🧠:[<think>Okay, so I need to find the length of CE in triangle ABC. Let me start by visualizing the problem. Triangle ABC has an altitude CD of length 12, and CE is the angle bisector of angle ACB. The sides AC and BC are given as 15 and 20, respectively. Hmm, okay. Let me sketch this in my mind: triangle ABC with AC=15, BC=20, and altitude CD=12. CE splits angle ACB into two equal parts. I need to find CE's length.First, maybe I should recall some theorems related to angle bisectors. The Angle Bisector Theorem states that an angle bisector in a triangle divides the opposite side into segments proportional to the adjacent sides. So, if CE is the angle bisector of angle C, then it divides AB into segments AE and EB such that AE/EB = AC/BC = 15/20 = 3/4. That ratio might come in handy later.But before that, maybe I need to find some more information about the triangle. We know AC=15, BC=20, CD=12 (the altitude). Let me see. Since CD is the altitude from C to AB, then CD is perpendicular to AB. So, triangle ABC is divided into two right triangles: ADC and BDC. Given CD=12, AC=15, BC=20, perhaps I can use the Pythagorean theorem to find AD and DB?Yes, let's try that. In triangle ADC, AC is the hypotenuse with length 15, and CD is the height with length 12. So, AD^2 + CD^2 = AC^2. Plugging in the numbers: AD^2 + 12^2 = 15^2. So, AD^2 = 225 - 144 = 81. Therefore, AD = 9. Similarly, in triangle BDC, BC=20, CD=12, so BD^2 + 12^2 = 20^2. BD^2 = 400 - 144 = 256. Therefore, BD = 16. So, AB = AD + DB = 9 + 16 = 25.Alright, so the triangle ABC has sides AC=15, BC=20, AB=25. Wait, 15, 20, 25. That's a Pythagorean triple: 15-20-25. So, triangle ABC is a right-angled triangle? Wait, because 15² + 20² = 225 + 400 = 625, which is 25². So, yes! Triangle ABC is a right-angled triangle at C? Wait, but then CD is the altitude from C to AB. In a right-angled triangle, the altitude from the right angle to the hypotenuse has some known properties.Wait, in a right-angled triangle, the altitude to the hypotenuse is equal to (product of the legs)/hypotenuse. So, if ABC is right-angled at C, then the legs are AC and BC, which are 15 and 20, hypotenuse AB=25. Then altitude CD should be (15*20)/25 = 300/25 = 12. Which matches the given CD=12. So, that confirms that triangle ABC is right-angled at C. Wait, but the problem didn't state that. But given the lengths, it must be. So, angle C is the right angle. So, angle ACB is 90 degrees, and CE is the bisector of this right angle. So, angle ECD and angle ECB would each be 45 degrees?Wait, but if angle C is 90 degrees and CE is the angle bisector, then it divides the 90-degree angle into two 45-degree angles. Therefore, CE makes a 45-degree angle with both AC and BC. Hmm. But in that case, maybe CE can be calculated using some trigonometric relations or perhaps using the angle bisector length formula.Alternatively, since the triangle is right-angled, maybe coordinate geometry can be helpful here. Let me try placing the triangle on a coordinate system. Let me place point C at the origin (0,0), since it's the right angle. Then, AC is 15 units long. Let me assume point A is along the x-axis. Wait, if C is at (0,0), then AC could be along the x-axis, so point A is at (15,0). Then BC is 20 units, and since angle C is the right angle, BC is along the y-axis. So point B is at (0,20). Wait, but then AB would be from (15,0) to (0,20). Let me check the length of AB in that case. The distance between (15,0) and (0,20) is sqrt(15² + 20²) = sqrt(225 + 400) = sqrt(625) = 25, which matches AB=25. So that works.Now, CD is the altitude from C to AB. Since ABC is a right-angled triangle at C, the altitude from C to hypotenuse AB is indeed CD=12. So, point D is the foot of the perpendicular from C to AB. But in coordinate terms, maybe I can find the coordinates of D. Alternatively, since we already know AD=9 and DB=16 from earlier, maybe the coordinates of D can be found by dividing AB in the ratio AD:DB = 9:16. Wait, coordinates of A (15,0) and B (0,20). So, point D divides AB into segments AD=9 and DB=16. So, the coordinates of D can be found using the section formula. If AD:DB = 9:16, then the coordinates of D would be [(16*15 + 9*0)/(9+16), (16*0 + 9*20)/(9+16)] = [(240 + 0)/25, (0 + 180)/25] = (240/25, 180/25) = (9.6, 7.2). Hmm, but maybe we don't need that right now.We need to find the length of CE, where E is on AB, and CE is the angle bisector of angle ACB (which is 90 degrees). So, CE splits AB into AE:EB = AC:CB = 15:20 = 3:4. Since AB is 25, AE = (3/7)*25 and EB = (4/7)*25. Wait, let me check that. The Angle Bisector Theorem states that AE/EB = AC/CB = 15/20 = 3/4. So, AE = (3/7)*AB and EB = (4/7)*AB. Since AB=25, then AE = (3/7)*25 = 75/7 ≈ 10.714, and EB = 100/7 ≈ 14.286.Therefore, point E divides AB in the ratio 3:4. So, coordinates of E can be found using the section formula again. Coordinates of A (15,0) and B (0,20). So, if E divides AB in the ratio AE:EB = 3:4, then coordinates of E would be [(4*15 + 3*0)/7, (4*0 + 3*20)/7] = [(60 + 0)/7, (0 + 60)/7] = (60/7, 60/7). So, coordinates of E are (60/7, 60/7). Then, CE is the distance from point C (0,0) to E (60/7, 60/7). So, CE = sqrt[(60/7)^2 + (60/7)^2] = sqrt[2*(60/7)^2] = (60/7)*sqrt(2) ≈ (60/7)*1.414 ≈ 8.571*1.414 ≈ 12.12. But the problem is probably expecting an exact value. So, CE = (60√2)/7. Let me check if this is correct.Alternatively, maybe there's a formula for the length of an angle bisector in a right-angled triangle. The formula for the angle bisector length in a triangle is given by:If in a triangle, the angle bisector from angle C meets AB at E, then the length of CE can be calculated by:CE = (2ab/(a + b)) * cos(theta/2)Where a and b are the sides adjacent to angle C, and theta is angle C. In this case, angle C is 90 degrees, so theta/2 is 45 degrees, and cos(45°) = √2/2.So, substituting a=15, b=20, theta=90°, we get:CE = (2*15*20)/(15 + 20) * (√2/2) = (600/35)*(√2/2) = (300/35)*√2 = (60/7)*√2. So, that's the same result as before. So, CE = (60√2)/7. So, that's the exact value. Therefore, the length of CE is 60√2 divided by 7. So, boxed, that's boxed{dfrac{60sqrt{2}}{7}}.Wait, but let me confirm using another method to be sure. Let's use coordinates again. Point C is at (0,0), E is at (60/7, 60/7). So, distance CE is sqrt[(60/7)^2 + (60/7)^2] = sqrt[2*(60/7)^2] = (60/7)*sqrt(2), which is the same as before. So that's correct.Alternatively, maybe using trigonometry. Since CE is the angle bisector of a 90-degree angle, it creates two 45-degree angles. So, in triangle CEB, angle at C is 45 degrees, side CB is 20, and angle at B can be calculated. Wait, maybe not straightforward. Alternatively, in triangle CEA, angle at C is 45 degrees, side AC=15. Maybe use the Law of Sines or Cosines.In triangle CEA, angle at C is 45 degrees, side AC=15, CE is the bisector. Let's denote CE = x. Then, in triangle CEA, using the Law of Sines: sin(angle at A)/CE = sin(45°)/EA. Wait, angle at A in triangle CEA? Maybe this is getting too complicated. Alternatively, in triangle CEA, angle at C is 45°, angle at A is the same as angle at A in triangle ABC. Since ABC is a right-angled triangle at C, angle at A is arctan(BC/AC) = arctan(20/15) = arctan(4/3). So angle at A is arctan(4/3). Then, in triangle CEA, angles are 45°, arctan(4/3), and the remaining angle. But this seems more complicated than the coordinate geometry approach.Alternatively, since we know coordinates, maybe using vectors. But coordinate method already gave the answer. Alternatively, using the formula for the angle bisector length. Wait, the formula for the length of an angle bisector in any triangle is:CE = (2ab/(a + b)) * cos(theta/2)Which we already used. So, that's another confirmation. Therefore, CE = (60√2)/7. Hence, the answer is 60√2 over 7. So, the box answer is boxed{dfrac{60sqrt{2}}{7}}.I think that's correct. Let me just verify once more. The key steps were recognizing that ABC is a right-angled triangle, then applying the Angle Bisector Theorem to find the ratio AE:EB = 3:4, calculating the coordinates of E, and then computing the distance from C to E. Alternatively, applying the angle bisector length formula directly. Both methods lead to the same result. So, I'm confident that this is the correct answer.Final AnswerThe length of C E is boxed{dfrac{60sqrt{2}}{7}}.</think>Given triangle ( triangle ABC ) with ( AC = 15 ), ( BC = 20 ), and altitude ( CD = 12 ). We need to find the length of the angle bisector ( CE ) of ( angle ACB ).1. Identify the triangle type and sides: - Using the Pythagorean theorem, we find: - ( AD = 9 ) (since ( AD^2 + 12^2 = 15^2 )) - ( BD = 16 ) (since ( BD^2 + 12^2 = 20^2 )) - Therefore, ( AB = AD + BD = 9 + 16 = 25 ). - Recognizing ( 15^2 + 20^2 = 25^2 ), triangle ( triangle ABC ) is a right-angled triangle at ( C ).2. Apply the Angle Bisector Theorem: - The angle bisector ( CE ) divides ( AB ) in the ratio ( AC : BC = 15 : 20 = 3 : 4 ). - Using the section formula, coordinates of ( E ) are found by dividing ( AB ) in the ratio 3:4: - Coordinates of ( A ) are ( (15, 0) ) and ( B ) are ( (0, 20) ). - Coordinates of ( E ) are ( left( frac{4 cdot 15 + 3 cdot 0}{3 + 4}, frac{4 cdot 0 + 3 cdot 20}{3 + 4} right) = left( frac{60}{7}, frac{60}{7} right) ).3. Calculate the length of ( CE ): - Using the distance formula from ( C(0, 0) ) to ( E left( frac{60}{7}, frac{60}{7} right) ): [ CE = sqrt{left( frac{60}{7} right)^2 + left( frac{60}{7} right)^2} = sqrt{2 left( frac{60}{7} right)^2} = frac{60}{7} sqrt{2} ]4. Verification using the angle bisector length formula: - For a right-angled triangle, the length of the angle bisector is: [ CE = frac{2ab}{a + b} cos left( frac{theta}{2} right) ] - Substituting ( a = 15 ), ( b = 20 ), ( theta = 90^circ ), and ( cos 45^circ = frac{sqrt{2}}{2} ): [ CE = frac{2 cdot 15 cdot 20}{15 + 20} cdot frac{sqrt{2}}{2} = frac{600}{35} cdot frac{sqrt{2}}{2} = frac{60 sqrt{2}}{7} ]Thus, the length of ( CE ) is (boxed{dfrac{60sqrt{2}}{7}}).,<think>Okay, so I have this geometry problem here. Let me try to visualize it first. There's triangle ABC, with CD being the altitude from C to AB, and CE being the angle bisector of angle ACB. The given lengths are AC = 15, BC = 20, and CD = 12. We need to find the length of CE. Hmm, alright.First, let me draw triangle ABC. Let me label the vertices: A, B, and C. CD is the altitude from C to AB, so D must be the foot of the perpendicular from C to AB. CE is the angle bisector of angle ACB, so E is a point on AB such that CE splits angle ACB into two equal angles. The sides AC and BC are given as 15 and 20 respectively, and the altitude CD is 12. I need to find CE.Hmm, okay. Let me recall some properties. Since CD is the altitude, triangle ABC is split into two right triangles: ADC and BDC. Maybe I can use the Pythagorean theorem here. Also, angle bisector theorem might come into play for finding CE.First, maybe I should find the length of AB. Wait, but to find AB, perhaps I can use the area of the triangle. The area can be calculated in two ways: using the base AB and height CD, or using the sides AC and BC with the sine of the included angle. But since we don't know the included angle, maybe using the altitude is better.The area of triangle ABC is (1/2)*AB*CD = (1/2)*AB*12 = 6*AB. Also, the area can be calculated using Heron's formula, but I need to know all three sides. Wait, but I only know AC and BC. Maybe I need to find AB first?Alternatively, since CD is the altitude, in right triangles ADC and BDC, we can use the Pythagorean theorem. Let me denote AD as x and DB as y. Then, AB = x + y. In triangle ADC, AC^2 = AD^2 + CD^2. So, 15^2 = x^2 + 12^2. Calculating that, 225 = x^2 + 144, so x^2 = 225 - 144 = 81. Therefore, x = 9. Similarly, in triangle BDC, BC^2 = BD^2 + CD^2. So, 20^2 = y^2 + 12^2. That's 400 = y^2 + 144, so y^2 = 400 - 144 = 256. Thus, y = 16. Therefore, AB = AD + DB = 9 + 16 = 25. So, AB is 25.Alright, so triangle ABC has sides AC = 15, BC = 20, and AB = 25. Let me check if this is a right triangle. 15^2 + 20^2 = 225 + 400 = 625, which is 25^2. Oh! So triangle ABC is a right triangle with right angle at C. Wait, but CD is the altitude from C to AB. In a right triangle, the altitude from the right angle to the hypotenuse is equal to the geometric mean of the segments into which the hypotenuse is divided. Wait, but here we found AD = 9 and DB = 16, and CD = 12. Let me verify: in a right triangle, the altitude to the hypotenuse should be equal to (AC * BC)/AB. So, (15*20)/25 = 300/25 = 12. Which matches CD = 12. So that checks out.So, triangle ABC is a right-angled triangle at C. Therefore, angle ACB is 90 degrees. Therefore, the angle bisector CE will split the 90-degree angle into two 45-degree angles. Hmm, okay. So CE is the angle bisector of the right angle at C, so it divides AB into segments proportional to the adjacent sides. Wait, angle bisector theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So, in triangle ABC, angle bisector from C to E on AB, so AE/EB = AC/BC = 15/20 = 3/4. Therefore, AE = (3/7)*AB and EB = (4/7)*AB. Since AB is 25, then AE = (3/7)*25 = 75/7 ≈ 10.714, and EB = 100/7 ≈ 14.286.But how do we find CE? Since we know the coordinates, maybe coordinate geometry would help here. Let me assign coordinates to the triangle. Let me place point C at the origin (0,0), since it's the right angle. Then, since AC = 15 and BC = 20, and angle at C is 90 degrees, point A can be at (15,0) and point B at (0,20). Wait, but then AB would be the hypotenuse from (15,0) to (0,20). Let me check the distance between (15,0) and (0,20). The distance is sqrt((15-0)^2 + (0-20)^2) = sqrt(225 + 400) = sqrt(625) = 25, which matches AB = 25. Okay, so coordinates:C: (0,0)A: (15,0)B: (0,20)Then, the altitude CD from C to AB. Wait, but in a right triangle, the altitude from the right angle to the hypotenuse is the same as the point C itself? Wait, no. Wait, in a right triangle, the altitude from the right angle to the hypotenuse is actually the same as the vertex itself. Wait, but in this problem, CD is given as 12. Wait, but in our coordinate system, the length from C to AB is 12? But in reality, in a right triangle, the distance from C to AB is zero because C is on AB? Wait, no. Wait, in a right triangle, the hypotenuse is AB, and the right angle is at C. So point C is not on AB. Wait, hold on, maybe my coordinate system is conflicting with the problem statement.Wait, in the problem statement, triangle ABC has CD as the altitude from C to AB, which is given as 12. But if triangle ABC is a right triangle at C, then AB is the hypotenuse, and the altitude from C to AB is indeed CD = 12. But in my coordinate system, if C is at (0,0), A at (15,0), and B at (0,20), then AB is from (15,0) to (0,20). The equation of AB can be found to compute the distance from C to AB, which should be 12. Let's check that.The equation of AB: Let's compute the equation. The coordinates are A(15,0) and B(0,20). The slope of AB is (20 - 0)/(0 - 15) = 20/(-15) = -4/3. So the equation is y - 0 = -4/3(x - 15), which simplifies to y = -4/3 x + 20. The distance from point C(0,0) to the line AB is given by the formula:Distance = |Ax + By + C| / sqrt(A^2 + B^2)First, let's write AB in standard form. From y = -4/3 x + 20, multiply both sides by 3: 3y = -4x + 60. Then, 4x + 3y - 60 = 0. So A=4, B=3, C=-60.Distance from (0,0) is |4*0 + 3*0 -60| / sqrt(16 + 9) = | -60 | / 5 = 60/5 = 12. Yes! So CD = 12, which matches the problem statement. So everything checks out.So in this coordinate system, point D is the foot of the altitude from C(0,0) to AB. We had already calculated AD = 9 and DB = 16, so point D divides AB into segments of 9 and 16. Since AB is from (15,0) to (0,20), the coordinates of D can be found by moving 9 units from A towards B, since AD = 9 and AB = 25.Using the section formula: if D divides AB in the ratio AD:DB = 9:16, then the coordinates of D are [(16*15 + 9*0)/(9+16), (16*0 + 9*20)/(9+16)] = [(240)/25, (180)/25] = (9.6, 7.2). Alternatively, since AD = 9, moving 9/25 of the way from A to B. The vector from A to B is (-15, 20). So starting at A(15,0), adding 9/25*(-15,20) gives:x-coordinate: 15 + (9/25)*(-15) = 15 - 135/25 = 15 - 5.4 = 9.6y-coordinate: 0 + (9/25)*(20) = 180/25 = 7.2So D is at (9.6, 7.2). That's correct.Now, we need to find point E on AB such that CE is the angle bisector of angle ACB (which is 90 degrees). As per the angle bisector theorem, AE/EB = AC/BC = 15/20 = 3/4. So as I mentioned before, AE = (3/7)*AB = (3/7)*25 = 75/7 ≈ 10.714, and EB = (4/7)*25 = 100/7 ≈14.286. Therefore, point E divides AB in the ratio AE:EB = 3:4.So coordinates of E can be found similarly. Since AB is from A(15,0) to B(0,20), and E divides AB in the ratio 3:4 (from A to B), the coordinates are:x-coordinate: (4*15 + 3*0)/(3+4) = (60 + 0)/7 = 60/7 ≈8.571y-coordinate: (4*0 + 3*20)/7 = (0 + 60)/7 = 60/7 ≈8.571Wait, so E is at (60/7, 60/7). Let me check this. Alternatively, since moving from A to B, which is 25 units, E is 3/7 from A and 4/7 from B.But let's use the section formula properly. If E divides AB in the ratio AE:EB = 3:4, then the coordinates of E are [(4*15 + 3*0)/7, (4*0 + 3*20)/7] = [60/7, 60/7]. So E is at (60/7, 60/7). Approximately (8.571, 8.571). Okay.Now, we need to find the length of CE. Since C is at (0,0) and E is at (60/7, 60/7), the distance CE is sqrt[(60/7 - 0)^2 + (60/7 - 0)^2] = sqrt[(60/7)^2 + (60/7)^2] = sqrt[2*(60/7)^2] = (60/7)*sqrt(2).So CE = (60√2)/7. Let me compute that. 60/7 is approximately 8.571, so CE is approximately 12.12. But the problem probably expects an exact value, which is (60√2)/7. Let me check if that's the correct answer.Alternatively, maybe there's a formula for the length of the angle bisector in a right triangle. Let me recall. The angle bisector length formula in a triangle is:If in a triangle, the lengths of the sides are a, b, c, opposite to angles A, B, C respectively, then the length of the angle bisector from angle C is given by:t_c = (2ab/(a + b)) * cos(C/2)But in this case, angle C is 90 degrees, so cos(45°) = √2/2. So t_c = (2*15*20)/(15 + 20) * (√2/2) = (600/35)*(√2/2) = (300/35)*√2 = (60/7)√2. Which matches the previous result. So CE = (60√2)/7. So that's the answer.Alternatively, another formula for the angle bisector length is:If in a triangle, the angle bisector from angle C meets AB at E, then CE can be calculated by:CE = (2ab/(a + b)) * cos(theta/2)Where theta is the angle at C. Which is exactly what we used.Alternatively, using coordinates was straightforward here because it's a right triangle. So both methods confirm that CE = (60√2)/7. Therefore, the answer should be 60√2/7. Let me check if there's another method.Alternatively, using the angle bisector theorem. The angle bisector theorem states that AE/EB = AC/BC = 15/20 = 3/4. So AE = 3k, EB = 4k, and AE + EB = AB = 25. Therefore, 3k + 4k = 7k = 25, so k = 25/7. Therefore, AE = 75/7, EB = 100/7. Then, coordinates of E as above.Then, using coordinates to compute CE. So CE is distance from (0,0) to (60/7, 60/7). Which is sqrt[(60/7)^2 + (60/7)^2] = sqrt[2*(3600/49)] = sqrt[7200/49] = (60√2)/7. Yep, same result.Alternatively, using vectors or trigonometry. But seems consistent.Alternatively, in the right triangle, since angle bisector divides the angle into two 45-degree angles, perhaps triangle CEB is a 45-45-90 triangle? Wait, no. Because CE is the angle bisector, but E is on AB. So angle ECD is 45 degrees, but not necessarily the triangle.Alternatively, maybe using the formula for the length of the angle bisector:In any triangle, the length of the angle bisector from angle C to side AB is given by:CE = (2ab/(a + b)) * cos(theta/2)Where a and b are the sides adjacent to angle C (i.e., AC and BC), and theta is angle C. So here, a = 15, b = 20, theta = 90 degrees. Therefore,CE = (2*15*20)/(15 + 20) * cos(45°) = (600/35)*(√2/2) = (300/35)*√2 = (60/7)√2. Same as before.So regardless of the method, the answer is 60√2 over 7. Therefore, the length of CE is 60√2/7.Final AnswerThe length of C E is boxed{dfrac{60sqrt{2}}{7}}.</think>Given triangle ( triangle ABC ) with ( CD ) as the altitude from ( C ) to ( AB ) and ( CE ) as the angle bisector of ( angle ACB ). The given lengths are ( AC = 15 ), ( BC = 20 ), and ( CD = 12 ). We need to find the length of ( CE ).First, we use the Pythagorean theorem to find the lengths of ( AD ) and ( BD ):- In ( triangle ADC ), ( AC^2 = AD^2 + CD^2 ): [ 15^2 = AD^2 + 12^2 implies 225 = AD^2 + 144 implies AD^2 = 81 implies AD = 9 ]- In ( triangle BDC ), ( BC^2 = BD^2 + CD^2 ): [ 20^2 = BD^2 + 12^2 implies 400 = BD^2 + 144 implies BD^2 = 256 implies BD = 16 ]Thus, the length of ( AB ) is:[AB = AD + BD = 9 + 16 = 25]Next, we verify that ( triangle ABC ) is a right triangle with ( angle ACB = 90^circ ):[AC^2 + BC^2 = 15^2 + 20^2 = 225 + 400 = 625 = 25^2 = AB^2]Using the angle bisector theorem, the angle bisector ( CE ) divides ( AB ) into segments proportional to ( AC ) and ( BC ):[frac{AE}{EB} = frac{AC}{BC} = frac{15}{20} = frac{3}{4}]Let ( AE = 3k ) and ( EB = 4k ). Since ( AB = 25 ):[3k + 4k = 25 implies 7k = 25 implies k = frac{25}{7}]Thus, the coordinates of ( E ) can be found using the section formula. Placing ( C ) at the origin ((0,0)), ( A ) at ((15,0)), and ( B ) at ((0,20)), the coordinates of ( E ) dividing ( AB ) in the ratio ( 3:4 ) are:[E = left( frac{4 cdot 15 + 3 cdot 0}{3 + 4}, frac{4 cdot 0 + 3 cdot 20}{3 + 4} right) = left( frac{60}{7}, frac{60}{7} right)]Finally, the length of ( CE ) is calculated using the distance formula:[CE = sqrt{left( frac{60}{7} - 0 right)^2 + left( frac{60}{7} - 0 right)^2} = sqrt{2 left( frac{60}{7} right)^2} = frac{60}{7} sqrt{2}]Thus, the length of ( CE ) is (boxed{dfrac{60sqrt{2}}{7}}).]

🔑:9. frac{60 sqrt{2}}{7}.By the Pythagorean theorem, we know A D=9, B D=16.Therefore, A B=A D+B D=25.Hence, by the converse of the Pythagorean theorem, triangle A C B is a right triangle, and angle A C B=90^{circ}.As shown in Figure 4, draw E F perp B C, with the foot of the perpendicular at F. Let E F=x.Since angle E C F=frac{1}{2} angle A C B=45^{circ}, we have C F=x.Thus, B F=20-x.Because E F parallel A C, we have frac{E F}{A C}=frac{B F}{B C}, which means frac{x}{15}=frac{20-x}{20} Rightarrow x=frac{60}{7}.Therefore, C E=sqrt{2} x=frac{60 sqrt{2}}{7}.

🧠:[<think>Okay, let me try to figure out this problem. So, the question is about comparing the accuracy of Soviet rocket tests to a student throwing a softball. The rockets had all hits within a circle of radius less than 1 km at a target distance of 12,500 km. We need to find out what the radius of the hit circle would be for a student throwing a softball 25 meters away with the same accuracy.Hmm, first, I need to understand what "same accuracy" means here. I think it's referring to the angular accuracy or maybe the relative error. Since rockets are hitting within a certain radius at a very long distance, and the student is doing the same but at a much shorter distance, we need to scale the radius proportionally.So, if the radius of the hit circle is proportional to the distance, then we can set up a proportion. Let me write that down.The rocket's distance is 12,500 km, and the radius is 1 km. The student's distance is 25 m, and we need to find the radius R. If the ratio of radius to distance is the same, then:Rocket's ratio: 1 km / 12,500 kmStudent's ratio: R / 25 mSetting them equal:1 / 12,500 = R / 25Solving for R:R = 25 * (1 / 12,500)Wait, let's check the units. The rocket's distance and radius are in kilometers, and the student's are in meters. So we need to convert units to make them consistent. Let me convert 12,500 km to meters to keep the same unit system.1 km = 1,000 meters, so 12,500 km = 12,500 * 1,000 m = 12,500,000 m.So the rocket's ratio is 1 km / 12,500 km, which is 1,000 m / 12,500,000 m.Simplifying that: 1,000 / 12,500,000 = 1 / 12,500.Wait, that's the same as before. So whether we convert to meters or not, the ratio remains the same because it's a unitless ratio. So maybe the units cancel out. Let me confirm:1 km / 12,500 km = (1/12,500) km/km = 1/12,500 (unitless)Similarly, R / 25 m = (R m) / (25 m) = R/25 (unitless)Therefore, setting them equal: 1/12,500 = R/25Solving for R:R = 25 / 12,500Calculating that: 25 divided by 12,500. Let me compute that.12,500 divided by 25 is 500, so 25 divided by 12,500 is 1/500.1/500 of a meter. Let's convert that to a more understandable unit. 1/500 m = 0.002 m = 2 mm.Wait, 0.002 meters is 2 millimeters? That seems really small. Is that right?Wait, if the rocket's error is 1 km over 12,500 km, that's a very small angular error. So scaling down to 25 meters, the linear error would be proportionally smaller. Let's see:Angular error in radians is approximately radius divided by distance (for small angles). So the angular error for the rocket is 1 km / 12,500 km = 1/12,500 radians.Then, for the student, the radius R would be angular error multiplied by distance:R = (1/12,500 radians) * 25 mSo R = 25 / 12,500 m = 0.002 m = 2 mm.That's correct. So if the angular accuracy is the same, then the linear error at 25 meters is 2 millimeters. That seems extremely precise for a student throwing a softball. Maybe even more precise than possible, but perhaps the question is just theoretical.But let me check again. Maybe there is a unit conversion mistake. Let's go step by step.Original problem:Rocket: Distance = 12,500 km, Radius = 1 kmStudent: Distance = 25 m, Radius = ?Assuming the same accuracy, meaning the same ratio of radius to distance.So, ratio = Radius / DistanceFor the rocket: 1 km / 12,500 km = 1 / 12,500For the student: R / 25 m = 1 / 12,500Therefore, R = 25 m / 12,500 = 0.002 m = 2 mm.Yes, that's correct. So the answer should be 2 millimeters. Although in real life, a student throwing a softball can't have that precision, but in terms of scaling the problem, that's the mathematical answer.Alternatively, maybe the question is considering area? Wait, the problem says "all hits were within a circle whose radius was less than 1 km." So it's a circular error probable or something. If they're talking about the same circular error scaled down, then it's linear scaling, not area. Because if you scale the area, the radius would scale with the square root, but I think in terms of accuracy, it's linear with distance. For example, if you have a certain angular error, the linear error at the target is proportional to the distance. So if you half the distance, the linear error is halved. So in this case, since 25 m is much less than 12,500 km, the linear error should be much less.Wait, 12,500 km is 12,500,000 meters. So 25 meters is 25 / 12,500,000 = 1 / 500,000 of the original distance. Therefore, scaling the radius by the same ratio: 1 km * (25 / 12,500,000) = 1,000 m * (25 / 12,500,000) = (25,000 / 12,500,000) m = 2 / 1,000 m = 0.002 m = 2 mm. Same result.Therefore, the answer is 2 millimeters. So the student's hits would be within a circle of radius 2 mm. That seems insanely precise, but mathematically, that's the answer. Maybe the question is testing understanding of scaling factors and ratios, regardless of real-world feasibility.I think that's it. The key is to recognize that the radius scales linearly with the distance if the angular accuracy is the same. So the calculation gives 2 mm. Let me just verify once more.1 km at 12,500 km. So 1 / 12,500 is the ratio. For 25 m, 25 * (1 / 12,500) = 25 / 12,500 = 0.002 km? Wait, no. Wait, no, units. Wait, if we take 1 km per 12,500 km, that's (1/12,500) km per km. So per kilometer of distance, the error is 1/12,500 km. Therefore, per meter of distance, the error would be (1/12,500) km / 1,000 m = (1/12,500,000) km/m, but this seems more complicated. Wait, perhaps better to convert everything to meters.Original error: 1,000 meters at 12,500,000 meters.Ratio: 1,000 / 12,500,000 = 1 / 12,500.So, for 25 meters, error = 25 * (1 / 12,500) = 25 / 12,500 = 0.002 meters = 2 millimeters. Yep, same answer.Therefore, the answer is 2 mm. So boxed, it would be boxed{2 text{ mm}}.Final AnswerThe radius of the hit circle for the student would be boxed{2 text{ mm}}.</think>To determine the radius of the hit circle for a student throwing a softball at a target 25 meters away with the same accuracy as the Soviet rocket tests, we need to use the concept of proportional scaling based on distance.1. Rocket Test Data: - Distance: (12,500 mathrm{~km}) - Radius of hit circle: (1 mathrm{~km})2. Student's Data: - Distance: (25 mathrm{~m}) - Radius of hit circle: (R) (to be determined)3. Proportional Relationship: The ratio of the radius to the distance should be the same for both the rocket and the student. [ frac{1 mathrm{~km}}{12,500 mathrm{~km}} = frac{R}{25 mathrm{~m}} ]4. Unit Conversion: Convert kilometers to meters to maintain consistent units: - (12,500 mathrm{~km} = 12,500,000 mathrm{~m}) - (1 mathrm{~km} = 1,000 mathrm{~m})5. Setting Up the Equation: [ frac{1,000 mathrm{~m}}{12,500,000 mathrm{~m}} = frac{R}{25 mathrm{~m}} ]6. Simplifying the Ratio: [ frac{1}{12,500} = frac{R}{25} ]7. Solving for (R): [ R = 25 mathrm{~m} times frac{1}{12,500} = frac{25}{12,500} mathrm{~m} = 0.002 mathrm{~m} = 2 mathrm{~mm} ]Thus, the radius of the hit circle for the student would be boxed{2 text{ mm}}.,<think>Okay, let me try to figure out this problem step by step. So, the question is about comparing the accuracy of Soviet rocket tests to a student throwing a softball. The rockets had all their hits within a circle of radius less than 1 km at a target distance of 12,500 km. Now, we need to find out what the radius of the hit circle would be for a student throwing a softball 25 meters away with the same accuracy. First, I need to understand what "same accuracy" means here. I think it's referring to the angular accuracy. That is, the angle between the intended trajectory and the actual trajectory is the same for both the rocket and the softball. If that's the case, then the error radius would be proportional to the distance to the target. So, if the distance is shorter, the radius should be smaller proportionally.Let me verify this idea. Angular accuracy is often measured in radians. The formula for the radius of the circle (which is the error) would be the tangent of the angular error multiplied by the distance. But for small angles, the tangent of the angle is approximately equal to the angle itself in radians. So, if the angular error is θ, then the radius r ≈ θ * d, where d is the distance.So, if both the rocket and the student have the same angular accuracy θ, then the ratio of their error radii should be the same as the ratio of their distances. That is, r_rocket / d_rocket = r_student / d_student. Therefore, r_student = (r_rocket / d_rocket) * d_student.Let me plug in the numbers. The rocket's error radius is 1 km, and the distance is 12,500 km. The student's distance is 25 meters. So,r_student = (1 km / 12,500 km) * 25 m.First, let's convert 1 km to meters to keep the units consistent. 1 km = 1000 meters. So,r_student = (1000 m / 12,500,000 m) * 25 m.Wait, hold on. 12,500 km is 12,500 * 1000 m = 12,500,000 meters. The student's distance is 25 meters. So,r_student = (1000 / 12,500,000) * 25.Let me compute that. Let's simplify the fraction 1000 / 12,500,000 first. Dividing numerator and denominator by 1000 gives 1 / 12,500. Then, 25 * (1 / 12,500) = 25 / 12,500.25 divided by 12,500. Let's see, 12,500 divided by 25 is 500. So, 25 / 12,500 = 1 / 500. Therefore, 1/500 meters.Convert that to meters: 1/500 m = 0.002 meters. Which is 2 millimeters.Wait, that seems really small. 2 millimeters for a student throwing a softball? That seems too precise. Maybe I made a mistake in unit conversions?Let me check again. The rocket's error radius is 1 km at 12,500 km. So the ratio is 1 km / 12,500 km = 1 / 12,500. Then, applying that ratio to the student's distance of 25 m: 25 m * (1 / 12,500) = 25 / 12,500 m.25 divided by 12,500. Let's do this division step by step. 25 ÷ 12.5 = 2, so 25 ÷ 12,500 = 2 ÷ 1000 = 0.002 meters. So 0.002 meters is 2 millimeters. Hmm. 2 millimeters is 0.2 centimeters, which is about the thickness of a pencil. For a student throwing a softball, expecting all hits to be within a circle of radius 2 mm is way too accurate. That doesn't make sense. Maybe the assumption about angular accuracy is wrong?Alternatively, perhaps it's a linear accuracy. But if it's linear, then the radius would be the same regardless of distance, which also doesn't make sense. So maybe there's a different way to interpret "same accuracy".Wait, maybe it's about the relative error. So, the rocket has an error of 1 km over 12,500 km. The relative error is 1/12,500. Then for the student, the relative error would be the same, so the radius would be 25 m * (1/12,500). Which again gives 0.002 meters or 2 millimeters. Same result. But that still seems too small.Alternatively, maybe the question is considering the absolute error, but scaled by distance. If you scale down the distance by a factor, the error radius scales by the same factor. Let's compute the scaling factor. The rocket's distance is 12,500 km, and the student's distance is 25 m. Let's convert both to the same unit. Let's use meters.12,500 km = 12,500 * 1000 m = 12,500,000 meters. The student's distance is 25 meters. So the scaling factor is 25 / 12,500,000 = 1 / 500,000. Then the error radius for the rocket is 1 km = 1000 meters. Scaling that by 1/500,000 gives 1000 / 500,000 = 0.002 meters = 2 millimeters. Again same answer.But a softball is about 30 cm in circumference, so diameter about 10 cm. A 2 mm radius circle is way smaller than the ball itself. So a student can't even hit a target with that precision. That seems unrealistic. So maybe the problem is in the way I'm interpreting "same accuracy". Let me check the problem statement again."In the Soviet rocket tests conducted in October 1961, at a target distance of about 12500 km, all hits were within a circle whose radius was less than 1 km. How large would the radius of the hit circle be for a student who, with the same accuracy, throws a softball at a target 25 m away?"So the key is "same accuracy". So the rockets have a certain accuracy, which results in a 1 km radius at 12500 km distance. The student has the same accuracy, so we need to compute the radius at 25 m.If "accuracy" here is defined as the ratio of error radius to distance, then it's 1/12500 for the rockets, so 25 m * 1/12500 = 0.002 m = 2 mm. But as I said, that's too small. However, maybe accuracy in terms of angle. Let me check using angles.The angular error θ can be calculated by θ = arctan(r / d). For small angles, θ ≈ r / d. So, for the rocket, θ ≈ 1 km / 12500 km = 1/12500 radians. Then, for the student, the radius r = θ * d = (1/12500) * 25 m = 25/12500 m = 0.002 m = 2 mm. So same answer.But this seems to suggest that if you have a certain angular accuracy, the linear error scales with distance. So, if the student's distance is much shorter, their error radius is proportionally smaller. However, 2 mm is indeed extremely precise. But maybe in the context of the problem, that's the answer they expect.Alternatively, maybe the problem is using a different measure of accuracy. For example, maybe precision in terms of the area. The area of the circle is πr². If the accuracy is the same in terms of area per unit distance or something. Let's see.But the problem states "with the same accuracy". If it's about the same absolute area, that would be different, but the distances are different. But the problem doesn't specify that, so probably not. The problem says "the same accuracy", which is most likely angular accuracy, leading to linear error proportional to distance. So 2 mm would be the answer.But just to think again, maybe I messed up the unit conversions. Let me check:12500 km is 12,500,000 meters. 1 km is 1000 meters. So the ratio is 1000 / 12,500,000 = 1 / 12,500. Then 25 meters * (1 / 12,500) = 25 / 12,500 = 0.002 meters. Yes, 0.002 meters is 2 millimeters. So that's correct.Therefore, even though 2 mm seems unrealistic for a student's throw, the mathematics point to that answer. Perhaps in the context of a physics problem, we are to ignore the practicality and just compute based on proportional scaling.Therefore, I think the answer is 2 millimeters, which is 0.002 meters. So the radius would be boxed{0.002 mathrm{~m}} or boxed{2 mathrm{~mm}}.But let me check if the problem expects units in meters or millimeters. The original problem uses kilometers and meters. The target distance for the student is 25 meters, so likely the answer should be in meters. 0.002 meters is 2 millimeters. So both are correct, but perhaps meters is preferable here. Let me confirm. If the question says "how large", and the original radius was given in kilometers, then for the student, probably meters is the appropriate unit. So 0.002 meters, which is 2 millimeters. So I can write it as boxed{0.002 mathrm{~m}} or boxed{2 times 10^{-3} mathrm{~m}}.Alternatively, if they want centimeters, it's 0.2 cm. But since the problem is in meters, I think 0.002 m is the right way to present it.Yes, I think that's the answer. The key is that the accuracy is angular, so the linear error scales linearly with distance. So the answer is 2 millimeters.Final AnswerThe radius of the hit circle for the student would be boxed{0.002 mathrm{~m}}.</think>To determine the radius of the hit circle for a student throwing a softball at a target 25 meters away with the same accuracy as the Soviet rocket tests, we start by noting that the accuracy is angular. This means the radius of the hit circle is proportional to the distance to the target.Given:- The rocket's hit circle radius is 1 km at a distance of 12,500 km.- The student's target distance is 25 meters.First, we calculate the ratio of the rocket's error radius to its distance:[text{Ratio} = frac{1 text{ km}}{12,500 text{ km}} = frac{1}{12,500}]Using this ratio, we apply it to the student's distance to find the corresponding radius:[text{Radius for student} = 25 text{ m} times frac{1}{12,500} = frac{25}{12,500} text{ m} = 0.002 text{ m}]Converting 0.002 meters to millimeters confirms it is 2 millimeters, but since the problem uses meters, we present the answer in meters.Thus, the radius of the hit circle for the student would be boxed{0.002 mathrm{~m}}.,<think>Okay, let's see. The problem is about comparing the accuracy of Soviet rocket tests to a student throwing a softball. The rockets were hitting within a circle of radius less than 1 km at a target distance of 12,500 km. We need to find out what the radius of the hit circle would be for a student throwing a softball 25 meters away with the same accuracy.Hmm, so I think this is a proportionality problem. The idea is that the accuracy, which is the radius of the hit circle, is proportional to the distance to the target. If the rockets can hit within 1 km at 12,500 km, then the student's accuracy at 25 m would be scaled down by the same ratio.Let me write that out. If R1 is the radius of the hit circle for the rockets, D1 is the distance for the rockets, and R2 and D2 are the corresponding values for the student, then:R1 / D1 = R2 / D2So solving for R2:R2 = (R1 * D2) / D1Plugging in the numbers. R1 is 1 km, which is 1000 meters. D1 is 12,500 km, which is 12,500,000 meters. D2 is 25 meters.So R2 = (1000 m * 25 m) / 12,500,000 mLet me compute that. First, 1000 * 25 = 25,000. Then, divide by 12,500,000.25,000 / 12,500,000 = 25 / 12,500 = 0.002Wait, 25 divided by 12,500 is 0.002. So 0.002 meters. Converting that to centimeters, since it's a small distance. 0.002 meters is 2 millimeters. But that seems really small. Did I do that right?Wait, maybe I messed up the units. Let's check again.Original R1 is 1 km, which is 1000 meters. D1 is 12,500 km, which is 12,500 * 1000 = 12,500,000 meters. D2 is 25 meters. So R2 = (1000 * 25) / 12,500,000 = 25,000 / 12,500,000 = 0.002 meters. 0.002 meters is 0.002 * 100 cm = 0.2 cm, which is 2 millimeters. Hmm, 2 millimeters is like the size of a small bead. That does seem really tiny for a student throwing a softball. Maybe I made a mistake.Alternatively, maybe the problem is in angular terms. Maybe the radius is proportional to the distance, which is linear. So if you have a certain angular error, then the linear error at the target is proportional to the distance. So if the angular accuracy is the same, then the linear accuracy scales with distance.So for example, if the rockets had an angular error that resulted in 1 km at 12,500 km, then the same angular error for the student at 25 m would result in (25 / 12,500,000) * 1 km? Wait, no. Wait, if angular error θ is R / D, so θ = R1 / D1 = R2 / D2. Therefore, R2 = θ * D2 = (R1 / D1) * D2. Which is the same as the previous equation.So, according to that, R2 = (1 km / 12,500 km) * 25 m. Wait, 1 / 12,500 is 0.00008. Then 0.00008 * 25 m = 0.002 meters, which is 2 millimeters. So same answer.But that still seems too small. A student throwing a softball 25 meters and having all hits within a circle of 2 mm radius? That's superhuman accuracy. Maybe the problem is using radius as in the circular error probable or something else?Wait, maybe the problem is just a straightforward scaling problem, regardless of the practical aspects. The question says "with the same accuracy," so we have to take that as given, even if the result seems unrealistic.Alternatively, maybe the question is referring to the ratio of the area? Wait, but the problem mentions the radius of the circle, not the area. So if the radius scales with distance, then the area would scale with the square of the distance. But the problem says "all hits were within a circle whose radius was less than 1 km," so the radius is the linear measure. Therefore, the radius should scale linearly with distance.So even though 2 mm seems unrealistic, maybe that's the answer.Wait, let me check the calculation again.Convert all distances to meters:R1 = 1 km = 1000 metersD1 = 12,500 km = 12,500,000 metersD2 = 25 metersR2 = (R1 / D1) * D2 = (1000 / 12,500,000) * 25Calculate 1000 divided by 12,500,000. Let's see, 12,500,000 divided by 1000 is 12,500. So 1 / 12,500 = 0.00008.0.00008 * 25 = 0.002 meters. 0.002 meters is 2 millimeters. Yeah, that's correct.So unless there's a different interpretation, the answer is 2 millimeters. But that seems way too precise for a student throwing a softball. For example, even Olympic athletes couldn't throw a softball into a 2 mm circle from 25 meters. But maybe the problem is just mathematical, not considering real-world feasibility.Alternatively, maybe the problem expects the answer in centimeters? Let me see. 0.002 meters is 0.2 centimeters, which is 2 millimeters. So if they want centimeters, it's 0.2 cm. Maybe they expect it in meters? 0.002 meters. But 0.002 meters is 2 mm. Hmm.Alternatively, maybe the problem is using radius as a ratio of the total distance? So like 1 km is 1/12500 of 12500 km. Then for 25 m, 25 m * (1/12500) = 0.002 m. Same result.So unless the question is tricking us, the answer is 0.002 meters, which is 2 millimeters.But maybe I made a unit conversion mistake? Let's see. If D1 is 12,500 km, R1 is 1 km. So the ratio is 1/12500. Then for D2 = 25 m, R2 = 25 m / 12500 = 0.002 m. So same.Alternatively, maybe it's 12500 km vs 25 m. Let's convert both distances to the same unit. Let's convert 12500 km to meters: 12,500,000 meters. 25 meters is 25 meters. So the ratio is 25 / 12,500,000 = 1 / 500,000. So scaling down the radius by the same ratio. 1 km is 1000 meters. So 1000 meters * (1 / 500,000) = 0.002 meters. Yeah, same answer.So I think that is the answer. Even though it's unrealistic, the math says 2 millimeters. Maybe in the context of the problem, it's just about the proportional calculation, not about real-world expectations.Alternatively, maybe the problem meant to say 25 meters instead of 25 km? If it were 25 km, then the radius would be 2 meters, which is more reasonable. But the problem says 25 meters. Hmm.Alternatively, maybe the original radius was 1 km, which is diameter 2 km, but the problem says radius less than 1 km. So radius is 1 km. So scaling down.Alternatively, maybe the question is in error, but assuming the numbers are correct, the answer is 0.002 meters, which is 2 millimeters.So I think the answer is boxed{0.002} meters or boxed{2} millimeters. But the problem says "radius was less than 1 km", so maybe the answer should be in kilometers? Wait, no, for the student it's 25 meters away, so the radius would be in meters or centimeters.Wait, the problem says "how large would the radius of the hit circle be", so probably in the same units as the target distance. The target distance for the student is 25 meters, so the radius would be in meters. So 0.002 meters, which is 0.2 centimeters or 2 millimeters.But maybe they expect the answer in centimeters? 0.002 meters is 0.2 cm. Hmm.Wait, let's check the original problem statement again. It says "all hits were within a circle whose radius was less than 1 km". So the answer for the student should be "less than X", where X is 0.002 km? No, the student's distance is 25 m, so the radius would be proportional. Since 1 km is the radius at 12500 km, so scaling factor is 25 / 12,500,000. So 25 / 12,500,000 = 1 / 500,000. So 1 km * 1 / 500,000 = 0.000002 km, which is 0.002 meters.Alternatively, if the original accuracy is 1 km per 12,500 km, that's an accuracy of 1/12500. So for any distance, the radius is distance * (1/12500). So 25 m * (1/12500) = 0.002 m.So yeah, I think that's correct. The answer is 0.002 meters, which is 2 millimeters. So I will go with that.Final AnswerThe radius of the hit circle for the student would be boxed{0.002} meters.</think>To determine the radius of the hit circle for a student throwing a softball 25 meters away with the same accuracy as the Soviet rocket tests, we use proportionality. The rockets had a hit circle radius of 1 km at a distance of 12,500 km. The key is to scale the radius proportionally to the target distance.1. Convert all distances to the same unit (meters): - Radius for rockets, ( R_1 = 1 ) km = 1000 meters - Distance for rockets, ( D_1 = 12,500 ) km = 12,500,000 meters - Distance for student, ( D_2 = 25 ) meters2. Using the proportionality relationship: [ frac{R_1}{D_1} = frac{R_2}{D_2} ] Solving for ( R_2 ): [ R_2 = frac{R_1 times D_2}{D_1} ]3. Plugging in the values: [ R_2 = frac{1000 , text{m} times 25 , text{m}}{12,500,000 , text{m}} = frac{25,000}{12,500,000} = 0.002 , text{m} ]Thus, the radius of the hit circle for the student would be (boxed{0.002}) meters.,<think>Okay, let's see. So the problem is about comparing the accuracy of Soviet rocket tests to a student throwing a softball. The rockets had a hit radius of less than 1 km at a target distance of 12,500 km. The question is, if a student throws a softball with the same accuracy at a target 25 meters away, what would be the radius of the hit circle?Hmm, first, I need to understand what "same accuracy" means here. I think it refers to the angular accuracy, right? Like, the angle from the target within which the projectile can land. Because if you have the same angular error, then the radius of the hit circle would scale linearly with the distance. So it's a proportional relationship.Let me verify. If the original problem is at 12,500 km distance with a 1 km radius, then the angular error can be calculated using trigonometry. The radius is the opposite side of a right triangle where the adjacent side is the distance. So the tangent of the angle would be radius divided by distance. But for very small angles, tangent of theta is approximately equal to theta in radians. So maybe they just used a linear approximation.So, angle theta ≈ radius / distance. Therefore, theta = 1 km / 12,500 km = 1 / 12,500 radians. Then, for the student's case, the radius would be theta multiplied by the new distance. Which is 25 meters * (1 / 12,500). Let me calculate that.First, 25 divided by 12,500. Let's see, 12,500 divided by 100 is 125, so 25 divided by 125 is 0.2, and then divide by 100 again, so 0.002 km. Wait, but 25 meters is 0.025 km. Wait, no, maybe I messed up the units.Wait, the rocket's distance was 12,500 km, radius 1 km. The student's distance is 25 meters. Let's convert everything to the same units. Let's do meters. So 12,500 km is 12,500,000 meters. Radius is 1 km, which is 1000 meters.So angle theta ≈ 1000 meters / 12,500,000 meters = 1 / 12,500 radians. Then, applying that angle to 25 meters: radius = 25 meters * (1 / 12,500) = 25 / 12,500 = 0.002 meters. That's 2 millimeters. Wait, that seems really small. 2 millimeters? For a student throwing a softball? That seems way too accurate. Maybe I made a mistake here.Alternatively, maybe "same accuracy" refers to the ratio of the radius to the distance. So, for the rocket, it's (1 km / 12,500 km) = 0.00008. Then, applying that ratio to 25 meters: 25 * 0.00008 = 0.002 meters, which is 2 millimeters. Hmm, same answer. But is that realistic?Wait, maybe the problem is just a simple proportion. If you scale down the distance by a factor, the radius scales by the same factor. So 25 meters is 25 / 12,500,000 km. Wait, 25 meters is 0.025 km. So 0.025 km / 12,500 km = 1 / 500,000. Then, the original radius was 1 km, so scaling by 1/500,000 would give 1 km * (1/500,000) = 0.000002 km = 0.002 meters = 2 millimeters. Same answer again. So according to the math, it's 2 mm. But that seems unrealistic for a student's throw. Maybe the problem is expecting this answer, even if it's not realistic? Or perhaps I misunderstood the concept.Alternatively, maybe it's about the absolute error in angle, but that would require calculating the actual angle and then applying it to the shorter distance, which also leads to the same result. Because if the angle is preserved, then the linear error is proportional to distance.Wait, so in essence, if the rockets have a certain angular precision, then at a shorter distance, the linear error (radius) would be smaller proportionally. So, the student's radius would be (25 m / 12,500,000 m) * 1,000 m. Let me check the units:25 m / 12,500,000 m = 25 / 12,500,000 = 0.000002. Multiply by 1,000 m gives 0.002 m, which is again 2 mm.Hmm, so all the methods lead to 2 millimeters. But 2 millimeters is extremely precise for a student throwing a softball. Even professional pitchers can't hit a target with 2 mm accuracy. So is this a trick question? Or maybe it's just a straightforward proportionality problem, regardless of real-world accuracy?The problem does state "with the same accuracy," so maybe we just have to go with the mathematical answer. So, 1 km at 12,500 km scales to 25 m as follows:(1 km / 12,500 km) * 25 m = (1 / 12,500) * 25,000 m = 2 m. Wait, wait, hold on. Wait, 1 km is 1000 meters. So 1000 m / 12,500,000 m = 0.00008. Then 25 m * 0.00008 = 0.002 m. So 0.002 m is 2 mm. Wait, but if I use km for 25 m, which is 0.025 km, then 0.025 km /12,500 km = 0.000002, and 0.000002 * 1 km = 0.000002 km = 0.002 m. Hmm, same result. So perhaps the answer is 2 millimeters. But is that correct?Wait, maybe the problem is in different units. Wait, the distance is 12,500 km, and the radius is 1 km. So the ratio is 1/12,500. Then, for 25 m, the radius would be 25 m * (1/12,500). 25 divided by 12,500 is 0.002 m, which is 2 millimeters. Yeah, that seems to check out. So even though it's unrealistic, the mathematical answer is 2 mm. Maybe the problem is testing the understanding of proportionality, not real-world feasibility. So the answer is 2 millimeters. But just to confirm, let's see.Alternatively, maybe the question is using the same linear accuracy, which would be 1 km regardless of distance, but that doesn't make sense. If you have the same linear accuracy, then the radius would be 1 km even at 25 m, which is impossible, so that can't be.Alternatively, same relative accuracy, which is the ratio of radius to distance. So for the rocket, it's 1/12500, and for the student, it's radius /25 m = 1/12500, so radius =25/12500 m = 0.002 m = 2 mm. So same ratio. That's the same answer. So regardless of the approach, it's 2 mm.But since the problem mentions "with the same accuracy," which in technical terms could refer to angular accuracy, which would lead to the same result. So yeah, 2 mm is the answer.But just to make sure, perhaps converting units again:Original CEP (Circular Error Probable) is 1 km at 12,500 km. So scaling factor is (25 m) / (12,500,000 m) = 1 / 500,000. So scaling the CEP by the same factor: 1 km * 1 / 500,000 = 0.000002 km = 0.002 m = 2 mm. Yep, same answer. So I think that's it.Final AnswerThe radius of the hit circle would be boxed{0.002} meters, or equivalently boxed{2} millimeters. Since the problem asks for the radius in a box, and the answer can be presented in meters, the appropriate boxed answer is boxed{0.002} meters.However, checking the problem statement again, it says "put your final answer within boxed{}". It might expect the answer in meters, so 0.002 meters, but sometimes problems prefer different units. Let's check units in the question. The target distance is 25 meters, so answer in meters is fine. Alternatively, 2 millimeters. But in the original problem, the radius was given in km, so the answer here should be in meters. So boxed{0.002} is correct. Alternatively, if converted to millimeters, boxed{2}, but the problem didn't specify. Let me check the original problem again. It says "radius was less than 1 km", and asks for the radius in the student's case. Since the student is throwing at 25 meters, answer in meters is more appropriate, so 0.002 meters. But 0.002 meters is 2 millimeters, so maybe both are correct. But to follow the units of the question: original radius in km, target distance in km, student's distance in meters. The answer would be in meters. So yeah, 0.002 meters.Final Answerboxed{0.002}</think>To determine the radius of the hit circle for a student throwing a softball at a target 25 meters away with the same accuracy as the Soviet rocket tests, we use proportionality based on angular accuracy.1. Calculate the angular accuracy for the rocket tests: - The radius of the hit circle for the rockets is 1 km at a distance of 12,500 km. - The angular accuracy (in radians) can be approximated as the ratio of the radius to the distance: [ theta approx frac{1 text{ km}}{12,500 text{ km}} = frac{1}{12,500} text{ radians} ]2. Apply the same angular accuracy to the student's throw: - Convert the student's target distance to kilometers (25 meters = 0.025 km). - Calculate the radius using the same angular accuracy: [ text{Radius} = theta times text{distance} = frac{1}{12,500} times 25 text{ meters} ] - Simplify the calculation: [ text{Radius} = frac{25}{12,500} text{ meters} = 0.002 text{ meters} ]Thus, the radius of the hit circle for the student throwing a softball at a target 25 meters away is boxed{0.002} meters.]

🔑:The ratio of the radius of the hit circle to the distance of the target corresponds to the accuracy of the hit. That is, according to the problem statement, the unknown ratio x: 25 mathrm{~m} should be equal to the ratio 1 mathrm{~km} : 12500 mathrm{km}.x=25 mathrm{~m} cdot frac{1 mathrm{~km}}{12500 mathrm{~km}}=0.002 mathrm{~m}=2 mathrm{~mm}The hit circle of the student would have a radius of 2 mathrm{~mm}.

🧠:[<think>Okay, let's see. The problem is about finding the sum of a and b, where the interval [a, b] contains the root of the function f(x) = ln x + 3x - 8. The key points here are that a and b are positive integers, and the length of the interval b - a is 1. So, we need to determine which integers a and b (with b = a + 1) bracket the root of the function, and then compute a + b.First, let me recall that finding a root in an interval typically involves checking the sign of the function at the endpoints. If the function changes sign between a and b, then by the Intermediate Value Theorem, there must be at least one root in that interval, provided the function is continuous. Since ln x is continuous for x > 0, and 3x - 8 is a linear function (also continuous), the sum f(x) is continuous on its domain, which is x > 0. Therefore, if f(a) and f(b) have opposite signs, there's a root in (a, b).Given that a and b are positive integers and b = a + 1, the possible intervals are [1, 2], [2, 3], [3, 4], etc. Our task is to check these intervals sequentially until we find the one where f(a) and f(b) have opposite signs.Let me compute the values of f(x) at these integers:Starting with x = 1:f(1) = ln(1) + 3*1 - 8 = 0 + 3 - 8 = -5. That's negative.x = 2:f(2) = ln(2) + 3*2 - 8 ≈ 0.6931 + 6 - 8 ≈ -1.3069. Still negative.x = 3:f(3) = ln(3) + 3*3 - 8 ≈ 1.0986 + 9 - 8 ≈ 2.0986. Positive.x = 4:f(4) = ln(4) + 3*4 - 8 ≈ 1.3863 + 12 - 8 ≈ 5.3863. Positive.So, let's check the intervals:Between 1 and 2: f(1) = -5, f(2) ≈ -1.3069. Both negative. No sign change. So no root here.Between 2 and 3: f(2) ≈ -1.3069, f(3) ≈ 2.0986. Here, the function changes from negative to positive. Therefore, by the Intermediate Value Theorem, there is a root in (2, 3). Therefore, a = 2 and b = 3. Hence, a + b = 5, which corresponds to option A.Wait, but let me make sure. Let's check x=2 and x=3 again. f(2) is approximately -1.3069, which is negative, and f(3) is approximately 2.0986, which is positive. So, indeed, the sign changes. Therefore, the root is between 2 and 3. The interval length is 1, so since a and b are integers, it's [2,3]. Therefore, a + b = 2 + 3 = 5, which is option A.But wait, the options given are A:5, B:4, C:3, D:2. So 5 is an option. But let me check once again because sometimes these problems can trick you.Wait, perhaps the problem is written in a way that the interval [a, b] is such that b - a =1, so the interval is of length 1, and we have to find a and b. So, if the root is between 2 and 3, then a=2, b=3, so a + b=5. That's correct.But just to be thorough, maybe I should check x=2.5 to see if the function is actually crossing there. Let's compute f(2.5):f(2.5) = ln(2.5) + 3*(2.5) - 8 ≈ 0.9163 + 7.5 - 8 ≈ 0.4163. Hmm, that's positive. Wait, but f(2) is negative and f(2.5) is positive. So the root is actually between 2 and 2.5. But since our interval has to be [a, b] where a and b are integers with b - a =1, then the interval [2,3] would contain the root even though the crossing is at 2.5. So the correct interval is [2,3], so a=2, b=3, a + b=5, which is option A.But hold on, looking at the answer options, option A is 5. But let me check if maybe there's a miscalculation somewhere. Let me recompute f(2):f(2) = ln(2) + 3*2 -8. ln(2) is approximately 0.693. 3*2=6. So 0.693 +6 -8 = -1.307. That's correct. And f(3) is ln(3)≈1.0986 +9-8=2.0986. Correct. So yes, the sign changes between 2 and3.Wait, but the problem states "its root x0 ∈ [a, b]", so it's within the closed interval [a, b]. However, strictly speaking, the root is in the open interval (a, b). But since a and b are integers, and the function is continuous, even if the root was exactly at an integer point, but since f(2) is negative and f(3) is positive, the root must be between 2 and 3, not at 2 or 3. So [2,3] is still the correct interval.Therefore, the answer should be A:5.But wait, the options are A:5, B:4, C:3, D:2. But in the original problem statement, the interval is [a,b], with b -a =1, and a,b are positive integers. So the possible intervals are [1,2], [2,3], [3,4], etc. We saw that the root is in [2,3], so a=2, b=3. So a +b=5. So option A. But wait, why is there an option B:4? Maybe there is a miscalculation?Wait, maybe I made a mistake in calculating f(3). Let me recalculate f(3):ln(3) is approximately 1.0986. 3*3=9. 9-8=1. So total f(3)=1.0986 +1≈2.0986. That's correct.Wait, maybe the question is in Chinese, and the translation might have some nuances? But I don't think so. The problem seems straightforward.Alternatively, maybe I need to check if there's another interval where the function crosses. Let's check x=1: f(1)= -5, x=2: -1.307, x=3:2.098, x=4:5.386. So the function crosses from negative to positive between 2 and3, so that's the only interval. So the answer is 2+3=5, option A.But let me check if maybe there's a trick here. The function is f(x) = ln x +3x -8. Let's analyze its behavior. The derivative f’(x) = 1/x +3, which is always positive for x>0. Therefore, the function is strictly increasing. Therefore, there can be only one root. So once we find an interval where the function crosses from negative to positive, that's the only interval. So since between 2 and3 it crosses, that's the only interval. Therefore, a=2, b=3, a +b=5, option A.Therefore, the answer should be A.But why does the option B:4 exist? Maybe a and b are 1 and3? But b -a has to be 1. So if a=1, b=2, but that interval doesn't contain the root. If a=2, b=3, which is the correct interval. So 2+3=5. So answer is A.Wait, maybe the problem is written differently? Let me check again."Given the function f(x) = ln x + 3x - 8, its root x0 ∈ [a, b], and b - a =1 (a, b ∈ N⁺), then a + b=?"Yes, so as per the problem, the root is in [a, b], which is of length 1, with a and b positive integers. So the possible pairs are (1,2), (2,3), (3,4), etc. Since the function is increasing, it crosses from negative to positive only once. Since at x=2, it's negative, at x=3, it's positive. Therefore, the interval is [2,3], so a=2, b=3, sum is 5. Answer A.Therefore, I think the correct answer is A:5.Final Answerboxed{A}</think>Given the function ( f(x) = ln x + 3x - 8 ), we need to find the interval ([a, b]) where the root ( x_0 ) lies, with ( b - a = 1 ) and ( a, b in mathbb{N}^+ ). We then determine ( a + b ).1. Evaluate the function at integer points: - ( f(1) = ln 1 + 3 cdot 1 - 8 = 0 + 3 - 8 = -5 ) (negative) - ( f(2) = ln 2 + 3 cdot 2 - 8 approx 0.6931 + 6 - 8 = -1.3069 ) (negative) - ( f(3) = ln 3 + 3 cdot 3 - 8 approx 1.0986 + 9 - 8 = 2.0986 ) (positive) - ( f(4) = ln 4 + 3 cdot 4 - 8 approx 1.3863 + 12 - 8 = 5.3863 ) (positive)2. Check for sign changes: - Between ( x = 1 ) and ( x = 2 ): Both ( f(1) ) and ( f(2) ) are negative. - Between ( x = 2 ) and ( x = 3 ): ( f(2) ) is negative and ( f(3) ) is positive, indicating a sign change.3. Conclusion: - The interval ([2, 3]) contains the root ( x_0 ) since the function changes sign from negative to positive. - Therefore, ( a = 2 ) and ( b = 3 ), leading to ( a + b = 2 + 3 = 5 ).The final answer is (boxed{A}).,<think>Okay, let's see. I need to find the value of a + b where the interval [a, b] contains a root of the function f(x) = ln x + 3x - 8. The problem states that b - a = 1 and both a and b are positive integers. The options are A:5, B:4, C:3, D:2. Hmm, let's break this down step by step.First, I remember that to find an interval where a root exists, we can use the Intermediate Value Theorem. That theorem says that if a function is continuous on [a, b] and takes on values of opposite sign at the endpoints, then there is at least one root in the interval. Since ln x is continuous for x > 0 and 3x -8 is a linear function, the entire function f(x) is continuous for x > 0. So, we just need to find two consecutive integers a and b (since b - a =1) where f(a) and f(b) have opposite signs.Given that a and b are positive integers, and their difference is 1, possible pairs could be (1,2), (2,3), (3,4), (4,5), etc. But since the answer choices are up to 5, maybe the root is somewhere in the lower integers.Let me start by testing some integer values.First, let's try x=1:f(1) = ln 1 + 3*1 -8 = 0 + 3 -8 = -5. That's negative.x=2:f(2) = ln 2 + 3*2 -8 ≈ 0.6931 + 6 -8 ≈ -1.3069. Still negative.x=3:f(3) = ln 3 + 9 -8 ≈ 1.0986 +1 ≈ 2.0986. Positive.So between x=2 and x=3, the function goes from negative to positive. Therefore, by the Intermediate Value Theorem, there must be a root in (2,3). So the interval [a, b] would be [2,3], so a=2, b=3, hence a + b =5. Wait, but option A is 5. But let me check if there might be another interval.Wait, maybe the problem is referring to a different interval? Let's check higher integers just in case.x=4: f(4) = ln4 +12 -8 ≈1.3863 +4 ≈5.3863. Positive. So between 3 and 4, the function stays positive. Similarly, x=5: ln5 +15 -8≈1.6094 +7≈8.6094. Still positive. So from x=3 onward, it's increasing but already positive. What about lower integers?x=1: f(1)= -5, as before. x=2: -1.3069. So between 1 and 2, both negative. So the only sign change is between 2 and 3. Therefore, the interval [2,3] must contain the root. Therefore a=2, b=3, a + b=5. So the answer should be A:5. But let me double-check.Wait, the problem says "its root x0 ∈ [a, b], and b - a=1 (a, b ∈ ℕ⁺)", so a and b must be positive integers with difference 1, and the root is in that interval. The answer is the sum a + b. So in our case, 2 +3=5, which is option A. But the answer options given are A:5, B:4, C:3, D:2.Wait, but wait a second. Maybe I made a mistake in calculating f(3). Let me check again.f(3) = ln3 +3*3 -8. ln3 is approximately 1.0986, 3*3=9, so 1.0986 +9 -8=1.0986 +1=2.0986. Yes, that's correct. So positive.f(2) = ln2 +6 -8≈0.6931 -2≈-1.3069. Negative. So yes, the sign changes between 2 and 3. Therefore, the interval [2,3] must contain the root. So a=2, b=3, sum is 5. So answer is A. But the options list A as 5, which would be correct.But wait, the problem says "b - a=1 (a, b ∈ ℕ⁺)". So since 3 -2=1, yes, that's correct. So then a + b=5. Therefore, the correct answer should be A:5.But let me check if the problem is perhaps in another interval. Maybe I made a miscalculation.Wait, let's test x=1.5, just to see how the function behaves. Maybe the root is between 1 and 2? But since at x=1, f(x)=-5; x=2, f≈-1.3069. Still negative. So between 1 and 2, function is negative. Then x=3 is positive. So the crossing is between 2 and3. So [2,3], sum 5. So answer is A.But the options include B:4. Let me check if maybe there's a miscalculation here.Wait, perhaps the problem is presented differently? Let me confirm:The function is f(x) = ln x + 3x -8. So, the root is where ln x +3x -8=0.Alternatively, maybe there's a mistake in my calculations. Let me check x=2 again:ln2 ≈0.6931, 3*2=6, so 0.6931+6-8≈-1.3069. Correct.x=3: ln3≈1.0986, 3*3=9, so 1.0986 +9 -8≈2.0986. Correct.So, yes, sign changes between 2 and3.But maybe the problem expects the interval to be [a, b] where the root is between a and b, but a and b are integers with b -a=1. So yes, 2 and3. So a=2, b=3. Sum is5. So answer A.But the options are:A:5B:4C:3D:2So unless I made a mistake, answer is A.Wait, but hold on. Let me check x=2 again. Maybe if I compute f(2) more accurately. Let's use more decimal places.ln2 is approximately 0.69314718056.So f(2)=0.69314718056 + 6 -8=0.69314718056 -2≈-1.30685281944. So that's correct.f(3)=ln3≈1.098612288668, 3*3=9, so 1.098612288668 +9 -8=1.098612288668 +1≈2.098612288668. So positive.Therefore, the root is between 2 and3, so interval [2,3], a+b=5.Hmm. Therefore, unless the question is tricking me, the answer should be A. But why is the first option A:5? Wait, maybe I need to check another interval. Wait, could there be another interval where the function crosses zero? For example, maybe between 1 and2? But as x approaches 0 from the right, ln x approaches negative infinity, and 3x -8 approaches -8, so f(x) approaches negative infinity. At x=1, f(1)=0 +3 -8=-5. At x=2, f(2)≈-1.3069. So between 1 and2, the function is still negative. From x=2 to x=3, it goes from negative to positive, so crosses zero once. Then, from x=3 onwards, f(x) is positive and increasing because the derivative f’(x)=1/x +3, which is always positive for x>0. Therefore, the function is strictly increasing. Therefore, only one root exists, between 2 and3.Therefore, the answer is A:5.But let me check the answer options again. The user lists the options as A:5, B:4, C:3, D:2. So A is 5. But maybe the original problem in Chinese had a different interval? Wait, no, the problem states that b -a=1, and a, b are positive integers. So 2 and3 are consecutive integers, difference1. Hence sum is5. So answer is A. But why is the option B:4 here? Maybe I'm missing something.Wait, let me re-express the function. Maybe I misread the function. The problem says "ln x +3x -8". Yes. So, let me check x=1, 2,3 again.Wait, maybe there's another root in a higher interval? Let's check x=4, even though the function is increasing.f(4)=ln4 +12 -8≈1.386 +4=5.386. Positive.x=5≈ln5≈1.609, 15-8=7, total≈8.609. Still positive.So after x=3, the function continues to increase. So only one root between2 and3.Alternatively, maybe the problem uses natural logarithm with base e? Well, ln x is natural logarithm. Yes. So calculations are correct.Alternatively, maybe the problem had a typo, but as given, the answer should be A.Wait, but let me check the possibility that the answer is 4. If a=1 and b=2, then a +b=3. If a=2 and b=3, sum=5. If a=3 and b=4, sum=7. So none of the other intervals give sum 4. Unless the interval is [1,2], but that would sum to3, which is option C. So 3 is C, which isn't our case. The only interval where the sign changes is [2,3], sum5. So answer is A.Wait, but let me check once more. Maybe my calculation at x=2 is wrong. Let's compute f(2) again:ln(2) is approximately 0.6931. 3*2=6. So 0.6931 +6 -8=0.6931 -2= -1.3069. Correct.Wait, if the problem had been, say, lnx +3x -5, then maybe the root would be between1 and2. But with -8, it's definitely higher.Wait, let me see. Let's try solving ln x +3x -8=0 numerically.Let’s define the function f(x)=lnx +3x -8.We know that at x=2, f(x)=~ -1.3069At x=3, f(x)=~2.0986We can use linear approximation to estimate the root.The difference between x=2 andx=3 is1. The function increases by approximately 2.0986 - (-1.3069)=3.4055 over that interval.We need to find c such that f(c)=0.Starting at x=2, f(x)=-1.3069. The amount needed to reach zero is1.3069.So the fraction of the interval needed is1.3069 /3.4055≈0.3836.Therefore, the root is approximately at x=2 +0.3836≈2.3836. So indeed, between2 and3.Thus, the interval is [2,3], sum5. So answer is A.But let me check why the options include B:4. Maybe the problem had a different function? Wait, no, the user provided the function as lnx +3x -8. Maybe in Chinese, the comma is a decimal point? Wait, no, the problem is written in English here. So no, probably not.Alternatively, maybe the problem is stating that b -a=1, but a and b are the floor and ceiling of the root? Wait, but the root is approximately2.3836, so floor is2, ceiling is3. So the interval [2,3], sum5. So answer is A.Therefore, I think the correct answer is A:5. So why is the option B:4 present? Maybe I made a miscalculation. Let me check once again.Wait, could the interval be [1,2], but with b -a=1, but since the function is negative at both ends, no. [2,3] is the only interval with a sign change. So unless the problem is considering the interval around the root where the function is closest to zero, but the problem states that the root is in [a,b], so by IVT, [2,3] is the correct interval.Alternatively, maybe the answer is different. Let me check other integer pairs:Suppose a=1, b=2. f(1)=-5, f(2)≈-1.3, both negative. No root.a=2, b=3: f(2)≈-1.3, f(3)≈2.09, so root here.a=3, b=4: both positive. No root.Thus, only possible interval is [2,3], so sum5.So the answer is A.Final Answerboxed{A}</think>Given the function ( f(x) = ln x + 3x - 8 ), we need to find the interval ([a, b]) where ( b - a = 1 ) (with ( a ) and ( b ) being positive integers) that contains the root ( x_0 ). We then need to determine ( a + b ).To find the interval, we use the Intermediate Value Theorem, which states that if a function is continuous on ([a, b]) and takes on values of opposite signs at the endpoints, there is a root in that interval. 1. Evaluate ( f(x) ) at consecutive integers: - ( f(1) = ln 1 + 3 cdot 1 - 8 = 0 + 3 - 8 = -5 ) (negative) - ( f(2) = ln 2 + 3 cdot 2 - 8 approx 0.6931 + 6 - 8 approx -1.3069 ) (negative) - ( f(3) = ln 3 + 3 cdot 3 - 8 approx 1.0986 + 9 - 8 approx 2.0986 ) (positive) - ( f(4) = ln 4 + 3 cdot 4 - 8 approx 1.3863 + 12 - 8 approx 5.3863 ) (positive)2. The function changes sign between ( x = 2 ) and ( x = 3 ), indicating a root in the interval ([2, 3]).3. Therefore, ( a = 2 ) and ( b = 3 ), leading to ( a + b = 2 + 3 = 5 ).The correct answer is (boxed{A}).]

🔑:The domain of the function f(x) = ln x + 3x - 8 is x > 0, and it is a continuous increasing function. We have f(2) = ln 2 + 6 - 8 0, thus f(2)f(3) < 0. Since the root x_0 of the function f(x) = ln x + 3x - 8 is in the interval [a, b] and b-a=1 (a, b in mathbb{N}^+), we can determine that a=2 and b=3, therefore, a+b=5. Hence, the correct answer is: boxed{A}. This problem involves applying the theorem of determining the roots of a function and tests computational skills.

🧠:[<think>Okay, let's see. I need to find the explicit expression for a linear function f(x) that satisfies f(2) = 1 and f(3) = -5. Hmm, linear function, so it should be in the form f(x) = mx + b, right? Where m is the slope and b is the y-intercept. First, maybe I should find the slope. The slope between two points (x1, y1) and (x2, y2) is given by (y2 - y1)/(x2 - x1). Here, the two points are x=2 with f(2)=1 and x=3 with f(3)=-5. So substituting those into the formula, that would be (-5 - 1)/(3 - 2). Let me compute that. The numerator is -5 - 1, which is -6, and the denominator is 3 - 2, which is 1. So the slope m is -6/1 = -6. Okay, so m = -6.Now that I have the slope, I need to find the y-intercept b. To do that, I can use one of the given points and plug it into the equation f(x) = mx + b. Let me use the point (2, 1). So substituting x=2 and f(x)=1 into the equation:1 = (-6)(2) + bCalculating that, -6 times 2 is -12. So:1 = -12 + bTo solve for b, add 12 to both sides:1 + 12 = bWhich gives b = 13. Hmm, let me check if that's right. If b is 13, then the equation is f(x) = -6x + 13. Let's verify this with the other point (3, -5). Plugging x=3 into the equation:f(3) = -6(3) + 13 = -18 + 13 = -5. Yep, that matches the given value. So that's correct.Alternatively, I could have used the other point (3, -5) to solve for b. Let me try that as a check. Using x=3 and f(x)=-5:-5 = (-6)(3) + bCalculating, -6*3 is -18. So:-5 = -18 + bAdding 18 to both sides:-5 + 18 = b => 13 = bSame result. So that's consistent. Therefore, the equation is f(x) = -6x + 13.Wait, but let me make sure I didn't make any calculation errors. Let's go through the steps again. Starting with slope: between (2,1) and (3,-5). The change in y is -5 - 1 = -6. Change in x is 3 - 2 = 1. So slope m = -6/1 = -6. Correct. Then using point (2,1):1 = (-6)(2) + b => 1 = -12 + b => b = 13. Correct. Then the function is f(x) = -6x +13. Testing for x=2: -6*2 +13 = -12 +13 =1. Good. For x=3: -6*3 +13 = -18 +13 = -5. Perfect. So all checks out. Therefore, the explicit expression is f(x) = -6x +13.I think that's it. Alternatively, could I have approached this problem differently? Maybe using point-slope form? Let's try that to see if it gives the same answer. The point-slope form is y - y1 = m(x - x1). Using point (2,1) and slope -6:y - 1 = -6(x - 2)Expanding that: y -1 = -6x +12. Then add 1 to both sides: y = -6x +13. Same result. So that works too. Similarly, if I use the other point (3,-5):y - (-5) = -6(x -3) => y +5 = -6x +18. Subtract 5: y = -6x +13. Same equation. So regardless of the method, the answer is f(x) = -6x +13. Just to ensure there's no alternative interpretation, the problem says it's a linear function. A linear function in this context is a first-degree polynomial, so there's no ambiguity. So, all steps seem correct. Therefore, the answer should be f(x) = -6x +13.Final AnswerThe explicit expression for ( f(x) ) is boxed{-6x + 13}.</think>Given a linear function ( f(x) ) satisfying ( f(2) = 1 ) and ( f(3) = -5 ), we need to find the explicit expression for ( f(x) ).A linear function is of the form ( f(x) = mx + b ), where ( m ) is the slope and ( b ) is the y-intercept.First, we calculate the slope ( m ) using the two points ((2, 1)) and ((3, -5)):[m = frac{f(3) - f(2)}{3 - 2} = frac{-5 - 1}{3 - 2} = frac{-6}{1} = -6]Next, we use the point-slope form of a linear equation to find ( b ). Using the point ((2, 1)):[1 = -6 cdot 2 + b]Solving for ( b ):[1 = -12 + b implies b = 13]We verify this result using the other point ((3, -5)):[-5 = -6 cdot 3 + b]Solving for ( b ):[-5 = -18 + b implies b = 13]Both methods confirm that ( b = 13 ). Therefore, the explicit expression for ( f(x) ) is:[f(x) = -6x + 13][boxed{-6x + 13}],<think>Okay, so I need to find the explicit expression for a linear function f(x) that satisfies f(2) = 1 and f(3) = -5. Hmm, let's think. Since it's a linear function, I remember that linear functions can be written in the form f(x) = mx + b, where m is the slope and b is the y-intercept. Right? So my goal is to find the values of m and b that make this function pass through the points (2, 1) and (3, -5).First, I need to find the slope, m. The formula for slope between two points (x1, y1) and (x2, y2) is m = (y2 - y1)/(x2 - x1). Let me apply that here. So here, x1 is 2, y1 is 1, x2 is 3, and y2 is -5. Plugging those into the formula: m = (-5 - 1)/(3 - 2) = (-6)/1 = -6. So the slope is -6. Got that part.Now that I have the slope, I need to find the y-intercept, b. To do this, I can use one of the given points and plug the values into the equation f(x) = mx + b. Let's use the point (2, 1). So when x is 2, f(x) is 1. Plugging into the equation: 1 = (-6)(2) + b. Let me calculate that: -6 times 2 is -12. So 1 = -12 + b. To solve for b, I can add 12 to both sides: 1 + 12 = b, which is 13. So b is 13.Wait, let me check that again. If the slope is -6 and we plug in x=2, then the equation should be 1 = (-6)(2) + b. So that's 1 = -12 + b. Adding 12 to both sides: 1 + 12 = 13 = b. Yeah, that seems right. Let me verify with the other point to make sure. Let's use x=3, f(x)= -5. Plugging into f(x) = -6x + 13. So -6*(3) + 13 = -18 + 13 = -5. Which matches the given value. Perfect, that checks out. So both points satisfy the equation. So the function should be f(x) = -6x + 13.But wait, let me go through the steps again to make sure I didn't make any mistakes. First, finding the slope. Between (2,1) and (3,-5), the difference in y is -5 -1 = -6, difference in x is 3 -2 =1. So slope m = -6/1 = -6. That's correct. Then using point-slope form or plugging into the equation. Alternatively, I could use the point-slope formula. Let me try that as another method.The point-slope form is y - y1 = m(x - x1). Let's use the point (2,1). So plugging in m=-6, x1=2, y1=1, we get y -1 = -6(x -2). Distribute the -6: y -1 = -6x +12. Then add 1 to both sides: y = -6x +13. Same result. So that's consistent. Alternatively, using the other point (3, -5): y - (-5) = -6(x -3). That's y +5 = -6x +18. Subtract 5: y = -6x +13. Again, same answer. So that seems solid.Alternatively, if I didn't remember the formulas, maybe I can think of it as a linear function changing by m each time x increases by 1. Since when x goes from 2 to 3 (an increase of 1), the function goes from 1 to -5 (a decrease of 6). So the slope is change in y over change in x, which is -6/1 = -6, same as before. Then starting from x=2, f(2)=1. So for each x less than 2, we would have to add 6 for each unit decrease in x. But since we need the equation in terms of x, maybe it's easier to use the formula.But either way, all methods are pointing to the same answer: f(x) = -6x +13. Let me just visualize the graph to confirm. If x=0, then f(0) = 13. So the y-intercept is at (0,13). Then with a slope of -6, each time x increases by 1, y decreases by 6. So from (0,13) to (1,7), then (2,1), (3,-5), which matches the given points. Perfect. So that seems correct.I think that's all. The key steps were calculating the slope using the two given points, then using one of the points to solve for the y-intercept. Both methods (slope-intercept form and point-slope form) gave the same equation, and checking with both points confirmed that the equation works. So I'm confident that the explicit expression is f(x) = -6x +13.Final AnswerThe explicit expression for ( f(x) ) is boxed{-6x + 13}.</think>Given a linear function ( f(x) ) satisfying ( f(2) = 1 ) and ( f(3) = -5 ), we need to find the explicit expression for ( f(x) ).1. Calculate the slope (m): The slope ( m ) between the points ((2, 1)) and ((3, -5)) is given by: [ m = frac{y_2 - y_1}{x_2 - x_1} = frac{-5 - 1}{3 - 2} = frac{-6}{1} = -6 ]2. Determine the y-intercept (b): Using the point-slope form with the point ((2, 1)): [ y - y_1 = m(x - x_1) ] Substituting ( m = -6 ), ( x_1 = 2 ), and ( y_1 = 1 ): [ y - 1 = -6(x - 2) ] Simplifying this equation: [ y - 1 = -6x + 12 implies y = -6x + 13 ]3. Verification with the other point: Using the point ((3, -5)) in the equation ( f(x) = -6x + 13 ): [ f(3) = -6(3) + 13 = -18 + 13 = -5 ] This confirms the equation is correct.Thus, the explicit expression for ( f(x) ) is (boxed{-6x + 13}).]

🔑:Let's assume the form of the function is f(x) = kx + b (where k neq 0).Since f(2) = 1 and f(3) = -5, we can set up the following system of equations:[begin{cases}2k + b = 1, 3k + b = -5.end{cases}]Subtracting the first equation from the second, we get:[3k + b - (2k + b) = -5 - 1,]which simplifies to:[k = -6.]Using k = -6 in the first equation:[2(-6) + b = 1,]we find:[b = 1 + 2 cdot 6 = 13.]Therefore, the explicit expression for f(x) is:[f(x) = -6x + 13. quad boxed{f(x) = -6x + 13}]

🧠:[<think>Okay, let's go through each statement one by one and see if they're correct or not. I'll start with statement ①: "Rational numbers can be divided into positive rational numbers and negative rational numbers." Hmm, so rational numbers are numbers that can be expressed as a fraction a/b where a and b are integers and b is not zero. But wait, zero is also a rational number, right? Because 0 can be written as 0/1. So if we're dividing all rational numbers into positive and negative, where does zero fit? It's neither positive nor negative. So statement ① is not entirely correct because it doesn't account for zero. Therefore, ① is incorrect.Moving on to statement ②: "If |a| = -a, then a is a negative number." Let's recall that the absolute value of a number is always non-negative. So |a| = -a implies that -a must be non-negative. Therefore, -a ≥ 0, which means that a ≤ 0. But does this mean a is negative? Well, if a is zero, then |a| = 0 and -a is also 0, so |a| = -a holds when a is zero. So the statement says that a is a negative number, but a could be zero. Therefore, the conclusion that a is negative is not necessarily true. So statement ② is incorrect.Statement ③: "The coefficient of the quadratic term of the polynomial 2x³ - 3xy + 3y is 3." First, let's identify the quadratic term in the polynomial. A quadratic term is a term with degree 2. The given polynomial is 2x³ which is cubic (degree 3), then -3xy which is degree 2 (since x has exponent 1 and y has exponent 1, adding to 2), and 3y which is degree 1. So the quadratic term is -3xy, and its coefficient is -3, not 3. Therefore, statement ③ is incorrect.Statement ④: "All rational numbers can be represented by points on the number line." Well, rational numbers are indeed real numbers, and every real number corresponds to a point on the number line. Since rational numbers are a subset of real numbers, they can all be represented on the number line. So this statement is correct. Therefore, ④ is correct.Statement ⑤: "A pentagonal prism has 7 faces, 10 vertices, and 15 edges." Let's check each part. A pentagonal prism has two pentagonal bases and five rectangular lateral faces. So total faces = 2 + 5 = 7. That's correct. For vertices, each pentagon has 5 vertices, so two pentagons would have 5*2 = 10 vertices. That's correct. Edges: each pentagon has 5 edges, so two pentagons contribute 10 edges, but since they are connected by the rectangular faces, there are also 5 edges connecting the two bases. So total edges = 10 + 5 = 15. Wait, hold on, actually, in a prism, the number of edges should be 3 times the number of sides of the base. For a pentagonal base, edges = 5 (from top base) + 5 (from bottom base) + 5 (connecting the bases) = 15. So that's correct. However, according to Euler's formula for polyhedrons: V - E + F = 2. Let's check with V=10, E=15, F=7. 10 - 15 + 7 = 2. 10 -15 is -5, -5 +7=2. Yes, that satisfies Euler's formula. So all parts of statement ⑤ are correct. Therefore, ⑤ is correct.So summarizing:① Incorrect② Incorrect③ Incorrect④ Correct⑤ CorrectSo there are two correct statements: ④ and ⑤. Therefore, the answer should be option B: 2 statements.Final Answerboxed{A}Wait, hold on, wait, I thought statements ④ and ⑤ were correct. That's two correct statements. So the answer should be B. But the initial summary in the Final Answer was boxed{A}, which is 1 statement. But according to the reasoning, ④ and ⑤ are correct, which is two. So why did I put A?Wait, let me check again.Statement ④: Correct. All rational numbers can be represented on the number line. Yes, because they are real numbers.Statement ⑤: Correct. Faces: 2 pentagons + 5 rectangles =7. Vertices: 5*2=10. Edges: Each pentagon has 5 edges, but in the prism, each edge of the pentagon is connected. Wait, actually, for a prism, the number of edges is 3n where n is the number of sides. For a pentagon, n=5, so edges=15. And Euler's formula holds: 10-15+7=2. So ⑤ is correct.So two correct statements. Therefore, the answer is B: 2 statements. But why did I write boxed{A}? That must have been a mistake when I was concluding. So I need to correct that. So the correct answer should be boxed{B}.But let me check once more to be sure.①: Incorrect (misses zero)②: Incorrect (a could be zero)③: Incorrect (coefficient is -3)④: Correct⑤: CorrectSo two correct statements. So answer is B. Therefore, the initial conclusion of A was an error. So the correct final answer is B.Final Answerboxed{B}</think>Let's analyze each statement one by one:1. Statement ①: "Rational numbers can be divided into positive rational numbers and negative rational numbers." - Rational numbers include zero, which is neither positive nor negative. Therefore, this statement is incorrect.2. Statement ②: "If ( |a| = -a ), then ( a ) is a negative number." - ( |a| = -a ) implies ( a leq 0 ). Since ( a ) could be zero, this statement is incorrect.3. Statement ③: "The coefficient of the quadratic term of the polynomial ( 2x^3 - 3xy + 3y ) is ( 3 )." - The quadratic term is ( -3xy ) (degree 2), so the coefficient is ( -3 ), not ( 3 ). This statement is incorrect.4. Statement ④: "All rational numbers can be represented by points on the number line." - All rational numbers are real numbers and can be represented on the number line. This statement is correct.5. Statement ⑤: "A pentagonal prism has ( 7 ) faces, ( 10 ) vertices, and ( 15 ) edges." - A pentagonal prism has 2 pentagonal bases and 5 rectangular faces (total 7 faces), 10 vertices (5 from each base), and 15 edges (5 edges on each base and 5 connecting edges). Euler's formula ( V - E + F = 2 ) holds (10 - 15 + 7 = 2). This statement is correct.Summarizing the results: - Incorrect statements: ①, ②, ③ - Correct statements: ④, ⑤ The number of correct statements is 2. Therefore, the answer is (boxed{B}).,<think>Okay, let's tackle this problem step by step. There are five statements here, and I need to figure out which ones are correct. The answer choices tell me that somewhere between 1 and 4 statements are correct, but I need to check each one carefully. Let me go through them one by one.Statement ①: Rational numbers can be divided into positive rational numbers and negative rational numbers.Hmm, rational numbers are numbers that can be expressed as a fraction a/b where a and b are integers and b ≠ 0. So, positive rationals would be those where a and b are both positive or both negative, and negative rationals would be where one is positive and the other is negative. But wait, does this cover all rational numbers? What about zero? Zero is a rational number (since 0 = 0/1), but it's neither positive nor negative. So, if we divide rational numbers into positive and negative, we're excluding zero. Therefore, this statement is incorrect because it doesn't account for zero. So ① is false.Statement ②: If |a| = -a, then a is a negative number.The absolute value of a, |a|, is always non-negative. The equation |a| = -a implies that -a is non-negative, right? So, if |a| = -a, then -a ≥ 0, which means a ≤ 0. But does that make a necessarily negative? Wait, if a is zero, then |a| = 0, and -a is also 0. So in that case, |a| = -a would hold true even when a is zero. But zero isn't a negative number. Therefore, the conclusion that a is a negative number isn't entirely correct because a could be zero. So ② is false.Statement ③: The coefficient of the quadratic term of the polynomial 2x³ - 3xy + 3y is 3.Let me look at the polynomial: 2x³ - 3xy + 3y. A quadratic term is a term where the sum of the exponents of the variables is 2. Let's check each term:- 2x³: The exponent is 3, so that's cubic, not quadratic.- -3xy: The exponents are 1 for x and 1 for y, so 1+1=2. That's a quadratic term. The coefficient here is -3.- 3y: The exponent is 1, so that's linear.Therefore, the quadratic term is -3xy, and its coefficient is -3, not 3. So the statement claims it's 3, which is wrong. Hence, ③ is false.Statement ④: All rational numbers can be represented by points on the number line.Yes, the number line includes all real numbers, and rational numbers are a subset of real numbers. Each rational number can be represented as a fraction, which corresponds to a specific point on the number line. For example, 1/2 is halfway between 0 and 1, -3/4 is three-quarters of the way from 0 to -1, etc. So every rational number has a unique position on the number line. Therefore, ④ is true.Statement ⑤: A pentagonal prism has 7 faces, 10 vertices, and 15 edges.Let me recall what a pentagonal prism is. It's a three-dimensional shape with two pentagonal bases connected by rectangular faces. Let's verify each part:- Faces: The two pentagonal bases and five rectangular sides. So 2 + 5 = 7 faces. That matches the statement.- Vertices: Each pentagon has 5 vertices. Since there are two pentagons, that's 5*2 = 10 vertices. That also matches.- Edges: Each pentagon has 5 edges. Since there are two pentagons, that's 5*2 = 10 edges. Additionally, the sides are connected by rectangles, which connect corresponding vertices of the two pentagons. There are 5 such vertical edges. So total edges are 10 + 5 = 15. That matches the statement as well.So all three attributes (faces, vertices, edges) given in statement ⑤ are correct. Therefore, ⑤ is true.Now, let's recap which statements are correct:- ①: False (doesn't include zero)- ②: False (could be zero)- ③: False (coefficient is -3)- ④: True- ⑤: TrueSo there are 2 correct statements: ④ and ⑤. Therefore, the answer should be B: 2 statements.Wait, just to double-check, especially on statement ⑤. Let me confirm Euler's formula for polyhedrons, which states that V - E + F = 2. For the pentagonal prism:V = 10, E = 15, F = 7.Plug into Euler's formula: 10 - 15 + 7 = 2. Indeed, 2 = 2. So Euler's formula holds, which supports that the counts are correct. So ⑤ is definitely correct.Yes, so only statements ④ and ⑤ are correct, so the number of correct statements is 2. The answer is option B.Final Answerboxed{A}Wait, hold on! Wait, this contradicts my previous conclusion. Wait, I thought ④ and ⑤ are correct, which would be 2 statements, so answer B. But when I looked at the original problem's options, let me check again.Wait, the options are:A: 1 statementB: 2 statementsC: 3 statementsD: 4 statementsBut in my reasoning, I concluded that ④ and ⑤ are correct, so that's 2 statements. So answer B. But the initial assistant thought process wrote Final Answer as boxed{A}. That must be a mistake. So the user is presenting the assistant's thought process, but in the hidden final answer, it's boxed{A}. That seems like an error. Wait, maybe I made a mistake in my reasoning?Wait, let me check again:Statement ④: All rational numbers can be represented on the number line. Yes, because rational numbers are real numbers, and every real number has a place on the number line. So ④ is correct.Statement ⑤: Pentagonal prism has 7 faces, 10 vertices, 15 edges. As per Euler's formula, V - E + F = 2: 10 -15 +7=2, which works. So ⑤ is correct.Therefore, two statements are correct, answer B. So why did I initially write boxed{A}? That must have been an error. Therefore, the correct answer should be B.But let me check once more. Wait, perhaps I made a mistake in counting?Wait, statement ①: Rational numbers divided into positive and negative. But zero is neither, so statement ① is false.Statement ②: |a| = -a. Then a ≤0. But since a can be zero, not necessarily negative. So ② is false.Statement ③: Coefficient of quadratic term is 3. But quadratic term is -3xy, coefficient -3. So ③ is false.④: Correct.⑤: Correct.Thus, 2 correct statements. Answer is B. Therefore, the final answer should be boxed{B}.But the original assistant's thought process had boxed{A} at the end. That's an error. Therefore, I need to correct that. Wait, maybe the user made a mistake when writing the problem?Wait, in the original problem statement, maybe statement ⑤ says 7 faces, 10 vertices, and 15 edges. Let me check again:"A pentagonal prism has 7 faces, 10 vertices, and 15 edges."Faces: 2 pentagons + 5 rectangles = 7. Correct.Vertices: 5*2=10. Correct.Edges: Each pentagon has 5 edges, so 10 edges from both, plus 5 edges connecting the two pentagons. Total 15. Correct.Euler's formula: 10 -15 +7=2. Correct.So ⑤ is correct. So two correct answers. Answer is B.Wait, but in the options given by the user, options are A, B, C, D. So unless the problem was presented differently... Wait, the user wrote:"⑤ A pentagonal prism has 7 faces, 10 vertices, and 15 edges."Wait, actually, in a pentagonal prism, the number of edges should be calculated as follows: each pentagon has 5 edges, but in the prism, each edge of the pentagon is connected to the corresponding edge of the other pentagon via a rectangle. Wait, actually, each original edge of the pentagon (5 per base) is connected by a rectangle, but the vertical edges are the ones connecting the vertices.Wait, perhaps there's confusion here. Let me re-verify:A pentagonal prism has:- Two pentagonal bases.- Each base has 5 edges, but in the prism, these are considered as the top and bottom edges. Then, the lateral edges (connecting corresponding vertices of the two pentagons) are 5.Thus, total edges: 5 (top) + 5 (bottom) + 5 (sides) = 15. Correct.Vertices: Each pentagon has 5 vertices, so 5*2=10. Correct.Faces: 2 pentagons + 5 rectangles = 7. Correct.Euler's formula: V - E + F = 10 -15 +7 = 2. So it satisfies Euler's formula. Therefore, the statement ⑤ is correct.Therefore, statements ④ and ⑤ are correct. So two correct statements. Answer B.But the user initially wrote in the problem statement that the answer is A? No, the user just provided the options. So in that case, the correct answer is B. So the Final Answer should be boxed{B}.But the initial thought process ended with boxed{A}, which is incorrect. So I need to correct that.Final Answerboxed{B}</think>Let's analyze each statement one by one:1. Statement ①: Rational numbers can be divided into positive rational numbers and negative rational numbers. - This statement is false because rational numbers also include zero, which is neither positive nor negative.2. Statement ②: If (|a| = -a), then (a) is a negative number. - This statement is false because (a) could be zero, which is not a negative number.3. Statement ③: The coefficient of the quadratic term of the polynomial (2x^3 - 3xy + 3y) is 3. - This statement is false. The quadratic term is (-3xy), so the coefficient is (-3), not 3.4. Statement ④: All rational numbers can be represented by points on the number line. - This statement is true because every rational number is a real number and can be represented on the number line.5. Statement ⑤: A pentagonal prism has 7 faces, 10 vertices, and 15 edges. - This statement is true. A pentagonal prism has 2 pentagonal faces and 5 rectangular faces (total 7 faces), 10 vertices (5 from each pentagon), and 15 edges (5 from each pentagon plus 5 connecting edges). Euler's formula (V - E + F = 2) is satisfied: (10 - 15 + 7 = 2).The correct statements are ④ and ⑤. Therefore, the number of correct statements is 2.[boxed{B}]]

🔑:To evaluate the correctness of each statement, we proceed as follows:- Statement ①: Rational numbers include positive rational numbers, negative rational numbers, and also 0. Therefore, the statement that rational numbers can only be divided into positive and negative rational numbers is incorrect because it omits 0.- Statement ②: The absolute value of a number, |a|, is defined as a if a geqslant 0 and -a if a < 0. Therefore, if |a| = -a, it implies that a must be less than or equal to 0 (i.e., a is non-positive). This means a could be negative or 0, making the statement incorrect since it asserts a must be a negative number without considering a could be 0.- Statement ③: The polynomial given is 2x^{3}-3xy+3y. The coefficient of the quadratic term would be associated with x^2, but there is no x^2 term in this polynomial. The term -3xy is not a quadratic term in x; it is a term of degree 2 when considering both variables, but not a quadratic term in the sense usually meant (which would be x^2). Therefore, saying the coefficient of the quadratic term is 3 is incorrect; there is no quadratic term in x alone in this polynomial.- Statement ④: It is a well-established fact that all rational numbers can be represented by points on the number line. This is because rational numbers are defined as the quotient of two integers, and every possible quotient can be located on the number line. Therefore, this statement is correct.- Statement ⑤: A pentagonal prism is a type of polyhedron with a pentagon base. It indeed has 7 faces (2 pentagonal faces and 5 rectangular faces), 10 vertices (5 vertices on each pentagonal base), and 15 edges (5 edges on each pentagon and 5 edges connecting the two bases). Therefore, this statement is correct.Given the above analysis, the correct statements are ④ and ⑤. This means there are 2 correct statements.Therefore, the answer is: boxed{B}.

🧠:[<think>Okay, so I need to find the equation of the tangent line to the curve C at x=0. The function is f(x) = sin x + e^x + 2. Hmm, tangent lines... I remember that the equation of a tangent line at a point involves the derivative of the function at that point. Let me recall the formula. I think it's something like y = f(a) + f'(a)(x - a), where a is the x-coordinate where the tangent is drawn. In this case, a is 0. So first, I need to find f(0) and f'(0).Let me compute f(0) first. Plugging x=0 into the function: sin(0) + e^0 + 2. Sin(0) is 0, e^0 is 1, so 0 + 1 + 2 = 3. So f(0) is 3. That gives me the point (0, 3) where the tangent line touches the curve.Now, I need to find the derivative f'(x) to get the slope of the tangent line at x=0. Let's differentiate term by term. The derivative of sin x is cos x. The derivative of e^x is e^x. The derivative of a constant, 2, is 0. So f'(x) = cos x + e^x. Then, plugging in x=0: cos(0) + e^0. Cos(0) is 1, e^0 is also 1. So f'(0) = 1 + 1 = 2. Therefore, the slope of the tangent line is 2.Now, using the point-slope form of a line. Wait, but the formula I mentioned earlier is already a version of that. So using y = f(a) + f'(a)(x - a). Here, a is 0, so that simplifies to y = f(0) + f'(0)(x - 0), which is y = 3 + 2x. So the equation of the tangent line should be y = 2x + 3.Let me double-check my steps. First, f(0) calculation: sin(0) is 0, e^0 is 1, plus 2 gives 3. Correct. Then derivative: cos x + e^x, evaluated at 0 gives 1 + 1 = 2. Correct. Then plugging into the equation: y = 3 + 2x. That seems right.Wait, just to make sure I didn't mix up any terms. The original function is sin x + e^x + 2. The derivative of that is indeed cos x + e^x. Yes, because the derivative of sin x is cos x, derivative of e^x is e^x, and derivative of constant 2 is 0. So f'(x) is correct. Then f'(0) is 1 + 1 = 2. Correct.And plugging into the tangent line formula: since a=0, x - a is just x, so the equation is y = 3 + 2x. So yes, that should be the right answer. Maybe write it as y = 2x + 3. Either way, it's the same equation.I think that's all. Let me see if there's another way to approach this, but I don't think so. The process is standard: find the point, find the slope via the derivative, plug into the line equation. Seems straightforward. No, I don't think I made any mistakes here. Let's confirm once again:Point: x=0, y=f(0)=3.Slope: f'(0)=2.Equation: y - 3 = 2(x - 0) => y = 2x + 3. Yep, that's the same. So I think that's correct.Final AnswerThe equation of the tangent line is boxed{y = 2x + 3}.</think>To find the equation of the tangent line to the curve ( C ) given by ( f(x) = sin x + e^x + 2 ) at ( x = 0 ):1. Calculate ( f(0) ): [ f(0) = sin(0) + e^0 + 2 = 0 + 1 + 2 = 3 ] This gives the point ((0, 3)) on the curve.2. Find the derivative ( f'(x) ): [ f'(x) = cos x + e^x ] Evaluate the derivative at ( x = 0 ): [ f'(0) = cos(0) + e^0 = 1 + 1 = 2 ] This gives the slope of the tangent line at ( x = 0 ).3. Form the equation of the tangent line using the point-slope form ( y = f(a) + f'(a)(x - a) ) with ( a = 0 ): [ y = 3 + 2(x - 0) = 3 + 2x ]Thus, the equation of the tangent line is (boxed{y = 2x + 3}).,<think>Okay, so I need to find the equation of the tangent line to the curve C at x=0. The function is f(x) = sin x + e^x + 2. Hmm, tangent lines... Right, the tangent line at a point is given by the formula y = f(a) + f’(a)(x - a), where a is the x-coordinate of the point. In this case, a is 0. So first, I should find f(0) and f’(0). Let me start with f(0).Calculating f(0) is straightforward. Plugging x=0 into the function: sin(0) + e^0 + 2. Sin of 0 is 0, e^0 is 1, so 0 + 1 + 2 equals 3. Therefore, f(0) = 3. So the point is (0, 3).Next, I need the derivative f’(x) to find the slope of the tangent line at x=0. Let's differentiate term by term. The derivative of sin x is cos x. The derivative of e^x is still e^x. The derivative of a constant, 2, is 0. So f’(x) = cos x + e^x. Now, plug in x=0 into the derivative: cos(0) + e^0. Cos(0) is 1, and e^0 is also 1. So f’(0) = 1 + 1 = 2. Therefore, the slope of the tangent line at x=0 is 2.Now that I have the point (0, 3) and the slope 2, I can plug these into the point-slope form of a line. Wait, the formula I mentioned earlier is essentially point-slope form. So using y = f(a) + f’(a)(x - a), which becomes y = 3 + 2*(x - 0). Simplifying that, it's just y = 2x + 3. Let me double-check my steps. Calculated f(0) correctly: sin(0) is 0, e^0 is 1, plus 2 gives 3. The derivative: d/dx sin x is cos x, d/dx e^x is e^x, derivative of 2 is 0. So f’(x) is cos x + e^x. Plugging in 0, cos(0)=1 and e^0=1, so 2. Then point-slope gives y = 2x + 3. That seems right.Wait, just to make sure I didn't mix up any terms. Sometimes when there's a constant term, like the +2 in f(x), people might forget that it doesn't affect the derivative. But in this case, I accounted for it by saying the derivative of 2 is 0, so that's correct. The slope only depends on the derivative of the variable terms. So no issues there.Another check: if I were to graph the original function, at x=0, the value is 3. The tangent line should touch the curve at that point and have a slope of 2. So if I go one unit to the right (x=1), the tangent line would be at y = 2(1) + 3 = 5. The actual function at x=1 is sin(1) + e^1 + 2. Which is approximately 0.8415 + 2.7183 + 2 ≈ 5.5598, which is higher than 5, but that's expected because the tangent line is just an approximation at x=0. So the slope seems reasonable. If I check closer to x=0, say x=0.1, the tangent line would give y ≈ 2*0.1 + 3 = 3.2. The actual function: sin(0.1) + e^0.1 + 2 ≈ 0.0998 + 1.1052 + 2 ≈ 3.205, which is very close. That seems consistent. So the tangent line equation is likely correct.Therefore, the equation of the tangent line is y = 2x + 3.Final AnswerThe equation of the tangent line is boxed{y = 2x + 3}.</think>To find the equation of the tangent line to the curve ( C ) given by ( f(x) = sin x + e^x + 2 ) at ( x = 0 ):1. Calculate ( f(0) ): [ f(0) = sin(0) + e^0 + 2 = 0 + 1 + 2 = 3 ] The point on the curve is ( (0, 3) ).2. Find the derivative ( f'(x) ): [ f'(x) = cos x + e^x ]3. Evaluate the derivative at ( x = 0 ): [ f'(0) = cos(0) + e^0 = 1 + 1 = 2 ] The slope of the tangent line is 2.4. Use the point-slope form of the tangent line equation ( y = f(a) + f'(a)(x - a) ) with ( a = 0 ): [ y = 3 + 2(x - 0) = 2x + 3 ]The equation of the tangent line is (boxed{y = 2x + 3}).]

🔑:Given f(x)=sin x+e^{x}+2, thus f'(x)=cos x+e^{x}, therefore, the slope of the tangent line to the curve f(x)=sin x+e^{x}+2 at point P(0,3) is: k=cos 0+e^{0}=2, hence, the equation of the tangent line to the curve f(x)=sin x+e^{x}+2 at point P(0,3) is: y=2x+3, so the answer is y=2x+3. To find the equation of the tangent line at x=0, it is only necessary to find its slope. Therefore, first use the derivative to find the value of the derivative function at x=0, then combine the geometric meaning of the derivative to find the slope of the tangent line, thus solving the problem. This question mainly examines the application of derivatives to study the equation of the tangent line at a certain point on the curve, the equation of a line, and other basic knowledge, focusing on computational problem-solving ability, and the integration of mathematical and geometric ideas, simplification, and transformation. It is a basic question.The final answer is boxed{y=2x+3}.