answer：First, I need to calculate the expression ((lg 2)^2 + (lg 5)^2 + lg 4 cdot lg 5).I notice that (lg 4) can be rewritten as (2lg 2) using the logarithmic identity (lg a^b = blg a).Substituting this back into the expression, it becomes:[(lg 2)^2 + (lg 5)^2 + 2lg 2 cdot lg 5]This expression resembles the expansion of a squared binomial. Specifically, it matches the form (a^2 + b^2 + 2ab), which is equal to ((a + b)^2).Applying this to our expression:[(lg 2 + lg 5)^2]Using the logarithmic property that (lg a + lg b = lg(ab)), we can simplify further:[(lg(2 times 5))^2 = (lg 10)^2]Since (lg 10 = 1), the expression simplifies to:[1^2 = 1]Therefore, the value of the expression is 1.

answer：First, I need to understand the relationship between the number of revolutions and the distance traveled by each wheel.I'll denote the number of revolutions made by the back wheel as ( R_b ) and the front wheel as ( R_f ). According to the problem, the front wheel makes 10 more revolutions than the back wheel, so:[ R_f = R_b + 10 ]Next, I'll express the distance traveled by each wheel in terms of their perimeters and the number of revolutions. The distance traveled by the back wheel is:[ D_b = R_b times 9 ]And the distance traveled by the front wheel is:[ D_f = R_f times 7 ]Since both wheels travel the same distance, I can set these two expressions equal to each other:[ R_b times 9 = (R_b + 10) times 7 ]Now, I'll solve this equation for ( R_b ):[ 9R_b = 7R_b + 70 ]Subtracting ( 7R_b ) from both sides:[ 2R_b = 70 ]Dividing both sides by 2:[ R_b = 35 ]With the number of revolutions for the back wheel determined, I can calculate the total distance traveled by multiplying the number of revolutions by the perimeter of the back wheel:[ D = 35 times 9 = 315 ]Therefore, the distance traveled is 315 feet.

answer：First, I need to determine the original length of the movie before any scenes were cut. I know that a 6-minute scene was removed, and the final length of the movie is 54 minutes.To find the original length, I can add the length of the cut scene to the final length of the movie. This is because the final length represents the movie after the scene was removed.So, the original length of the movie would be the final length plus the length of the cut scene.By performing this addition, I can calculate the original length of the movie.

answer：Okay, so I have this problem about a right triangle, and I need to find the least possible value of one of its acute angles, which is a prime number. Let me try to break this down step by step.First, I remember that in a right triangle, one angle is 90 degrees, and the other two angles are acute, meaning they are less than 90 degrees. Also, the sum of all three angles in any triangle is 180 degrees. So, if one angle is 90 degrees, the other two must add up to 90 degrees. That means if the two acute angles are a degrees and b degrees, then a + b = 90.The problem says that both a and b are prime numbers, and a is greater than b. I need to find the smallest possible value of b. The options given are 2, 3, 5, 7, and 11.Let me list out the prime numbers less than 90 because both a and b have to be prime and add up to 90. The prime numbers less than 90 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89.Now, I need to find pairs of these primes that add up to 90. Since a is greater than b, I can start checking from the smallest prime for b and see if 90 - b is also a prime.Let's start with the smallest option, which is 2. If b is 2, then a would be 90 - 2 = 88. Is 88 a prime number? No, because 88 is divisible by 2, 4, 8, 11, etc. So, 88 is not prime. Therefore, b cannot be 2.Next, let's try b = 3. Then a would be 90 - 3 = 87. Is 87 a prime number? Hmm, 87 divided by 3 is 29, so 87 = 3 * 29. That means 87 is not prime either. So, b can't be 3.Moving on to b = 5. Then a would be 90 - 5 = 85. Is 85 a prime number? 85 divided by 5 is 17, so 85 = 5 * 17. Not prime. So, b can't be 5.Next, b = 7. Then a would be 90 - 7 = 83. Is 83 a prime number? Let me think. 83 is a well-known prime number because it's only divisible by 1 and itself. So, yes, 83 is prime. Therefore, b = 7 is a possible value.Just to make sure there isn't a smaller prime that works, let me check the next option, which is b = 11. Then a would be 90 - 11 = 79. 79 is also a prime number. But since 7 is smaller than 11, and we already found that b = 7 works, 7 is the smaller one.Wait, but just to be thorough, let me check if there are any primes smaller than 7 that could work. We already checked 2, 3, and 5, and none of them worked because their corresponding a values weren't prime. So, 7 is indeed the smallest possible value for b.Therefore, the least possible value of b is 7.

answer：Alright, so I have this problem here about a cubic function, f(x) = ax³ + bx² + cx, where a is not zero. The problem says that this function has extremum values at x = ±1, and also that f(-1) = 1. There are two parts: first, I need to find the constants a, b, and c, and second, I need to find the extremum values of f(x).Okay, let's start with part (1). I remember that extremum points of a function occur where the first derivative is zero. So, I should find the derivative of f(x) and set it equal to zero at x = 1 and x = -1. That should give me some equations to solve for a, b, and c.First, let's find the derivative of f(x). The derivative of ax³ is 3ax², the derivative of bx² is 2bx, and the derivative of cx is c. So, f'(x) = 3ax² + 2bx + c.Since the function has extremum values at x = 1 and x = -1, that means f'(1) = 0 and f'(-1) = 0. Let me write those equations down:1. f'(1) = 3a(1)² + 2b(1) + c = 3a + 2b + c = 02. f'(-1) = 3a(-1)² + 2b(-1) + c = 3a - 2b + c = 0So now I have two equations:3a + 2b + c = 0 ...(1)3a - 2b + c = 0 ...(2)Hmm, if I subtract equation (2) from equation (1), I can eliminate a and c:(3a + 2b + c) - (3a - 2b + c) = 0 - 03a + 2b + c - 3a + 2b - c = 0(3a - 3a) + (2b + 2b) + (c - c) = 00 + 4b + 0 = 04b = 0So, b = 0Okay, that's helpful. So, b is zero. Now, let's plug b = 0 back into equation (1):3a + 2(0) + c = 03a + c = 0So, c = -3aAlright, now we have expressions for b and c in terms of a. But we need another equation to find the value of a. The problem gives us f(-1) = 1. Let's use that.f(-1) = a(-1)³ + b(-1)² + c(-1) = -a + b - c = 1We already know that b = 0 and c = -3a, so let's substitute those in:-a + 0 - (-3a) = 1-a + 3a = 12a = 1a = 1/2Great, so a is 1/2. Now, since c = -3a, c = -3*(1/2) = -3/2.So, summarizing:a = 1/2b = 0c = -3/2Let me double-check these values to make sure they satisfy the original conditions.First, f'(x) = 3*(1/2)x² + 2*0*x + (-3/2) = (3/2)x² - 3/2Let's check f'(1):(3/2)(1)² - 3/2 = 3/2 - 3/2 = 0. Good.f'(-1):(3/2)(-1)² - 3/2 = 3/2 - 3/2 = 0. Also good.Now, f(-1):(1/2)(-1)³ + 0*(-1)² + (-3/2)(-1) = (-1/2) + 0 + (3/2) = (-1/2 + 3/2) = 1. Perfect.Okay, so part (1) seems solved. Now, moving on to part (2): finding the extremum values of f(x).We already know that the function has extremum points at x = 1 and x = -1. To find the extremum values, we just need to compute f(1) and f(-1).Wait, we already computed f(-1) as 1. Let me compute f(1):f(1) = (1/2)(1)³ + 0*(1)² + (-3/2)(1) = (1/2) + 0 - 3/2 = (1/2 - 3/2) = -1So, f(1) = -1.Therefore, the function has a maximum value of 1 at x = -1 and a minimum value of -1 at x = 1.But just to be thorough, maybe I should check the second derivative to confirm whether these points are maxima or minima.The second derivative f''(x) is the derivative of f'(x). So, f'(x) = (3/2)x² - 3/2, so f''(x) = 3x.At x = -1, f''(-1) = 3*(-1) = -3, which is negative, so the function is concave down there, meaning it's a local maximum.At x = 1, f''(1) = 3*(1) = 3, which is positive, so the function is concave up there, meaning it's a local minimum.So, that confirms our earlier conclusion.Just to visualize, the function f(x) = (1/2)x³ - (3/2)x is a cubic function. Since the coefficient of x³ is positive, it goes from negative infinity to positive infinity. It has a local maximum at x = -1 and a local minimum at x = 1.I think that's all for this problem. I've found the constants a, b, c, and determined the extremum values with their respective points.**Final Answer**(1) The constants are a = boxed{dfrac{1}{2}}, b = boxed{0}, and c = boxed{-dfrac{3}{2}}.(2) The extremum values of f(x) are a maximum of boxed{1} at x = -1 and a minimum of boxed{-1} at x = 1.

answer：Okay, so I need to find the value of g(h(2)) where g(x) is 3x² + 4 and h(x) is -2x³ + 2. Hmm, let me break this down step by step because I'm still getting used to function composition.First, I remember that g(h(2)) means I have to evaluate h at x=2 first, and then plug that result into g. So, let me start by finding h(2).Looking at h(x), it's -2x³ + 2. So, plugging in x=2, I get:h(2) = -2*(2)³ + 2.Wait, what's 2 cubed? That's 2*2*2, which is 8. So, substituting that back in:h(2) = -2*8 + 2.Now, -2*8 is -16, so:h(2) = -16 + 2.Adding those together gives me:h(2) = -14.Okay, so h(2) is -14. Now, I need to find g(-14) because g(h(2)) is g(-14).Looking at g(x), it's 3x² + 4. So, plugging in x = -14:g(-14) = 3*(-14)² + 4.Wait, what's (-14) squared? That's positive because a negative times a negative is positive. So, (-14)*(-14) is 196.So, substituting that back in:g(-14) = 3*196 + 4.Now, 3*196 is... let me calculate that. 196*3: 100*3 is 300, 90*3 is 270, 6*3 is 18. So, 300 + 270 is 570, plus 18 is 588.So, g(-14) = 588 + 4.Adding those together gives me:g(-14) = 592.Therefore, the value of g(h(2)) is 592.Wait, let me double-check to make sure I didn't make any mistakes. Starting with h(2):h(2) = -2*(2)^3 + 2 = -2*8 + 2 = -16 + 2 = -14. That seems right.Then, g(-14) = 3*(-14)^2 + 4 = 3*196 + 4 = 588 + 4 = 592. Yep, that checks out.I think I did everything correctly. So, the final answer should be 592.