answer:This problem involves finding the minimum value of a function, which means I need to use some calculus. I want to find the critical points of the function, which are where the derivative is either zero or undefined. So, I need to find the derivative of the function and set it equal to zero. To find the derivative, I use the power rule and the constant multiple rule. The derivative of 4x^5 is 20x^4, and the derivative of 5x^{-4} is -20x^{-5}. So, the derivative of the function is 20x^4 - 20x^{-5}. Now, I set this equal to zero and solve for x. I multiply both sides by x^5 to clear the fractions, and get 20x^9 - 20 = 0. I add 20 to both sides and divide by 20, and get x^9 = 1. I take the ninth root of both sides and get x = 1. This is the only critical point of the function, since the derivative is defined for all positive x. To check if this is a minimum or a maximum, I can use the second derivative test. I find the second derivative of the function by differentiating the first derivative. The second derivative is 80x^3 + 100x^{-6}. I plug in x = 1 and get 80 + 100 = 180. This is positive, which means the function is concave up at x = 1, and therefore x = 1 is a minimum point. To find the minimum value of the function, I plug in x = 1 into the original function and get 4 + 5 = 9.

answer:The expression inside the square root must be greater than 0 because the denominator cannot be equal to 0. Therefore, x-1>0, so x>1. The expression inside the logarithm must be greater than 0, so 3-x>0, which gives x<3. Therefore, the interval of x for which the expression frac{log{(3-x)}}{sqrt{x-1}} is defined is 1<x<3, which is (1,3).

answer:Let [f(t) = x_1(t+1)^2 + x_2(t+2)^2 + cdots + x_7(t+7)^2.]Then the three given equations say f(0) = 1, f(1) = 12, and f(2) = 123, and we want to find f(3). Since f(t) is a quadratic, we may let f(t) = At^2 + Bt + C, where A, B, C are constant. Then we have the equations [begin{aligned} C &= 1, A+B+C &= 12, 4A+2B+C &= 123. end{aligned} ]Substituting C=1 into the second and third equations gives A+B=11 and 4A+2B=122. Then 2A+B=61, so A = (2A+B)-(A+B) = 61-11=50. Then B=11-A=-39, and so [f(3) = 9A+3B+C=9(50)+3(-39)+1= 334.]

answer:I notice that the given polynomial has a constant term of 35, which is divisible by 7. That means that if 2x+7 is a factor, then the other factor must also have a constant term that is a multiple of 5. So I can try to write the given polynomial as (2x+7)(3x^2+ax+5), where a is some constant. To find a and c, I can multiply out the two factors and compare the coefficients of each term with the given polynomial. I get (2x+7)(3x^2+ax+5) = 6x^3+(2a+21)x^2+(7a+10)x+35. Comparing the coefficients of x^2, I get 2a+21=19, so a=-1. Comparing the coefficients of x, I get 7a+10=c, so c=7(-1)+10=3.