answer:To multiply the two polynomials, we can use the FOIL method (First, Outer, Inner, Last). First, we multiply the first terms of each polynomial: (-frac{24 x^2}{7})(9 x^2) = -frac{216 x^4}{7} Next, we multiply the outer terms of each polynomial: (-frac{24 x^2}{7})(frac{90 x}{7}) = -frac{2160 x^3}{49} Then, we multiply the inner terms of each polynomial: (frac{20 x}{7})(9 x^2) = frac{180 x^3}{7} Finally, we multiply the last terms of each polynomial: (frac{20 x}{7})(frac{1}{7}) = frac{20 x}{49} Now, we add up the like terms: (-frac{216 x^4}{7}) + (-frac{2160 x^3}{49}) + (frac{180 x^3}{7}) + (frac{20 x}{49}) + (-frac{95}{7}) = -frac{216 x^4}{7}-frac{900 x^3}{49}-frac{4209 x^2}{49}-frac{8530 x}{49}-frac{95}{49} Therefore, the product of p(x) and q(x) is -frac{216 x^4}{7}-frac{900 x^3}{49}-frac{4209 x^2}{49}-frac{8530 x}{49}-frac{95}{49}. The answer is -frac{216 x^4}{7}-frac{900 x^3}{49}-frac{4209 x^2}{49}-frac{8530 x}{49}-frac{95}{49}

answer:The Jacobian matrix of the given vector-valued function mathbf{r}(x, y, z) is: frac{partial mathbf{r}}{partial (x, y, z)} = begin{bmatrix} frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z} frac{partial g}{partial x} & frac{partial g}{partial y} & frac{partial g}{partial z} frac{partial h}{partial x} & frac{partial h}{partial y} & frac{partial h}{partial z} end{bmatrix} = begin{bmatrix} -frac{3}{2}(y - x)^{-frac{1}{2}} & frac{3}{2}(y - x)^{-frac{1}{2}} & 0 0 & 5y^4 & 0 0 & 5(y - z)^4 & -5(y - z)^4 end{bmatrix}

answer:The Euler totient function, phi(n), counts the number of positive integers less than n that are relatively prime to n. Since 48991 is a prime number, phi(48991) = 48991 - 1 = 48990.

answer:Given the following: - Future value of the investment (A) = 23,000 - Investment period (T) = 7 years - Continuous annual interest rate (r) = 3.5% = 0.035 Let R be the constant, continuous rate at which money must be deposited. Using the formula for the future value of a continuous deposit, we can find R: [ A = intlimits_0^T R e^{rt} dt ] Substitute the given values: [ 23000 = R intlimits_0^7 e^{0.035t} dt ] Evaluate the integral: [ 23000 = R left[ dfrac{e^{0.035t}}{0.035} right]_0^7 ] [ 23000 = R left( dfrac{e^{0.035 times 7}}{0.035} - dfrac{e^{0.035 times 0}}{0.035} right) ] [ 23000 = R times 7.93 ] [ R = dfrac{23000}{7.93} ] [ R approx 2900.3 text{ per year} ] Therefore, the constant, annual deposit rate needed to reach 23,000 in 7 years with a 3.5% compounded continuous interest rate is approximately 2900.3.