Question

Which shows how to find the value of this expression when x=-2 and y=5? (3x^3y^-2)2

Answer 1

Put x = -2 and y = 5 to the expression.

$(3x^3y^{-2})^2$

$\left[3\cdot(-2)^3(5)^{-2}\right]^2=\left[3(-8)\cdot\dfrac{1}{5^2}\right]^2=\left(-24\cdot\dfrac{1}{25}\right)^2=\left(-\dfrac{24}{25}\right)^2\\\\=\dfrac{24^2}{25^2}=\dfrac{576}{625}$

explanation

$(-2)^3=(-2)(-2)(-2)=-8\\\\a^{-n}=\dfrac{1}{a^n}\to5^{-2}=\dfrac{1}{5^2}=\dfrac{1}{(5)(5)}=\dfrac{1}{25}\\\\a^2\geq0\ \text{for any real number}\to\left(-\dfrac{24}{25}\right)^2=\left(\dfrac{24}{25}\right)^2\\\\\left(\dfrac{a}{b}\right)^2=\dfrac{a^2}{b^2}\to\left(\dfrac{24}{25}\right)^2=\dfrac{24^2}{25^2}=\dfrac{(24)(24)}{(25)(25)}=\dfrac{576}{625}$