Question

Simplify the following questions using the rules for exponents.

Answer 1

Answer:

$\frac{9x^{-1}y^6}{3x^{-2}y{-9}}=3xy^{15}\\\\\frac{-6a^7b^8}{-3a^5b^7}=2a^2b\\\\(\frac{a^{-2}b^{-8}}{a^6b^4})^{\frac{1}{2}}=\frac{1}{a^4b^6}$

Step-by-step explanation:

$\frac{9x^{-1}y^6}{3x^{-2}y^{-9}}$

Multiply numerator and denominator by $x^2y^9$

$\frac{9x^{-1}y^6}{3x^{-2}y^{-9}}\times \frac{x^2y^2}{x^2y^2}=\frac{9xy^{15}}{3x^0y^o}=3xy^{15}$

$\frac{-6a^7b^8}{-3a^5b^7}$

multiply numerator and denominator by $a^{-5}b^{-7}$

$\frac{-6a^7b^8}{-3a^5b^7}\times \frac{a^{-5}b^{-7}}{a^{-5}b^{-7}}=\frac{-6a^{7-5}b^{8-7}}{-3a^{5-5}b^{7-7}}=2a^2b\\\\\\\\(\frac{a^{-2}b^{-8}}{a^6b^4})^{\frac{1}{2}}=\frac{a^{\frac{-2}{2}}b^{\frac{-8}{2}}}{a^{\frac{6}{2}}b^{\frac{4}{2}}}=\frac{a^{-1}b^{-4}}{a^3b^2}$

Multiply numerator and denominator by $a^1b^4$

$\frac{a^{-1}b^{-4}}{a^3b^2} \times \frac{ab^4}{ab^4}=\frac{a^0b^0}{a^4b^6}=\frac{1}{a^2b^6}}$