Question

Please help me with this Algebra I question.

Answer 1

Answer:

$-\dfrac{b^2}{16a^4}$

Step-by-step explanation:

Given expression:

$\dfrac{(2a)^{-2}b^5}{-4a^2b^3}$

$\textsf{Apply the exponent rule}\quad (ab)^c=a^c \cdot b^c:$

$\implies \dfrac{2^{-2}a^{-2}b^5}{-4a^2b^3}$

Separate the variables:

$\implies \dfrac{2^{-2}}{-4}\cdot \dfrac{a^{-2}}{a^2}\cdot \dfrac{b^5}{b^3}$

$\textsf{Apply the exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:$

$\implies \dfrac{2^{-2}}{-4}\cdot a^{-2-2}\cdot b^{5-3}$

$\implies \dfrac{2^{-2}}{-4}\cdot a^{-4}\cdot b^{2}$

$\textsf{Apply the exponent rule}\quad a^{-n}=\dfrac{1}{a^n}:$

$\implies \dfrac{1}{-4(2^2)}\cdot \dfrac{1}{a^4}\cdot b^{2}$

$\implies -\dfrac{1}{16}\cdot \dfrac{1}{a^4}\cdot b^{2}$

Simplify:

$\implies -\dfrac{b^2}{16a^4}$