Answer:
[tex]-\dfrac{b^2}{16a^4}[/tex]
Step-by-step explanation:
Given expression:
[tex]\dfrac{(2a)^{-2}b^5}{-4a^2b^3}[/tex]
[tex]\textsf{Apply the exponent rule}\quad (ab)^c=a^c \cdot b^c:[/tex]
[tex]\implies \dfrac{2^{-2}a^{-2}b^5}{-4a^2b^3}[/tex]
Separate the variables:
[tex]\implies \dfrac{2^{-2}}{-4}\cdot \dfrac{a^{-2}}{a^2}\cdot \dfrac{b^5}{b^3}[/tex]
[tex]\textsf{Apply the exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:[/tex]
[tex]\implies \dfrac{2^{-2}}{-4}\cdot a^{-2-2}\cdot b^{5-3}[/tex]
[tex]\implies \dfrac{2^{-2}}{-4}\cdot a^{-4}\cdot b^{2}[/tex]
[tex]\textsf{Apply the exponent rule}\quad a^{-n}=\dfrac{1}{a^n}:[/tex]
[tex]\implies \dfrac{1}{-4(2^2)}\cdot \dfrac{1}{a^4}\cdot b^{2}[/tex]
[tex]\implies -\dfrac{1}{16}\cdot \dfrac{1}{a^4}\cdot b^{2}[/tex]
Simplify:
[tex]\implies -\dfrac{b^2}{16a^4}[/tex]
