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Answer ya so 1 is the answer
Step-by-step explanation:
Answer:
1
Step-by-step explanation:
Given expression:
[tex]\sf \left(\dfrac{3a^{-2}b^6}{2a^{-1}b^5} \right)^2[/tex]
To find the value of the expression when a = 3 and b = -2, substitute these values into the expression:
[tex]\implies \sf \left(\dfrac{3(3)^{-2}(-2)^6}{2(3)^{-1}(-2)^5} \right)^2[/tex]
[tex]\textsf{Apply exponent rule} \quad a^{-n}=\dfrac{1}{a^n}[/tex]
[tex]\sf \implies \left(\dfrac{3\left(\dfrac{1}{3^2}\right)(-2)^6}{2\left(\dfrac{1}{3^1} \right)(-2)^5} \right)^2[/tex]
[tex]\sf \implies \left(\dfrac{3\left(\dfrac{1}{9}\right)(-2)^6}{2\left(\dfrac{1}{3} \right)(-2)^5} \right)^2[/tex]
[tex]\sf \implies \left(\dfrac{\left(\dfrac{3}{9}\right)(-2)^6}{\left(\dfrac{2}{3} \right)(-2)^5} \right)^2[/tex]
[tex]\sf \implies \left(\dfrac{\left(\dfrac{1}{3}\right)(-2)^6}{\left(\dfrac{2}{3} \right)(-2)^5} \right)^2[/tex]
[tex]\textsf{Apply exponent rule} \quad (-a)^n=a^n,\:\: \textsf{ if }n \textsf{ is even}[/tex]
[tex]\textsf{Apply exponent rule} \quad (-a)^n=-a^n,\:\: \textsf{ if }n \textsf{ is odd}[/tex]
[tex]\sf \implies \left(\dfrac{\left(\dfrac{1}{3}\right) (2^6)}{\left(\dfrac{2}{3} \right) (-(2^5))} \right)^2[/tex]
[tex]\sf \implies \left(\dfrac{\left(\dfrac{1}{3}\right) (64)}{\left(\dfrac{2}{3} \right) (-32)} \right)^2[/tex]
[tex]\sf \implies \left(\dfrac{\left(\dfrac{1 \times 64}{3}\right)}{\left(\dfrac{2 \times -32}{3} \right)} \right)^2[/tex]
[tex]\sf \implies \left(\dfrac{\dfrac{64}{3}}{\dfrac{-64}{3}} \right)^2[/tex]
When dividing fractions, flip the second fraction and multiply it by the first:
[tex]\implies \sf \left( \dfrac{64}{3} \times \dfrac {3}{-64} \right)^2[/tex]
[tex]\implies \sf \left( \dfrac{64 \times 3}{3 \times (-64)}\right)^2[/tex]
[tex]\implies \sf \left( \dfrac{192}{-192}\right)^2[/tex]
[tex]\implies \sf \left(-1\right)^2[/tex]
[tex]\textsf{Apply exponent rule} \quad (-a)^n=a^n,\:\: \textsf{ if }n \textsf{ is even}[/tex]
[tex]\sf \implies 1^2=1[/tex]