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2 years ago

Help please! this is due tomorrow so please help out! will mark as brainiest!!

Mathematics

1 answer:

lesya692 [45]2 years ago

5 0

Answer:

$\large\boxed{\bold{Q1}\ -\dfrac{128}{9}}\\\boxed{\bold{Q2}\ 2^{25}}$

Step-by-step explanation:

$\bold{Q1}\\\\\dfrac{-2(3^2\cdot3^3)^2}{3^{12}\cdot2^{-6}}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=-\dfrac{2(3^{2+3})^2}{3^{12}\cdot2^{-6}}=-\dfrac{2(3^5)^2}{3^{12}\cdot2^{-6}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=-\dfrac{2\cdot3^{5\cdot2}}{3^{12}\cdot2^{-6}}=-\dfrac{2\cdot3^{10}}{3^{12}\cdot2^{-6}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=-2^{1-(-6)}\cdot3^{10-12}=-2^{7}\cdot3^{-2}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=-2^7\cdot\dfrac{1}{3^2}=-\dfrac{2^7}{3^2}=-\dfrac{128}{9}$

$\bold{Q2}\\\\\dfrac{4^2}{4^{-7}}\cdot(2^4\cdot2^3)\cdot(3^{-2})^0\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m},\ a^n\cdot a^m=a^{n+m},\ a^0=1\\\\=4^{2-(-7)}\cdot2^{4+3}\cdot1=4^{9}\cdot2^7=(2^2)^{9}\cdot2^7\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{2\cdot9}\cdot2^7=2^{18}\cdot2^7\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{18+7}=2^{25}$

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