Ignore my work but solve the problem and check. Thank you!!

Mathematics

1 answer:

Free_Kalibri [48]1 year ago

8 0

Yes good job you are correct

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Answer:

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

Step-by-step explanation:

$\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228$

$=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}$

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

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