3 years ago

474 ( ) 548 42. - 8 how do I arrive at this answer.

Mathematics

1 answer:

dexar [7]3 years ago

5 0

9514 1404 393

Answer:

(64x^28)/(5y^22)

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)^c = a^(bc)

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

First, eliminate the outside exponent on the left factor. Then use the above rules to combine factors with the same base.

$\displaystyle\left(\frac{4x^4}{5y^6}\right)^5\cdot\left(\frac{5^4y^8}{4^2x^{-8}}\right)=\frac{(4^5x^{20})(5^4y^8)}{(5^5y^{30})(4^2x^{-8})}=4^{5-2}5^{4-5}x^{20-(-8)}y^{8-30}\\\\=4^35^{-1}x^{28}y^{-22}=\boxed{\frac{64x^{28}}{5y^{22}}}$

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474 ( ) 548 42. - 8 how do I arrive at this answer.​

474 ( ) 548 42. - 8 how do I arrive at this answer.