How do you simplify $(- 3 r^{2} s^{- 4}) {(2 r^{- 3} s^{7})}^{2}$ $(-3r^2s^-4)(2r^-3s^7)^2$ ?

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How do you simplify $(- 3 r^{2} s^{- 4}) {(2 r^{- 3} s^{7})}^{2}$ $(-3r^2s^-4)(2r^-3s^7)^2$ ?

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Answer 1 · 2018-06-20T11:32:23

$\Rightarrow - 12 r^{- 4} \cdot s^{10} = \frac{- 12 s^{10}}{r^{4}}$ $color(indigo)(=> -12 r^-4 * s^10 = (-12 s^10)/r^4$

Explanation:

![http://quantbasics.blogspot.com/2010/04/http://indices-and-surds.html](https://useruploads.socratic.org/AAJCnfJYRLOF0YhMuGcR_theory%20of%20indices.jpg)

$(- 3 r^{2} s^{- 4}) ({(2 r^{- 3} s^{7})}^{2}$ $(-3 r^2 s^-4)((2r^-3s^7)^2$

$\Rightarrow (- 3 r^{2} s^{- 4}) (2^{2} r^{- 6} s^{14})$ $=> (-3 r^2 s^-4) (2^2 r^-6 s^14)$

$\Rightarrow (- 3 \cdot 4 \cdot r^{2} \cdot r^{- 6} \cdot s^{- 4} \cdot s^{14})$ $=> (-3 * 4 * r^2 * r^-6 * s^-4 * s^14)$

$\Rightarrow - 12 r^{- 4} \cdot s^{10} = \frac{- 12 s^{10}}{r^{4}}$ $color(indigo)(=> -12 r^-4 * s^10 = (-12 s^10)/r^4$

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How do you simplify $(- 3 r^{2} s^{- 4}) {(2 r^{- 3} s^{7})}^{2}$ $(-3r^2s^-4)(2r^-3s^7)^2$ ?

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How do you simplify (-3r^2s^-4)(2r^-3s^7)^2(−3r2s−4)(2r−3s7)2(-3r^2s^-4)(2r^-3s^7)^2?

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How do you simplify $(- 3 r^{2} s^{- 4}) {(2 r^{- 3} s^{7})}^{2}$ $(-3r^2s^-4)(2r^-3s^7)^2$ ?