How do you simplify $(7 d^{- 6}) (- 4 d^{- 2})$ $(7d^-6)(-4d^-2)$ ?

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How do you simplify $(7 d^{- 6}) (- 4 d^{- 2})$ $(7d^-6)(-4d^-2)$ ?

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Answer 1 · 2015-09-24T14:03:09

$- \frac{28}{d^{8}}$ $-28/d^8$

Explanation:

To multiply $d^{- 6}$ $d^-6$ and $d^{- 2}$ $d^-2$ , you only have to add their exponents together (Law of Indices). Also, to remove the negative sign from the exponent, you simply have to get the reciprocal of the base.

$(7 d^{- 6}) (- 4 d^{- 2})$ $(7d^(−6))(−4d^(−2))$

$= (7) \cdot (- 4) \cdot (d^{- 6 + (- 2)})$ $=(7)*(-4)⋅(d^(-6+(-2)))$

$= (- 28) \cdot (d^{- 8})$ $=(-28)*(d^-8)$

$= (- 28) \cdot (\frac{1}{d^{8}})$ $=(-28)*(1/d^8)$

$= - \frac{28}{d^{8}}$ $=color(blue)(-28/d^8)$

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How do you simplify $(7 d^{- 6}) (- 4 d^{- 2})$ $(7d^-6)(-4d^-2)$ ?

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How do you simplify $(7 d^{- 6}) (- 4 d^{- 2})$ $(7d^-6)(-4d^-2)$ ?