How do you evaluate $(5 x^{- 3} y) (6 x^{7} y^{- 4})$ $(5x ^ { - 3} y ) ( 6x ^ { 7} y ^ { - 4} )$ ?

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Answer 1 · 2017-10-30T02:29:19

$(5 x^{- 3} y) (6 x^{7} y^{- 4}) = \frac{30 x^{4}}{y^{3}}$ $(5x^(-3)y)(6x^7y^(-4))=color(blue)((30x^4)/(y^3)$

Simplify:

$(5 x^{- 3} y) (6 x^{7} y^{- 4})$ $(5x^(-3)y)(6x^7y^(-4))$

Simplify the parentheses.

$5 x^{- 3} y \times 6 x^{7} y^{- 4}$ $5x^(-3)yxx6x^7y^(-4)$

Multiply $5$ $5$ and $6$ $6$ .

$30 x^{- 3} y x^{7} y^{- 4}$ $30x^(-3)yx^7y^(-4)$

Gather like variables.

$30 x^{- 3} x^{7} y y^{- 4}$ $30x^(-3)x^(7)yy^(-4)$

Apply the product rule of exponents: $a^{m} a^{n} = a^{m + n}$ $a^ma^n=a^(m+n)$ . (No exponent is understood to be $1$ $1$ .

$30 x^{- 3 + 7} y^{1 - 4}$ $30x^(-3+7)y^(1-4)$

Simplify.

$30 x^{4} y^{- 3}$ $30x^(4)y^(-3)$

Apply the negative exponent rule: $a^{- m} = \frac{1}{a^{m}}$ $a^(-m)=1/a^m$ .

$\frac{30 x^{4}}{y^{3}}$ $(30x^4)/(y^3)$ $\leftarrow$ $larr$ Answer

How do you evaluate (5x−3y)(6x7y−4)(5x ^ { - 3} y ) ( 6x ^ { 7} y ^ { - 4} )?