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$\begingroup$

$$(-11a^2)(-4a^{-7})$$

Can someone reformat, $a$ is second set of parenthesis is to the $-7$ power.

Change to reciprocal so we get $$\left(\frac{1}{-4a}\right)^7 * \frac{11a}{1} = $$ confused

Answer should be $$\frac{44}{a^5}$$

Unless,

$$(-11a^2)(-4a^{-7}) = 44a^{-5} = \frac{1}{44a^5}$$

but I guess, only $a$ was moved to the bottom because 44 is not negative.

Confused.

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1 Answers 1

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$(-11a^2)(-4a^{-7})$ means $(-11) \times a^2 \times (-4) \times a^{-7}$. (No real math there, just convention about how the notation works).

We can reorder the factors such that the two constants are multiplied together first and then use the rule $a^n a^k=a^{n+k}$ to get $$\Bigl((-11)\times(-4)\Bigr)\times a^{2-7} = 44\times a^{-5} = \frac{44}{a^5}$$