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nignag [31]
3 years ago
7

14. Simplify (4a^-4)^-3 a. –64x^9 b. x^12/64 c. -8x^12 d. -1/2x^4

Mathematics
1 answer:
Anvisha [2.4K]3 years ago
7 0
Remember
(ab)^c=(a^c)(b^c) and
x^{-m}= \frac{1}{x^m}

so
(4a^{-4})^{-3}=(4^{-3})((a^{-4}^{-3})=( \frac{1}{4^3})(a^{12})= \frac{a^{12}}{64}

answer would be B if we were using x


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Reil [10]
<h2>Answer:</h2>

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<h2>Step-by-step explanation:</h2>

The question in this problem is:

<em>The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger. What are the numbers?</em>

<em />

First of all, let's name the first variable x which is the smaller number. Accordingly, the lager number will be x+1 given that those numbers are consecutive. On the other hand<em> at most </em>conveys the idea of an inequality, which is:

\leq \\ which \ means \ less \ than

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The sum of 2 consecutive integers is at most the difference between nine times the smaller and 5 times the larger:

x+(x+1) \leq 9x-5(x+1) \\ \\ x+x+1\leq 9x-5x-5 \\ \\ 2x+1 \leq 4x-5 \\ \\ 6 \leq 2x \\ \\

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The two numbers are:

x\geq 3 \ and \ x+1 \geq 4

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