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Alenkinab [10]
3 years ago
10

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mathematics
2 answers:
beks73 [17]3 years ago
8 0

3\sqrt{32x^4z}=3\sqrt{32}\cdot\sqrt{x^4}\cdot\sqrt{z}\\\\=3\sqrt{16\cdot2}\cdot\sqrt{(x^2)^2}\cdot\sqrt{z}\\\\=3\sqrt{16}\cdot\sqrt2\cdot x^2\cdot\sqrt{z}\\\\\=3\cdot4\cdot\sqrt2\cdot x^2\cdot\sqrt{z}\\\\=12x^2\sqrt{2z}\\\\Used:\\\\\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{a^2}=a

Semmy [17]3 years ago
7 0
3√32x^4z

= 3√(2^5)* x^4 *z

(Square rooting the terms)

= 3*2*2*x*x√2*z

=12x^2√2z

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Y = x - 6<br> 5x - 2y = -30
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xeze [42]
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3 years ago
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3 more than the product of 7 and a number x is less than 26 Find the solution set for x. Write the solution using a fraction or
Tresset [83]

Answer: The solution set is x\:


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4 0
3 years ago
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Oliga [24]
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6 0
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