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nataly862011 [7]
3 years ago
10

Find the measure of x.

Mathematics
1 answer:
krok68 [10]3 years ago
3 0

Answer:

Step-by-step explanation:

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KatRina [158]

1.

(x^2y^8)^\frac{2}{3}=x^{2\cdot\frac{2}{3}}y^{8\cdot\frac{2}{3}}=x^{\frac{4}{3}}y^{\frac{16}{3}}=x^{1\frac{1}{3}}y^{5\frac{1}{3}}=x^1x^\frac{1}{3}y^5y^\frac{1}{3}=xy^5\sqrt[3]{x}\sqrt[3]{y}=\boxed{xy^5\sqrt[3]{xy}}

Answer B)

2.

\frac{1}{3}\left(\frac{2}{15}x-\frac{2}{3}\right)>x+\frac{1}{5}\quad|\cdot3\\\\\\\frac{2}{15}x-\frac{2}{3}>3x+\frac{3}{5}\quad|\cdot15\\\\\\2x-\frac{30}{3}>45x+\frac{45}{5}\\\\2x-10>45x+9\\\\-10-9>45x-2x\\\\-19>43x\quad|:43\\\\\boxed{x

Answer B)

3.

\sqrt{14xy}\cdot\sqrt{12}\cdot\sqrt{30y}=\sqrt{14\cdot12\cdot30}\cdot\sqrt{xy^2}=\sqrt{2\cdot7\cdot3\cdot4\cdot2\cdot3\cdot5}\cdot\sqrt{xy^2}=\\\\=\sqrt{2^2\cdot3^2\cdot2^2\cdot5\cdot7}\cdot\sqrt{xy^2}=2\cdot3\cdot2\cdot\sqrt{35}\cdot y\cdot\sqrt{x}=\boxed{12y\sqrt{35x}}

3 0
3 years ago
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