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nlexa [21]
3 years ago
11

What is the common denominator of 1/8 and 9/5​

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
4 0

Answer:

40, they both are 40, because multiplied you get 40

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Help with Algebra 2: 1. <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B7%5D%7Bx%5E5%7D%20%7D%7B%5Csqrt%5B4%5D%7Bx%5E2%7D%
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See explanation

Step-by-step explanation:

1. Given the expression

\dfrac{\sqrt[7]{x^5} }{\sqrt[4]{x^2} }

Note that

\sqrt[7]{x^5}=x^{\frac{5}{7}} \\ \\\sqrt[4]{x^2}=x^{\frac{2}{4}}=x^{\frac{1}{2}}

When dividing \sqrt[7]{x^5} by \sqrt[4]{x^2}, we have to subtract powers (we cannot subtract 4 from 7, because then we get another expression), so

\dfrac{5}{7}-\dfrac{2}{4}=\dfrac{5}{7}-\dfrac{1}{2}=\dfrac{5\cdot 2-1\cdot 7}{14}=\dfrac{3}{14}

and the result is x^{\frac{3}{14}}=\sqrt[14]{x^3}

2. Given equation 3\sqrt[4]{(x-2)^3} -4=20

Add 4:

3\sqrt[4]{(x-2)^3} -4+4=20+4\\ \\3\sqrt[4]{(x-2)^3}=24

Divide by 3:

\sqrt[4]{(x-2)^3} =8

Rewrite the equation as:

(x-2)^{\frac{3}{4}}=8\\ \\(x-2)^{\frac{3}{4}}=2^3

Hence,

\left((x-2)^{\frac{3}{4}}\right)^{\frac{4}{3}}=(2^3)^{\frac{4}{3}}\\ \\x-2=2^{3\cdot \frac{4}{3}}\\ \\x-2=2^4\\ \\x-2=16\\ \\x-2+2=16+2\\ \\x=18

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3 years ago
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