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I am Lyosha [343]
3 years ago
5

Identify next three numbers in this sequence: 3, 12, 6, 24, 18, … 27, 13.5, 22.5 72, 66, 264 72, 36, 144 27, 21, 30

Mathematics
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

27

Step-by-step explanation:

sorry try and use google, google never lies

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Which of the following is equal to the expression below? (160x243)^1/5
Kipish [7]
160 = 2 x 2 x 2 x 2 x 2 x 5 = 2^5 x 5
243 = 3 x 3 x 3 x 3 x 3 = 3^5
So
(160 * 243)^1/5
= 5th root of (160 * 243)
= 5th root of (2^5 * 5 * 3^5)
= 2 * 3 * (5th root of 5)
= 6 * (5th root of  5)
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To calculate h evaluate inside the square root before taking the square root

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Answer:

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Step-by-step explanation:

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Solve this x²+8/10 -x = 10/2
AysviL [449]
x^2+\frac{8}{10}-x=\frac{10}{2}\\
x^2-x+\frac{8}{10}-\frac{10}{2}=0\\
x^2-x+\frac{8}{10}-\frac{50}{10}=0\\
x^2-x-\frac{42}{10}=0\\
x^2-x+\frac{1}{4}-\frac{1}{4}-\frac{42}{10}=0\\
(x-\frac{1}{2})^2=\frac{1}{4}+\frac{42}{10}\\
(x-\frac{1}{2})^2=\frac{5}{20}+\frac{84}{20}\\
(x-\frac{1}{2})^2=\frac{89}{20}\\
x-\frac{1}{2}=\sqrt{\frac{89}{20}} \vee x-\frac{1}{2}=-\sqrt{\frac{89}{20}}\\
x=\frac{1}{2}+\frac{\sqrt{89}}{\sqrt{20}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{\sqrt{20}}\\



x=\frac{1}{2}+\frac{\sqrt{89}}{2\sqrt{5}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{2\sqrt5}}\\
x=\frac{1}{2}+\frac{\sqrt{445}}{10} \vee x=\frac{1}{2}-\frac{\sqrt{445}}{10}}\\
x=\frac{5+\sqrt{445}}{10} \vee x=\frac{5-\sqrt{445}}{10}}\\
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