The value of $\sqrt[3]{8^2}$ is [#1694]

$\sqrt[3]{8^2}$ = $\sqrt[3]{64}$ = $\sqrt[3]{4^3}$ = 4

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$\sqrt[3]{8^2}$ = $\sqrt[3]{64}$ = $\sqrt[3]{4^3}$ = 4

$\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}-\sqrt{\frac{16}{81}}$ = $\sqrt{\frac{15*15}{27*27}}-\sqrt{\frac{5*5}{12*12}}-\sqrt{\frac{4*4}{9*9}}$ = $\frac{15}{27}-\frac{5}{12}-\frac{4}{9}$ = $\frac{60-45-48}{108}$ = $\frac{-33}{108}$ = $\frac{-11}{36}$

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$\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}-\sqrt{\frac{16}{81}}$ = $\sqrt{\frac{15*15}{27*27}}-\sqrt{\frac{5*5}{12*12}}-\sqrt{\frac{4*4}{9*9}}$ = $\frac{15}{27}-\frac{5}{12}-\frac{4}{9}$ = $\frac{60-45-48}{108}$ = $\frac{-33}{108}$ = $\frac{-11}{36}$

In ascending order = -2, -1, 1, 2, 3, 5, 6 median = $\frac{7+1}{2}$th term = 4th term = 2

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In ascending order = -2, -1, 1, 2, 3, 5, 6 median = $\frac{7+1}{2}$th term = 4th term = 2

3.5 m

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3.5 m

The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.

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The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.

Integers

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Integers

43

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43

1250 = 2*5*5*5*5 1250*2 = 2*2*5*5*5*5 = 2500 $\sqrt{2500}$ = 2*5*5 = 50

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1250 = 2*5*5*5*5 1250*2 = 2*2*5*5*5*5 = 2500 $\sqrt{2500}$ = 2*5*5 = 50

60

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60

To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.

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To solve for x, subtract 5 from both sides of the equation, resulting in 2x = 6. Then, divide both sides by 2, giving x = 3.