The smallest among $\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{3}}, \sqrt{\frac{1}{4}} and \sqrt{\frac{2}{3}}$ is [#1691]

X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23

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X + (X+2) + (X+4) + (X+6) + (X+8) = 95 => 5X + 20 = 95 => 5X = 95-20 => 5X = 75 => X = 15 Hence (X+8) = 15+8 = 23

40

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40

(360°/15) * (7-1) = 144°

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(360°/15) * (7-1) = 144°

18

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18

12

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12

$\frac{45}{100}*X = \frac{30}{100} * Y$ = $\frac{X}{Y} = \frac{30*100}{100*45}$ = $\frac{X}{Y} = \frac{2}{3}$ = X:Y = 2:3

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$\frac{45}{100}*X = \frac{30}{100} * Y$ = $\frac{X}{Y} = \frac{30*100}{100*45}$ = $\frac{X}{Y} = \frac{2}{3}$ = X:Y = 2:3

376 -> 367 629 -> 269 921 -> 129 * 397 -> 379

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376 -> 367 629 -> 269 921 -> 129 * 397 -> 379

A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km

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A starts journey before B at a speed of 1 m/s. Hence A will be ahead of B (10*60)s * 1m/s = 600m Let A covers a distance of X after starting of B, Then X + 600m = 2X => 2X - X = 600m => X = 600m Hence B will cover a distance of 2X = 2 * 600m = 1200m = 1.2 km

2 * $\frac{22}{7}$ * 7 = 2 * 22 = 44

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2 * $\frac{22}{7}$ * 7 = 2 * 22 = 44

The result of multiplying a number by itself is called a square, such as 4 × 4 = 16, which is denoted as 4² (four squared).

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The result of multiplying a number by itself is called a square, such as 4 × 4 = 16, which is denoted as 4² (four squared).