Find the least number by which 1250 must be multiplied to make it a perfect square. [#1700]

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

2⁵ = 2*2*2*2*2 = 32

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2⁵ = 2*2*2*2*2 = 32

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

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

4

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4

60

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60

36

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36

2

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2

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

An obtuse angle is an angle greater than 90 degrees but less than 180 degrees, like the angle formed by two walls that meet at a corner.

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An obtuse angle is an angle greater than 90 degrees but less than 180 degrees, like the angle formed by two walls that meet at a corner.

20% = $\frac{25}{100+25} * 100$% $\frac{25}{125} * 100$% = 20%

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20% = $\frac{25}{100+25} * 100$% $\frac{25}{125} * 100$% = 20%

5 : (5² + 2) = 5 : 27 Hence 9 : (9² + 2) = 9 : 83

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5 : (5² + 2) = 5 : 27 Hence 9 : (9² + 2) = 9 : 83