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Andrej [43]
3 years ago
13

When 1,250^3/4 is written in simplest radical form, which value remains under the radical?

Physics
2 answers:
GaryK [48]3 years ago
7 0

Answer:

125\sqrt[4]{8}

Explanation:

A number of the form

a^{\frac{m}{n}}

can be re-written in the radical form as follows:

\sqrt[n]{a^m}

In this problem, we have:

a = 1,250

m = 3

n = 4

So, if we apply the formula, we get

1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}

Then, we can rewrite 1250 as

1250 = 2\cdot 5^4

So we can rewrite the expression as

=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}

seraphim [82]3 years ago
7 0

Answer: The value remains under the radical is 8.

Explanation:

Given that,

1250^{\frac{3}{4}}

We know that,

a^{\frac{x}{n}}=n\sqrt{a^x}

Here, n = integer number

n is greater than x.

a = real number

Therefore,

1250^{\frac{3}{4}}=\sqrt[4]{1250}^3

1250^{\frac{3}{4}}=\sqrt[4]{2.5^{4}}^3

1250^{\frac{3}{4}}=5^3\sqrt[4]{2}^3

1250^{\frac{3}{4}}=125\sqrt[4]{8}

Hence, The value remains under the radical is 8.

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A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth
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The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

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Distance = 16 m

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Using formula of angle

\tan\theta=\dfrac{8}{3}

\thta=\tan^{-1}\dfrac{8}{3}

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Using formula of tension

mg=2T\cos\theta

Put the value into the formula

3\times9.8=2T\times\cos69.44

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4 0
4 years ago
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Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = 20.6\times 10^3\ m

\theta = The angle subtended by moon = 9.6\times 10^{-3}\ rad

Distance traveled is given by

s=r\times\theta

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Answer:

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According to Snell's law,

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Sin r = 0.375

r = 22 degree

Thus, the angle of refraction is 22 degree.

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3 years ago
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