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Vladimir79 [104]
3 years ago
8

The weight of an object on the moon varies directly with its weight on the earth. If an object weighing 95 lbs on the moon weigh

s 570 lb on the earth, how much douse an object weigh on the moon if it weighs 2700 lb on the earth?
Mathematics
1 answer:
Luda [366]3 years ago
8 0
ANSWER

450lb

EXPLANATION

If the weight, m of an object on the moon varies directly as the weight of an object e, on the earth, then we can write the mathematical statement,.
m \propto \: e
We introduce the constant of proportionality to obtain,

m = ke

When, m=95, e=570,

This implies that,

95 = 570k
k = \frac{95}{570}

k = \frac{1}{6}

The equation now becomes,

m = \frac{1}{6} e

We want to find e, when m=2700,

m = \frac{1}{6} \times 2700

m =450lb
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Given Rhombus ADEF is inscribed into a triangle ABC so that they share angle A and the vertex E lies on the side BC. We have to find the length of side of rhombus.

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What is the square root of m^6?
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Answer:
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If m is allowed to be negative, then the answer is either |m^3| or |m|^3

==============================

Explanation:

There are two ways to get this answer. The quickest is to simply divide the exponent 6 by 2 to get 6/2 = 3. This value of 3 is the final exponent over the base m. Why do we divide by 2? Because the square root is the same as having an exponent of 1/2 = 0.5, so
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---------------------------

A slightly longer method is to break up the square root into factors of m^2 each and then apply the rule that sqrt(x^2) = x, where x is nonnegative

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------------------------------

If we allow m to be negative, then the final result would be either |m^3| or |m|^3. 

The reason for the absolute value is to ensure that the expression m^3 is nonnegative. Keep in mind that m^6 is always nonnegative, so sqrt(m^6) is also always nonnegative. In order for sqrt(m^6) = m^3 to be true, the right side must be nonnegative.

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Without the absolute value, sqrt(m^6) = m^3 is false when m = -2

4 0
3 years ago
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