The weight needed to balance a lever varies inversely with the distance

Mathematics

1 answer:

Andrews [41]3 years ago

5 0

Let w represents the weight and d represents the distance.

It is given that

The weight needed to balance a lever varies inversely with the distance from the fulcrum to the weight. A 120-lb weight is placed on a lever, 5 ft from the fulcrum.

$w = \frac{k}{d}$

Substituting the values of w and d, we will get

$120= \frac{k}{5}
\\
k =600lb \ ft$

Using the value of k which is 600, with the above equation and the given value of d which is 8, we will get

$w = \frac{600}{8} = 75lb$

Therefore the correct option is B .

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Quick pls

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First, let's see how 23 compares with the squares of the positive whole numbers on the number line.

1² = 1

2² = 4

3² = 9

4² = 16

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The value of 23 is right between the square of 4 and the square of 5. Thus, the value √23 will be between 4 and 5.

Since 23 is much, much closer to the square of 5 than the square of 4, we can assume that the value √23 will be closer to 5 on the number line than 4.

Look at the attached image to see where I plotted the approximate location of √23.

You will realize that this approximation is pretty close since the actual value is roughly 4.80.

Let me know if you need any clarifications, thanks!