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3 years ago

15

the point-slope form of the question of the line that passes through (-4,-3) and (12,1) is y-1=1/4(x-12). what is the standard f

orm of the equation for this line?

1 answer:

Alexeev081 [22]3 years ago

5 0

Well first off you multiply the parenthesis 1/4(x-12) you get -1=1/4x-3 then you subtract the one so now the y can be by itself the you get Y=1/4x-2

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2 years ago

For each of the transactions in items 1 through 5, indicate the effects on the accounting equation elements.

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2 years ago

Approximate the change in the volume of a sphere when its radius changes from r = 40 ft to r equals 40.05 ft (Upper V (r )equal

alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere ( $V$ ), measured in cubic feet, is represented by the following formula:

$V = \frac{4\pi}{3}\cdot r^{3}$

Where $r$ is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

$\Delta V = \frac{\partial V}{\partial r}\Delta r$

The derivative of the volume as a function of radius is:

$\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}$

Then, the change in volume is expanded:

$\Delta V = 4\pi \cdot r^{2}\cdot \Delta r$

If $r = 40\,ft$ and $\Delta r = 40\,ft-40.05\,ft = 0.05\,ft$ , the change in the volume of the sphere is approximately:

$\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)$

$\Delta V \approx 1005.310\,ft^{3}$

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

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