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3 years ago

What is the decimal equivalent to the fraction 80/32

Mathematics

1 answer:

Dafna1 [17]3 years ago

5 0

Answer:

2.5

Step-by-step explanation:

$\frac{80 \div 16}{32 \div 16} = \frac{5}{2}$

first you have to decide by a number that goes into both the numator and denominator in this case you dive the top and bottom by 32. witch equals

$\frac{5}{2}$

then you divide 5 and 2

$5 \div 2$

which equals

$5 \div 2 = 2.5$

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Step-by-step explanation:

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2 years ago

the width of a rectangle is 2/3 length, The perimeter of the rectangle is 100 ft. What is the width, in feet, of the rectangle.

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Step-by-step explanation:

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3 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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3 years ago