1/5 divided by 3 is what

Mathematics

1 answer:

Vsevolod [243]2 years ago

8 0

Answer:

the answer would be 1/15, or approximately 0.067.

Step-by-step explanation:

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When considering the normal power demand as reported by a utility company, the power demand at midnight is 1040 MW. During the f

anastassius [24]

Answer:

120 MW/hour

Step-by-step explanation:

The formula of the average rate of change between two points in a function is:

Average rate of change (ARC) = f(x2) -f(x1)/(x2-x1)

Let's think the power demad as a function d(x) depending of the hour of the day, so the variable x= hour of the day.

Now we have:

d(5)= 1600

d(8)= 1960

If we apply the mentioned ARC formula = [d(8)-d(5)] MW/(8-5)hour= (1960-1600)MW/3hour= 360/3=120 MW/Hour

3 0

3 years ago

Plot √0.5 on the number line.

Paul [167]

It is roughly 0.707 so it should be very close to .7

3 0

3 years ago

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Find the definite integral from 0 to 1/16 of arcsin(8x)/sqrt(1-64x^2)dx

kolezko [41]

$\displaystyle\int_0^{1/16}\frac{\arcsin8x}{\sqrt{1-64x^2}}\,\mathrm dx$

First let $y=8x$ , so that $\mathrm dx=\dfrac{\mathrm dy}8$ to write the integral as

$\displaystyle\frac18\int_0^{1/2}\frac{\arcsin y}{\sqrt{1-y^2}}\,\mathrm dy$

Now recall that $(\arcsin y)'=\dfrac1{\sqrt{1-y^2}}$ , so substituting $z=\arcsin y$ should do the trick. The integral then becomes

$\displaystyle\frac18\int_0^{\pi/6}z\,\mathrm dz=\frac1{16}z^2\bigg|_{z=0}^{z=\pi/6}=\frac{\pi^2}{576}$

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2 years ago

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Anuta_ua [19.1K]

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3 years ago

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Answer: 4

Step-by-step explanation: