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xenn [34]
2 years ago
13

When completely filled, a spherical storage tank

Mathematics
1 answer:
mamaluj [8]2 years ago
6 0

Answer:

radius: 12

Given the following:

Volume of sphere: 7,240

Formula of volume of sphere:

Volume = 4/3 π (radius)³

Find radius:

  • 4/3 π (radius)³ = 7,240

<u>change sides</u>

  • (radius)³ = 7,240/(4π/3)

<u>simplify</u>

  • (radius)³ = 1728.422682

<u>cube root on both sides</u>

  • radius = ∛1728.422682

<u>simplify</u>

  • radius = 12
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The power generated by an electrical circuit (in watts) as a function of its current x xx (in amperes) is modeled by P ( x ) = −
marysya [2.9K]

Given:

The power generated by an electrical circuit (in watts) as a function of its current x (in amperes) is modeled by

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To find:

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Solution:

We have,

P(x)=-15x(x-8)

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Differentiate with respect to x.

P'(x)=-15(2x)+120(1)

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To find the extreme point equate P'(x)=0.

-30x+120=0

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Divide both sides by -30.

x=4

Differentiate (i) with respect to x.

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P'(x)=-30 (Maximum)

It means, the given function is maximum at x=4.

Therefore, the current of 4 amperes will produce the maximum power.

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4 years ago
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Vera_Pavlovna [14]

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Find derivative problem<br> Find B’(6)
dalvyx [7]

Answer:

B^\prime(6) \approx -28.17

Step-by-step explanation:

We have:

\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]

We can move the constant outside:

\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]

Now, we will utilize the product rule. The product rule is:

(uv)^\prime=u^\prime v+u v^\prime

We will let:

\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t

Then:

\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1

(The derivative of u was determined using the chain rule.)

Then it follows that:

\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}

Therefore:

\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]

By simplification:

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So, the slope of the tangent line to the point (6, B(6)) is -28.17.

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