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3 years ago

Which choice could represent the volume of a cake pan

Mathematics

1 answer:

Sav [38]3 years ago

5 0

Answer:

3150^3

Step-by-step explanation:

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Find derivative problem 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:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

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

5 0

3 years ago

Every 10 seconds, a printing company prints 50 books. How many books are printed in 1 minute?

Svetlanka [38]

Answer:

300 books

Step-by-step explanation:

10 x 6 = 1 min

50 x 6 = 300

8 0

3 years ago

Read 2 more answers

Let f(x) = 1/x . Find the number b such that the average rate of change of f on the interval [2, b] is − 1/8

vekshin1

Answer:

b=4

Step-by-step explanation:

So, we have the function $f(x)=1/x$ . We need to find b such that the average rate of change or the slope is -1/8 between the intervel [2, b]. First, let's find f(2).

f(2) = 1/(2) = 1/2

So, we have the point (2, 1/2)

At point b, f(b) = 1/b.

Let's plug this into the slope formula:

$\frac{y_2-y_1}{x_2-x_1}=\frac{.5-\frac{1}{b} }{2-b} =-1/8$

Now, we just need to solve for b. First, let's multiply both the numerator and denominator by b (to get rid of the annoying fraction in the numerator).

$\frac{.5b-1}{2b-b^2} =\frac{-1}{8}$