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adelina 88 [10]
3 years ago
13

Helpppp pleaseeee , thank you .

Mathematics
1 answer:
34kurt3 years ago
4 0

Answer:

20 yds

Step-by-step explanation:

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marshall27 [118]

Answer:

IM preeetty sure the answer is x=4 and y=26 so the solution would be (4,26)

Step-by-step explanation:

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3 years ago
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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:

\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.

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3 years ago
12
ruslelena [56]
I need more information to answer this question.
8 0
3 years ago
Round .859 to nearest tenth
Nikolay [14]
0.9. The reason why is because the 5 brings the 8 to a 9.
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3 years ago
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Bateman's jar of marbles contains 8 blue marbles. 6 yellow marbles, and 9 tiger eyes. Bateman pulls a blue marble without puttin
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Answer: no

Step-by-step explanation:

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