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Alika [10]
3 years ago
13

Jeanetta collects antique picture frames. In her first year of collecting, she collects 10 frames. Each year, she hopes to colle

ct
three more frames than the previous year. The number of frames Jeanetta will have after 10 years of collecting is represented
by the expression below.
10
2 [10+3(n-1)]
=1
How many frames will Jeanetta have after 10 years of collecting?
•235
•255
•1,640
•1,649
Mathematics
1 answer:
GREYUIT [131]3 years ago
5 0

Answer:

235

Step-by-step explanation:

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Frederick designed an experiment in which he spun a spinner 20 times and recorded the results of each spin.
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Answer:

the statements given above are true.

Step-by-step explanation:

Given that Frederick designed an experiment in which he spun a spinner 20 times and recorded the results of each spin.  

He spun a 4 five times.

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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:

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\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:

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Then:

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(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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