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3 years ago

HELP HELL HELP HELP HELP HELP HELP HELL

Mathematics

2 answers:

atroni [7]3 years ago

5 0

J is your answer to the question

Paraphin [41]3 years ago

3 0

Answer:

Step-by-step explanation:

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Speed of a 747 airplane is 9,944 inches per second. What is the nearest foot per second

laiz [17]

Answer:

12 inches = 1 foot

9,944 inches = 9,944/12 = 826.67 ft

3 0

2 years ago

Solve for x. 7x−9<12 or 14x+8>11

Vitek1552 [10]

Answer:

For 7x-9<12, the answer is x<3

For 14x+8>11, the answer is x>3/14

Step-by-step explanation: Isolate the variables by dividing each side by factors that do not contain the variable.

I hope this helps you out! :)

4 0

2 years ago

Home help plz really appreciate it!!!!!

Schach [20]

Answer:

angle AGE=113.5

Step-by-step explanation:

3x-28=66-x

4x=94

x=23.5

23.5+90=113.5

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3 years ago

Read 2 more answers

4th grade work help my brotherrr please!!

exis [7]

4 x 20 and that is 160

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2 years ago

Read 2 more answers

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.

5 0

3 years ago