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3 years ago

Someone please help me asap!!!!

Mathematics

1 answer:

sergejj [24]3 years ago

3 0

21.) is clue
I don't know the others I'm sorry...

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What is the exact circumference of a circle with a radius of 12 inches?

kvv77 [185]

The radius of a circle is half the length of the circumference.
So multiply 12 by itself and you get 24.
The answer is C.

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4 years ago

Read 2 more answers

If possible, find AB. & State the dimension of the result.

Nikitich [7]

Answer:

Step-by-step explanation:

$A=\begin{bmatrix}0 &0 &5 \\ 0 &0 &-3 \\ 0 &0 &3 \end{bmatrix}$

$B=\begin{bmatrix}8 &-12 &5 \\ 7 &19 &5 \\ 0 &0 &0 \end{bmatrix}$

A.B = A × B

$A.B=\begin{bmatrix}0 &0 &0 \\ 0 &0 &0 \\ 0 &0 &0 \end{bmatrix}$

Dimension of the resultant matrix is (3 × 3)

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3 years ago

Surface area of a sphere is 615.752 m2, what is the volume of the sphere?

const2013 [10]

the volume (V) of a sphere = $\frac{4}{3}$ πr³

We require to know the radius r to calculate the volume

using surface area of a sphere = 4πr² = 615.752

divide both sides by 4π

r² = $\frac{615.752}{12.56}$ = 49.0248...

take the square root of both sides

r = $\sqrt{49.0248..}$ ≈ 7, hence

V= $\frac{4}{3}$ × π × 7³ = 1436.76 m³ → B

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4 years ago

a store is selling uniform shirts for 18.95 each. if samuel has a coupon for 1/4 off the price, how much would he pay for the sh

balandron [24]

First take the original price and multiple by 1/4 or .25. You should get 4.73 then you subtract that from 18.95 and your answer is 14.22

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3 years ago

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