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4 years ago

Which pairs of points would result in a line with a rate of change of 2:1? select all that apply

Mathematics

1 answer:

Reil [10]4 years ago

4 0

Answer:

A,C,D

Step-by-step explanation:

Complete Question:

"Which pairs of points would result in a line with a rate of change of 2:1? select all that apply"

A.(−4,−11) and (−2−7)

B.(5,6) and (−5, 12)

C.(0,−3)and (−4,−11)

D. (0,−3) and (5, 7)

Rate of change is the slope. 2:1 would mean a slope of simply, 2.

We need to find slope of each of the choices from A - D and see which one has 2. The formula would be:

$m=\frac{y_2-y_1}{x_2-x_1}$

Where m is the slope and

x_1 is first x point

x_2 is second x point

y_1 is first y point

y_2 is second y point

Let's calculate slope of each of the choices.

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{-7+11}{-2+4}\\m=2$

THis one is okay.

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{12-6}{-5-5}\\m=-\frac{3}{5}$

THis one is not 2:1.

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{-11+3}{-4-0}\\m=2$

THis one is okay.

$m=\frac{y_2-y_1}{x_2-x_1}\\m=\frac{7+3}{5-0}\\m=2$

This one is okay.

Hence, A, C, and D all have rate of change of 2:1

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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

Please help!

agasfer [191]

Ok im not sure the answer yet so ima work on it at the same time while im explaining it. (The answer will probably be near the end)

We can use the elimination method to eliminate y out. To do that we multiply the first equation by 3.

6x+3y=-12

Now just subtract it from the other equation.

6x+3y=-12
5x+3y=-6

***x=-6***

Usually after doing the elimination method you will have to solve for x but in this case its already solved for you. If you want to find y now you just take the first equation and fill x with -6 and solve for y.

2(-6)+y=-4
-12+y=-4
y=8

Brainliest my answer if it helps you out?

4 0

4 years ago

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