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2 years ago

Please help me i'm not sure of this question i've bin on it for a while now.

Mathematics

2 answers:

photoshop1234 [79]2 years ago

5 0

Answer:

it is so easy

Step-by-step explanation:

the answer is is -5x+60

marin [14]2 years ago

4 0

Answer:

D. y= -5x + 60

Step-by-step explanation:

Have a nice day/night :)

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The coordinates of trapezoid ABCD are A(−4, 3), B(2, 3), C(4, −1) and D(−4, −1). A line segment runs through the trapezoid with

madam [21]

The distance from point Y to the y-axis is 4 units and the distance from point Z to the y-axis is 3 units, then the lelgth of the segment YZ is 4+3=7 units.
If a scale factor is 3, then the length of Y'Z' will be 7·3=21 units.

P.S. In the added picture you can see trapezoid ABCD that was dilated by a scale factor 3 about the origin. This may help to understand that all linear values after dilation become multiplied by scale factor.

5 0

3 years ago

Read 2 more answers

Find the slope of the line that passes through these two points: (-12, -1) and (-3,-4)

ziro4ka [17]

-1/3 is the slope of the line

7 0

3 years ago

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

Explain an example of how taking an amount (for example, $20), decreasing it by 40% and then increasing that amount by 40% does

Likurg_2 [28]

Answer:

See below

Step-by-step explanation:

What we can do is try it out.

20*0.6=12

since 0.6 is left

now we increase 12 by 40, or multiply it by 1.4

12*1.4= 16.8

See what is happening here is that at first when we are decreasing it by 40 percent, we are taking 40 percent of 20. However, when we increase that amount by 40 percent, we are now increasing 40 percent of 20 by 40 percent, or increasing 16 by 40 percent. Therefore, we will not get the original amount.

4 0

3 years ago

Please help with this I am completely stuck on it

vaieri [72.5K]

Answer:

$f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4$