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3 years ago

The figures are similar. Give the ratio of the perimeters and the ratio of the areas of the first figure to the second.

Mathematics

1 answer:

pickupchik [31]3 years ago

5 0

Answer: first option.

Step-by-step explanation:

The ratio of the area of the triangles can be calculated as following:

$ratio_{(area)}=(\frac{l_1}{l_2})^2$

Where $l_1$ is the lenght of the given side of the smaller triangle and $l_2$ is the lenght of the given side of the larger triangle.

Therefore:

$ratio_{(area)}=(\frac{28}{32})^2=\frac{49}{64}$

It can be written as following:

$ratio_{(area)}=49:64$

The ratio of the perimeter is:

$ratio_{(perimeter)}=\frac{l_1}{l_2}$

Where $l_1$ is the lenght of the given side of the smaller triangle and $l_2$ is the lenght of the given side of the larger triangle.

Therefore:

$ratio_{(perimeter)}=\frac{28}{32}=\frac{7}{8}$

It can be written as following:

$ratio_{(perimeter)}=7:8$

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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:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

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$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

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We can setup an inequality as,

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