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3 years ago

12

Help plzzz 60 points

2 answers:

Rudiy273 years ago

8 0

Answer:

Neither lines r and s nor lines t and u must be parallel from the information given

Step-by-step explanation:

<2 and <4 are vertical angles so they are always equal

<5 and <7 are vertical angles so they are always equal

We are not given any information to tell if any of the lines are parallel

Neither lines r and s nor lines t and u must be parallel from the information given

Debora [2.8K]3 years ago

7 0

I made a comment about this but I mean, free 30 points.

I think the answer is the last one. It only mentions <2 = <4 and <5 = <7. Nowhere does it say <2 =< 5 for example. So it can be any two lines of which don't have to be parallel to each other.

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The volume of one cloud, in cubic kilometers, grows according to the expression 43x – 1, where x is time in hours. Another cloud

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Step-by-step explanation:

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3 years ago

Pam has 90 m of fencing to enclose an area in a petting zoo with two dividers to separate three types of young animals. The thre

andrew-mc [135]

Answer:

The area function is

$A=\frac{135}{2}x-\frac{9}{2}x^2$ .

The domain and range of A is $(0,15m)$ and $(0, 253.125 m^2]$ .

Step-by-step explanation:

The given length of fencing is $90 m$ .

Let the length and width of each pen be $x$ and $y$ respectively as shown in the figure.

As there are 3 pens, so, the total area,

$A= 3 xy \;\cdots (i)$

From the figure the total length of fencing is $6x+4y$ .

Here, for a significant area for the animals, $x>0$ as well as $y>0$ as $x$ and $y$ are the sides of ben.

From the given value:

$6x+4y=90\;\cdots (ii)$

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Now, from equation (i)

$A=3x\left(\frac {45}{2}-\frac{3x}{2}\right)$

$\Rightarrow A=\frac{135}{2}x-\frac{9}{2}x^2\;\cdots (iii)$

This is the required area function in the terms of variable $x$ .

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$x=15-\frac{2y}{3}$

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For the range of area function:

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$\Rightarrow A>0$

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Equate $\frac {dA}{dx}$ to zero to get the extremum point.

$\frac {dA}{dx}=0$

$\Rightarrow \frac{135}{2}-9x=0$

$\Rightarrow x=\frac{15}{2}$

Check this point by double differentiation

$\frac {d^2A}{dx^2}=-9$

As, $\frac {d^2A}{dx^2}$ , so, point $x=\frac{15}{2}$ is corresponding to maxima.

Put this value back to equation (iii) to get the maximum value of area function. We have

$A=\frac{135}{2}\times \frac {15}{2}-\frac{9}{2}\times \left(\frac {15}{2}\right)^2$

$\Rightarrow A=253.125 m^2$

Hence, the range of area function is $(0, 253.125 m^2]$ .

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