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notsponge [240]

3 years ago

10

In the polygon to the right, what is the value of x?

1 answer:

olganol [36]3 years ago

5 0

Answer:

14

Step-by-step explanation:

im ur teacher what are u doing

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30 pts.!! Which expression is equivalent to (4√5^3)^1/2? <br>5^3/8 <br>5^6 5^16/3 <br>5^2/3

vampirchik [111]

it would be 5^6 5^16/

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4 years ago

What are the coordinates of the origin?<br> A) (0, 1) <br> B) (x, y) <br> C) (0, 0) <br> D) (1, 0)

ratelena [41]

Answer:

i believe it's c

Step-by-step explanation:

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4 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:

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

$\Rightarrow y=\frac {45}{2}-\frac{3x}{2}$

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 $x$ .

For the domain of area function, from equation (ii)

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

$\Rightarrow x$ [as y>0]

So, the domain of area function is $(0,15m)$ .

For the range of area function:

As $x \rightarrow 0$ or $y\rightarrow 0$ , then $A\rightarrow 0$ [from equation (i)]

$\Rightarrow A>0$

Now, differentiate the area function with respect to $x$ .

$\frac {dA}{dx}=\frac{135}{2}-9x$

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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