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3 years ago

9

A spotlight can be adjusted to effectively light a circular area of up to 6 meters in diameter. To the nearest tenth, what is th

e maximum area that can be effectively lit by the spotlight?

1 answer:

levacccp [35]3 years ago

3 0

You would find the radius by halving the diameter. Then you use the equation
$\pi \: r { }^{2} =$
and you get about 28.2 meters.

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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 $(0, 253.125 m^2]$ .

4 0

3 years ago

HELP ME I NEED HELLLPPP I AM A 6th grater if u ask HELLLPP ME

Ulleksa [173]

Answer:

Multiply by 3;9

Step-by-step explanation:

9/3=3

12/4=3

18/6=3

27/n=3

N=9

8 0

3 years ago

What equation is graphed in this figure? HELP PLEASE

UkoKoshka [18]

I believe that would be b

5 0

3 years ago

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I need help with this qustion ?

emmainna [20.7K]

The answer should be 32.

4 0

3 years ago

Solve for x: 4 over 5 x + 4 over 3 = 2x x = __________________ Write your answer as a fraction in simplest form. Use the "/" sym

Vesna [10]

4/5x + 4/3 = 2x....there are 2 ways to do this...one with fractions, one without.

with :
4/5x + 4/3 = 2x
4/3 = 2x - 4/5x
4/3 = 10/5x - 4/5x
4/3 = 6/5x
4/3 * 5/6 = x
20/18 = x
10/9 = x

without :
4/5x + 4/3 = 2x....multiply by common denominator of 15
12x + 20 = 30x
20 = 30x - 12x
20 = 18x
20/18 = x
10/9 = x

5 0

3 years ago

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