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Masja [62]
3 years ago
9

Find the dimensions of the rectangular corral split into two pens (sharing a fence) of the same size producing the greatest encl

osed area given 630 630 feet of fencing.
Mathematics
1 answer:
dimaraw [331]3 years ago
8 0
             L
--------------------------
|                             |    w
|                             |
--------------------------
|                             |
|                             |  w
--------------------------

Area = l* 2w

fence length: 3l + 4w = 630 =>  3l = 630 - 4w => l = 630 / 3 - (4w/3)

=> l = 210 -4w/3


=> area = (210 - 4w/3) * (2w) = 420w - 8(w^2)/3

The maximum or minimum of a parabola of the form ax^2 + bx + c is at x = -b/2a


So, the maximum of - 8(w^2) / 3 + 420 w is at w = - 420 / ( - 2*8/3) = 78.75

and l = 210 - 4(78.75)/3 = 105.

Answer: l = 105 feet and 2* w = 157.5 feet

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luda_lava [24]

Answer:

hello and thank u for asking questions

the correct answers are...

A. is correct because an obtuse angle is an angle that measures over 90°

B. is incorrect because the acute angle is less than 90°

C. is correct because a reflex is "Reflex angles are angles measuring greater than 180 degrees and less than 360 degrees. The measure of a reflex angle is added to an acute or obtuse angle to make a full 360 degree circle"

D. is incorrect because if it is less than 2 but greater than 4 that does not make any sence

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4 years ago
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nignag [31]

Answer:

  • 3 ft, 144 ft³

Step-by-step explanation:

Find the width as below

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Substitute the values and solve for w

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  • w² + 160 = 169
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Now find the volume

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2 years ago
Solve:-<br><br> 7^5<br><br> Thanks!!!!!!!!!!!!!
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These are exponents. To solve an exponent you need to multiply the base it self how many times its says in the power.

a² 

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<span>2 = power
</span>
7^5 = 7 × 7 × 7 × 7 × 7 = <span>16807
7^5 = </span><span>16807</span>
3 0
3 years ago
Read 2 more answers
A parabola can be drawn given a focus of (-11, -2) and a directrix of x= -3
fgiga [73]

Check the picture below, so the parabola looks more or less like that.

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\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7

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