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2 years ago

15

How long is JK?

1 answer:

-Dominant- [34]2 years ago

5 0

Answer:

C) 3 times the sq root of 3

Step-by-step explanation:

In a right angle triangle, the COS is definned as the ajacent side over the hypotenuse. So

COS 60 = JK/JL

COS 60 = JK/(6*sq root 3)

.5 = JK/(6*3**1/2)

JK = .5 *(6*3**1/2)

JK = 3*3**1/2

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Which number line shows the solution 1/2x-2>0

Ymorist [56]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$\frac{1}{2}x-2>0$

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3 years ago

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3 years ago

BARSIC [14]

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3 years ago

Read 2 more answers

A rancher wishes to build a fence to enclose a rectangular pen having area 24 square yards. Along one side the fence is to be ma

Gnoma [55]

Answer:

The value is $C \approx \$76$

Step-by-step explanation:

From the question we are told that

The area of the rectangular pen is $A = 24 \ yard^2$

The cost of material used to make one side is $z = \$ 6$

The cost of material used to make the other sides is $r = \$ 3$

Now , the fence to be build around the rectangular pen has four sides, the first opposite sides are equal, let assume each of the to be x yard and the other opposite sides are also equal as well let assume of the to be y yard

So the cost is mathematically represented as

$C = zx + r (x + 2y )$

=> $C = 6x + 3(x + 2y)$

=> $C = 9x + 6y$

Now the area of the fence is mathematically represented as

$A = x* y = 24$

=> $y = \frac{24}{x}$

=> $C = 9x + 6[\frac{24}{x} ]$

=> $C = 9x + [\frac{144}{x} ]$

Now differentiating

$C' = 9 + 144* (-2) x^{-2}$

$C' = 9 - 288x^{-2}$

At minimum $C' = 0$

So

$9 - 288x^{-2} = 0$

$x^{-2} = 0.03125$

$x = \sqrt{\frac{1}{0.03125} }$

$x = 5.66$

Now substituting for x in the equation above to obtain minimum cost

$C = 9(5.66) + [\frac{144}{5.66} ]$

$C \approx \$76$

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3 years ago

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qaws [65]

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