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2 years ago

Please help fast!!!!

Mathematics

1 answer:

Damm [24]2 years ago

8 0

Answer:

BGHjf, it skf skoad 78, then mutli 0867, then you'll get your answer!! 657.

Step-by-step explanation:

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Determine the constant of varlation for the direct variation glven. Rvaries directly with S. When S is 16, R Is 80.

navik [9.2K]

Answer:

The constant of variation is 5.

Step-by-step explanation:

In direct variation, as y varies directly with x, the standard equation is

y = kx,

where k is the constant of variation.

In your case, we use R and S. R varies directly with S, so we have

R = kS

We know that when S = 16, R is 80, so we plug in those values and solve for k, the constant of variation.

R = kS

80 = k(16)

k = 80/16

k = 5

4 0

2 years ago

You are designing a rectangular poster to contain 7575 in2 of printing with a 33-in margin at the top and bottom and a 11-in m

erma4kov [3.2K]

Answer:

The dimensions of the rectangular poster is 15 in by 5 in.

Step-by-step explanation:

Given that, the area of the rectangular poster is 75 in².

Let the length of the rectangular poster be x and the width of the rectangular poster be y.

The area of the poster = xy in².

$\therefore xy=75$

$\Rightarrow y=\frac{75}{x}$ ....(1)

1 in margin at each sides and 3 in margin at top and bottom.

Then the length of printing space is= (x-2.3) in

=(x-6) in

The width of printing space is = (y-2.1) in

=(y-2) in

The area of the printing space is A =(x-6)(y-2) in²

∴ A =(x-6)(y-2)

Putting the value of y

$\Rightarrow A =(x-6)(\frac{75}{x}-2)$

$\Rightarrow A = 87-\frac{450}{x}-2x$

Differentiating with respect to x

$A '= \frac{450}{x^2}-2$

Again differentiating with respect to x

$A''=-\frac{900}{x^3}$

To find the minimum area of printing space, we set A' = 0

$\therefore \frac{450}{x^2}-2=0$

$\Rightarrow 450 =2x^2$

$\Rightarrow x^2=225$

$\Rightarrow x=\pm 15$

Now putting x=±15 in A''

$A''|_{x=15}=-\frac{900}{15^3}$

$A''|_{x=-15}=-\frac{900}{(-15)^3}=\frac{900}{(15)^3}>0$

Since at x=15 , A"<0 Therefore at x=15 , the area will be minimize.

From (1) we get

$y=\frac{75}{x}$

Putting the value of x

$y=\frac{75}{15}$

=5 in

The dimensions of the rectangular poster is 15 in by 5 in.

4 0

2 years ago

ollegr [7]

Answer:

Lower quartile (xL): 16.75

Median (xm): 24.5

Upper quartile (xU): 33.5

4 0

2 years ago

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How do you use equivalent rates in the real world?

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Cross multiplication
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The answer to this is 8

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1 year ago