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3 years ago

10

The volume of a cylinder is 8.478 cubic inches, and the radius is 15 inches. Which measurement is closest to the height of this

cylinder in cubic inches?
A. 90 in
B. 12 in
C. 48
D. 8

1 answer:

Harrizon [31]3 years ago

6 0

Answer:

No correct answer

Step-by-step explanation:

volume of a cylinder= pi*r2*height

making height the subject of formula

height = volume/pi*r2

height= 8.478/pi*15*15 = 0.012in

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Step-by-step explanation:

there are 6 squares with area of 4 so 4 x 6 is 24

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Stephen and Alice are both reading the same book for a book club. Stephen reads 278 pages in 7 weeks. Alice reads 31 pages each

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A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0

3 years ago

6) If sec theta+tan theta = P. PT sin theta=P^2-1/P^2+1 ...?

jarptica [38.1K]

I will use the letter x instead of theta.

Then the problem is, given sec(x) + tan(x) = P, show that

sin(x) = [P^2 - 1] / [P^2 + 1]

I am going to take a non regular path.

First, develop a little the left side of the first equation:

sec(x) + tan(x) = 1 / cos(x) + sin(x) / cos(x) = [1 + sin(x)] / cos(x)

and that is equal to P.

Second, develop the rigth side of the second equation:

[p^2 - 1] / [p^2 + 1] =

= [ { [1 + sin(x)] / cos(x) }^2 - 1] / [ { [1 + sin(x)] / cos(x)}^2 +1 ] =

= { [1 + sin(x)]^2 - [cos(x)]^2 } / { [1 + sin(x)]^2 + [cos(x)]^2 } =

= {1 + 2sin(x) + [sin(x)^2] - [cos(x)^2] } / {1 + 2sin(x) + [sin(x)^2] + [cos(x)^2] }

= {2sin(x) + [sin(x)]^2 + [sin(x)]^2 } / { 1 + 2 sin(x) + 1} =

= {2sin(x) + 2 [sin(x)]^2 } / {2 + 2sin(x)} = {2sin(x) ( 1 + sin(x)} / {2(1+sin(x)} =

= sin(x)

Then, working with the first equation, we have proved that [p^2 - 1] / [p^2 + 1] = sin(x), the second equation.

4 0

4 years ago