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vodomira [7]
2 years ago
13

What is the ratio of the volume of a cube with edge length six inches to the volume of a cube with edge length one foot? Express

your answer as a common fraction.
Mathematics
1 answer:
gladu [14]2 years ago
3 0

Answer:

The ratio is \frac{1}{8}

Step-by-step explanation:

The volume of a cube with edge length a is:

V = a^{3}

We have two cubes:

Cube 1: The one with edge length of six inches.

Cubs 2: The one with edge length of one foot.

The ratio is:

R = \frac{V_{C1}}{V_{C2}}

In which V_{C1} is the volume of the first cube and V_{C2} is the volume of the second cube.

Volume of the first cube

The edge length is 6 inches, so a = 6.

V_{C1} = 6^{3} = 216 inches^{3}

Volume of the second cube

To express the ratio as a common fraction, both volumes must be in the same unit. So we must pass the edge length of the second cube to inches.

The second cube has edge length of one foot. Each foot has twelve inches. So

V_{C2} = 12^{3} = 1720 inches^{3}

Ratio:

R = \frac{V_{C1}}{V_{C2}} = \frac{216}{1720} = \frac{1}{8}

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Step-by-step explanation:

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3 years ago
Thomas Vega offers to pay $59 of the March cell phone bill. Each of the other 4 members of the family agrees to split the rest o
bija089 [108]
59×2=118 cost of bill.
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or...59 ×5=295
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3 years ago
Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?
cluponka [151]

Under the given transformation, the Jacobian and its determinant are

\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2

so that

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv

where D' is the region D transformed into the u-v plane. The remaining integral is the twice the area of D'.

Now, the integral over D is

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y

but through the given transformation, the boundary of D' is the set of equations,

\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}

We require that u>0, and the last equation tells us that we would also need v>0. This means 1\le v\le\sqrt2 and 0, so that the integral over D' is

\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6

4 0
2 years ago
Shaneeka is saving $5.75 of her allowance each week to buy a new camera that costs $51.75. How many weeks will she have to save
nika2105 [10]

Answer:

a. 9 weeks

Step-by-step explanation:

$51.75÷$5.75= 9 weeks

She will have to save 9 weeks to be able to afford the camera.

6 0
3 years ago
Help its due in 5 mins lol
inessss [21]
Y= 26 & x=35 hope I did it on time!
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