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Lana71 [14]
3 years ago
10

The area of the base of a rectangular juice box is 4 1/2 sq.in. If the volume of the box is 18 cubic inches, how tall is the box

?
Mathematics
1 answer:
Ksju [112]3 years ago
6 0
Volume = Area x Height. We are given Volume and Area, so plug in and solve for Height (Height = Volume/Area)

18in^3 = 4.5 in^2 x Height (divide both sides by 4.5) -> 4in = Height
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Can anyone tell the answer for the 3rd qouestion
N76 [4]

Answer:

Your first 2 questions are correctly answered.

I think the answer to the third question is NO.

Step-by-step explanation:

You can see that the sides with 3 and 4 are adjacent, which means they cannot possibly be opposite. For a valid dice, they should be opposite because they add up to 7. Hence this net is not valid.

8 0
3 years ago
10 pens cost £6.50.<br> Find the cost of 4 pens
Setler [38]
10 pens=£6.50
1 pen=£0.65
4 pens=£0.65×4
4 pens=£2.6
3 0
3 years ago
How do you find the volume of the solid generated by revolving the region bounded by the graphs
d1i1m1o1n [39]

Answer:

About the x axis

V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}

About the y axis

V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}

About the line y=8

V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}

About the line x=2

V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi

Step-by-step explanation:

For this case we have the following functions:

y = 2x^2 , y=0, X=2

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on  y from 0 to 8.

We can find the area like this:

A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4

And we can find the volume with this formula:

V = \int_{a}^b A(x) dx

V= 4\pi \int_{0}^2 x^4 dx

V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}

About the y axis

For this case we need to find the function in terms of x like this:

x^2 = \frac{y}{2}

x = \pm \sqrt{\frac{y}{2}} but on this case we are just interested on the + part x=\sqrt{\frac{y}{2}} as we can see on the second figure attached.

We can find the area like this:

A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})

And we can find the volume with this formula:

V = \int_{a}^b A(y) dy

V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy

V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)

And we can find the volume with this formula:

V = \int_{a}^b A(x) dx

V= \pi \int_{0}^2 64-32x^2 +4x^4 dx

V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}

And we can find the volume with this formula:

V = \int_{a}^b A(y) dy

V= \frac{\pi}{2} \int_{0}^8 y dy

V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi

6 0
3 years ago
before conducting some experiments, a scientist mixes 1/2 gram of Substance A with 3/4 gram of Substance B. If the scientist use
horrorfan [7]

Mixing the two ingredients would be adding them together:

1/2 gram + 3/4 gram = 1 1/4 total grams


Divide the total grams by 1/8 to find the number of experiments they can do:

1 1/4 / 1/8 = 5/4 / 1/8 = 5/4 * 8/1 = 40/4 = 10


They can conduct 10 experiments.


5 0
4 years ago
Translate this sentence into an equation. 59 is the sum of 11 and Mai’s score
trapecia [35]

Answer:

11 + Mai's Score = 59

Step-by-step explanation:

You need to add 11 and Mai's score together to get 59, so with the values given we can make the equation 11 + Mai's Score = 59.

*depending on the question, Mai's score may need to be said as a letter variable, so:

If m = mai's score,

11 + m = 59

I hope this helped! :)

3 0
3 years ago
Read 2 more answers
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