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sveticcg [70]
3 years ago
9

Solve for x. x^3=8/125 Enter your answer in the box as a fraction in simplest form.

Mathematics
1 answer:
Dahasolnce [82]3 years ago
8 0
First you divide 8 by 125, with a new equation of:
x^3=0.064
You then take the cube root of 0.064, which happens to be 0.4
x=0.4
To make sure you can do 0.4*0.4*0.4.
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Find the volume of the figure
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MAXImum [283]
37.1
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98 POINTS! PLEASE HELP! I been stuck on this assignment for over 2 weeks. My teacher gave me a simple explanation when I asked f
Colt1911 [192]

See the attached picture.

  • Create an equation for the volume of the box, find the zeroes, and sketch the graph of the function.

The resulting box has a volume

V(x)=x(8-x)(12-x)=x^3-20x^2+96x

because the volume of a box is the product of its width (12-x), length (8-x), and height (x).

  • find the zeroes

You know right away from the factored form of V(x) that the zeroes are x=0,8,12. (zero product property)

  • sketch the graph of the function

Easy to plot by hand. You know the zeroes, and you can check the sign of V(x) for any values of x between these zeros to get an idea of what the graph of V(x) looks like. See the second attached picture.

Here's what I mean by "check the sign" in case you don't follow. We know V(x)=0 when x=0 and x=8. So we pick some value of x between them, say x=1, and find that

V(1)=1(8-1)(12-1)=7\cdot11=77

which is positive, so V(x) will be positive for any other x between 0 and 8. Similarly we would find that V(x) for x between 8 and 12, and so on.

  • What is the size of the cutout he needs to make so that he can fit the most marbles in the box?

It's impossible to answer this without knowing the volume of each marble...

  • If Thomas wants a volume of 12 cubic inches, what size does the cutout need to be?

Thomas wants V(x)=12, so you solve

x^3-20x^2+96x=12

While this is possible to do by hand, the procedure is tedious (look up "solving the cubic equation"). With a calculator, you'd find three approximate solutions

x\approx0.1284

x\approx7.6398

x\approx12.2318

but you throw out the third solution because, realistically, the cutout length can't be greater than either of the sheet's dimensions.

  • What would be the dimensions of this box?

The box's dimensions are (x in) x (8-x in) x (12-x in).

If x\approx0.1284, then 8-x\approx7.8716 and 12-x\approx11.8716.

If x\approx7.6398, then 8-x\approx0.3602 and 12-x\approx4.3602.

8 0
2 years ago
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