Question 165522
volume = length times width times height.
let V = volume
let L = length
let W = width
let H = height
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V = L * W * H (equation 1)
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The length of a rectangular box is 1 inch more than twice the height of the box.
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L = 2 * H + 1 (equation 2)
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the width is 3 inches more than the height.
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W = H + 3 (equation 3)
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the volume of the box is 126 cubic inches.
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V = 126 cubic inches.
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you start with equation 1
V = L * W * H (equation 1)
you use equation 2 and equation 3 to substitute for W and L.
L = 2 * H + 1 (equation 2)
W = H + 3 (equation 3)
equation 1 becomes:
V = (2 * H + 1) * (H + 3) * H
126 = (2*H+1)*(H+3)*H
this becomes
2h^3 + 7h^2 + 3h - 126 = 0
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i couldn't solve it so i went to an online calculator and asked it to do the dirty work for me.
first calculator i went to couldn't handle it, but the second calculator i went to did.
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the real solution (there were others but they were complex (real + imaginary)) was h = 3.
i plugged that into the equation and it proved to be good.
i used H = 3 and went back to equations 2 and 3 to get the values for L and W.
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H = 3
L = 2 * H + 1 (equation 2)
W = H + 3 (equation 3)
from equation 2:
L = 7
from equation 3:
W = 6
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dimension are:
L = 7
W = 6
H = 3
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V = 126
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V = L*W*H
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L*W*H = 126
7*6*3 = 126
126=126
answer is good.
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sorry i can't help you with the TI-83.
if you want to us an online calculator, this is the one that solved the equation for me.
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http://www.ifigure.com/math/algebra/algebra.htm
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i clicked on "equations" in the solving equations section.
then i clicked on "go to the solve page" in the solve section.
then i entered my equation and i told it the letter i wanted it to solve for (h in my case), and it gave me the solutions.
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