The height, width and length of a small box are consecutive integers with the height being the smallest of the three dimensions. If the length and width are increased by 2 cm each and the height is tripled,
then the volume of the box is increased by 552 cm3
. Use algebraic techniques to find the dimensions of
the original small box.

9514 1404 393

  4 cm, 5 cm, 6 cm

Step-by-step explanation:

If x represents the height of the original box, the other two dimensions are ...

  width = x+1

  length = x+2

and the volume is ...

  V = HWL = x(x +1)(x +2) = x³ +3x² +2x

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After the change in dimensions, the height is 3x, the width is x+3, and the length is x+4. The new volume is ...

  V = HWL = 3x(x+3)(x+4) = 3x³ +21x² +36x

The difference in volumes is ...

  (3x³ +21x² +36x) -(x³ +3x² +2x) = 552

In standard form, this equation is ...

  2x³ +18x² +34x -552 = 0

Factoring out 2 gives ...

  x³ +9x² +17x -276 = 0

Possible rational roots are 1, 2, 3, 4, 6, and other divisors of 276. The largest possibility worth considering is less than ∛276 ≈ 6.5. Trial and error, or a graphing calculator can show us that x=4 is the only real solution to this equation.

The dimensions of the original small box are 4 cm × 5 cm × 6 cm.

step-by-step explanation:

once you graph it hit 2nd (the blue button) and graph

that takes you to a table of coordinates

on the left is x coordinates and on the right is y coordinates and an empty box

(you may have to scroll up/down in order to get ( x, 0)

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !