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emmasim [6.3K]
3 years ago
8

The volume of a box is 80 cubic feet with length x-3 and width x-1 and height x+5. What are the possible values of x? What are t

he possible dimensions?

Mathematics
1 answer:
Kryger [21]3 years ago
4 0

Answer:

  • the only possible value of x is 5
  • the dimensions are 2 × 4 × 10

Step-by-step explanation:

The cubic equation ...

  (x -3)(x -1)(x +5) = 80

has one real root: x = 5. Using that value for x, the dimensions become ...

  length = 5 - 3 = 2

  width = 5 - 1 = 4

  height = 5 + 5 = 10

The dimensions are (length, width, height) = (2, 4, 10).

_____

We cannot tell the thrust of the problem, since it has only one solution. Perhaps you're supposed to write the cubic in standard form and use the <em>Rational Root theorem</em> to find <em>possible values of x</em>. That form can be found to be ...

  (x -3)(x -1)(x +5) -80 = 0

  x³ +x² -17x -65 = 0

Descartes' rule of signs tells you there is one positive real root. The rational root theorem tells you possible rational roots are factors of 65:

  1, 5, 13, 65

We know that x must be greater than 3 (so all dimensions are positive). Thus <em>possible values of x are 5, 13, 65</em>, and we're pretty sure that 65 is way too large.

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Answer:

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Step-by-step explanation:

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(1). Row one = (10 + 12 + 12 + 14 ) = 48/4 = 12.

(2). Row Two: (12 + 11 + 13 + 16 ) = 52/4 = 13.

(3). Row three : (11 + 13 + 14 + 14)/4 = 13.

(4). Row four: (11 + 10 + 7 + 8)/4 = 36/4 = 9.

(5). Row five: (13 +12 + 14 + 13)/4 = 52/4 = 13.

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Minimum value for each row:

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