Respuesta :
Volume of box has a formula of: length × width × height
We know from the question that the length is 2 times of the width, and the height is 2inches greater than the width.
Since all 3 sides are relative to the width, let x be the width,
Width = x inches
Length = 2x inches
Height = (x + 2) inches
We know that Volume = Length × Width × Height:
192 inches^3 = 2x × x × (x + 2) inches
192 = 2x × x × (x + 2)
192 = 2x^2 × (x + 2)
192 =2x^3 + 4x^2
{Divide both sides of equation by 2}
96 = x^3 + 2x^2
{Bring all values to one side of the equation by subtracting 96 from both sides}
0 = x^3 + 2x^2 - 96
{Factorise}
(x - 4) (x^2 + 6x + 24) = 0
{Divide both sides by (x^2 + 6x + 24). 0 divided by (x^2 + 6x + 24) will give 0}
(x - 4) = 0
x = 4
{x^2 + 6x + 24 has no real x-value, so leave it. Do let me know if you'd like me to prove that (x^2 + 6x + 24) has no real x-value :)
Recall:
Width = x inches
Width = 4 inches
Length = 2x inches
Length = 2 × 4 inches
= 8 inches
Height = x + 2 inches
Height = 4 + 2 inches
= 6 inches
Hope this helps! :)
We know from the question that the length is 2 times of the width, and the height is 2inches greater than the width.
Since all 3 sides are relative to the width, let x be the width,
Width = x inches
Length = 2x inches
Height = (x + 2) inches
We know that Volume = Length × Width × Height:
192 inches^3 = 2x × x × (x + 2) inches
192 = 2x × x × (x + 2)
192 = 2x^2 × (x + 2)
192 =2x^3 + 4x^2
{Divide both sides of equation by 2}
96 = x^3 + 2x^2
{Bring all values to one side of the equation by subtracting 96 from both sides}
0 = x^3 + 2x^2 - 96
{Factorise}
(x - 4) (x^2 + 6x + 24) = 0
{Divide both sides by (x^2 + 6x + 24). 0 divided by (x^2 + 6x + 24) will give 0}
(x - 4) = 0
x = 4
{x^2 + 6x + 24 has no real x-value, so leave it. Do let me know if you'd like me to prove that (x^2 + 6x + 24) has no real x-value :)
Recall:
Width = x inches
Width = 4 inches
Length = 2x inches
Length = 2 × 4 inches
= 8 inches
Height = x + 2 inches
Height = 4 + 2 inches
= 6 inches
Hope this helps! :)
The dimensions of the box whose volume is 192 inches³ are as follows;
- Length = 2 × 4 = 8 inches
- Width = 4 inches
- Height = 4 + 2 = 6 inches
.According to the question;
The volume of the box is; V = 192 inches³
- The length is twice as long as it's width, 2x
- A height that is 2 inches greater than the width, (x + 2)
Let the width of the box be represented by, x.
The volume, V of the box is therefore;
- V = l × w × h.
- V = 2x × x × (x+2)
- 192 = 2x³ + 4x²
- 2x³ + 4x² - 192 = 0
- x³ + 2x² - 96 = 0
By factorization;
- (x-4)(x² + 6x + 24) = 0
The roots of (x² + 6x -24) = 0 are complex roots.
As such, we can consider x = 4 as the width of the box.
In essence, the dimensions of the box are as follows;
- Length = 2 × 4 = 8 inches
- Width = 4 inches
- Height = 4 + 2 = 6 inches.
Read more:
https://brainly.com/question/10865653