After filling the first sandbox to a height of 6 inches, the groundskeeper had 24 cubic feet of sand left over. She used all the leftover sand to fill the second sandbox.
. The second sandbox has a base in the shape of a rectangle.
The base of the second sandbox has a perimeter that is less than 35 feet.
The groundskeeper filled the second sandbox to a height of 6 inches.
What could be the length and width, in feet, of the second sandbox?

Length could be 14.05 ft and width 3.41 ft.

Step-by-step explanation:

The volume of the sand in the second box = 24 ft^3.

Let its base be x feet wide and y feet long.

Then  2(x + y) < 35.

The volume of sand = 0.5xy ft^3.

So 0.5xy = 24

xy = 48.

x = 48/y

Substituting ih the above inequality:

2(48/y + y) < 35

96/y + 2y < 35

2y^2 - 35y + 96 < 0

Solving 2y^2 - 35y + 96 = 0 gives y = 3.41, 14.09.

So the length  could be 14.05 ft and the width 48/14 = 3.41 ft.

length 8ft width 6ft

Step-by-step explanation:

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0