a rectangle has a length of 8 and a width of 4. four equal rectangles are drawn within that rectangle, creating the longest lengths possible for the rectangles. what is the sum of the perimeters of two of these equal rectangles?

Given:

A rectangle has a length of 8 and a width of 4.

Four equal rectangles are drawn within that rectangle, creating the longest lengths possible for the rectangles.

To find:

The sum of the perimeters of two of these equal rectangles.

Solution:

We have to draw 2 lines which are bisecting the length and width respectively, to divide a rectangle in 4 equal parts with longest lengths.

Half of length 8 = 4

Half of width 4 = 2

It means each rectangle have a length of 4 and a width of 2.

Perimeter of a rectangle is

So, perimeter of each equal rectangle is 12 units.

Sum of the perimeters of two of these equal rectangles is

Therefore, sum of the perimeters of two of these equal rectangles is 24.

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.

it is true

step-by-step explanation:

all the combinations add up i used   calculator