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.
Answer:
There is no solution
Step-by-step explanation:
Let's use distributive property to solve this equation.
4(7+8m) = 8(-1+4m)
28+32m = -8+32m
-32m -32m
28= -8
There is no solution
Hope this helps!
Here are 3 equivalent expressions that would represent the total number of pencils and pens in Karen's backpack.
1. 4p + 6
2. p+ p + p + p + 6
3. 6 + 4(p)
Each of these has four groups of p +6 represented.
The simplified form of this polynomial is (n(20n + 1))/5
Answer:
0.800 or 0.008
Step-by-step explanation: