Answer:
1)The dimensions you are solving for
length of the rectangular room
width of the rectangular room
2)Now let the length be L and the width be w then
the perimeter is 2L+2W
2L+ 2W = 24------------------------------(1)
Also, "Twice the length decreased by three times the width is 4 feet" can be written as
2L - 3W = 4-------------------------------------(2)
Solving (1) and (2) by elimination method, by subtraction
2L+ 2W = 24
2L - 3W = 4
(-) (-) (-)
----------------------------
0L +5W = 20
-----------------------------
5W = 20
W = 4--------------------------------------------(3)
Now by substitution method,substituting (3) in(1)
2L+ 2(4) = 24
2L + 8 = 24
2L = 24 -8
2L = 16
L = 8
Substituting in the original equation and rechecking the perimeter
=2(8) + 2(4)
=16 + 8
= 24
Thus the found dimensions are correct
3)The length of the room is 8 feet and the width of the room is 4 feet.
This is the formula of finding the hypotenuse: a² + b² = c².
So the legs would be a and b.
4² + 4² = c²
c² = 16 + 16
c² = 32
c = √32 (Option C)
As a mixed number, that would be 4 2/4, or 4 1/2 simplified.
I hope this answer helped you! Feel free to ask me any more questions if you need help! :)
The answer is 150.
(5)^2(9-3)
25(6)
=150
(5x^2-15x) (-2x+6)
5x (x-3) -2 (x-3)
(5x-2) (x-3)