Answer:
6.1
Step-by-step explanation:
hope it helps<3
Answer:
- 2(L +W) ≤ 600
- W ≤ 200
- L ≥ 2W
Step-by-step explanation:
We assume the problem wording means the length is to be at least 2 times <em>as long as</em> the width. (<em>Longer than</em> usually refers to a difference, not a scale factor.)
If we let "W" and "L" represent the width and length, respectively, then we can translate the problem statement to ...
2(L + W) ≤ 600 . . . . . . the perimeter is twice the sum of length and width
W ≤ 200 . . . . . . . . . . . . the width is at most 200 inches
L ≥ 2W . . . . . . . . . . . . . the length is at least twice the width
BD = 2x-4
But it's also BE - CE + CD
BE is 3x-1
CE is 2x-3
CD is 2
so
BD = 3x-1 -(2x-3) +2
simplified
BD = 1x +4
since BD = BD (obviously), we can also say that the righthand sides of the two equation must be equal
2x -4 = 1x +4
let's solve for x and that put it into either of the 2 equations.
2x -4 = 1x +4
subtract 1x on both sides
x -4 = 4
add 4 on both sides
x = 8
substitute x for its value in 2x -4 = 1x +4
(this way we double check BD in two expressions at once. the equation should come out true, meaning same value for BD in each expression. both sides should be equal).
2*8 -4 = 1*8 +4
12 = 12
seems true
BD is safely 12
