SOLUTION: A rectangle has length ( x + 11) in width 2x in . the perimeter of the rectangle is at least 47 inches and at most 52 inches .find the greatest value of x that satisfies the cond
Algebra ->
Quadratic Equations and Parabolas
-> SOLUTION: A rectangle has length ( x + 11) in width 2x in . the perimeter of the rectangle is at least 47 inches and at most 52 inches .find the greatest value of x that satisfies the cond
Log On
Question 323595: A rectangle has length ( x + 11) in width 2x in . the perimeter of the rectangle is at least 47 inches and at most 52 inches .find the greatest value of x that satisfies the condition Answer by JBarnum(2146) (Show Source):
You can put this solution on YOUR website! x+11+x+11+2x+2x
6x+22=47 = 6x=25 =
6x+22=48 = 6x=26 =
6x+22=49 = 6x=27 =
6x+22=50 = 6x=28 =
6x+22=51 = 6x=29 =
6x+22=52 = 6x=30 = x=5
all the other answers would be less than 5 so x=5 is greatest
