The length of a rectangle is 3 cm more than 5 times its width. If the area of the rectangle is 71 cm2, find the dimensions of the rectangle to the nearest thousandth.
2007-02-25 06:25:21 · 5 answers · asked by Specter 3 in Science & Mathematics ➔ Mathematics
length= 3+5w
width= w
So, length X width= Area
(3+5w)(w)=71
3w+5w^2=71
5w^2+3w-71=0
(5w-1)(w+71)=0
Using a calc, the width is around 3.480211...cm
Substitute it into the length equation:
3+5(3.480211...)= 20.4011cm
2007-02-25 06:36:42 · answer #1 · answered by bluefairy421 4 · 0⤊ 0⤋
I would set this up as an algebraic equation with x= width of the rectangle. Let 5x+3= the length of the rectangle. Now since the area of the rectangle is equal to length x width you would multiply 5x+3 times x giving you 5x^2+3x=71cm^2. Subtract 71 and your final equation is 5x^2+3x-71=0. Now I would use the quadratic equation to solve for x.
2007-02-25 06:34:19 · answer #2 · answered by frank3444 2 · 0⤊ 0⤋
Just set up the problem, first assign variable to length = l and width = w
Then l = 3 + 5w
and lw = 71
Now solve these two equations for l and w.
2007-02-25 06:30:30 · answer #3 · answered by rscanner 6 · 0⤊ 0⤋
w : x
L : 5x+3
A = WL
71=x(5x+3)
71 = 5x^2 + 3x
5x^2 + 3x - 71
now just solve for x, don't take nagative numbers
2007-02-25 06:32:49 · answer #4 · answered by 7 · 0⤊ 0⤋
Well I do not know the problem but I hope you can finish up so ......
2007-02-25 10:39:03 · answer #5 · answered by j13 3 · 0⤊ 1⤋