Solve the following application problem (show your equation and then your solution). Round answers to nearest thousandths and properly label your answer as well.
The length of a rectagle is 5 cm. more than its width. If the area of the rectange is 90 cm^2, find the dimensions.
2006-12-29 08:03:13 · 11 answers · asked by Big Daddy 2 in Education & Reference ➔ Homework Help
7.31 * 12.31
2006-12-29 08:10:26 · answer #1 · answered by -J- 3 · 0⤊ 0⤋
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2016-10-19 04:32:41 · answer #2 · answered by valda 4 · 0⤊ 0⤋
w=width L=length
w+5=L
L*w=90 so w=90/L now if you substitute this in the first equation you get 90/L+5=L and if you multiply the whole thing by L you get
90+5L=L^2 which is equal to L^2-5L-90=0 and now if you use the equation (b+rad b^2-4ac)/2a you get 12.31 for L
and 12.31=5+w by the first equation so you get 7.31for w
L=12.31 w=7.31
2006-12-29 08:27:52 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Area = length x width
90 = (w + 5) x w
or 90 = 2w + 5
Solve the equation for w to get your answer.
2006-12-29 08:12:27 · answer #4 · answered by bugjrmom 3 · 0⤊ 0⤋
x=width
x+5=lenth
x(x+5)=90,
x squared + 5x =90
x squared + 5x -90 =0
From here you need to use the quadratic formula, after which you'll find that x(width)= 7.3107 and x+5(length) = 12.3107
2006-12-29 08:19:46 · answer #5 · answered by waldon l 2 · 0⤊ 0⤋
Let L=length and W=width.
L=W+5
L*W=A
(W+5)W=90
W^2+5W=90
W^2+5W-90=0
This will get you started.
2006-12-29 08:14:44 · answer #6 · answered by slobberknocker_usa 7 · 0⤊ 0⤋
Sorry, not that good in Math but I do understand +2
Maybe check your ? in adult catagories & ask nicely.
2006-12-29 08:06:32 · answer #7 · answered by North of Heaven 3 · 0⤊ 0⤋
w=width, l=5w, 5w * w =90, you do the rest
2006-12-29 08:09:37 · answer #8 · answered by gggjoob 5 · 0⤊ 0⤋
no way im not doing your homework
2006-12-29 08:04:49 · answer #9 · answered by love100 2 · 0⤊ 0⤋
http://www.webmath.com/
2006-12-29 08:11:31 · answer #10 · answered by Anonymous · 0⤊ 0⤋