then solve by factoring
another question : a rectangle is 5 ft longer than its wide and its area is 50 square ft. find the dimensions?
2007-11-09 08:34:20 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
sneaking in another questioins
according to the book of records,stuntman fell a distance of 365ft into airbag after jumping from vegas Hotel. The distance d in ft traveled by free falling object in t seconds is given by the formula d=16t^2 to the nearest tenth of a second, how long did the stuntman free fall last ?????
answers 3.6sec
4.8 sec
6.3 sec
5.9 sec
or none of these
2007-11-09 08:59:14 · update #1
Your first question was answered correctly. you just have to distribute the x into the (x-10) by multiplying both terms by x and then combine like terms.
For your second question, the area of a rectangle is length times width.
Let x = width
Let x+5 = length.
(x) * (x+5) = 50
x^2 + 5x = 50
Set the equation equal to zero so that you can factor it.
x^2 + 5x - 50 = 0
(x + 10) (x - 5) = 0
x = -10 and 5
Since this is an area question, you can disregard the -10 because you can't have a negative distance. So your x is 5.
This means that the width is 5 and the length is 10. Hope that helped.
2007-11-09 08:43:42 · answer #1 · answered by npontello12 2 · 0⤊ 0⤋
x(x -10) + sixteen = 0 x^2 -10x + sixteen = 0 x^2 - 8x - 2x + sixteen = 0 (x - 8)(x - 2) = 0 x = 2, 8 enable L = the dimensions of the rectangle. Then its width, W = L - 5 (length is 5 greater advantageous than width) the section is basically A = L*W = 50 L*(L-5) = 50 L^2 - 5L - 50 = 0 L^2 -10L + 5L - 50 = 0 (L -10)(L + 5) = 0 L = 10, -5 as a results of fact the dimensions is obviously beneficial, then L = 10 ft, and W = L - 5 = 5 ft
2016-10-15 22:39:52 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Tom has the correct answer for the first part. For the rectangle...
L = W + 5
LW = 50
(W+5)W = 50
W^2 + 5W - 50 = 0
(W+10)(W-5) = 0
Width = -10 or 5 since you can't have a negitive width
W = 5 and L = 10.
2007-11-09 08:42:42 · answer #3 · answered by J D 5 · 0⤊ 0⤋
x^2 -10x+16 = 0
(x-2)(x-8) = 0
x-2 = 0 or x- 8 = 0
x = 2, 8
h = w + 5
h*w=50
(w+5)*w = 50
w^2+5w = 50
w^2+5w-50 = 0
(w-5)(w+10)=0
w = 5, -10 . . . w = 5 since width cannot be negative
If w = 5, then h = 10
2007-11-09 08:43:44 · answer #4 · answered by mortgagelns 3 · 0⤊ 0⤋
x(x-10)+16=0
xsquared - 10x + 16 = 0
(x-8)(x-2)=0
either, x = 8, or x = 2
rectangle, width = 5 ft, Length = 10ft.
Hope tht helps!
2007-11-09 08:41:07 · answer #5 · answered by Devil Boy 2 · 0⤊ 0⤋
x squared - 10x +16 = 0
2007-11-09 08:55:15 · answer #6 · answered by skitta922 1 · 0⤊ 0⤋
x^2-10+16=0
(x-8)(x-2) = 0
x = 2 or 8
x = width
x+5 = length
x^2+5x = 50
x^2+5x-50 = 0
(x+10)(x-5) = 0
x = 5
x+5 = 10
Area = 5*10 = 50ft^2
2007-11-09 08:44:25 · answer #7 · answered by ironduke8159 7 · 0⤊ 0⤋
x^2 -10x+16
(x-2)(x-8)
2007-11-09 08:38:35 · answer #8 · answered by Tom M 3 · 0⤊ 0⤋