SOLUTION: the length of a rectanlge is twice the width. The area is 128yd^2. I know that A=2(L)+2(w). But with this equation wouldn't it be calcutaed differently? given the equation -0.4x^2+

Question 329425: the length of a rectanlge is twice the width. The area is 128yd^2. I know that A=2(L)+2(w). But with this equation wouldn't it be calcutaed differently? given the equation -0.4x^2+11.2x+14 for an event selling tickets. I know that the graph would open downward because of the negative 0.4x. Now it is asking the# if days since the 1st. day of tickets sold, when the sales peaked and how many sold that day. I'm assuming that the quadratic equation should apply here?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
the length of a rectanlge is twice the width. The area is 128yd^2. I know that A=2(L)+2(w). But with this equation wouldn't it be calcutaed differently?
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2L + 2W is the perimeter
Area = L*W
L = 2W
A = 2W*W = 128
W^2 = 64
W = 8
L = 16
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given the equation -0.4x^2+11.2x+14 for an event selling tickets. I know that the graph would open downward because of the negative 0.4x. Now it is asking the# if days since the 1st. day of tickets sold, when the sales peaked and how many sold that day. I'm assuming that the quadratic equation should apply here?
That's not an equation, there's no equal sign.

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