SOLUTION: A rectangle has an area of 2x^2 + 2x - 112.
A) if one side is (x-7) cm, show that the perimeter is 6(x+3) cm.
B) hence it otherwise, find the range of possible values for x
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Rectangles
-> SOLUTION: A rectangle has an area of 2x^2 + 2x - 112.
A) if one side is (x-7) cm, show that the perimeter is 6(x+3) cm.
B) hence it otherwise, find the range of possible values for x
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Question 1118941: A rectangle has an area of 2x^2 + 2x - 112.
A) if one side is (x-7) cm, show that the perimeter is 6(x+3) cm.
B) hence it otherwise, find the range of possible values for x Answer by solver91311(24713) (Show Source):
It is impossible for the given function to represent the area of a rectangle. The area of a rectangle has a lower limit of zero (make the positive length as long as you like and the positive width as close to zero as you like, or vice versa) and an upper limit of the area of the rectangle when the length equals the width, i.e. a square.
The graph of the given function which has a positive lead coefficient is a convex UP parabola. It has an absolute minimum function value of -112. This means that when the length and width of the rectangle are equal, namely negative(?) one-half centimeters, then the area is negative(?) 112 square centimeters. Such a result is ludicrous in the extreme.
Furthermore, since the value of the function increases without bound as the value of either increases or decreases without bound, you could make the area of the rectangle as large as you want by taking as large as you like or as small as you like. The very idea is an affront to every Geometer since and including Pythagoras.
The entire problem is nonsense.
John
My calculator said it, I believe it, that settles it
