How does a focus differ from an epicenter?
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Lets calculate the parameters of the problem. Suppose that the width is x feet. THen, the length is x+30 feet. The area of this parallelogram is given by the product of length and width, hence x(x+30). This has to be equal to 600.
Hence, x is the solution to the equation x(x+30)=600.
x*x+30x=600 hence x*x+30x-600=0. This is a second degree equation and we can calculate Δ=b*b-4ac=3300. The general solution is given by:
.
Substituting, we get that x=-30/4+
or x=-7.5+25*
. This is the width; the length is given by x+30.
Hence, x is the solution to the equation x(x+30)=600.
x*x+30x=600 hence x*x+30x-600=0. This is a second degree equation and we can calculate Δ=b*b-4ac=3300. The general solution is given by:
Substituting, we get that x=-30/4+