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8 < x(7 – x)
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start with \[8<7x-x^2\] then \[x^2-7x+8<0\]
you are going to have so finish by solving \[x^2-7x+8=0\] using the quadratic formula once you know the zeros, since this is a parabola that opens up, it will be negative between the zeros, which is what you are looking for
HELP
i think its 0
x=(-b+-sqrt(b^2-4ac))/2a x=(-(-7)+-sqrt((-7)^2-4*1*8))/2*1 x=(7+-sqrt(49-32))/2 x=(7+-sqrt(17))/2 x1=(7+sqrt(17))/2 x2=(7-sqrt(17))/2 now, plug these points to line number
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