The area of a rectangle can be calculated using the formula A = length x width. Given that the length is x and the width is 8 - x, the area can be expressed as A = x(8 - x). Expanding this equation results in A = 8x - x². The discussion highlights confusion regarding the initial problem statement, indicating that additional information may be needed for clarity. Ultimately, the relationship between length, width, and area is confirmed through algebraic expansion.
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tayross
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if the length of a rectangle = x,
show that the area of the rectangle is given by A = 8x - x[2]
Neither do I, there must be some information you are not giving us.
Please post the entire question.
Other than that, let us assume for a moment that we know that the length of a rectangle is x and its width is 8 - x.
Then do you agree that its area must be x(8 - x)?
Do you see how opening the brackets gives 8x - x2?
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense.
Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities?
Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots.
Let ##f(x)=27x^{18}+bx^9+70##, then:
$$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$
$$b=27 \frac{q}{p}+70 \frac{p}{q}$$
$$b=\frac{27q^2+70p^2}{pq}$$
From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##.
How to find the greatest value of ##b##?
Thanks