What is the constant rate of change between (3.6,5.2) and (-0.6,7)

Mathematics

1 answer:

Marina86 [1]3 years ago

7 0

Answer:

Investigation 7 introduces a method for computing the ... at a constant rate with respect to the measure of another, then the changes in the ... –5.2 = –3.1∆x ... y = 3.6 + (5.1/–3.5)(x – 12.2) or y = –1.5 + (5.1/–3.5)(x – 8.7).

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Step-by-step explanation:

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$\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill$

$\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}$