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andrezito [222]
3 years ago
9

If x = 1/2 what is the value of the following expression?

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
4 0

Answer:

=-2x+14

Step-by-step explanation:

\left(\frac{1}{2}\right)^2\cdot \:8-2x+\left(5+7\right)\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=\left(\frac{1}{2}\right)^2\cdot \:8-2x+5+7\\\left(\frac{1}{2}\right)^2\cdot \:8=2\\\left(\frac{1}{2}\right)^2\cdot \:8\\\left(\frac{1}{2}\right)^2=\frac{1}{2^2}\\\left(\frac{1}{2}\right)^2\\\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}\\=\frac{1^2}{2^2}\\\mathrm{Apply\:rule}\:1^a=1\\1^2=1\\=\frac{1}{2^2}\\=8\cdot \frac{1}{2^2}

\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}\\=\frac{1\cdot \:8}{2^2}\\\mathrm{Multiply\:the\:numbers:}\:1\cdot \:8=8\\=\frac{8}{2^2}\\\mathrm{Factor}\:8:\quad 2^3\\\mathrm{Factor\:}8=2^3\\=\frac{2^3}{2^2}\\\mathrm{Cancel\:}\frac{2^3}{2^2}:\quad 2\\\frac{2^3}{2^2}\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}\\\frac{2^3}{2^2}=2^{3-2}\\=2^{3-2}\\\mathrm{Subtract\:the\:numbers:}\:3-2=1\\=2\\=2-2x+5+7\\Group\:like\:terms\\=-2x+2+5+7

\mathrm{Add\:the\:numbers:}\:2+5+7=14\\=-2x+14

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<u>Step-by-step explanation:</u>

Here we have , A manufacturer finds that the revenue generated by selling of cortan commodity is given by function R(x)=80x-0.2x^ 2 , where he maximum reveremany should be manufactured to obtain this maximum units .Let's find out:

We have following function as R(x)=80x-0.2x^ 2 . Let's differentiate this and equate it to zero to find value of x for which the function is maximum!

⇒ R(x)=80x-0.2x^ 2

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