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3 years ago

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Miguel draws △JKL with vertices J(−12, −1) , K(12, 9) , and L(12, −1) . Then he dilates the triangle with a scale factor of 6 to

create △J′K′L′. How are the areas of the triangles related?
Enter your answer in the box.
The area of △J′K′L′ is [__?__] times the area of △JKL.

1 answer:

Zina [86]3 years ago

3 0

Ik I'm late but for my homies in k12: 36

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The total cost (in hundreds of dollars) to produce x units of a product is c(x) = (3x-2) / (8x+1), find the average cost for eac

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Answer:

a) $\frac{74}{10025}$

b) $\frac{3x-2}{x(8x+1)}$

c) $\frac{-24x^2+32x-2}{(8x^2+x)^2}$

Step-by-step explanation:

For total cost function $c(x)$ , average cost is given by $\frac{c(x)}{x}$ i.e., total cost divided by number of units produced.

Marginal average cost function refers to derivative of the average cost function i.e., $\left ( \frac{c(x)}{x} \right )'$

Given: $c(x)=\frac{3x-2}{8x+1}$

Average cost = $\frac{c(x)}{x}=\frac{3x-2}{x(8x+1)}$

a)

At x = 50 units,

$\frac{c(50)}{50}=\frac{150-2}{50(400+1)}=\frac{148}{50(401)}=\frac{74}{10025}$

b)

Average cost = $\frac{c(x)}{x}=\frac{3x-2}{x(8x+1)}$

c)

Marginal average cost:

Differentiate average cost with respect to $x$

Take $f=3x-2\,,\,g=8x^2+x$

using quotient rule, $\left ( \frac{f}{g} \right )'=\frac{f'g-fg'}{g^2}$

Therefore,

$\left ( \frac{c(x)}{x} \right )'=\left ( \frac{3x-2}{8x^2+x} \right )'\\=\left ( \frac{3(8x^2+x)-(16x+1)(3x-2)}{(8x^2+x)^2} \right )\\=\frac{24x^2+3x-48x^2-3x+32x+2}{(8x^2+x)^2}\\=\frac{-24x^2+32x-2}{(8x^2+x)^2}$

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3 years ago