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Mumz [18]
2 years ago
9

Find the value of x. The figure is not drawn to scale. LOOK AT THE PICTURE ASAP!

Mathematics
2 answers:
MArishka [77]2 years ago
7 0

Answer:

from the graph I think it's 16

BaLLatris [955]2 years ago
6 0

Step-by-step explanation:

I m pretty sure that the correct answer is 24

16/2=8

16+8=24

x=24

I hope it helps

can I be brainliest

have a nice day

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The answer is B i just took the test m∠C < ∠S
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What is 2.33 written as a mixed number in simplest form?
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To help racecar drivers safely negotiate turns, racetracks are banked at the curves. At one location, a 58 ft track is banked su
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The correct answer is Choice B: 53.0 feet.

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3 years ago
Match the angle with it's corresponding classification​
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∠TPR = right angle

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Step-by-step explanation:

Right angles are 90°, acute angles are angles less than 90°, obtuse angles are angles more than 90°

8 0
3 years ago
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
olya-2409 [2.1K]

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}

3 0
3 years ago
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