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2 years ago

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The cost of a cab ride is $10 for the first mile or part of a mile. It costs $3 for each additional mile or part of a mile and a

ll cab rides are less than six miles. The cost is modeled by the function C(m), where "m" is the number of miles. The driver's pay "P" is a function of the cost "C" and is modeled by P(C) = 0.60C. What is the domain of P(C)? A) {1, 2, 3, 4, 5} B) {10, 13, 16, 19, 22, 25} C) {any positive real number less than 6} D) {any positive real number less than 25}

2 answers:

Nostrana [21]2 years ago

7 0

Answer:

B) {10, 13, 16, 19, 22, 25}

Step-by-step explanation:

The cost of the cab ride will be one of 6 possible values. Those values constitute the domain of the function P(C).

ioda2 years ago

7 0

Answer:

B) {10, 13, 16, 19, 22, 25}

Step-by-step explanation:

Cost of cab ride for the first mile or part of a mile = $10

Cost for each additional mile or part of a mile = $3

Therefore,

Cost of cab ride for the two miles = $10 + $3 = $13

Cost of cab ride for the three miles = $10 + $3 +$3= $16

Cost of cab ride for the four miles = $10 + $3 +$3 + $3= $19

Cost of cab ride for the five miles = $10 + $3 +$3 + $3 +$3 = $22

Cost of cab ride for the six miles = $10 + $3 +$3 + $3 +$3 + $3 = $25

So, domain of P(C) is equal to {10, 13, 16, 19, 22, 25}

Therefore, option B) is correct .

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

The correct option is;

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Answer:

The total revenue is $TR=580x-10x^2$ .

The marginal revenue is $MR=580-20x$ .

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The marginal cost function is $MC=50x+300$ .

Step-by-step explanation:

The Total Revenue ( $TR$ ) received from the sale of $x$ goods at price $p$ is given by

$TR=p\cdot x$

The Marginal Revenue ( $MR$ ) is the derivative of total revenue with respect to demand and is given by

$MR=\frac{d(TR)}{dx}$

From the information given we know that the price they can sell cakes is given by the function $p=580-10x$ , where $x$ is the number of cakes sold per day.

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$TR=(580-10x)\cdot x\\TR=580x-10x^2$

And the marginal revenue is

$MR=\frac{d}{dx}(580x-10x^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\MR=\frac{d}{dx}\left(580x\right)-\frac{d}{dx}\left(10x^2\right)\\\\MR=580-20x$

The Fixed Cost ( $FC$ ) is the amount of money you have to spend regardless of how many items you produce.

The Marginal Cost ( $MC$ ) function is the derivative of the cost function and is given by

$MC=\frac{d(TC)}{dx}$

We know that the total cost function of the company is given by $C=(30+5x)^2$ , which it is equal to

$\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a+b\right)^2=a^2+2ab+b^2\\a=30,\:\:b=5x\\\\\left(30+5x\right)^2=30^2+2\cdot \:30\cdot \:5x+\left(5x\right)^2=25x^2+300x+900\\\\C=25x^2+300x+900$

From the total cost function and applying the definition of fixed cost, the fixed cost is $900.

And the marginal cost function is

$MC=\frac{d}{\:dx}\left(25x^2+300x+900\right)\\\\MC=\frac{d}{dx}\left(25x^2\right)+\frac{d}{dx}\left(300x\right)+\frac{d}{dx}\left(900\right)\\\\MC=50x+300+0=50x+300$

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