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2 years ago

Did I do this right?

Mathematics

1 answer:

kupik [55]2 years ago

5 0

Answer:

Yes, you did.

Step-by-step explanation:

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Answer:

2x-3+8v

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3 years ago

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What is the slope of the line shown?

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Answer:

The slope is 45.

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Using the slope formula by two points:

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Find the value of the X to the nearest tenth 57* 6

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3 years ago

QUESTION 3 [10 MARKS] A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the nu

Harman [31]

Answer:

(a)Revenue function, $R(x)=580x-x^2$

Marginal Revenue function, R'(x)=580-2x

(b)Fixed cost =900 .

Marginal Cost Function=300+50x

(c)Profit, $P(x)=-35x^2+280x-900$

(d)x=4

Step-by-step explanation:

Part A

Price Function $= 580 - 10x$

The revenue function

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

The marginal revenue function

$\dfrac{dR}{dx}= \dfrac{d}{dx}(R(x))=\dfrac{d}{dx}(580x-x^2)=580-2x\\R'(x)=580-2x$

Part B

(Fixed Cost)

The total cost function of the company is given by $c=(30+5x)^2$

We expand the expression

$(30+5x)^2=(30+5x)(30+5x)=900+300x+25x^2$

Therefore, the fixed cost is 900 .

Marginal Cost Function

If $c=900+300x+25x^2$

Marginal Cost Function, $\frac{dc}{dx}= (900+300x+25x^2)'=300+50x$

Part C

Profit Function

Profit=Revenue -Total cost

$580x-10x^2-(900+300x+25x^2)\\580x-10x^2-900-300x-25x^2\\$Profit,P(x)=-35x^2+280x-900$

Part D

To maximize profit, we find the derivative of the profit function, equate it to zero and solve for x.

$P(x)=-35x^2+280x-900\\P'(x)=-70x+280\\-70x+280=0\\-70x=-280\\$Divide both sides by -70\\x=4$

The number of cakes that maximizes profit is 4.

6 0

3 years ago

Solve. -7x=x^2+12 help

Nat2105 [25]

Hey There!

The answer to your problem is $x=-3$ or $x=-4$ both answers are correct!

Step 1: Subtract x²+12 from both sides.

−7x−(x²+12)=x²+12−(x²+12)

−x²−7x−12=0

Step 2: Factor left side of equation.

$(-x-3)(x+4)=0$

Step 3: Set factors equal to 0.

$-x-3=0$ or $x+4=0$

Get your answer:

$x=-3$ or $x=-4$

3 0

3 years ago