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3 years ago

The 18th term of an arithmetic sequence is 49 and the first term is 15.

Mathematics

1 answer:

Tju [1.3M]3 years ago

5 0

Answer:

Step-by-step explanation:

49-15=34

34/17=2

For this sequence, you can write this function:

f(x)=15+2(x-1)

Check:

f(18)=15+2(18-1)

f(18)=15+2(17)

f(18)=15+34

f(18)=49

So:

f(10)=15+2(10-1)

f(10)=15+2(9)

f(10)=15+18

f(10)=33

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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

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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.

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Divide 29.99 by .55 (55% to a decimal is .55)

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Therefore, there's a relation that aroun Unlimited Meals tickets and Adults-Seniors costumers: they tend to get these tickest more than children.

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