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3 years ago

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the table below shoes the rates for two duplicating companies. quick copy copy place initial fee of $5.00 initial fee of $10.00

$0.06 per page $0.04 per page the total cost, c, is dependent on the number of pages, p, that are duplicated. which of these systems of equations represents this situation? p = 5 + 6c p = 10 + 4c ⊝ c = 5 + 6p c = 10 + 4p ⊝ p = 5 + 0.06c p = 10 + 0.04c ⊝ c = 5 + 0.06p c = 10 + 0.04p

1 answer:

Mariana [72]3 years ago

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Answer:

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How would the average rate of change over years 1 to 5 and years 6 to 10 be affected if the population increased at a rate of 8%

Sever21 [200]

Answer:

(A): As year increases the number of pikas reduces.

(B): As year increases the number of pikas increases as opposed to when the rate reduces.

Step-by-step explanation:

<em>See comment for complete question</em>

<u>Given</u>

$a = 144$ --- Initial Population

$r = 8\%$ --- rate

<u>(A) WHEN THE RATE DECREASES</u>

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1-8\%)^x$

Convert % to decimal

$f(x) = 144(1-0.08)^x$

$f(x) = 144(0.92)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(0.92)^x$

$f(1) = 144*0.92^1 =144*0.92=132.48$

$f(5) = 144*0.92^5 =144*0.66=95.04$

$Rate = \frac{95.04 - 132.48 }{4}$

$Rate = \frac{-37.44}{4}$

$Rate = -9.36$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(0.92)^x$

$f(6) = 144*0.92^6 =144*0.61=87.84$

$f(10) = 144*0.92^{10} =144*0.43=61.92$

$Rate = \frac{61.92-87.84}{4}$

$Rate = \frac{-25.92}{4}$

$Rate = -6.48$

So, we have:

$Rate = -9.36$ for year 1 to 5

This means that the number of pikas reduces by 9.36 yearly

$Rate = -6.48$ for year 6 to 10

This means that the number of pikas reduces by 6.48 yearly

So, we can say that, as year increases the number of pikas reduces.

<u>(B) WHEN THE RATE INCREASES</u>

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1+8\%)^x$

Convert % to decimal

$f(x) = 144(1+0.08)^x$

$f(x) = 144(1.08)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(1.08)^x$

$f(1) = 144(1.08)^1 = 144*1.08= 155.52$

$f(5) = 144(1.08)^5 = 144*1.47= 211.68$

$Rate = \frac{211.68 - 155.52}{4}$

$Rate = \frac{56.16}{4}$

$Rate = 14.04$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(1.08)^x$

$f(6) = 144(1.08)^6 = 228.52$

$f(10) = 144(1.08)^{10} = 310.89$

$Rate = \frac{310.89-228.52}{4}$

$Rate = \frac{82.37}{4}$

$Rate = 20.59$

So, we have:

$Rate = 14.04$ for year 1 to 5

This means that the number of pikas increases by 14.04 yearly

$Rate = 20.59$ for year 6 to 10

This means that the number of pikas increases by 20.59 yearly

So, we can say that, as year increases the number of pikas increases as opposed to when the rate reduces.

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Answer:

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Step-by-step explanation:

The given information are;

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The two equations used to set up the problem are;

The y-coordinate value, y₁ = -40×sin(30°) = -20 feet

The x-coordinate value, x₁ = 40×cos(30°) = 34.64 feet

The coordinates of the point location of Jack's seat = (20×√3, -20)

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