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3 years ago

Select the expressions that are equivalent to 5(6x+3)+4x.

Mathematics

1 answer:

bekas [8.4K]3 years ago

4 0

Answer:

the first 2

Step-by-step explanation:

5(6x + 3) + 4x = (6x + 3)5 + 4x due to the commutative rule of the multiplication (a×b = b×a).

and when you simplify the expression we get

5(6x + 3) + 4x = 30x + 15 + 4x = 34x + 15

so, the first two answer options are correct

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He sold 15% of them. "of" means multiply.

80 x 15% is the same as 80 x .15

80 x .15 = 12

He sold 12 coins

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3 years ago

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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:

See comment for complete question

Given

$a = 144$ --- Initial Population

$r = 8\%$ --- rate

(A) WHEN THE RATE DECREASES

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.

(B) WHEN THE RATE INCREASES

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:

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Step-by-step explanation:

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