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ch4aika [34]
3 years ago
13

Which statement describes the end behavior of the function f(x) = 3|x − 7| − 7? A. As x approaches negative infinity, f(x) appro

aches negative infinity. B. As x approaches negative infinity, f(x) approaches positive infinity. C. As x approaches positive infinity, f(x) approaches negative infinity. D. As x approaches positive infinity, f(x) is no longer continuous.
Mathematics
1 answer:
Firlakuza [10]3 years ago
4 0

Answer:

Our expressions is f(x) = 3|x-7|-7

  • in positive values |x-7| is x-7 ⇒f(x) = 3x-28
  • in negative ones |x-7| is 7-x wich is the opposite⇒f(x) = 14-3x

Let's calculate the limits in +∞ and -∞

  • \lim_{x \to +\infty} (3x-28)=\lim_{x \to+ \infty} 3x = +∞
  • \lim_{x \to- \infty} (14-3x) = \lim_{x \to- \infty} -3x =+∞

So the right staement is B

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If we assume the population of Grand Rapids is growing at a rate of approximately 4% per decade, we can model the population fun
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Answer:

The average rate of change of the population on the intervals [ 5 , 10 ] , [ 5 , 9 ] , [ 5 , 8 ] , [ 5 , 7 ] , and [ 5 , 6 ]  are 734.504, 733.06, 731.62, 730.185 and 728.75 respectively.

Step-by-step explanation:

The given function is

P(t)=181843(1.04)^{(\frac{t}{10})}

where, P(t) is population after t years.

At t=5,

P(5)=181843(1.04)^{(\frac{5}{10})}=185444.20

At t=6,

P(6)=181843(1.04)^{(\frac{6}{10})}=186172.95

At t=7,

P(7)=181843(1.04)^{(\frac{7}{10})}=186904.57

At t=8,

P(8)=181843(1.04)^{(\frac{8}{10})}=187639.06

At t=9,

P(9)=181843(1.04)^{(\frac{9}{10})}=188376.44

At t=10,

P(10)=181843(1.04)^{(\frac{10}{10})}=189116.72

The rate of change of P(t) on the interval [x_1,x_2] is

m=\frac{P(x_2)-P(x_1)}{x_2-x_1}

Using the above formula, the average rate of change of the population on the intervals [ 5 , 10 ] is

m=\frac{P(10)-P(5)}{10-5}=\frac{189116.72-185444.20}{5}=734.504

The average rate of change of the population on the intervals [ 5 , 9 ] is

m=\frac{P(9)-P(5)}{9-5}=\frac{188376.44-185444.20}{4}=733.06

The average rate of change of the population on the intervals [ 5 , 8 ] is

m=\frac{P(8)-P(5)}{8-5}=\frac{187639.06-185444.20}{3}=731.62

The average rate of change of the population on the intervals [ 5 , 7 ] is

m=\frac{P(7)-P(5)}{7-5}=\frac{186904.57-185444.20}{2}=730.185

The average rate of change of the population on the intervals [ 5 , 6 ] is

m=\frac{P(6)-P(5)}{6-5}=\frac{186172.95-185444.20}{1}=728.75

Therefore the average rate of change of the population on the intervals [ 5 , 10 ] , [ 5 , 9 ] , [ 5 , 8 ] , [ 5 , 7 ] , and [ 5 , 6 ]  are 734.504, 733.06, 731.62, 730.185 and 728.75 respectively.

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percent means out of a hundred, hence

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(3.14)(16)(12)

602.88

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