Question

Please need help!

A population of 100 rabbits can be modeled by the equation y = 100e^0.12x, where y is the final population of rabbits and x is time in years.

Which graph models this population correctly and what is the approximate number of rabbits in the population after 5 years?

Graph 1 and there would be approximately 54 rabbits
Graph 1 and there would be approximately 100 rabbits
Graph 2 and there would be approximately 182 rabbits
Graph 2 and there would be approximately 100 rabbits

Answer 1

http://learnft.flvs.net/webdav/assessment_images/educator_lam2_v14/q142368-la_exp_graph_5.jpg

Answer 2

the graph is in the link

Answer 3

Plug in 5 for x to get the number

Answer 4

link doesn't work.

Answer 5

So the answer would be the third answer

Answer 6

182

Answer 7

yes.

Answer 8

Ok I have a few more, would you be willing to help me?

Answer 9

@peachpi

Answer 10

ok

Answer 11

The depreciating value of a semitruck can be modeled by y = Ao(0.87)^x, where y is the remaining value of the semi and x is time in years. The truck is depreciating at 13% per year.

What is the value of the truck initially, Ao, and how does the graph change if the initial value was only $67,000?

$90,000, and the graph has a y-intercept at 20,000
$87,000, and the graph has a y-intercept at 67,000
$87,000, and the graph falls at a slower rate to the right
$87,000, and the graph falls at a faster rate to the right

Answer 12

@peachpi

Answer 13

Is this the whole question? it seems like something's missing. They're asking you to find \(A_0\), but there's nothing there that allows you to solve for it.

Answer 14

Would we use each answer choice as a kind of a eliminating factor maybe? @peachpi

Answer 15

I guess you could. There's nothing to say whether 87k or 90k is the correct initial value, so you'd have to look at the effect on the graph. Initial value doesn't affect the rising/falling rate. It does change the y-intercept.

y -intercept = initial value, so the choice with 67k as the initial is a good guess

Answer 16

Okay thank you! I just have one more @peachpi

Answer 17

When their child was born, her parents invested $40,000 in a plan guaranteed to increase by 3% per year. The logarithmic function y equals 77.9 log x/40,000 can be used to determine the number of years, y, it takes for the investment to be worth x dollars.

After how many years will the investment be worth approximately 50% more than it was originally and how old will their child be when the investment is worth about $100,000?

The investment is worth approximately 50% more after 25 years, and the investment is worth $100,000 when their child is about 31 years old.
The investment is worth approximately 50% more after 5 years, and the investment is worth $100,000 when their child is about 25 years old.
The investment is worth approximately 50% more after 14 years, and the investment is worth $100,000 when their child is about 31 years old.
The investment is worth approximately 50% more after 31 years, and the investment is worth $100,000 when their child is about 25 years old.

Answer 18

you have to plug in the given values for y then solve for x.

y = 77.9 log x/40,000

For the 1st question
1.5(40,000) = 77.9 log x/40,000

for the 2nd
100,000 = 77.9 log x/40,000

Answer 19

Okay hold on, will you check my answer?

Answer 20

k

Answer 21

How do I set this up? @peachpi

Answer 22

I am kind of lost with this one

Answer 23

oh wait y is the year, x is the amount. so it's

y = 77.9 log x/40,000

For the 1st question
y = 77.9 log (1.5*40,000)/40,000

for the 2nd
y = 77.9 log 100,000/40,000

Answer 24

Okay one sec :) @peachpi

Answer 25

Is it the second choice?

Answer 26

@peachpi

Answer 27

i got 13.7 and 31.9

Answer 28

oops 30.9

Answer 29

So the the third one would be the answer? I am not sure what I did wrong

Answer 30

yes the third. It might have been an issue with division/parenthes on your calculator

Answer 31

Okay thank you so much! I really appreciate all your help!!!!!!

Answer 32

you're welcome