- anonymous

HELP?!

- katieb

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- mathstudent55

How do you show growth?

- mathstudent55

For example, if the growth is a certain percentage each year, what equation shows growth?

- mathstudent55

Let's say the growth is 10%. Since 100% + 10% = 110% = 1.1, then each year, multiply the previous year's amount by 1.1
Year 0 a
Year 1 1.1a
Year 2 1.1(1.1a) = 1.1^2 * a
Year 3 1.1(1.1^2)a) = 1.1^3 * a
Year x 1.1^x * a

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## More answers

- mathstudent55

In your case, the growth is 20%, so we use 1.2
Neighborhood A
\(y = 30 \times 1.2^x\)

- anonymous

and I would just replace x with 5?

- mathstudent55

In Neighborhood B, you start with 45 and the growth is linear, so its:
Year 0 45 houses
Year 1 45 + 3 * 1
Year 2 45 + 3 * 2
Year 4 45 + 3 * 3
etc.
Year x 45 + 3 * x
\(y = 45 + 3x\)

- mathstudent55

Part A.
Neighborhood A. \(y = 30 \times 1.2^x\)
Neighborhood B. \(y = 45 + 3x\)

- mathstudent55

Part B.
Let x equal 5 in each of the two functions above to find the number of homes in the two neighborhoods after 5 years.

- mathstudent55

Part C.
Set the two functions equal and solve for x.

- anonymous

so y=30*1.2^5 and y=45+3(5) for B?

- mathstudent55

Yes, and evaluate each expression.

- anonymous

y=74.6496
y=60

- mathstudent55

Good.
75 & 60

- mathstudent55

Notice that Neighborhood A started with 30 homes and B with 45.
By year 5, Neighborhood A already has more homes than B.
This gives you a hint as to the solution of part C.

- anonymous

what would I set my function equal to?

- anonymous

would I make a function table and use the formula until I get the same number of houses?

- mathstudent55

For Part C?

- anonymous

yes

- mathstudent55

\(30 \times 1.2^x = 45 + 3x\)

- anonymous

ohhh ok... and so that's all I need for part c, is to solve that?

- mathstudent55

You can use a table.
The hint I mentioned above is that since by year 5, Neighborhood A already has more homes than Neighborhood B, that means the year they have the same number of homes is between years 0 and 5.

- mathstudent55

Yes, up need to solve the equation, but your idea of letting x = 1, 2, 3, 4, 5 is good.

- mathstudent55

Make a table and see when Neighborhood A overtakes B.

- anonymous

awee yes I get it now

- mathstudent55

Are you given choices?

- anonymous

no it's open response

- anonymous

I'm making a table rn

- mathstudent55

OK. Then do what you mentioned above. Make a table.

- mathstudent55

We already know this:
|dw:1438567229776:dw|

- mathstudent55

It happens between year 3 and year 4.
|dw:1438567359681:dw|

- anonymous

ok thanks for that because I was plugging in the different numbers and didn't see when they were equal, I didn't know that it could be in between years. So it would be after approximately 3 and a half yrs?

- mathstudent55

The smallest difference between A and B is at 3.3 years, so that is a close approximation.
|dw:1438567705900:dw|

- mathstudent55

Your guess of approximately 3.5 years is also good.

- anonymous

Oh my gosh! You were more than helpful and I honestly understand and appreciate everything you've told me. Thank you sooooo much!!! If I were there I'd give you a hug! Absolute lifesaver!!!

- mathstudent55

You are very welcome.
I just want to mention one point.
I don't know if you've learned logarithms yet.
Usually, when you have an equation with a variable in the exponent, you use logarithms to find the solution.
In this case, the equation we had does not lend itself to taking logarithms of both sides and simplifying easily. That is why I recommended you use a table with values (which was your idea) to see when the functions are of equal value to then see at which time the number of homes are the same.

- anonymous

You are correct, I have not learned logarithms yet, but I will remember your advice.

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