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How do you show growth?
For example, if the growth is a certain percentage each year, what equation shows growth?
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
In your case, the growth is 20%, so we use 1.2 Neighborhood A \(y = 30 \times 1.2^x\)
and I would just replace x with 5?
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 \(y = 45 + 3x\)
Part A. Neighborhood A. \(y = 30 \times 1.2^x\) Neighborhood B. \(y = 45 + 3x\)
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.
Part C. Set the two functions equal and solve for x.
so y=30*1.2^5 and y=45+3(5) for B?
Yes, and evaluate each expression.
Good. 75 & 60
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.
what would I set my function equal to?
would I make a function table and use the formula until I get the same number of houses?
For Part C?
\(30 \times 1.2^x = 45 + 3x\)
ohhh ok... and so that's all I need for part c, is to solve that?
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.
Yes, up need to solve the equation, but your idea of letting x = 1, 2, 3, 4, 5 is good.
Make a table and see when Neighborhood A overtakes B.
awee yes I get it now
Are you given choices?
no it's open response
I'm making a table rn
OK. Then do what you mentioned above. Make a table.
We already know this: |dw:1438567229776:dw|
It happens between year 3 and year 4. |dw:1438567359681:dw|
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?
The smallest difference between A and B is at 3.3 years, so that is a close approximation. |dw:1438567705900:dw|
Your guess of approximately 3.5 years is also good.
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!!!
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.
You are correct, I have not learned logarithms yet, but I will remember your advice.