HELP?!
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.
y=74.6496 y=60
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?
yes
\(30 \times 1.2^x = 45 + 3x\)
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.
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