Suppose you take a job teaching college math, for instance, at an annual salary of $20,000, with promised annual raises of $2000. a) Find a formula for the salary as it depends on the year. b) Use your formula to predict the salary after 10 years
This is effectively the formula for interest. Do you know how to do that?
Sorry, to be clear, it's *compound* interest.
we haven't used that in my class (i'm taking math for liberal arts). we used the linear formula y=mx+b
Oh, wait. Yes, never mind. It's not compound interest, because it's not a percentage raise. You're just increasing by $2000 each year.
So, let's use \(y = mx + b\).
\(y\) is your wage after \(m\) years. \(x\) is your guaranteed raise each year. \(b\) is your starting wage. Does that help get you started?
we were solving this in class but my notes are confusing: y=20000 + 2000(x-1) y = 2000x+18000 I don't know remember how I got 18000 (or if that's correct)
I don't see a need to use x - 1. You can start by charting a table. Year Salary 0 $20,000 1 $22,000 2 $24,000 3 $26,000
okay... but would my formula be: y= 20000 + 2000 (x-1), if x equals the number of years?
I already answered that, but okay ... Plug in your numbers, using x - 1. Does it give the result you expect? For example, "now" is year 0. Putting that in your equation, you get a current salary of $18,000. Does that seem right?
i see... the way we were doing this in class was year 1 = 20000, then to find the raise for an additional salary, you subtract the first year using that formula
Okay, in that case we are doing the same thing. However, you will probably find later on in economics, math, statistics, physics, etc. that "right now" is assigned time 0. For now do what you need to do to get it right in class, but keep that in mind.
Ah, I understand now.... I have a question about formulas for calculating supply rates... Would I add one amount to the rate increase if the formula is in linear form (like you would do in exponential form )?
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