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MaryJaneMescudi
01-16-12 11:54 PM
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crazycatpup
01-18-12 02:32 PM

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Do my homework dudes!

 

01-16-12 11:54 PM
MaryJaneMescudi is Offline
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College Algebra sucks.
I have a program I use to do equations that shows me the answers and the work.
However it doesn't do word problems.
One question is:
"phyllis invested 12000$, a portion earning a simple interest rate at 4.5% per year and the rest earning 4% per year. After 1 year the total interest earned on these investments was 525$. how much money did she invest at each rate?"
who names their kid phyllis for starters?
and second, when will i need this in my life?
but if you could please legitly show me the equation for this word problem, i might be able to solve it.
(:
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01-17-12 12:10 AM
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I'll try to explain it as best as I can.

First off, you know that there are two numbers that add up to 12,000.
So: x + y = 12000

Next you know that the interest rates per year, and the total interest after one year.
So: .045*x + .04*y = 525
NOTE: It doesn't matter which variable is multiplied by which rate.

Next you play with the first equation to get it in terms of one variable.
So: y = 12000 - x.
NOTE: It doesn't matter which variable you subtract, I just tend to prefer working with x's.

Next you substitute that into the interest equation.
So: .045*x + .04*(12000-x) = 525

Play with the numbers a little to make it easier to see what's going on.
So: .045*x + (.04*12000) - (.04*x) = 525

Work that out a little.
So: .045x - .04x + 480 = 525
Then: .005x = 45
Finally: x = 9000
Because x = 12000 - y, that means that y = 3000.

Plug both numbers into the equation that has the interest rates to check the answers and you'll see that it works. I hope that helps you, and just let me know if there was anything unclear or if you still don't understand something.
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01-17-12 01:07 AM
MaryJaneMescudi is Offline
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thanks!
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01-18-12 02:32 PM
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Well..I guess you already got your answer.

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