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anonymous

  • 5 years ago

eresa invested $1425.00 in a savings account at her local bank. the bank pays simple interest of 3.5% annually. how much money would she have in 4 years?

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  1. bahrom7893
    • 5 years ago
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    is this D.E?

  2. bahrom7893
    • 5 years ago
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    differential equations?

  3. anonymous
    • 5 years ago
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    simple interest i guess

  4. anonymous
    • 5 years ago
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    i=prt?

  5. bahrom7893
    • 5 years ago
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    oh i dunno i never used that stuff

  6. bahrom7893
    • 5 years ago
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    i think i can do it another way

  7. anonymous
    • 5 years ago
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    please show

  8. bahrom7893
    • 5 years ago
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    okay workin on it, afk

  9. bahrom7893
    • 5 years ago
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    let me find a similar problem, i was solving these a couple of weeks ago, will get back to u asap

  10. bahrom7893
    • 5 years ago
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    can u check the answer or do u have the answer?

  11. anonymous
    • 5 years ago
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    nopes

  12. bahrom7893
    • 5 years ago
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    k ill do my best

  13. bahrom7893
    • 5 years ago
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    So here I am going to use the following differential equation: dN/dt - k*N = 0

  14. bahrom7893
    • 5 years ago
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    Where N is the amount of money and k is the percent at which it grows. You also know that she invested $1425.00 at some point, so let's call that point the beginning, or time is 0 t = 0

  15. bahrom7893
    • 5 years ago
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    k is given to be 3.5%; so this is what you know: N(0) = 1425.00 (initial value - initial investment is $1425) dN/dt - 3.5N = 0

  16. bahrom7893
    • 5 years ago
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    Move -3.5N to the right: dN/dt = 3.5N Multiply both sides by dt: dN = 3.5N*dt Divide both sides by N: dN/n = 3.5dt

  17. bahrom7893
    • 5 years ago
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    Integrate both sides: \[\int\limits_{}^{}dN/N = \int\limits_{}^{}3.5dt\]

  18. bahrom7893
    • 5 years ago
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    on the left side you get: ln|N| + C1= 3.5t + C2, where C1 and C2 are arbitrary constants.

  19. bahrom7893
    • 5 years ago
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    Ln|N| = 3.5t + C2 - C1 (some constant minus another constant is still a constant so ima replace the diff by C3)

  20. bahrom7893
    • 5 years ago
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    Ln|N| = 3.5t + C3. Now I have to get rid of Ln, so we raise e to the both sides of equation: \[e^{Ln \left| N \right|} = e^{3.5t+C _{3}}\]

  21. bahrom7893
    • 5 years ago
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    are you following me?

  22. anonymous
    • 5 years ago
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    i dont think its that complicated. i am in 6th grade

  23. bahrom7893
    • 5 years ago
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    okay im sure its not that complicated

  24. bahrom7893
    • 5 years ago
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    oh well do u want me to finish it? at least get the answer? I think there def is an easier way using logic or smth, but i have no idea how to solve it that way lol. im in college now

  25. anonymous
    • 5 years ago
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    :-) thanks for your help. you could try

  26. bahrom7893
    • 5 years ago
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    wait i think i messed up somewehere in between

  27. bahrom7893
    • 5 years ago
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    im gettin a ridiculous number

  28. anonymous
    • 5 years ago
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    thats ok...i think it would be: Amount= 1425*(3.5/100)*4

  29. bahrom7893
    • 5 years ago
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    oh crap im using the wrong method. Mine is for changing amount of money, urs is the ez way, they'r assuming that the APR won't change. Like I mean 3.5% of 1425 is not the same as 3.5 of 20k. That's what im solving for, in ur case the 3.5% stays the same.

  30. bahrom7893
    • 5 years ago
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    which means my solution is totally wrong for this problem lol sorry off to help others, but i really cant do this any other way

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