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Homework Help: Math Help needed

  1. Jan 22, 2006 #1
    I am new to the site but I need some math help. I am a business major and I am taking Business Finance. He wants us to do a bonus quiz to see how well our math skills are for the class. Well I cant seem to figure them out.

    Someone plz help.

    (1 + r)^2 = 10
    (1+r)^2=10
    (1+r)(1+r)=10
    1+1r+1r+r^2
    1+2r+r^2=10
    2r+r^2=9
    r^2=9-2x/x
    r=9-2 =7?


    60= [400/ (1 + r)^3]




    If PV=110 r=.09 and t=4 find fv when fv= pv*(1+r)^t.


    Solve for r directly when FVt=PVo *(1+r)^t
     
    Last edited: Jan 22, 2006
  2. jcsd
  3. Jan 22, 2006 #2

    benorin

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    Up to this point, namely [tex]2r+r^2=9,[/tex] your work is ok, but the next line isn't right. There is, however, a more direct solution:

    [tex](1+r)^2=10[/tex]
    [tex]\sqrt{(1+r)^2}=\sqrt{10}[/tex]
    [tex]1+r=\pm\sqrt{10}[/tex]
    [tex]r=-1\pm\sqrt{10}[/tex]
     
  4. Jan 22, 2006 #3

    benorin

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    As you are new here, I should advise you that most poeple will not reply to your posts unless you show some work.
     
  5. Jan 22, 2006 #4

    benorin

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    [tex]60= \frac{400}{(1 + r)^3}[/tex] assume that [tex]r\neq -1[/tex] to get [tex](1 + r)^3= \frac{400}{60}=\frac{20}{3}[/tex]
    and then take the cube root of both sides to get... you try.
     
  6. Jan 22, 2006 #5

    so does r = 295

    (20/3)^3 = 296
    r=296-1
     
  7. Jan 22, 2006 #6
    Cube root, one third power. You're cubing it when you should be taking the cube root.
     
  8. Jan 22, 2006 #7

    Pyrrhus

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    Use

    [tex] r = \sqrt[3]{\frac{20}{3}} -1 [/tex]
     
  9. Jan 22, 2006 #8
    how do i put that in my calculator
     
  10. Jan 22, 2006 #9
    This is just substitution. [tex]fv=(110)(1+.09)^4[/tex]
    Remember to use the order of operations.

    Edit: Type this in you calculator: [tex](\frac{20}{3})^{\frac{1}{3}}[/tex], then subtract one from that answer.

    2nd Edit: I'm using LaTeX for the first time, and I was wondering if/how you can make the parentheses larger. That looks weird. Could someone please clean that up and make it look better so I can look at the proper code :)?
     
    Last edited: Jan 22, 2006
  11. Jan 22, 2006 #10
    type (20/3)^(1/3) - 1
     
  12. Jan 22, 2006 #11
    Here's another

    Solve for r directly when FVt =PVo *(1+r)^t
     
  13. Jan 22, 2006 #12

    Pyrrhus

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    Well what comes first to your mind?

    Start by isolationg the (1+r)^t term.
     
  14. Jan 22, 2006 #13
    If r=8%, t=5 and m=4 find the timve value factors given below:

    FVIF r%/m,t*m=[(1+(r/m))^t*m]

    I got 1 as the answer I just canceled out and got 1 left over.
     
  15. Jan 22, 2006 #14

    Pyrrhus

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    Well actually is basic algebra.

    For example:

    [tex] a = \frac{bc}{d \sqrt{e}} [/tex]

    Now if you want to isolate let's say e what must you do?

    first off you could multiply both sides of the equation by [itex] \sqrt{e} [/itex]

    [tex] \sqrt{e} a = \frac{ \sqrt{e} bc}{d \sqrt{e}} [/tex]

    Now you can see you have

    [tex] \sqrt{e} a = \frac{bc}{d} \frac{\sqrt{e}}{\sqrt{e}} [/tex]

    Thus

    [tex] \sqrt{e} a = \frac{bc}{d}[/tex]

    Now you should divide both members by a

    [tex] \sqrt{e} \frac{a}{a} = \frac{bc}{da}[/tex]

    thus

    [tex] \sqrt{e} = \frac{bc}{da}[/tex]

    Now using the fact that [itex] \sqrt[n]{e} = e^{\frac{1}{n}} [/itex]

    [tex] e^{\frac{1}{2}} = \frac{bc}{da}[/tex]

    So if we square both sides

    [tex] (e^{\frac{1}{2}})^2 = (\frac{bc}{da})^2[/tex]

    Using the fact that [itex] (e^n)^m = e^{nm} [/itex]

    [tex] e^{\frac{2}{2}} = (\frac{bc}{da})^2[/tex]

    [tex] e = (\frac{bc}{da})^2[/tex]

    We succesfully isolated e.

    Now try it with your equation.
     
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