Math Help needed

  • Thread starter Grits
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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:

Answers and Replies

  • #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]
 
  • #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.
 
  • #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.
 
  • #5
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benorin said:
[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.


so does r = 295

(20/3)^3 = 296
r=296-1
 
  • #6
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Grits said:
so does r = 295
(20/3)^3 = 296
r=296-1

Cube root, one third power. You're cubing it when you should be taking the cube root.
 
  • #7
Pyrrhus
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Use

[tex] r = \sqrt[3]{\frac{20}{3}} -1 [/tex]
 
  • #8
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Cyclovenom said:
Use
[tex] r = \sqrt[3]{\frac{20}{3}} -1 [/tex]

how do i put that in my calculator
 
  • #9
Grits said:
If PV=110 r=.09 and t=4 find fv when fv= pv*(1+r)^t.
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:
  • #10
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type (20/3)^(1/3) - 1
 
  • #11
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Here's another

Solve for r directly when FVt =PVo *(1+r)^t
 
  • #12
Pyrrhus
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Grits said:
Here's another
Solve for r directly when FVt =PVo *(1+r)^t

Well what comes first to your mind?

Start by isolationg the (1+r)^t term.
 
  • #13
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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.
 
  • #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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