Math Help Needed: Solving Business Finance Problems

[Grits]

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 can't seem to figure them out.

Someone please 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

 

[benorin]

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

(1+r)^2=10
\sqrt{(1+r)^2}=\sqrt{10}
1+r=\pm\sqrt{10}
r=-1\pm\sqrt{10}

 

[benorin]

As you are new here, I should advise you that most poeple will not reply to your posts unless you show some work.

 

[benorin]

60= \frac{400}{(1 + r)^3} assume that r\neq -1 to get (1 + r)^3= \frac{400}{60}=\frac{20}{3}
and then take the cube root of both sides to get... you try.

 

[Grits]

60= \frac{400}{(1 + r)^3} assume that r\neq -1 to get (1 + r)^3= \frac{400}{60}=\frac{20}{3}
and then take the cube root of both sides to get... you try.

so does r = 295

(20/3)^3 = 296
r=296-1

 

[d_leet]

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.

 

[Pyrrhus]

Use

r = \sqrt[3]{\frac{20}{3}} -1

 

[Grits]

Use
r = \sqrt[3]{\frac{20}{3}} -1

how do i put that in my calculator

 

[ksinclair13]

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

This is just substitution. fv=(110)(1+.09)^4
Remember to use the order of operations.

Edit: Type this in you calculator: (\frac{20}{3})^{\frac{1}{3}}, 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 :)?

 

[Jameson]

type (20/3)^(1/3) - 1

 

[Grits]

Here's another

Solve for r directly when FVt =PVo *(1+r)^t

 

[Pyrrhus]

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.

 

[Grits]

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.

 

[Pyrrhus]

Well actually is basic algebra.

For example:

a = \frac{bc}{d \sqrt{e}}

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 \sqrt{e}

\sqrt{e} a = \frac{ \sqrt{e} bc}{d \sqrt{e}}

Now you can see you have

\sqrt{e} a = \frac{bc}{d} \frac{\sqrt{e}}{\sqrt{e}}

Thus

\sqrt{e} a = \frac{bc}{d}

Now you should divide both members by a

\sqrt{e} \frac{a}{a} = \frac{bc}{da}

thus

\sqrt{e} = \frac{bc}{da}

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

e^{\frac{1}{2}} = \frac{bc}{da}

So if we square both sides

(e^{\frac{1}{2}})^2 = (\frac{bc}{da})^2

Using the fact that (e^n)^m = e^{nm}

e^{\frac{2}{2}} = (\frac{bc}{da})^2

e = (\frac{bc}{da})^2

We succesfully isolated e.

Now try it with your equation.