Math Help needed

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

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

benorin
Homework Helper
$$60= \frac{400}{(1 + r)^3}$$ 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.

benorin said:
$$60= \frac{400}{(1 + r)^3}$$ 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

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.

Pyrrhus
Homework Helper
Use

$$r = \sqrt{\frac{20}{3}} -1$$

Cyclovenom said:
Use
$$r = \sqrt{\frac{20}{3}} -1$$

how do i put that in my calculator

Grits said:
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 :)?

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

Here's another

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

Pyrrhus
Homework Helper
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.

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
Homework Helper
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.