Hi I'm new here and I need some help please.

1. S=3/1-r if r= -3/4 , S=?

2. t=16(-1/2) ^x-1 if n=8

3. T= -1/2(2)^n-1 if t=-32 , n=?

4. write the quadratic equation whose roots are 2/3 and -1

5.What are the restrictions on the graph of y= x+2/ (x-a)(bx-1)

6. How many solutions are there for the follow non-linear system?

y= 2/x and y= (x+1)^2 -2

If anyone could me with these it would be great! Thanks a lot!

2. There are quite a few questions here. I believe the moderation team likes members to limit the number of questions to around 3 per post. However, here is assistance with the first four:

1. S=3/1-r if r= -3/4 , S=?
$S=\frac{3}{1}-r$

By subbing in the value of r:
$S=\frac{3}{1}+\frac{3}{4}$

Can you solve it from there?

2. $t=16(-1/2)^x-1$
Actually, this question makes no sense. No n value exists in the original equation, so you're unable to substitute n=8 anywhere.

3.
$T= -1/2(2)^n-1$ - I will assume the -1 is not part of the exponent.
$-32= -1/2(2)^n-1$
$-31= -1/2(2)^n$
$62=(2)^n$
$ln(62)=n*ln(2)$

Can you take it from there?

4. write the quadratic equation whose roots are 2/3 and -1
By using null factor rule:
$\frac{2}{3}=x$
$3x-2=0$

$-1=x$
$x + 1=0$

Therefore,
$0=(x+1)(3x-2)$

Can you solve it from there?

3. Thanks for the help, but for #1-3 I'll try to explain the question better.

1. S= 3
________ if r= -3/4 , S=?
1-r

2. the x is the n sorry, so its like:

t= 16(-1/2)^n-1 (yes the -1 is part of n)

3. the -1 is part of the exponent, but that wouldn't make a difference right? since your using ln anyway. Couldn't I use log?

4. I get 4 but that's the answer right? Because you cant really do anything more to it.

Thanks again!

4. Bump! someone please help lol I need to finish this tonight.

5. 1. S= 3
________ if r= -3/4 , S=?
1-r

$S=\frac{3}{1-r}$
$S=\frac{3}{1-(3/4)}$
$S=\frac{3}{.25}$.... Can you solve that?

2. $t=16(-1/2)^{(n-1)}$
$t=16(-1/2)^{(8-1)}$
$t=16(-1/2)^7$
$t=16 \times -\frac{1}{128}$.. can you finish it?

3. $T= -1/2(2)^{(n-1)}$
$-32= -1/2(2)^{(n-1)}$
$64= (2)^{(n-1)}$
$2^6= (2)^{(n-1)}$
$6= n-1$... sooo n=??

4. It depends what form you want the answer in. I'd suggest expanding out into 'standard form'.