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Thans for helping. I think, I have leads for the second one.

The solution shows this:

3^(2c+1)=28*(3^c)-9

3^(2c)*3^1 = 28*3^c - 9

3(3^c)^2 - 28(3^c) + 9 = 0

let 3^c = x

3x^2 - 28x + 9 = 0

(3x - 1)(x - 9) = 0

x = 1/3 or x = 9

if x = 1/3

3^c= 1/3 = 3^-1

c = -1

if x = 9

3^c = 9 = 3^2

c = 2

That's fine. Sorry I couldn't be any help!

3a. Do you know slope-intercept form? It is the form of a linear equation y=mx+ b. The b is a constant, which is also the y-intercept. So, can you take it from here?

3b. Think of a vertex form equation: y= a(x-h)^2 + k. k determines the y-intercept, while h determines the horizontal shift. I believe you should be able to take it from here.

1. The first statement shows that 5000 + x/100 * 5000 + (x/100)^2 * 5000 ... (x/100)^6 *5000 = 10000. I believe you can solve that. Then, plug it on the 300 dollars, which will look like 300 + x/100 * 300 + (x/100)^2 * 300 .... (x/100)^n * 300 = 9600, where n is number you are looking for.

2. Unclear of what you are asking for. If you are asking for \(3^{2c+1} = 28\cdot 3^{c-9}\), the answer is in some really ugly ln form. If you are asking for \(3^2 c + 1 = 28 \cdot 3^c -9,\) then it is also really ugly. Please specify what your equation is asking for.

So we have 2x - y = 9 and 3y + x = 1. Take the first one, and solving for y gets us y = 2x - 9. Substitute this into the second equation, and you have 3(2x - 9) + x = 1. Simplifying the left hand side gets us 6x - 27 + x = 1. 7x = 28. x = 4. Put this into either one, (I'm going to use the first) and you get 2(4) - y = 9,

8 - y = 9, -y = 1, so y = -1 and you can check this by putting both values of x and y into either equation. 

x = 4

y = -1

Hope this helps!

Thank you so much!

Pleaseeee...im studying for my exams in a few hours its ljke 12 am lol bht ehh

NVM i used 2 for ub in my e equation instead of -2 in all 4 of my attempts..

\(y=\log_2\left(\dfrac{x}{x-1}\right) = \dfrac{1}{\ln(2)}\ln\left(\dfrac{x}{x-1}\right)\\ \dfrac{dy}{dx} = \dfrac{1}{\ln(2)}\cdot \dfrac{1}{\dfrac{x}{x-1}}\cdot \left(\dfrac{(x-1)-x}{(x-1)^2}\right)= \\ -\dfrac{1}{\ln(2)}\cdot \dfrac{x-1}{x}\cdot \dfrac{1}{(x-1)^2} = \\ -\dfrac{1}{\ln(2)}\dfrac{1}{x(x-1)} =\\ \dfrac{1}{\ln(2)}\dfrac{1}{x(1-x)}\)

Thanks bunches!

\(\displaystyle \int_3^4 e^{5-x}~dx = \\ \left . -e^{5-x} \right |_3^4 = \\ e^2 - e\)

We have that

y^2  = k /x^3       and we know that

15^2 = k / 2^3

k =  15^2 / 2^3  =   225/8

So....

y^3 = (225/8) / 1^3

y^3 = (225/8)          

y = (225/8)^(1/3)  =  ∛225/ 2

I like asdf's method, as well.....sometimes....we try to over-complicate things.......the reasoning is very simple and logical.....

Well I got that 126 is right

Sorry LagTho, but I believe Chpill is right

What you coudl do is draw an octagon and find the intersection points. When I did this with on octagon, it made more points forming another octagon, and there were 49 of them. So I think both answers are off. :)

1.

1000 ( 1.05)^6  ≈  $1340.10

Total possible subcommittees =  C(14,5)   

Total possible subcommittees  that contain no Repubs = C(6,5)   [choose any 5 Dems from 6 ]

Total possible subcommittees that contain no Dems = C(8,5)   [ choose any 5 Repubs from 8 ]

So...the total number of subcommittees that contain at least one member of either party =

C (14,5) - C(6,5)  - C(8,5)   =   1940

1. There's definitely a formula for this, but I don't know it. So I just did 1000 * 1.05, and then timesed that by 1.05, and I just did that 6 times and I got 1340.09564063.

2. \(\frac {-5}{1176}\) I think

Seeing how we have low numbers, it's just trial and error. RRRRD, RRRDD, RRDDD, RDDDD, where R is Republicans and D is Democrats. It doesn't matter whether you do RRDRR or RRRRD, they are the same thing. So counting them, we have 4 possible scenarios. 

Hope this helps!

I'm not sure  what is wanted in the first one....the average value  ???

For the second, the average value is 

 6

 ∫  5  ( sqrt (t) + 30 )   dt  /  6   =

0

5 [ (2/3)(6)^(3/2) + 30(6) ]  / 6   ≈  158.16   (thousands  of   dollars )

First one

P = 422000 ( 1 + .12)^t       where P is the population  after t  years

Second one

A = 50 (1/2)^(t/16)       where A is the number of grams remaining after t minutes

You need to put 2 and 7 into the original function to find the heights....sorry that I didn't make that clear...