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OK.....welcome !!!

Thanks so much CPhill!!!

First one...the graph at the bottm right is correct

Second one....the graph at the bottom left is correct....

Your last answer is good....

That's ok! I appreciate you doing your best to help out :)

How to use nCr to find the 4th term of (3c+4d)^7

(3c+4d)^7

I know this is common wording for a question like this but I never know what end of the expansion that they want you to start from

anyway...

From one end it is 

    \(^7C_3\; (3c)^3(4d)^{(7-3)}\\\)

I used 3 because the first term is the 0th term.    So r=3 is the 4th term

If you started at the other end it would be

\(^7C_3\; (4d)^3(3c)^{(7-3)}\\\)

I suppose there is some convention about what end you are supposed to start at but i don't know what it is.  

I'm not sure how to explain how to graph it, but the correct graph is the bottom left one. My best suggestion for graphing would be to test it out on www.desmos.com/calculator.

Then I guess I can't remember. I just created a new one.

Thanks a lot!

So for this question, how would I graph this, and what graph would match the function?

function:

I do not believe accounts will be removed after inactivity.

thank you guest. that was very helpful

to simplify is to make terms look simpler. For example, we could have 5x-2+27x-42-3x+10=24

Simplifying would give 29x-34=24

simplifying does not have to mean solve.

Thanks Alan.

In other works Guest the identity you got is true except that you forgot to include the constant.

Yes I know you already said that Alan :)

the first one is A

Part 1:

When shifting an equation, if the value shifting is inside the equation, you go the opposite direction. So, for example, if it was:

\(f(x)=|x+7|\), then you would shift 7 to the left.

So, for your problem of \(f(x)= |x-3|+2\), when you replace x with x+2, the graph will shift 2 units to the left.

Part 2:

\(f(x)=\frac{1}{2}\sqrt{x+5}\)

So, square roots have specific domain restrictions, such that the value inside can't be negative because no number, when squared, can be negative, so when doing the opposite it can't be negative. 

So, with that in mind, x can be -5 but no lower.

So the domain would be \([-5,\infty)\)

The range is all possible y values. Since the square root can put out either positive or negative numbers, the range is \((-\infty,\infty)\).

\(\sqrt[3]{250x^6}\)

So, in this case, a cube root is the same as that value to the power of 1/3.

\((250x^6)^{\frac{1}{3}}\)

Then, you just multiply the exponents by each other. 

\(6*\frac{1}{3}=\frac{6}{3}=2\)

So that makes it:

\(250x^2\)

I apologize for not fully understanding your question :/

I see your actual question format, but I am unable to help you with this... sorry!

Yes. I couldnt figure out how to make it look like the question on here.

Is your question supposed to be formatted like this?: (250x^6)^3

( - inf, -4 ] U [ 3, 5)

y = -8 + 8x^3

y + 8 = 8x^3

[ y + 8 ] / 8 = x^3       take the cube root

 ∛ [ y + 8 ] / 2 =  x         swap x and y

 ∛ [ x + 8 ] / 2 = y = the inverse

lol i posted that too

Remember that some x value led to -8

So.....we can solve this to find f^-1(-8)

(x + 4) / ( x + 6) = -8

x + 4 = -8 (x + 6)

x + 4 = -8x - 48

9x = - 52

x = -52/9    =  f^-1(-8)

We have that

y = a (x - 5)^2 + 4

And if the y intercept is - 4, then (0, -4) is on the graph, we have

- 4 = a(0 - 5)^2 + 4

- 4 = 25a + 4     subtract 4 from both sides

-8 = 25a      divide both sides by 25

a = -8/25

So......the function is

y = (-8/25) ( x - 5)^2 + 4     simplify

y = (-8/25) ( x ^2 - 10x + 25) + 4

y = (-8/25)x^2 + (16/5)x - 8 + 4

y = (-25/8)x^2 + (16/5)x - 4

OK>>>>no prob !!!!