Melody

Because you asked 3 questions.

Answerers are here to help. They are not here to do all your homework for you.

Plus asking more than one question on a post causes confusion.

Hi Shaezy,

It is good to see you interacting here.

Your substitution is perfect, EP taught you well,  it is your simplifying that is incorrect.

\(If \;\;f(x) = (1 − x)^2 \qquad g(x) = 9 − x,\\  find\;\; f(g(x)) \;\;and ;\;g(f(x)). \\\\~\\ g(f(x))= 9-(1-x)^2\)

So far so good.  BUT

\(\text{This is NOT }(8-x)^2\\ \text{it is  }   \quad9-[(1-x)^2] \)

You have to expand the bracket and then take away ALL the terms from 9.

Have another go at it    

Thanks Alan,

It really is a strange one :)  

It is happening relatively often.

Maybe it is something to do with copyright that Admin is trying to fix? 

I have no idea.

Hi Ava and Geno,

I don't remember ever seeing that circle property before but yes Ava, you have done what Geno described.

My pic says blocked by moderator.

It has not been done by a human moderator, it is blocked automatically and I cannot unblock it.

I assume it has been picked up as copyrighted material.

What pic do you, or did you, see Helpmeeee ?

AELN is being sarcastic.

I Just thought I would point that out since it will go over the top of guests head,   

sounds reasonable to me.

Like this

\(f(x)=-\sec(x/4)+3\ \left\{-2\pi <=x<=6\pi\right\}\)

in the sampler I got an example 22 ^ 10 / (2 ^ 7 * 11 ^ 9) = 88 I want to understand the solution, how did it happen?

\(\frac{22^{10}}{2^7*11^9}\\~\\ =\frac{(2*11)^{10}}{2^7*11^7*11^{2}}\\~\\ =\frac{(2*11)^{10}}{(2*11)^7*11^{2}}\\~\\ =\frac{(2*11)^{10-7}}{11^{2}}\\~\\ =\frac{(2*11)^{3}}{11^{2}}\\~\\ =\frac{2^3*11^3}{11^{2}}\\~\\ =\frac{2^3*11^{(3-2)}}{1}\\~\\ =8*11\\~\\ =88 \)

Thanks Ginger, that is very interesting :)