Math Problems... help!?

Question

1. Given k(x): evaluate k(x+h)
6-x
___
x-3

2.Given h(x)=x^2 and p(x)= 2x-3, evaluate (p o h)(-3)
(btw, the o is a composite. )

3. For the group given in the chart at the right find the identity and inverse of each element. As well as evaluate (a@b)@c.
(There's a chart....

@ I a I b I c I d I
a I d I a I b I c I
b I a I b I c I d I
c I b I c I d I a I
d I c I d I a I b I
)

Can you explain the answers and the way to do it step by step. I don't understand the first two and I don't remember how to do the last one.

Thank you so much!

Answer 1

I can only help you on the first 2 I'm not sure I have enough info on he last one but...

function k of x k(x+x) that is too many varibles not enough #s the best I can tell you to do is turn 6-x/x-3 -> 6-(x+h)/(x+h)-3 or 6-x-h/x+h-3 which I did by placing x+h into everywhere there was an x

p(h(x))set up poh as
p(h(-3)) plug in your x
p(-3^2)plug in equation h using -3 as x h first because it is listed 2nd
p(9) evaluattion of h
2(9)-3 using 9 the evaluation of h as your x plug it into p
18-3 solve
15 final answer

sorry I can't be much help but I'm going on what I have

Answer 2

For the first problem, substitute (x+h) for the x's you have in your rational expression. After a bit of distributing in the numerator, your answer is: 6-x-h over x+h-3.

Answer 3

dear kala as with kelsey I only know the answer of ur second question but with another way.I thought it might be simpler for u to underestand:
we know that (poh)is a combination of two functions and u can write it like p(h(x)).it means that u have changed the parameter x with h(x).therefore u can write:
p(h(x))=2h(x)-3 so : p(h(x))=2(x^2)-3

now if u put -3 instead of x :p(h(x))=15

the same number of kelsey

Answer 4

Coming from someone pretty good at math, that's hard math.

Answer 5

...