for question 9a am not so good with my f(f(x) so would appreciate any help with this one please.

question 11

in a hexagon ave presumed tht it is 6 equilateral triangles..is this correct?

therefore a.(b+c) I multiplied out and got a.b +a.c so i did -2 times 2 cos 60 + a.c but i am unsure wot 2 substitue in for the a.c part!!

Again any help appreciated

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# winter diet 2002 non calc

Started by hssc11045, May 19 2004 07:47 PM

2 replies to this topic

### #1

Posted 19 May 2004 - 07:47 PM

### #2

Posted 19 May 2004 - 08:01 PM

Do you understand the idea of f(f(x))

so you would have f(f(x))

f(3/x+1)

3 / (3/x+1) + 1

from here on you just give in simplest form

for question 11 yes you do do as you say

a.b + a.c /a.b = |a|*|b| * cos 120 = 2*2*-0.5

a.c = |a| * |b| *cos 60 = 2*2* 0.5

so the final line is 2-2 = 0

so you would have f(f(x))

f(3/x+1)

3 / (3/x+1) + 1

from here on you just give in simplest form

for question 11 yes you do do as you say

a.b + a.c /a.b = |a|*|b| * cos 120 = 2*2*-0.5

a.c = |a| * |b| *cos 60 = 2*2* 0.5

so the final line is 2-2 = 0

If i am not here i am somewhere else

### #3

Posted 19 May 2004 - 08:56 PM

yes i understand tha concept of f of f of x meaning from this example 3 divided by whatevr x is and then add 1 but when i do this i get 3(x+1)/4.....which aint right!!

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