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Because repetition is allowed in that question. It would be 6*5*4 if repetition was not allowed.wait could you elaborate please? why did you take 6^3?
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Because repetition is allowed in that question. It would be 6*5*4 if repetition was not allowed.wait could you elaborate please? why did you take 6^3?
Sorry but I still don't understandBecause repetition is allowed in that question. It would be 6*5*4 if repetition was not allowed.
h(x) = 6x−x^2http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf
The last question's last part , I'm getting one of the signs wrong for some reason. They changed the equation when taking the inverse? :\
If someone could do it with steps included I'd love them forever .
h(x) = 6x−x^2
= - (x^2 - 6x)
= - [ (x-3)^2 - 9]
= 9 - (x-3)^2
To find h*inverse*
Let y = h(x)
y = 9 - (x-3)^2
(x-3)^2 = 9 - y
x - 3 = (+or -) *root* 9 - y
x = (+or -) [*root* 9 - y ] + 3
Since the domain is positive, we take the positive root.
Therefore h*inverse* = [*root* 9 - x ] + 3
Read thisSorry but I still don't understand
I've never encountered a question like this before
I didnt move the 9. I switched - (x-3)^2 with y. Got it now?The part I highlighted, how come you took 9 to the other side but changed the sign of 'y' ?
I didnt move the 9. I switched - (x-3)^2 with y. Got it now?
Not exactly, why are you equating the equation while finding the inverse?
This is what I did.
To find H inverse
Let y = h(x)
h(x)= 9-(x-3)^2
h(x)-9= (x-3)^2
*root*h(x)-9= x-3
[*root*h(x)-9]+3 = x
H inverse = [*root*h(x)-9]+3 where as yours and the MS's is "9-x" instead of "x-9"
EDIT: I undestand what you did, but why is my method wrong?
"To find H inverse
Let y = h(x)
h(x)= 9-(x-3)^2
h(x)-9= - (x-3)^2
The '-'... U omitted it! Correction in red
No prob!Ah thanks
can someone pls help me with question 5 and 6bii) nd 6c
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_63.pdf
thnx in advance
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