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Mathematics: Post your doubts here!

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Here: sin^2 (2x) (cosec^2 (x)-sec^2 (x))= 4cos2x
Sin^2 (2x) è4sin^2x cos^2 x
cosec^2 (x)-sec^2 (x) è 1/sin^2 x – 1/ cos^2 x = (cos^2 x – sin^2 x ) / (sin^2 x cos^2 X)
cos^2x – sin^2 x = cos 2x

now just simplify...
P.S. kinda difficult to put each and every step...that’s why did this way..

You are great ! Do you do M1 & S1 ?
 
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Assalamoalaikum
could i get help with these:
1. In october/november 2012 paper 2 question 1 i got the correct answer but my way was wrong could you tell me the correct way to go about doing this question.
2. Same paper question 5 same problem i got the same answer but i am not sure about my method.
3 .Same paper question 7 . I've got seriously no idea how to do it.
4.Same paper question 9 ,actually Im really bad at sequences not just this question .
5.Same paper question 15 part (a)

6.Same paper question 19 both parts


Ya they are a lot,(i am bad at maths ,.. you might have figured that out by now XP ) but i look forward to your help.

i dont think october / november 2011 papers are at xtremepapers you can see them here
http://www.mediafire.com/?nxir7u4uroyw8#cfgjat49f8s73
 
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hey guys need help with this question.
The variables x and y satisfy the equation x^ny = C, where n and C are constants. When x = 1.10,
y = 5.20, and when x = 3.20, y = 1.05. (i) Find the values of n and C. [5]
(ii) Explain why the graph of ln y against ln x is a straight line. [1]
This is from 9709 mj 10 31 q3.
Have understood the first step but cant go further,need some help with it.
thanks
 
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Can anyone help me out... ??
M/J 2005

Q2) b) It is given that
ln  = ln(y + 2) − 2 ln y,
where y > 0. Express  in terms of y in a form not involving logarithms
 
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thanx a lot but how did you know that u must convert t to radians not keep as degrees in the second part?
as it was 0.6:p
we use degree when angle is like 90, 10, 20 but it was 0.6 i do agree that we ussualy take something with pi as to radiance but still how can we use 0.6 for degree.......
 
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hey guys need help with this question.
The variables x and y satisfy the equation x^ny = C, where n and C are constants. When x = 1.10,
y = 5.20, and when x = 3.20, y = 1.05. (i) Find the values of n and C. [5]
(ii) Explain why the graph of ln y against ln x is a straight line. [1]
This is from 9709 mj 10 31 q3.
Have understood the first step but cant go further,need some help with it.
thanks
i) ln(x^ny) = lnC
ln(x^n) + lny = lnC
nlnx + lny = lnC
nln1.10 + ln5.20 = lnc
nln3.20 + ln1.05 = lnC
solve them simultaneously. u will get the value of n and C
 
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need help in 9709 mj 31 q7.
Have tried doing it but can't seem to arrive at the final answer.
thanks
 
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i) ln(x^ny) = lnC
ln(x^n) + lny = lnC
nlnx + lny = lnC
nln1.10 + ln5.20 = lnc
nln3.20 + ln1.05 = lnC
solve them simultaneously. u will get the value of n and C
Could please solve it simulatioeously?can't seem to get the correct answer.
 
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Could please solve it simulatioeously?can't seem to get the correct answer.
0.095n + 1.65 = lnC
1.16n + 0.05 = lnC
0.095n + 1.65 = 1.16n + 0.05
1.65 - 0.05 = 1.16n - 0.095n
1.6 = 1.065n
n = 1.5
0.095(1.5) + 1.65 = lnC
lnC= 1.79
C = e^1.79
C = 6
 
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s1 guyz !! plz help me asap !!
idk when to use in a frequency table the barrier if its ( a number <x<= a number) or ( a number =<x< a number)
 
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