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ln(x+5) - lnx =1Assalamo'alaikum, I need help with this question:
Solve the equation
ln(x + 5) = 1 + lnx,
giving your answer in terms of e.
i'm home studying and no one around can help me right now.
ln(x+5) - lnx =1
thanks. how about this one, how do you differentiate x = 4t/(2t + 3)?
Question 7 (ii)Which Question?
Its quite simple for the question 7 part 2 just do it like
3/(1 + 2x)^2 < 1/3 and then solve for x in the next step you will get like this
(1 + 2x)^2 (>)9
then
once when you solve it using the (a+b)^2 formulae you will get it like 4x^2 + 4x +1>9
you will get like
4x^2 +4x -8 and then solve it quadratically......ok get it like x > 1, x < –2 ok gt it?
Question 7 (ii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
In qs 5i , the qs says 94% of the letters are within 12g of mean, so isnt it supposed to be (20+12<x<20-12)?? my answer's not matching with the mark scheme......
pleaseee help Dug syed1995 Ashique
also tell me how to find the p is 0.97
This is very tricky question. Took me time to understand, anyway,
This is very tricky question. Took me time to understand, anyway,
Alice123
you are right about (20+12<x<20-12)
now you have to imagine the graph in your head, and you have to come into a conclusion that it is symmetrical.
Now imagine the z of 20+12 that is 32 is x
So the z of 20-12 that is 8 will be a negative number -x
Both of them are x but one is negative, because they are the same distance from the mean
Let the probability of x is p
so the probability of -x is 1-p
we know that difference of the two probability is 0.94
so p - (1-p) = 0.94
p - 1 + p = 0.94
2p = 0.95 + 1 = 1.94
p = 0.97
Now is p is 0.97 z is from the table, 1.882 (or 1.881)
now continue normally
1.882 = 32-20 / sd
sd = 12/1.882
=6.38
thanx...wat bout the oda 3qs?oct/nov 08
2. 1+sinx/cosx = cosx/1+sinx take common denominator
= [(1+sinx)^2 + (cos^2)x]/cosx(1+sinx)
= (1 + 2sinx + (sin^2)x + (cos^2)x)/cosx(1+sinx) here an identity needs to be used : (cos^2)x +(sin^2)x = 1, so (sin^2)x = 1 - (cos^2)x
= (1 + 2sinx + 1 - (cos^2)x + (cos^2)x )/ cosx(1+sinx) , the '(cos^2)x' cancels out
= 2 + 2sinx/cosx(1+sinx)
= [2(1 + sinx)]/cosx(1+sinx) , the '1+sinx' cancels out
= 2/cosx hence equal to right hand side
hope you understood
Help on this one please?
On a certain road 20% of the vehicles are trucks, 16% are buses and the remainder are cars.
(i) A random sample of 11 vehicles is taken. Find the probability that fewer than 3 are buses. [3]
(ii) A random sample of 125 vehicles is now taken. Using a suitable approximation, find the
probability that more than 73 are cars.
dnt have s1
p1,m1you goin for m1,m2 .. or just have p1,m1 this year?
LHS = 1/sin2Θ - cos2Θ/sin2ΘProve the identity: cosec 2Ɵ - cot 2Ɵ ≡ tan Ɵ
LHS = 1/sin2Θ - cos2Θ/sin2Θ
= (1 - cos2Θ)/sin2Θ
= [1 - (1 - 2sin²Θ)]/2sinΘcosΘ
= tanΘ
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