its p1? and i wish i had aritting like u . mine is likea chinese script
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Mathematics: Post your doubts here!
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jun2010 v11 no.1 pure 1
http://www.novapapers.com/images/PDF/CIE/A-LEVEL/maths/2011nov/9709_w11_qp_33.pdf
P3 q6 part 2. Its an argand diagram question! Please help
P3 q6 part 2. Its an argand diagram question! Please help
link?
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well we had couplr of formulaes in addmath which i remember they were tan(180-x)=-tanx , tan(90-x)=1/tanx
apply first formulae for part i you will get -k
apply second formulae for part two tan(pie/2 which is 90 - x) = 1/tanx means 1/k
for third we can use tan=perp/base to see perp=k base=1 use puthagorus theorum and you will get hyp= underroot(k^2+1)
so sinx=pep/hyp you wil get the answe
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Thank youuwell we had couplr of formulaes in addmath which i remember they were tan(180-x)=-tanx , tan(90-x)=1/tanx
apply first formulae for part i you will get -k
apply second formulae for part two tan(pie/2 which is 90 - x) = 1/tanx means 1/k
for third we can use tan=perp/base to see perp=k base=1 use puthagorus theorum and you will get hyp= underroot(k^2+1)
so sinx=pep/hyp you wil get the answe
Someone help!!!!! http://www.novapapers.com/images/PDF/CIE/A-LEVEL/maths/2011nov/9709_w11_qp_33.pdf
P3 q6 part 2. Its an argand diagram question! Please help
P3 q6 part 2. Its an argand diagram question! Please help
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http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_33.pdf
help needed in Q1o...Dug thanks in advance
help needed in Q1o...Dug thanks in advance
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- Messages
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you first plot points A and B on the diagram and i am assuming you know how to do that
next you find the mid point between A and B
you get -1/2-1/2i
this will be the centre of the circle
no you have to find the radius of the circle
find the distance between the two points using sqrt[(x2-x1)^2+(y2-y1)^2 and you get an answer
square root of answer will give you the radius
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I = ⌡(tan^(n+2) x + tan^n x) dx
take tan^n out as a common factor
I = ⌡[tan^n x(tan^2 x + 1)] dx
u = tan x
dx = du/sec^2 x
I = ⌡[tan^n x(tan^2 x + 1)] du/sec^2 x
tan^2 x + 1=sec^x so you can cancel sec^2 x
I = ⌡u^n du
I = u^n+1/(n + 1)
put in the new limits
u = tan x
tan(π/4) = 1
tan(0) = 0
I = 1/(n + 1)
can someone please help , w09_qp_51 question no 1 , i got the X of 0.05 and multiplied it by 2 due to the 2 forces acting on both sides why didin't the answer do that ?!! http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_51.pdf
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part 2 is easy i hope you can do it now
part 3 ) (tan9 + tan7) + (tan5+tan3) + 4(tan7+tan5)
tan n + tan (n+2) = 1/(n+1)
so for tan9 + tan7 it will be 1/(7+1)
for tan3 + tan5 = 1/(3+1)
for tan5 + tan7 = 1/(5+1)
1/(7+1) + 1/(3+1) + 4*1/(5+1)
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please pure 1 jun2010 v11 no.10 and n0.8 iii .
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a better dialogue p3 p3 every where xD. anyway in s1 heres a good one there are 9 pies to be divided between 3 people so that each person gets a odd no of pies. in how many no of ways can they be divided. i know the answerso i am posting it so u guys can solve it and others can get help.
Ways to be divided
3 5 1 x 3!
3 3 3 x 3!/3!
1 7 1 x 3!/2!
9C3 * 6C5 * 1C1 * 3! = 3024
9C3 * 6C3 * 3C3 * 3!/3! = 1680
9C1 * 8C7 * 1C1 * 3!/2! = 216
Total No. Of Ways = 4920..
If the pies are identical then it would be a whole different story..
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i ) OB = OA + OC
OB = 4i + 2j + 4k
Unit Vector of OB = (4i + 2j + 4k) / (√4^2+2^2+4^2)
= (4i + 2j + 4k)/6
ii) AC = OC - OA
AC = 2i -4j - 2k
Now use the scalar product formulae
OB = 4i + 2j + 4k
AC = 2i -4j - 2k
AC * OB = |AC| * |OB| * Cos(Theta)
(4*2)+(2*-4)+(4*-2) = 2√6 * 6 * Cos(Theta)
-8/12√6 = Cos(Theta)
Theta = Cos^-1(-8/12√6)
Theta = 105.8 Degrees
They asked for acute angle = 180 - Ans
= 74.2 Degrees Answer
iii) 2 * (Magnitude of OA + Magnitude of OC)
2 * (|OA| + |OC|)
2* (√19 + √11)
= 15.35 Answer
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8iii) The perpendicular bisector of AB meets BC at D.
Perpendicular bisector of a line is always from the middle of the line (Midpoint of the line) and is 90 Degrees to the line.
So find the co-ordinates of midpoint.. then find the equation of normal on the midpoint (normal gradient = -m/2 as [2m * grad = -1])
Then simultaneously solve equation of Normal (of AB) and equation of BC.. to get the co-ordinates of D.