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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w07_qp_3.pdf
question 10 part 2
anyone ?
PhyZac
question 10 part 2
anyone ?
PhyZac
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From the previous part, you have obtained a simplified form of the expression as p(x) = (x-2)(x+2)(x-4)
Thanks! : )From the previous part, you have obtained a simplified form of the expression as p(x) = (x-2)(x+2)(x-4)
Since the question now calls for evaluating p(x squared), you just substitute the x in the expression for x squared.
Your expression would become somewhat like this
(I'll be using x2 for x squared)
p(x2) = (x2 -2)(x2 +2)(x2 -4)
Now, you equate them to zero, each of them. You'll obtain six roots, two of which will be imaginary.
Okay, now u have to get a first which is
for stationary point, take the dy/dx equation and equate to 0.What equation did we take in Q10, part 2 ?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
AOA
can someone plz help me with 10c
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_12.pdf
Thank youWe take tan inverses of the gradient and subtract them.
tan^-1(m1) - tan^-1(m2) should give you the angle between the lines..
tan^-1(2) - tan^-1(1/2)
gives 36.9 Degrees Angle.
See, can u help me in (a) , i cant use the r-a.n equation , sorry.
share link next time plz.can you please help me in how to answer questions such as exam 9709/32/O/N10 the question 8 part b
its basically after making partial fractions...expanding it to ascending powers upto x^3 etc
how do you do these? plz help
i sorry about that....actually i have tried to solve it....but in Q8(i),my answer is C=-1/3,but the ms is 1/3...and i can‘t find any way to get “-”share link next time plz.
In the formula book u have the formula under binomial expansion
and use this video and then try resolving hope u will get it.
http://www.examsolutions.net/maths-...uences-series/binomial/formula/tutorial-3.php
I am afraid i am confused.i sorry about that....actually i have tried to solve it....but in Q8(i),my answer is C=-1/3,but the ms is 1/3...and i can‘t find any way to get “-”
SURESee, can u help me in (a) , i cant use the r-a.n equation , sorry.
for finding angle,
u take normal,
2i−3j+6k
and direction vector,
i − 2j + 2k
take scalar product
2+6+12 = 20
now divide with the mod of both vectors
and u get
20/21
20/21 normally =cos (angle)
but when plane and line
it is=sin (angle)
so sin^-1 (20/21) = 72.2
Yes part a.....thanks! Jazaki Allah khairan.SURE
part a right ?
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