Mathematics: Post your doubts here!

GorgeousEyes

GorgeousEyes

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Maz said:
June-12/12
7) Using the formula given first find the sum of the first 1 term. (sounds funny)
Sn= 1^2 + 8(1)
= 9
Then find of sum of the first two terms
Sn= 2^2 + 8(2)
= 20
subtract the two terms to get the second term
20-9 = 11
and then 11-9 = 2 is the common difference

[i cant seem to be able to solve the second part]

June-12/13
9(iii) Since BC = AD [since its a square], the translation of D from A is the same that of C from B.
i.e. the y-step is 6 and x-step is 2
D= (-3+2, 2+6)
= (-1, 8)

|AD|= use the formula that's all. (x2-x1)^2 - (y2-y1)^2 The whole square root. [i dont know how to type that in]

Hope that helps.
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Thaaanks a millioooon :)
 
Esme

Esme

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ahmed abdulla said:
by using the discriminant rule
ie) b^2-4ac
if b^2-4ac >0 (2 point intersection)
if b^2-4ac=0 (1 point intersection)
if b^2-4ac<0 (no intersection) thats for P1 and i dont now which paper ur talking about
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Thanks but you're telling me about coordinate geometry. I asked for vectors in P3. :)
It's ok, I got the explanation though.
 
iKhaled

iKhaled

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GorgeousEyes said:
How to prepare myself 4 the exam ? :)
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do loaaadsss of past papers and the most important thing is to make ur answer to 3 sig. figures because they r so strict on giving final answers to 3 sig figures so imagine u have 11 questions and in every question u lost that A1 mark because ur answer is not to 3 sig figures that means u lost 11 easy marks excluding the mistakes u have made...u need to be able to do every question in the past paper like literally every question..
 
P

PhyZac

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sma786 said:
Hii, can anyone solve this question in detail for me please?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_1.pdf Question 5, thanks, :love:
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You are lucky, i just finished this paper!


They said to find the distance between two points of intersection.

First we should find what are the points.

To do this we should do simultaneous equation.

3y= 4x+6 -----(1)
y^2 = 12x ----(2)

In equation (2)
x= y^2/12

Sub this to equation (1)

3y = 4(y^2/12) + 6
3y = y^2/3 +6
9y = y^2 + 18
y^2 - 9y +18 = 0
(y-3) or (y-6) = 0

y = 3 or y = 6

when y = 3 [sub to (1)]
3(3) = 4 x + 6
x = 3/4

so first point (3/4 , 3)

when y = 6
3(6) = 4x + 6
x = 3

so second point (3, 6)

Distance √(x1-x2)^2 + (y1-y2)^2
√(3-(3/4))^2 + (6-3)^2
= 3.75
 
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