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Qualification => Subject Doubts => IGCSE/ GCSE => Math => Topic started by: HUSH1994 on September 28, 2010, 01:57:09 pm

Title: Math P2 MJ 2010 doubts
Post by: HUSH1994 on September 28, 2010, 01:57:09 pm: i have a few doubts,paper 21 math MJ 2010:
Questions:
8.b
10
15.a+b
17.c
Title: Re: Math P2 MJ 2010 doubts
Post by: elemis on September 28, 2010, 02:01:20 pm: This was the paper I did for my IGCSEs :P PAWNED it :D Give me a minute to write the explanations ;)
Title: Re: Math P2 MJ 2010 doubts
Post by: HUSH1994 on September 28, 2010, 02:04:12 pm: good,im waitin and Thanks for it dude ;)
Title: Re: Math P2 MJ 2010 doubts
Post by: elemis on September 28, 2010, 02:14:16 pm: Oh, actually speaking, this is a different variant from the one I did :P

For 8 b) look at the first link its an excellent powerpoint (http://www.google.co.uk/#hl=en&source=hp&biw=1280&bih=610&q=planes+of+symmetry+in+3d+shapes&aq=1&aqi=g10&aql=&oq=planes+of+symmetr&gs_rfai=&fp=f1232bdfd13d2256)

For 10

3.5 *5 + 1.5*2 + $\sqrt{3.5^2+1.5^2}$

I basically used Pythagoras's Theorem to find the length of the diagonal wooden thingy.

For 15
a) $\frac{3-5}{3-2} = -2$

Hence, $\frac{5-1}{2-k}=-2$

b) We've already found the gradient from (a) hence : y=-2x + c.

Inputing one of the coordinates into this equation : 5 = -2*2 +c

We solve for c. Hence c = 9

I'm doing the rest.
Title: Re: Math P2 MJ 2010 doubts
Post by: elemis on September 28, 2010, 02:25:34 pm: Angle COD is subtended at the centre of the circle by the arc CD.

The angle CED is subtended by the same arc at the CIRCUMFERENCE of the circle.

Hence, you should know that angles subtended by the same arc at the centre are twice those subtended at the circumference.

Hence, CED = 19. And then EDO = 71
Title: Re: Math P2 MJ 2010 doubts
Post by: Arthur Bon Zavi on September 28, 2010, 05:33:32 pm: For The 15th one:

(a). See the pattern (sequence):

Y co-ordinates are 5, 3 and 1. Means -2 every time, i.e 5-2=3, then 3-2=1.
X co-ordinates are 2, 3 and k. This also has the sequence, i.e 2+1=3, then 3+1=4, so k(x co-ordinate)=4.

(b).
For the Gradient remember the formula:

y₂-y₁
-------
x₂-x₁

The Points are, (2,5), (3,3) and (4,1).

take any two points, e.g (2,5) and (3,3).

then:

3 - 5 -2
------ = ----- = -2
3 - 2 1

Gradient (Slope)= -2.

m = gradient = slope = -2.

so:

y=mx+c.
put any of the values.

3=-2*3+c

so,
c=9.

so equation goes:

y=-2*x+9

Or a simpler method if u have gradient, then:

y-y₁=m(x-x₁), where m is gradient (slope).

Example from this:

y-3=-2(x-3)
y-3=-2x+6
therefore: y=-2x+9. [It's Easy].

Another Method is when u don't have a gradient and have atleast two co-ordinates, so here u have 3 (more than necessary). :D

take any two:

like (2,5), and (3,3).

Equation is :

y-y₁ x-x₁
----- = ------
y₂-y₁ x₂-x₁

so, by substituting:

y -5 x - 2
----- = ------
3 -5 3 - 2

It gives:

y -5 = -2 (x - 2)
y - 5 = -2x+4
y = -2x + 9
Title: Re: Math P2 MJ 2010 doubts
Post by: Arthur Bon Zavi on September 28, 2010, 05:43:57 pm: This answer I just explained briefly with the help of alternative known methods, and other answers are explained by Ari.
;)