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Abid357 said:One question... in the elimination method, does it really matter whether I solve x or y the first? Will that have any effect on the final answer?
Math_angel said:Assalamoalaikum!!
ok so i've spotted ur first mistake!! well when u multiply 3 and 1/2 u get 3/2 ...right??
similarly 3[(11 – 3y) / 2] = (33-9y) /2
denominator is 2 not 6!!
let me see if there's any other mistake!!
post the question here and we will guide you !!WayneRooney10 said:Oh! okay...but i only need help in solving functions...like...f(x)= 2x-5.....i always get confused.
CaptainDanger said:Pernee26 said:Please can you solve October/November 2008 paper 4 Q3 Q8 and Q10 ?
Oh and can you give me useful tips on Loci and how Cambridge usually gives questions on this topic ?
Your help will be highly appreciated !
Q 8
(a)
x= 78° Answer (alternate angles)
y=180°-36°=144°Answer (Here you have to use the property that sum of opposite sides of CYCLIC quadrilateral = 180°)
z= 180°-78°=102° Answer again same property as above...
(b) The some of x+y needs to be 180° if they are parallel but its not the case so they are not... Taking alternate angles again... But that is only possible if they are parallel...
(c) Angle subtended at center is always double of the angle at the circumference from the same CHORD... So for E0C draw line EC now apply the property...
E0C=144/2=72° Answer
(d) Two sides equal so two angles equal... Its a triangle you can see... So 180°-78°=102° As both the angles are same so divide 102° by 2 = 51° Answer....
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