Qualification > Math

MATHS HELP!! (PAPER 4)

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examsaresoon:
Hi! I was just wondering if some one could be nice enough and help me solve a maths question?
Paper 4, Oct/Nov 2004, Question 7. I'm realllly bad at this topic...I sloved the first part....but got stuck after that.  :(
Help, Please?

Ascer Zayan:

--- Quote from: examsaresoon on May 17, 2009, 05:54:19 am ---Hi! I was just wondering if some one could be nice enough and help me solve a maths question?
Paper 4, Oct/Nov 2004, Question 7. I'm realllly bad at this topic...I sloved the first part....but got stuck after that.  :(
Help, Please?

--- End quote ---
Nov 2004 Q7)a)i)
1x(4)^2+(-2)(4)+(-3)
= 5
ii)Solve the Equation (x^2-2x-3)
Answer = (x-3)(x+1)
              X=3 or X=-1
So cordinates for K (-1,0) and for L it is (3,0)
III) -b divided by 2a
     -(-2)/(2x1)
          X=1                   Subsitite X in the Equation and find Y = -4
Hence Cordinates is (1,-4)
B)it will become upside down simple enough
ii)if there is no Y-intercept then the GRAPH SHOULD PASS THROUGH THE ORIGIN
C)C=0
 II)(3,0) and (4,8) Equation is Y=ax^2+Bx     
0=A(3^2)+3B
8=A(4^2)+4b                 THEN SOLVE SITMIELOUSLY
                                      A=2
                                      B=-6

examsaresoon:
Thanks! I finally understood that!!

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