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Home » Tag Archives: Waec 2016/2017 FURTHER MATHEMATICS Obj And Theory Answers

Tag Archives: Waec 2016/2017 FURTHER MATHEMATICS Obj And Theory Answers

Waec 2016/2017 FURTHER MATHEMATICS Obj And Theory Answers

FURTHER MATHEMATICS  ANSWERS FURTHER MATHS OBJ: 1-10 ABACCDCCCD 11-20 CDDDDABDBC 21-30 BDADABACBC 31-40 CCBBABABDD (2) (5,2)(-4,k)(2,1) (y3-y2)/(x3-x2)=(y2-y1)/(x2-x1) (1-k)/2-(-4)=(k-2)/(-4-5) (1-k)/(2+4)=(k-2)/-9 (1-k)/6=(k-2)/-9 -9(1-k)=6(k-2) -9+9k=6k-12 9k-6k=-12+9 3k=-3 k=-1 ==================================== (3a) If f(x+2)=6x^2+5x-8) To find f(5) Therefore f(x+2)=f(5) where x+2=5 x=5-2 x=3 therefore f(5)=6(3)^2+5(3)-8 =6(9)+15-8 =54+7 =61 (3b) (7root2+3root3)/(4root2-2roo3)*(4root2+2root3)/(4root2+2root3) (24*2+14root6+12root6+6*3)/(16*2+8root6-8root6-4*3) (48+26root6+18)/(32-12) =(66+26root6)/20 =66/20+(26root6/20) =33/10+(13root6/10) =3.3+1.3root6 ====================================== (4) (x^2+5x+1)sqroot(2x^3+mx^2+nx+11)=(2x-5) remainder:30x+16 (x^2+5x+1)(2x-5) =2x^3+10x^2+2x-5x^2-25x-5 =2x^3+10x^2-5x^2-25x-5 =2x^3+5x^2-23x+30x+16-5 =2x^3+5x^2+7x+11 ... Read More »