2010 - JAMB Mathematics Past Questions and Answers - page 7
3rd term : a + 2d = -9 .......(1)
7th term : a + 6d = -29 ......(2)
(2) - (1): 4d = -20
:. d = -20/4 = -5
From (1) : a + 2(-5) = -9
a - 10 = -9<br />
:. a = -9 + 10 = 1
:. 10th term of A.P is a + 9d = 1 + 9 (-5)
= 1 - 45 = -44
Users' Answers & CommentsGiven that y = -3 - 2x + X2
then, (\frac{dy}{dx}) = -2 + 2x
At maximum value, (\frac{dy}{dx}) = O
therefore, -2 + 2x
2x = 2
x = 2/2 = 1
Users' Answers & CommentsBy comparing y = mx + c
with y = -4x + 2,
the gradient of y = -4x + 2 is m1 = -4
let the gradient of the line parallel to the given line be m2,
then, m2 = m1 = -4
(condition for parallelism)
using, y - y1 = m2(x - x1)
Hence the equation of the parallel line is
y - 3 = -4(x-2)
y - 3 = -4 x + 8
y + 4x = 8 + 3
y + 4x = 11
y + 4x - 11 = 0
Users' Answers & Commentsp = (\frac{M}{5})(X + Q) + 1
P - 1 = (\frac{M}{5})(X + Q)
(\frac{5}{M})(p - 1) = X + Q
(\frac{5}{M})(p - 1)- x = Q
Q = (\frac{5(p -1) - Mx}{M})
= (\frac{5p - 5 - Mx}{M})
= (\frac{5p - Mx - 5}{M})
Users' Answers & Comments