2014 - JAMB Mathematics Past Questions and Answers - page 3
-x & 12 \\ -1 & 4
\end{vmatrix} = - 12, \text{ find x}\)
(\begin{vmatrix}
-x & 12 \
-1 & 4
\end{vmatrix} = - 12)
-4x - (-1)12 = -12
-4x + 12 = -12
-4x = -12 - 12
-4x = - 24
x = 6
Users' Answers & Comments0 & 3 & 2 \\ 1 & 7 & 8 \\ 0 & 5 & 4
\end{vmatrix}\)
(0 \begin{vmatrix}
7 & 8 \
5 & 4\end{vmatrix}-3 \begin{vmatrix}
1 & 8 \
0 & 4\end{vmatrix}+2 \begin{vmatrix}
1 & 7 \
0 & 5\end{vmatrix})
= 0(28 - 40) - 3(4 - 0) + 2(5 - 0)
= 0(-12) - 3(4) + 2(5)
= 0 - 12 + 10
= -2
Users' Answers & CommentsIf each interior angle of the polygon is 135o, then each exterior angle is 180o - 135o = 45o. Hence, number of sides =
(\frac{360^o}{\text{one exterior angle}})
(\frac{360^o}{45^o})
= 8
Users' Answers & CommentsUsing (V = \pi r^2 h)
6160 = 22/7 x 28 x 28 x h
(h = \frac{6160}{22 \times 4 \times 28} )
(h = 2.5m )
Users' Answers & CommentsMid point of S(-5, 4) and T(-3, -2) is
([\frac{1}{2}(-5 + -3), \frac{1}{2}(4 + 2)])
([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)])
([\frac{1}{2}(-8), \frac{1}{2}(2)])
([-4, 1])
Users' Answers & Comments(\text{Gradient m} = \frac{y_2 - y_1}{x_2 - x_1})
(\frac{1}{2} = \frac{2 - 4}{1 - x})
1 - x = 2(2 - 4)
1 - x = 4 - 8
1 - x = -4
-x = -4 - 1
x = 5
Users' Answers & Commentsy - 4x + 3 = 0
When y = 0, 0 - 4x + 3 = 0
Then -4x = -3
x = 3/4
So the line cuts the x-axis at point (3/4, 0).
When x = 0, y - 4(0) + 3 = 0
Then y + 3 = 0
y = -3
So the line cuts the y-axis at the point (0, 3)
Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;
([\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)])
([\frac{1}{2}(\frac{3}{4} + 0), \frac{1}{2}(0 + -3)])
([\frac{1}{2}(\frac{3}{4}), \frac{1}{2}(-3)])
([\frac{3}{8}, \frac{-3}{2}])
Users' Answers & Comments4x + 3y - 5 = 0 (given)
The equation of the line perpendicular to the given line takes the form 3x - 4y = k
Thus, substitution x = -2 and y = 3 in 3x - 4y = k gives;
3(-2) - 4(3) = k
-6 - 12 = k
k = -18
Hence the required equation is 3x - 4y = -18
3x - 4y + 18 = 0
Users' Answers & Comments