JAMB Mathematics Past Questions & Answers - Page 313

2y = 5x + 4 (4, 2)

y = $\frac{5x}{2}$ + 4 comparing with

y = mx + e

m = $\frac{5}{2}$

Since they are perpendicular

m1m2 = -1

m2 = $\frac{-1}{m_1}$ = -1

$\frac{5}{2}$ = -1 x $\frac{2}{5}$

The equator of the line is thus

y = mn + c (4, 2)

2 = -$\frac{2}{5}$(4) + c

$\frac{2}{1}$ + $\frac{8}{5}$ = c

c = $\frac{18}{5}$

y = -$\frac{2}{5}$x + $\frac{18}{5}$

5y = -2x + 18

or 5y + 2x - 18 = 0

P(x, y) Q(8, 6)

midpoint = (5, 8)

x + 8 = 5

$\frac{y + 6}{2}$ = 8

x + 8 = 10

x = 10 - 8 = 2

y + 6 = 16

y + 16 - 6 = 10

therefore, P(2, 10)

No explanation has been provided for this answer.

Volume of cube = L³

3³ = 27cm³

volume of rectangular tank = L x B X h

= 3 x 4 x 5

= 60cm³

volume of H₂O the tank can now hold

= volume of rectangular tank - volume of cube

= 60 - 27

= 33cm³

From Pythagoras theorem

|OA|2 = |AN|2 + |ON|2

72 = |AN|2 + (5)2

49 = |AN|2 + 25

|AN|2 = 49 - 25 = 24

|AN| = $\sqrt {24}$

= $\sqrt {4 \times 6}$

= 2√6 cm

|AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts)

|AN| + |NB| = |AB|

2√6 + 2√6 = |AB|

|AB| = 4√6 cm