JAMB Mathematics Past Questions & Answers - Page 306

Let (x1, y1) = (1, 1) and (x2, y2)= (2, 5)

then gradient m of $\bar{PQ}$ is

m = $\frac{y_2 - y_1}{x_2 - x_1}$ = $\frac{5 - 1}{2 - 1}$

= $\frac{4}{1}$

= 4

Let D denote the distance between ($\frac{1}{2}$, -$\frac{1}{2}$) then using

D = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

= $\sqrt{(-{\frac{1}{2} - \frac{1}{2}})^2 + (-{\frac{1}{2} - \frac{1}{2}})^2}$

= $\sqrt{(-1)^2 + (-1)^2}$

= $\sqrt{1 + 1}$

= √2

Total surface area of the cylindrical pipe = area of circular base + curved surface area

= $\pi$r$^2$ + 2$\pi$rh

= $\pi$ x 7$^2$ + 2$\pi$ x 7 x 50

= 49$\pi$ + 700$\pi$

= 749$\pi$m$^2$

(x + 15)° + (2x - 45)° + (x + 10)° = (2n - 4)90°

when n = 4

x + 15° + 2x - 45° + x - 30° + x + 10° = (2 x 4 - 4) 90°

5x - 50° = (8 - 4)90°

5x - 50° = 4 x 90° = 360°

5x = 360° + 50°

5x = 410°

x = $\frac{410^o}{5}$

= 82°

Hence, the value of the least interior angle is (x - 30°)

= (82 - 30)°

= 52°

P = $\begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}$

|P| = 2 - 1 x -3 = 5

P-1 = $\frac{1}{5}$$\begin{pmatrix} 1 & 3 \\ -1 & 2 \end{pmatrix}$

= $\begin{pmatrix} {\frac{1}{5}} & {\frac{3}{5}} \\ -{\frac{1}{5}} & {\frac{2}{5}} \end{pmatrix}$