Mathematics Questions and Answers

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

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