WAEC Further Mathematics Past Questions & Answers - Page 7

\(u = 27 ms^{-1}; a = -9 ms^{-2}; v = 0; t = ?\)

\(v = u + at; t = \frac{v - u}{a}\)

\(t = \frac{0 - 27}{-9} = \frac{-27}{-9}\)

\(\therefore t = 3 sec\)

\(|MQ| = \frac{1}{2} |TQ|\)

\(|TQ| = √((y2 - y1)^2 + (x2 - x1)^2)\)

\(|TQ| = √((6 - 4)^2 + (-8 - 2)^2)\)

\(|TQ| = √(2^2 + (-10)^2)\)

\(|TQ| = √(4 + 100) = √104\)

\(|TQ| = 2√26 units\)

\(|MQ| = \frac{1}{2} |TQ| = 2 \times 2√26\)

∴ \(|MQ| = √26 units\)

Standard Form equation of a circle (Center-Radius Form): \((x − a)^2 + (y − b)^2 = r^2\)

Where "a" and "b" are the coordinates of the center and "r" is the radius of the circle

\(2x^2+2y^2-4x+5y+1=0\)

Divide through by 2

= \(x^2+y^2-2x+\frac{5}{2}y+\frac{1}{2}= 0\)

=\(x^2-2x+y^2+\frac{5}{ 2}y=-\frac{1}{ 2}\)

=\(x^2-2x+1^2+y^2+\frac{5}{ 2}y+(\frac{5}{ 4})^2-1-\frac{25}{16}=-\frac{1}{2}\)

=\((x-1)^2+(y+\frac{5}{4})^2=-\frac{1}{2}+1+\frac{25}{16}\)

=\((x-1)^2+(y-(-\frac{5}{4}))^2=\frac{33}{16}\)

=\((x-1)^2+(y-(-\frac{5}{4}))^2=(\frac{\sqrt33}{4})^2\)

\(\therefore a = 1, b = - \frac{5}{4} and\) \(r \frac{\sqrt33}{4} (answer)\)

From the table, x * z = x, y * z = y, z * z = z and w * z = w

∴ z is the identity element

\((\frac{1}{9})^{2x-1} = (\frac{1}{81})^{2-3x}\)

\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{2(2-3x)}\)

\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{4-6x}\)

Since the bases are equal, powers can be equated

= 2x - 1 = 4 - 6x

= 2x + 6x = 4 + 1

= 8x = 5

\(\therefore x = \frac{5}{8}\)