WAEC Past Questions and Answers - Page 2235

Let p = \(\frac{2}{3}\) and q =  \(\frac{- 3}{4}\)

using (y - p)(y - q) = 0

= ( y -  \(\frac{2}{3})\)( y - (\(\frac{- 3}{4})) = 0\)

 =  (\( y -  \frac{2}{3})( y + \frac{3}{4})\) = 0

\( y^2 + \frac{3}{4}y - \frac{2}{3}y - \frac{6}{12} = 0 \)

\( y^2 + \frac{1}{12}y - \frac{1}{2}\) = 0

= multiply through by the l. c. m of 3 and 4 = 12

∴ the quadratic equation is  \(12y^2 + y - 6 = 0\)

\(log_3 x + log_3 (x - 8) = 2\)

\(log_3 x(x - 8) = log_39\) since 2 =   \(log_39\)

\(log_3\) cancels out

⇒ x(x - 8) = 9

⇒ \(x^2 - 8x = 9\)

⇒ \(x^2 - 8x - 9 = 0\)

⇒ \(x^2 - 9x + x - 9 = 0\)

⇒ x(x - 9) + 1(x - 9) = 0

⇒ (x - 9)(x + 1) = 0

⇒ x = 9 or x = -1

Since we can't have a log of negative numbers,

∴ x = 9.

∠RMN = 90° (angles in a semicircle is a right angle)
∠SRN = ∠RMN + ∠MNR (ext. angle of a ∆ is equal to the sum of two opp. int. angles)
⇒ 5x + 20 = 90 + x
⇒ 5x - x = 90 - 20
⇒ 4x = 70

x = \(\frac{70}{4}\)

 = 17.4°

therefore, 2x = 2\(\times17.5\) = 35°

Let x = original price

A decrease of 22% = (100-22)% = 78% of the original price.

78% of x = 27.30

= \(\frac{78x}{100} = 27.30\)

78x = 2730

then, x = \(\frac{2730}{78}\)

x =  35

Therefore, the original price of the shoe was  $35.00.

 Let A = (1, -3), B = (6, 2) and C = (0,4)

\(d =\sqrt{ (y_2 - y_1)^2 + (x_2 - x_1)^2}\)
|AB| = \(\sqrt{(2 - (-3))^2 + (6 - 1)^2}\) 
= \(\sqrt{(2 + 3)^2 + (6 - 1)^2}\) 
= \(\sqrt{5^2 + 5^2}\) 
= \(\sqrt{25 + 25}\)
= \(\sqrt50\)
= 5\(\sqrt2\) units

|BC| = \(\sqrt{(4 - 2)^2 + (0 - 6)^2}\)
= \(\sqrt{2^2 + (-6)^2}\)
= \(\sqrt{4 + 36}\)
= \(\sqrt40\) 
= 2\(\sqrt10\)  units

|AC| = \(\sqrt{(4 - (-3)^2 + (0 - 1)^2}\)
= \(\sqrt{(4 + 3)^2 + (0 - 1)^2}\)
= \(\sqrt{7^2 + (-1)^2}\) 
= \(\sqrt{49 + 1}\) 
= \(\sqrt50\)
= 5\(\sqrt{2}\)  units

Since two sides are equal (|AB| = |AC|), then the triangle is Isosceles.