2009 - JAMB Mathematics Past Questions and Answers - page 8

13² = 5² + b²

169 = 25 + b²

b² = 144

b = (\sqrt{144})

= 12 cm

Surface area of a sphere = 4(\pi)r²

4(\pi)r² = 154 cm²

(\pi)r² = (\frac{154}{4})

= 38.5

(\pi)r² = 38.5

r² = 38.5 + (\frac{22}{7}) (\to) r² = 38.5 x (\frac{7}{22})

r² 12.25 (\to) r = (\sqrt{12.25})

= 3.5 cm

(\frac{y_1 - y_o}{x_1 - x_o}) = m

(\frac{4 - p}{p + 1}) = (\frac{2}{3})

2p + 2 = 12 - 3p

5p = 10

p = (\frac{10}{5})

= 2

Applying the formula

(x - a)² + (y - b)² = r²

(4 - 1)² + (2 - r)² = (3)²

(3)² + (2 - r)² = (3)²

9 + ( 4 - 4r + r²) = 9

13 - 4r + r² = 9

r² - 4r + 13 - 9 = 0

(r² - 2r) - (2r + 4) = 0

r(r - 2) -2(r - 2) = 0

(r - 2) (r - 2)

r = 2 or -2

sin 45^o - cos 30^o

(\frac{1}{\sqrt{2}}) - (\frac{\sqrt{3}}{2})

= (\frac{2 - \sqrt{2} \times \sqrt{3}}{2\sqrt{2}}))

= (\frac{2 - \sqrt{6} \times \sqrt{2}}{2\sqrt{2} \sqrt{2}})

= (\frac{2\sqrt{2} - \sqrt{12}}{4})

= (\frac{2\sqrt{2} - 2\sqrt{3}}{4})

= (\frac{2(\sqrt{2} - \sqrt{3})}{4})

= (\frac{\sqrt{2} - \sqrt{3}}{2})

Tan 60^o = (\frac{300}{x})

x Tan 60^o = 300

x√3 = 300

x = (\frac{300}{\sqrt{3}}) x (\frac{\sqrt{3}}{\sqrt{3}})

x = (\frac{300\sqrt{3}}{3}) x 100(\sqrt{3})

If y = 3 cos 4x, (\frac{dy}{dx}) equals

(\frac{dy}{dx}) = 3(4 - sin 4x)

= -12 sin 4x

s = (2 + 3t)(5t - 4), find (\frac{dy}{dx}) when t = (\frac{4}{5}) sec

= 10t - 8 + 15t² - 12t

(\frac{ds}{dt}) = 15t² - 2t - 8

= 30t - 2 ; t = (\frac{4}{5})

= 30t - 2

= 30 x (\frac{4}{5}) - 2 (\to) 24 - 2

= 22 sec