2008 - JAMB Mathematics Past Questions and Answers - page 3

mΔn = m+n+mn

Let e be the identity element

∴mΔe = eΔm = m

m+e+me = m

e+me = m-m

e+me = 0

e(1+m) = 0

e = 0 / (1+m)

e = 0

Exterior angle = sum of two interior opposite angles

y = 50^o (vert. opp ∠s are equal)

X = y + 30

X = 50 + 30

X = 80^o

Exterior angle = 360 / n

= 360 / 12

= 30^o

a = a(base ∠s of Iss Δ)

∴ a+a+74 = 180

2a + 74 = 180

2a = 180-74

2a = 106

a = 53

∴∠OPQ = 53^o

Area of the figure = Area of rect + area of semi circle

(=L \times h + \frac{1}{2}\pi r^2\

5 \times 15 + \frac{1}{2} \times \frac{22}{7} \times \left(\frac{5}{2}\right)^2\

=75+\frac{(22\times 25)}{2\times7}\

=75 + 9\frac{23}{28}\

=84.8cm)

If a length of a chord is equal to the length of a radius in the same circle, the triangle formed y the chord and the radii is an equilateral triangle

∴each angle = 60^o

V = πr²h

1m = 100cm

14cm = 1400cm

(∴V = \frac{22}{7} \times \frac{100x100x1400}{1000}\

= 44, 000 liters)

(X = \frac{6+2}{2}=\frac{10}{2}=5\

Y = \frac{2+2}{2}=\frac{4}{2}=2\

∴X,Y = (5,2))

3X + 2Y + 1 = 0

2Y = -3X - 1

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

Gradient of 3X + 2Y +1 = 0 is -³/₂

Gradient of a line perpendicular to 3X + 2Y + 1 = 0

(=-1 \div \frac{3}{2}\

=-1 \times \frac{-2}{3}=\frac{2}{3})

sinθ = ³/₅

x² = 5² - 3²

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