2015 - JAMB Mathematics Past Questions and Answers - page 3

tan x = (\frac{2}{3})(given), is illustrated in a right-angled (\Delta)

thus m² = 2² + 3²

= 4 + 9 = 13

m = (\sqrt{13})

Hence, 2sin x = 2 x (\frac{2}{m})

2 x(\frac{2}{\sqrt{13}})

= (\frac{4}{\sqrt{13}})

= (\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}})

= (\frac{4\sqrt{13}}{13})

Percentage birthrate = 1.0%

1¼ ÷ [ 2 ÷ ¼] of 28

Apply BODMAS rules

( \frac{5}{4}) ÷ [ 2 ÷ ¼ × 28 ]

( \frac{5}{4} ÷ \frac{2}{7} )

( \frac{5}{4} × \frac{7}{2} )

( \frac{35}{8})

= (4 \frac{3}{8})

( x^2 + 9x + 20 )

Find the two numbers whose product is 20 and its sum is 9}

( 5x \times 4x = 20x^2)

(5x + 4x = 9x)

((x^2 + 5x) + (4x + 20))

(

x(x + 5)+ 4(x + 5)

= (x + 5)(x + 4))

n(X ∩ Y) = nx + ny − n(xX ∩ Y)

= 15 + 12 − 7

∴ n(X ∩ Y) = 20

Starting from 3x − z (\leq) 9

x = 0, substitute the value of x (i.e. x = 0) into the equation 3x − z (\leq) 9

3(0) − z (\leq) − 9

z (\leq) − 9

If z = 0

Then, 3x − 0 (\leq) 9

3x (\leq) 9

x (\leq) 3

Intercept of x and z i.e (x,z)

= (3, -9)

log₁₀1.5 + 3 log₁₀2 − log₁₀0.3

log₁₀1.5 + log₁₀23 − log₁₀0.3

[log₁₀(1.5 × 23) ÷ log₁₀0.3]

[log₁₀(15/10 × 8 ) ÷ log₁₀( 3/10)]

log₁₀(15/10 × 8 × 10/3 )

log₁₀40

The total surface area of a cylinder = 2πrl + 2πr2

= 2πr(l + r)

= 2 × 3.14(7+%=5)

2 × 3.14 × 12

= 75.4cm (1DP)