JAMB Mathematics Past Questions & Answers - Page 20

Using the expansion method on (10110.011)\(_2\)

(1 * 2\(^4\)) + (0 * 2\(^3\)) + (1 * 2\(^2\)) + (1 * 2\(^1\)) (0 * 2\(^0\)) + (0 * 2\(^{-1}\)) + (1 * 2\(^{-2}\))  + (1 * 2\(^{-3}\))

(1*16) + 0 (1*4) + (1*2) + 0 + 0 + (1*\(\frac{1}{4}\)) + (1*\(\frac{1}{8}\)) 

16 + 4 + 2 + (0.25 + 0.125)

22 + 0.375

(10110.011)\(_2\) = 22.375\(_{10}\)

When rearranged: 10, 10, 13, 14, 17, 18

median = \(\frac{13+14}{2}\)

= \(\frac{27}{2}\) or 13.5

Condition for Parallelism means that their gradient value is same

line 2x - y = 5 is rearranged

As y = 2x - 5 from y = mx + c

: line 2x - y = 5 has the gradient of 2

A parallel line with gradient of 2 and intercept of 5

→ 2x - y = -5

Rearranged as y = 2x + 5

\(\frac{x^3 + 6x - 7}{x-1}\):

When numerator is differentiated → 3x\(^2\) + 6  

When denominator is differentiated → 1

: \(\frac{3x^2 + 6}{1}\)

substitute x for 1

 \(\frac{3 * 1^2 + 6}{1}\) =  \(\frac{3 + 6}{1}\) 

=  \(\frac{9}{1}\)

= 9

Given that sec\(^2\)θ + tan\(^2\)θ = 3

Where sec\(^2\)θ = 1  + tan\(^2\)θ

: 1 + tan\(^2\)θ + tan\(^2\)θ = 3

2tan\(^2\)θ = 3 - 1

2tan\(^2\)θ = 2

divide both sides by 2

tan\(^2\)θ = 1

tanθ =  √1

tanθ = 1

θ = tan\(^{-1}\) (1)

θ = 45º