Correct Answer: A. 620

Explanation

d = -3 - (-7) = 4

S\(_{20}\) = \(\frac{20}{2}\){2(-7) + (20 - 1) 4}

= 10(- 14 + 76)

= 620

Correct Answer: D. 4.0cm

Explanation

Distance(d) = d² = 5² - 3²

= 25 - 9

= 16

d = \(\sqrt{16}\)

= 4cm

Explanation

(b) Median = \(L_{1} + \frac{(\frac{N}{2} - \sum f_{1}) \times c}{f_{med}}\)

= \(54.5 + \frac{(\frac{70}{2} - 31) \times 5}{23}\)

= \(54.5 + 0.9 = 55.4\)

Correct Answer: D. 13³/₅cm

Explanation

Perimeter of sector = \(\frac{\theta}{360°} \times 2\pi r + 2r\)

= \(\frac{108}{360} \times 2 \times \frac{22}{7} \times \frac{7}{2} + 2(\frac{7}{2})\)

= \(6 \frac{3}{5} + 7\)

= \(13 \frac{3}{5} cm\)

Correct Answer: B. 2:1

Explanation

From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A₁, the area of \(\bigtriangleup\)PXY

= Area of \(\bigtriangleup\)YZR = A₂

Area of \(\bigtriangleup\)PQR = A = A₁ + 2A₂

But from similarity of triangles

\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)

\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)

A = 4A₂ But, A = A₁ + 2A

A₁ = 4A₂ - 2A₂

A₁ = 2A₂

\(\frac{A_1}{A_2}\) = 2

A₁:A₂ = 2:1

Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1

Explanation

a.

b. the graph above, solution to question 8b.

ci. The roots of the equation from the graph are -1.2 and 1.7.

ii. Let y = \(2x^2 - x - 4\)

\(\frac{dy}{dx}\) = 4x - 1

At critical point \(\frac{dy}{dx}\) = 0

so that 4x - 1 = 0;

x = \(\frac{1}{4}\)

Using test points:

1. we use a test point less than \(\frac{1}{4}\), e.g x = 0

\(\frac{dy}{dx}\)(0) = 4(0) - 1 = -1

at this point, y = \(2x^2 - x - 4\) is decreasing

2. we use a test point greater than \(\frac{1}{4}\) e.g x = 1

\(\frac{dy}{dx}\)(1) = 4(1) - 1 = 4 - 1 = 3.

at this point, y = \(2x^2 - x - 4\) is increasing.

so, the function y = \(2x^2 - x - 4\) increases as x increases when x is greater than \(\frac{1}{4}\)

Therefore, the value of x for which y increases are x is greater than \(\frac{1}{4}\)

iii. The minimum point of y = -4.1.

Explanation

(a)

Let A and B be the reactions at P and Q respectively.

A + B = 80 + 350 + 1000 = 1430N

We take moments about Q.

Clockwise moments = 2.2A

Anticlockwise moments = \(80 \times 3 + 350 \times 1 + 1000 \times 0.6\)

= \(1190N\)

Both moments are equal

\(\therefore 2.2A = 1190 \implies A = \frac{1190}{2.2} = 540.9N\)

\(B = 1190N - 540.9N = 649.1N\)

The reactions at P and Q are 540.9N and 649.1N respectively.

(b)

Reaction at P : \((80 + 350 + 1000)N = 540.9N\)

\(\therefore \text{The new reaction at P} = (80 + 350 + 1200)N = \frac{1630}{1430} \times 540.9\)

= \(616.6N\)

Take moments at Q :

\(2.2 \times 616.6 = 80 \times 3 + 350 \times 1 + 1200x\)

\(1200x = 1356.5 - 590\)

\(1200x = 766.5\)

\(x = \frac{766.5}{1200}\)

= \(0.63875 \approxeq 0.64m\)

The 1200N should be placed 0.64m from Q.

Correct Answer: B. -1

Correct Answer: C. -1

Explanation

6 - x - x² = 0

a = -1; b = -1; c = 6

Sum of roots = c + k = -\(\frac{-b}{a}\)

= \(\frac{-(-1)}{-1}\)

= -1

Correct Answer: C. {1, 2, 4, 5}

Explanation

S = {1, 2, 3, 4, 5, 6}; T = {2, 4, 5, 7}; R = {1, 4, 5}

(S∩T) ∪ R = {2, 4, 5} ∪ {1, 4, 5}

= {1, 2, 4, 5}