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Question 1

(a) Solve the equation  \(\frac{\left ( 2x\: + \: 1 \right )}{\left ( 3\: -\: 4x \right )} = \frac{2}{3}\)

Solution

cross multiply

3(2x + 1) = 2(3 – 4x)

remove the brackets

6x + 3 = 6 – 8x

collect like terms

6x + 8x = 6 – 3

14x = 3

divide both sides by 14

\( \frac{14x}{14} = \frac{3}{14}\)

x = \( \frac{3}{14} \)

 

b(i) Make v the subject of the  formula

E = \(\frac{mv^{2}}{2}\)

Solution

cross multiply

mv² = 2E

divide both sides by m

\( \frac{mv^2}{m} = \frac{2E}{m}\)

Taking the square root of both sides

v = \( \normalsize \sqrt{\frac{2E}{m}} \)

 

b(ii) Find the value of v when m = 2 and E = 64

Solution

Using v = \( \normalsize \sqrt{\frac{2E}{m}} \)

v = \( \normalsize \sqrt{\frac{2\: \times \: 64}{2}} \)

v = \( \scriptsize \sqrt{64} \)

v = \( \scriptsize 8\)

Question 2

(a) Find the value of y in the figure below.

Solution

number of sides n = 5

sum of interior angles = (n – 2) x 180º

y + y + 80º + 90º + 90º = (n – 2) x 180º

2y + 260º = (5 – 2) x 180º

2y + 260 = 3 x 180º

2y + 260º = 540º

2y = 540º – 260º

2y = 280º

\( \frac{2y}{2} = \frac{280}{2} \)

y = 140º

 

(b) In the diagram below calculate

(i) /AC/

Solution

Using Pythagoras theorem

Hyp² = Opp² + Adj²

AC² = 4² + 3²

AC² = 16 + 9

AC² = 25

AC = √25

AC = 5cm

 

(ii) the area of ABC

Solution

Area of triangle = ½bh

 = ½ x 3 x 4

= ½ x 12

= 6cm²

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