Question 1:
(a) Convert 15026 to a number in base ten. [2 marks]
(b) It is given that 51079 – X9 = 3489.
Find the value of X.
[2 marks]
Answer:
(a)
$$ \begin{aligned} & 1\left(6^3\right)+5\left(6^2\right)+0\left(6^1\right)+2\left(6^0\right) \\ & =216+180+0+2 \\ & =398 \end{aligned} $$
(b)
$$ \begin{aligned} & 5107_9-348_9 \\ & =\left[5\left(9^3\right)+1\left(9^2\right)+0\left(9^1\right)+7\left(9^0\right)\right]-\left[3\left(9^2\right)+4\left(9^1\right)+8\left(9^0\right)\right] \\ & =3733_{10}-287_{10} \\ & =3446_{10} \end{aligned} $$
X = 4648
(a) Convert 15026 to a number in base ten. [2 marks]
(b) It is given that 51079 – X9 = 3489.
Find the value of X.
[2 marks]
Answer:
(a)
$$ \begin{aligned} & 1\left(6^3\right)+5\left(6^2\right)+0\left(6^1\right)+2\left(6^0\right) \\ & =216+180+0+2 \\ & =398 \end{aligned} $$
(b)
$$ \begin{aligned} & 5107_9-348_9 \\ & =\left[5\left(9^3\right)+1\left(9^2\right)+0\left(9^1\right)+7\left(9^0\right)\right]-\left[3\left(9^2\right)+4\left(9^1\right)+8\left(9^0\right)\right] \\ & =3733_{10}-287_{10} \\ & =3446_{10} \end{aligned} $$
X = 4648
Question 2:
Table 1 shows the values for the equation y = x3 + 1 for -3 ≤ x ≤ 2.
Draw the graph of y = x3 + 1 for -3 ≤ x ≤ 2.
[3 marks]
Answer:
Answer:
All 6 points are plotted correctly for -3 ≤ x ≤ 2 and -26 ≤ y ≤ 9.
The curve is smooth and passes through all 6 points.
Table 1 shows the values for the equation y = x3 + 1 for -3 ≤ x ≤ 2.
Draw the graph of y = x3 + 1 for -3 ≤ x ≤ 2.
[3 marks]
Answer:
Answer:
All 6 points are plotted correctly for -3 ≤ x ≤ 2 and -26 ≤ y ≤ 9.
The curve is smooth and passes through all 6 points.
Question 3:
Diagram 1 shows a cuboid, A B C D E F G H. It is given that P is the midpoint of line E H.
(a) State the length of GP, in cm. [1 mark]
(b) Hence, calculate the angle of elevation of B from P. [2 marks]
Answer:
(a) 50 cm
(b)
$$ \begin{aligned} \tan \theta & =\frac{30}{50} \\ \theta & =30^{\circ} 58^{\prime} \end{aligned} $$
Diagram 1 shows a cuboid, A B C D E F G H. It is given that P is the midpoint of line E H.
(a) State the length of GP, in cm. [1 mark](b) Hence, calculate the angle of elevation of B from P. [2 marks]
Answer:
(a) 50 cm
(b)
$$ \begin{aligned} \tan \theta & =\frac{30}{50} \\ \theta & =30^{\circ} 58^{\prime} \end{aligned} $$
Question 4:
(a) State whether the following statement is true or false.
(i) A circle is a polygon.
(ii) $$ a^{\frac{3}{2}}=\sqrt[3]{a^2} $$
[2 marks]
(b) Write down the converse of the following implication:
[1 mark]
(c) It is given that the volume of a sphere is 4/3 πr3, such that r is the radius.
If a hemisphere has a radius of 3 cm , make one conclusion by deduction for the volume of this hemisphere. [2 marks]
Answer:
(a)(i) False
(a)(ii) False
(b) If c is the y-intercept then y = mx + c is the equation of a straight line.
(c)
$$ \begin{aligned} &\text { Volume of hemisphere }\\ &\begin{aligned} & \frac{\frac{4}{3}(\pi)\left(3^3\right)}{2} \\ = & 18 \pi \text { or } \frac{396}{7} \end{aligned} \end{aligned} $$
(a) State whether the following statement is true or false.
(i) A circle is a polygon.
(ii) $$ a^{\frac{3}{2}}=\sqrt[3]{a^2} $$
[2 marks]
(b) Write down the converse of the following implication:
| If y = mx + c is the equation of a straight line then c is the y-intercept. |
(c) It is given that the volume of a sphere is 4/3 πr3, such that r is the radius.
If a hemisphere has a radius of 3 cm , make one conclusion by deduction for the volume of this hemisphere. [2 marks]
Answer:
(a)(i) False
(a)(ii) False
(b) If c is the y-intercept then y = mx + c is the equation of a straight line.
(c)
$$ \begin{aligned} &\text { Volume of hemisphere }\\ &\begin{aligned} & \frac{\frac{4}{3}(\pi)\left(3^3\right)}{2} \\ = & 18 \pi \text { or } \frac{396}{7} \end{aligned} \end{aligned} $$