A rally car travelled for 2 hours 40 minutes at an average speed of 120 km/h. The car consumes an average of 1 litre of fuel for every 4 kilometers. A litre costs Ksh. 189. Calculate the amount of mon

The car consumes 1 litre for every 4 km.

The amount of money spent on fuel is $\boxed{\text{Ksh. } 15120}$.

2. One interior angle of a polygon is equal to $80^\circ$ and each of the other interior angles are $128^\circ$. Find the number of sides of the polygon.

The sum of the interior angles of an $n$-sided polygon is given by the formula:

One angle is $80^\circ$. The remaining $(n-1)$ angles are each $128^\circ$.

The number of sides of the polygon is $\boxed{6}$.

3. (a) Using a pair of compasses and a ruler only, construct a triangle ABC such that AB = 4 cm, BC = 6 cm and angle ABC = $135^\circ$.

a) Construction steps:

1. Draw a line segment AB of length 4 cm.
2. At point B, construct an angle of $135^\circ$. To do this, construct a $90^\circ$ angle, then bisect the adjacent $90^\circ$ angle to get $45^\circ$, and add it to $90^\circ$ ($90^\circ + 45^\circ = 135^\circ$). Alternatively, construct a straight line through B, then construct a $45^\circ$ angle from the extension of AB, which will give $135^\circ$ on the other side.
3. Along the $135^\circ$ line from B, measure and mark a point C such that BC = 6 cm.
4. Join points A and C to complete triangle ABC.

(b) Construct the height of triangle ABC in (a) above taking AB as the base, hence calculate the area of triangle ABC.*

b) Construction steps for height:

1. Extend the line segment AB beyond B.
2. From point C, drop a perpendicular to the extended line AB. To do this, place the compass at C, draw an arc that intersects the extended line AB at two points.
3. From these two intersection points, draw two more arcs that intersect below (or above) the line.
4. Draw a straight line from C through the intersection of these two arcs. This line is the height of the triangle from C to the base AB (extended). Let the intersection point on the extended line AB be D. CD is the height.

Calculation of the area:
Given AB = 4 cm, BC = 6 cm, and $\angle ABC = 135^\circ$.
The area of a triangle can be calculated using the formula:

Here, $a = \text{BC} = 6 \text{ cm}$, $b = \text{AB} = 4 \text{ cm}$, and the included angle is $\angle ABC = 135^\circ$.

We know that $\sin(135^\circ) = \sin(180^\circ - 45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2}$.

Using $\sqrt{2} \approx 1.414$:

The area of triangle ABC is $\boxed{6\sqrt{2} \text{ cm}^2 \text{ or approximately } 8.484 \text{ cm}^2}$.

4. Solve the following inequalities and present your solution on a number line:

Inequality 1: $2x - 2 \le 3x + 1$
Inequality 2: $3x + 1 < x + 11$

We need values of $x$ that satisfy both $x \ge -3$ and $x < 5$.
The combined solution is $-3 \le x < 5$.

Draw a number line. Place a closed circle at -3 (because $x$ can be equal to -3) and an open circle at 5 (because $x$ cannot be equal to 5). Draw a line connecting these two circles.

<-----|-----|-----|-----|-----|-----|-----|-----|-----|----->
     -4    -3    -2    -1     0     1     2     3     4     5
            [------------------------------------------------)

The solution is $\boxed{-3 \le x < 5}$.

5. Without using mathematical tables or a calculator, evaluate leaving your answer as a simplified fraction:
The expression to evaluate is missing from the image. Please provide the expression for question 5.

6. Two similar solids have surface areas $48 \text{ cm}^2$ and $108 \text{ cm}^2$ respectively. Find the volume of the smaller solid if the bigger one has a volume of 162 milliliters.

Let $A_1 = 48 \text{ cm}^2$ (smaller solid) and $A_2 = 108 \text{ cm}^2$ (bigger solid).
The ratio of surface areas is $\frac{A_1}{A_2} = \frac{48}{108}$.
Simplify the ratio:

For similar solids, the ratio of surface areas is the square of the ratio of corresponding linear dimensions ($L_1/L_2$).

The ratio of volumes is the cube of the ratio of corresponding linear dimensions.

Given $V_2 = 162 \text{ milliliters}$. Since $1 \text{ ml} = 1 \text{ cm}^3$, $V_2 = 162 \text{ cm}^3$.

The volume of the smaller solid is $\boxed{48 \text{ cm}^3}$.

7. A triangular flower garden has an area of $28 \text{ cm}^2$. Two of its edges are 14 m and 8 m. find the angle between the two edges.

Given Area $= 28 \text{ m}^2$, $a = 14 \text{ m}$, $b = 8 \text{ m}$. Let $\theta$ be the angle between the two edges.

The angle whose sine is $\frac{1}{2}$ is $30^\circ$.

The angle between the two edges is $\boxed{30^\circ}$.

8. Simplify the expression: