Find the inequalities that satisfy the region labeled R in the figure below.

Line L1 passes through $(4, 0)$ and $(0, 4)$.
The slope $m_1 = \frac{4-0}{0-4} = \frac{4}{-4} = -1$.
The y-intercept is $c_1 = 4$.
The equation of L1 is $y = -x + 4$, which can be rewritten as $x + y = 4$.
The region R is below L1. Testing a point in R, e.g., $(2, 1)$: $2 + 1 = 3 < 4$. Since L1 is a dashed line, the inequality is $x + y < 4$.

Line L2 passes through $(4, 0)$ and $(0, -2)$.
The slope $m_2 = \frac{-2-0}{0-4} = \frac{-2}{-4} = \frac{1}{2}$.
The y-intercept is $c_2 = -2$.
The equation of L2 is $y = \frac{1}{2}x - 2$.
Multiply by 2: $2y = x - 4$, which can be rewritten as $x - 2y = 4$.
The region R is above L2. Testing a point in R, e.g., $(2, 1)$: $2 - 2(1) = 0 < 4$. This means the region is $x - 2y < 4$. If we want the region above the line, we need to consider the form $y > \frac{1}{2}x - 2$. So $2y > x - 4$, or $x - 2y < 4$. Since L2 is a dashed line, the inequality is $x - 2y < 4$.

Line L3 is a vertical line passing through $x = 0$.
The region R is to the right of L3. Testing a point in R, e.g., $(2, 1)$: $2 > 0$. Since L3 is a dashed line, the inequality is $x > 0$.

The inequalities that satisfy the region R are:

6. Rodi walked from Kisian to Otonglo, a distance of 8 km for $1\frac{1}{2}$ hours. He then took a motorbike to Rabuor that travel at a speed of 80 km/h and took 15 minutes. From Rabuor, he boarded a car to Kendu Bay, 60 km away and took 45 minutes to arrive in Kendu Bay. Calculate his average velocity for the whole journey.

• Segment 1 (Walk): Distance $d_1 = 8$ km.
• Segment 2 (Motorbike): Speed $v_2 = 80$ km/h, Time $t_2 = 15$ minutes $= \frac{15}{60}$ hours $= 0.25$ hours. Distance $d_2 = v_2 \times t_2 = 80 \text{ km/h} \times 0.25 \text{ h} = 20$ km.
• Segment 3 (Car): Distance $d_3 = 60$ km. Total distance $D_{total} = d_1 + d_2 + d_3 = 8 \text{ km} + 20 \text{ km} + 60 \text{ km} = 88$ km.

• Segment 1 (Walk): Time $t_1 = 1\frac{1}{2}$ hours $= 1.5$ hours.
• Segment 2 (Motorbike): Time $t_2 = 15$ minutes $= 0.25$ hours.
• Segment 3 (Car): Time $t_3 = 45$ minutes $= \frac{45}{60}$ hours $= 0.75$ hours. Total time $T_{total} = t_1 + t_2 + t_3 = 1.5 \text{ h} + 0.25 \text{ h} + 0.75 \text{ h} = 2.5$ hours.

Average velocity is defined as total displacement divided by total time. Assuming the journey is in a straight line, total displacement equals total distance.

The average velocity for the whole journey is $\boxed{35.2 \text{ km/h}}$.

7. Simplify the expression $\frac{(-4b + 3a)^2 - (3a + 4b)^2}{12a}$

Let $X = (-4b + 3a)$ and $Y = (3a + 4b)$.

The numerator becomes $(X-Y)(X+Y) = (-8b)(6a) = -48ab$.

Assuming $a \neq 0$:

The simplified expression is $\boxed{-4b}$.

*8. A saleslady earns a commission of 3% and 5% for sale of goods up to Kshs. 100,000 and above Kshs. 100,000 respectively. In a certain month, Sarah's total commission was Kshs. 15,