Show and mark the region R defined by the inequalities y >= x, x/8 + y/4 >= 1, and x < 5.

here are the solutions to the problems.

8. On the grid provided below, show and mark the region R defined by the following inequalities.
$L_1: y \ge x$
$L_2: \frac{x}{8} + \frac{y}{4} \ge 1$
$L_3: x < 5$

Step 1: Graph the boundary line for each inequality.

• For $L_1: y \ge x$, draw the solid line $y = x$. This line passes through points like $(0,0)$, $(1,1)$, $(2,2)$, etc.
• For $L_2: \frac{x}{8} + \frac{y}{4} \ge 1$, draw the solid line $\frac{x}{8} + \frac{y}{4} = 1$. This line has x-intercept $(8,0)$ and y-intercept $(0,4)$.
• For $L_3: x < 5$, draw the dashed line $x = 5$. This is a vertical line passing through $x=5$.

Step 2: Determine the region for each inequality.

• For $L_1: y \ge x$, the region is above or on the line $y=x$.
• For $L_2: \frac{x}{8} + \frac{y}{4} \ge 1$, test the origin $(0,0)$: $\frac{0}{8} + \frac{0}{4} = 0$, which is not $\ge 1$. So, the region is above or on the line (away from the origin).
• For $L_3: x < 5$, the region is to the left of the dashed line $x=5$.

Step 3: Mark the feasible region R.
The feasible region R is the area where all three shaded regions overlap. It is bounded by the line $y=x$, the line $\frac{x}{8} + \frac{y}{4} = 1$, and the dashed line $x=5$. The region R is an unbounded triangular region.

9. A flower garden is in the shape of a circle. It is surrounded by a circular path of 2m wide. If the area of the path is $\frac{9}{16}$ of the garden, find the radius of the garden.

Step 1: Define the radii and areas.
Let $r$ be the radius of the garden.
The area of the garden is $A_{garden} = \pi r^2$.
The path is 2m wide, so the outer radius (garden + path) is $R = r + 2$.
The area of the garden and path combined is $A_{total} = \pi R^2 = \pi (r+2)^2$.
The area of the path is $A_{path} = A_{total} - A_{garden} = \pi (r+2)^2 - \pi r^2$.

Step 2: Set up the equation based on the given condition.
The area of the path is $\frac{9}{16}$ of the garden's area:

$$\pi (r+2)^2 - \pi r^2 = \frac{9}{16} \pi r^2$$

Step 3: Solve the equation for $r$.
Divide both sides by $\pi$:

$$(r+2)^2 - r^2 = \frac{9}{16} r^2$$

Expand $(r+2)^2$:

$$(r^2 + 4r + 4) - r^2 = \frac{9}{16} r^2$$ $$4r + 4 = \frac{9}{16} r^2$$

Multiply by 16 to eliminate the fraction:

$$16(4r + 4) = 9r^2$$ $$64r + 64 = 9r^2$$

Rearrange into a quadratic equation:

$$9r^2 - 64r - 64 = 0$$

Use the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$r = \frac{-(-64) \pm \sqrt{(-64)^2 - 4(9)(-64)}}{2(9)}$$ $$r = \frac{64 \pm \sqrt{4096 + 2304}}{18}$$ $$r = \frac{64 \pm \sqrt{6400}}{18}$$ $$r = \frac{64 \pm 80}{18}$$

Two possible solutions:

$$r_1 = \frac{64 + 80}{18} = \frac{144}{18} = 8$$ $$r_2 = \frac{64 - 80}{18} = \frac{-16}{18} = -\frac{8}{9}$$

Since the radius cannot be negative, we take the positive value.
The radius of the garden is $\boxed{8 \text{ m}}$.

10. A foreign exchange bureau gives the rates shown in the table below:

Currency	Buying (Ksh)	Selling (Ksh)
1 South African Rand	7.50	7.80
1 Chinese Yuan	12.50	12.80