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

You're on a roll — 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 x L_2: (x)/(8) + (y)/(4) 1 L_3: x < 5 Step 1: Graph the boundary line for each inequality. For L_1: y x, draw the solid line y = x. This line passes through points like (0,0), (1,1), (2,2), etc. For L_2: (x)/(8) + (y)/(4) 1, draw the solid line (x)/(8) + (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 x, the region is above or on the line y=x. For L_2: (x)/(8) + (y)/(4) 1, test the origin (0,0): (0)/(8) + (0)/(4) = 0, which is not 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 (x)/(8) + (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 (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 = 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 = R^2 = (r+2)^2. The area of the path is A_path = A_total - A_garden = (r+2)^2 - r^2. Step 2: Set up the equation based on the given condition. The area of the path is (9)/(16) of the garden's area: (r+2)^2 - r^2 = (9)/(16) r^2 Step 3: Solve the equation for r. Divide both sides by : (r+2)^2 - r^2 = (9)/(16) r^2 Expand (r+2)^2: (r^2 + 4r + 4) - r^2 = (9)/(16) r^2 4r + 4 = (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 = -b ± sqrt(b^2 - 4ac)2a: r = -(-64) ± sqrt((-64)^2 - 4(9)(-64))2(9) r = 64 ± sqrt(4096 + 2304)18 r = 64 ± sqrt(6400)18 r = (64 ± 80)/(18) Two possible solutions: r_1 = (64 + 80)/(18) = (144)/(18) = 8 r_2 = (64 - 80)/(18) = (-16)/(18) = -(8)/(9) Since the radius cannot be negative, we take the positive value. The radius of the garden is 8 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 | a) A tourist arrived in Kenya with 10,000 South African Rand. He converted the whole amount to Kenya Shillings. How much did he get? Step 1: Identify the correct exchange rate. The tourist is selling Rand to the bureau, so the bureau is buying Rand. The buying rate for 1 South African Rand is Ksh 7.50. Step 2: Calculate the amount in Kenya Shillings. Amount in Ksh = 10,000 Rand × 7.50 Ksh/Rand = 75,000 Ksh The tourist got 75,000 Ksh. b) While in Kenya, he spent Ksh 25,000 and converted the rest to Chinese Yuan. Calculate the amount of Chinese Yuan he got. (Give your answer to 2.d.p) Step 1: Calculate the remaining amount in Kenya Shillings. Remaining Ksh = 75,000 Ksh - 25,000 Ksh = 50,000 Ksh Step 2: Identify the correct exchange rate for converting Ksh to Chinese Yuan. The tourist is buying Chinese Yuan from the bureau, so the bureau is selling Yuan. The selling rate for 1 Chinese Yuan is Ksh 12.80. Step 3: Calculate the amount in Chinese Yuan. Amount in Chinese Yuan = 50,000 Ksh12.80 Ksh/Yuan = 3906.25 Yuan The tourist got 3906.25 Chinese Yuan. What's next?