Find two consecutive even numbers that sum to 270. Also, describe the region R defined by x >= -1, 3y - x >= -5, and 3x + 5y < 15.

Hey dory, good to see you again. Problem 13: Step 1: Define the numbers. Let the first even number be x. Since the numbers are consecutive even numbers, the next even number is x+2. Step 2: Set up the equation. The sum of the two numbers is 270: x + (x+2) = 270 Step 3: Solve for x. 2x + 2 = 270 2x = 270 - 2 2x = 268 x = (268)/(2) x = 134 Step 4: Find the two numbers. The first number is x = 134. The second number is x+2 = 134+2 = 136. The two consecutive even numbers are 134 and 136. --- Problem 14: To draw and label the region R, we first need to draw the boundary lines for each inequality. Step 1: Draw the line for L_1: x -1. The boundary line is x = -1. This is a solid vertical line passing through x=-1. The region x -1 is to the right of this line. Step 2: Draw the line for L_2: 3y - x -5. The boundary line is 3y - x = -5. This is a solid line. To draw it, find two points: • If x=0, 3y = -5 y = -(5)/(3) ≈ -1.67. Point: (0, -(5)/(3)). • If y=0, -x = -5 x = 5. Point: (5, 0). Plot these points and draw a solid line through them. To determine the region, test a point, e.g., (0,0): 3(0) - 0 -5 0 -5, which is true. So, the region 3y - x -5 is above the line (containing the origin). Step 3: Draw the line for L_3: 3x + 5y < 15. The boundary line is 3x + 5y = 15. This is a dashed line because of the strict inequality (<). To draw it, find two points: • If x=0, 5y = 15 y = 3. Point: (0, 3). • If y=0, 3x = 15 x = 5. Point: (5, 0). Plot these points and draw a dashed line through them. To determine the region, test a point, e.g., (0,0): 3(0) + 5(0) < 15 0 < 15, which is true. So, the region 3x + 5y < 15 is below the line (containing the origin). Step 4: Identify and label region R. Region R is the area where all three shaded regions overlap. It will be a triangular region bounded by the lines x=-1, 3y-x=-5, and 3x+5y=15. • The region is to the right of x=-1. • The region is above 3y-x=-5. • The region is below 3x+5y=15. Shade this common region and label it R. --- Problem 15: Step 1: Set up initial number of members. Let the original number of males be M and females be F. The ratio of males to females is 2:3. (M)/(F) = (2)/(3) This means M = 2k and F = 3k for some constant k. Step 2: Account for changes on Tuesday. On Tuesday: • Ten males were absent, so the number of males present is M - 10 = 2k - 10. • Six new female members joined, so the number of females present is F + 6 = 3k + 6. Step 3: Set up the new ratio equation. On Tuesday, the ratio of males to females was 1:3. (2k - 10)/(3k + 6) = (1)/(3) Step 4: Solve for k. Cross-multiply: 3(2k - 10) = 1(3k + 6) 6k - 30 = 3k + 6 Subtract 3k from both sides: 3k - 30 = 6 Add 30 to both sides: 3k = 36 Divide by 3: k = (36)/(3) k = 12 Step 5: Calculate the original number of regular members. Original number of males M = 2k = 2(12) = 24. Original number of females F = 3k = 3(12) = 36. The total number of regular members is M + F = 24 + 36 = 60. The group has 60 regular members. Send me the next one 📸