Solve the equation: 2x - 4 2x = 0 for 0^ x 180^.

Here are the solutions to the problems. 3. Solve the equation: 2x - 4 2x = 0 for 0^ x 180^. Step 1: Rearrange the equation. 2x = 4 2x Step 2: Divide both sides by 2x (assuming 2x ≠ 0). 1 = 4 ( 2x)/( 2x) 1 = 4 2x 2x = (1)/(4) Step 3: Find the principal value for 2x. 2x = ((1)/(4)) ≈ 14.036^ Step 4: Find the general solutions for 2x. Since 2x is positive, 2x can be in the first or third quadrant. 2x = 14.036^ + n · 180^ where n is an integer Step 5: Solve for x. x = (14.036^)/(2) + n · (180^)/(2) x = 7.018^ + n · 90^ Step 6: Find values of x in the range 0^ x 180^. For n=0: x = 7.018^ + 0 · 90^ = 7.018^ For n=1: x = 7.018^ + 1 · 90^ = 97.018^ For n=2: x = 7.018^ + 2 · 90^ = 187.018^ (outside the range) The final answers are 7.02^, 97.02^ (rounded to two decimal places). 4. A team of 5 players is to be selected from 7 boys and 6 girls. If the team must have more girls than boys, in how many ways can the team be selected? Let B be the number of boys and G be the number of girls. Total players to be selected = 5. Total boys available = 7. Total girls available = 6. Condition: G > B and B+G=5. We list the possible combinations of boys and girls that satisfy the conditions: Case 1: 2 boys and 3 girls (2+3=5, 3>2) Number of ways = 72 × 63 72 = (7!)/(2!(7-2)!) = (7 × 6)/(2 × 1) = 21 63 = (6!)/(3!(6-3)!) = (6 × 5 × 4)/(3 × 2 × 1) = 20 Ways for Case 1 = 21 × 20 = 420 Case 2: 1 boy and 4 girls (1+4=5, 4>1) Number of ways = 71 × 64 71 = 7 64 = (6!)/(4!(6-4)!) = (6 × 5)/(2 × 1) = 15 Ways for Case 2 = 7 × 15 = 105 Case 3: 0 boys and 5 girls (0+5=5, 5>0) Number of ways = 70 × 65 70 = 1 65 = (6!)/(5!(6-5)!) = (6)/(1) = 6 Ways for Case 3 = 1 × 6 = 6 Total number of ways = Sum of ways for all valid cases. Total ways = $42