Save for theta given that sin2 theta - 3/2 sin theta - 1 = 0 for 0 degrees <= theta <= 360 degrees. Determine the average rate of cooling between second and tenth minutes.

Step 1: Let k = . Substitute this into the given equation. ^2 - (3)/(2) - 1 = 0 k^2 - (3)/(2)k - 1 = 0 Step 2: Multiply the entire equation by 2 to eliminate the fraction. 2k^2 - 3k - 2 = 0 Step 3: Solve the quadratic equation for k by factoring. 2k^2 - 4k + k - 2 = 0 2k(k - 2) + 1(k - 2) = 0 (2k + 1)(k - 2) = 0 This gives two possible values for k: 2k + 1 = 0 k = -(1)/(2) k - 2 = 0 k = 2 Step 4: Substitute back for k. = -(1)/(2) or = 2 Step 5: Evaluate the solutions for . The range of the sine function is [-1, 1]. Therefore, = 2 has no solution. For = -(1)/(2): The reference angle for = (1)/(2) is = 30^. Since is negative, lies in the third and fourth quadrants. In the third quadrant: = 180^ + 30^ = 210^. In the fourth quadrant: = 360^ - 30^ = 330^. Both angles are within the given range 0^ < 360^. The solutions for are 210^ or 330^. --- Step 1: Identify the temperature values at the given times from the graph. At the second minute (t_1 = 2 min), the temperature is T_1 = 70^ C. At the tenth minute (t_2 = 10 min), the temperature is T_2 = 20^ C. Step 2: Calculate the change in temperature and the change in time. Change in temperature ( T) = T_2 - T_1 = 20^ C - 70^ C = -50^ C. Change in time ( t) = t_2 - t_1 = 10 min - 2 min = 8 min. Step 3: Calculate the average rate of cooling using the formula: Rate = ( T)/( t). Rate = -50^ C8 min Rate = -(25)/(4) ^ Cmin Rate = -6.25 ^ Cmin The average rate of cooling is -6.25 ^ Cmin. That's 2 down. 3 left today — send the next one.