Given the curve y = x2 - 2x + 6, find the coordinate of the point on the curve at which the gradient is 4. Also, Mary has some money in two denominations only. Fifty shilling notes and twenty shilling

You're on a roll — Here are the solutions to the problems: Given the curve y = x^2 - 2x + 6, find the coordinate of the point on the curve at which the gradient is 4. Step 1: Find the derivative of the curve to get the gradient function. The gradient, (dy)/(dx), is found by differentiating the equation y = x^2 - 2x + 6 with respect to x. (dy)/(dx) = (d)/(dx)(x^2 - 2x + 6) (dy)/(dx) = 2x - 2 Step 2: Set the gradient equal to 4 and solve for x. 2x - 2 = 4 2x = 4 + 2 2x = 6 x = (6)/(2) x = 3 Step 3: Substitute the value of x back into the original equation to find the corresponding y-coordinate. y = (3)^2 - 2(3) + 6 y = 9 - 6 + 6 y = 9 The coordinate of the point is (3, 9). --- 13. Mary has some money in two denominations only. Fifty shilling notes and twenty shilling coins. She has three times as many fifty shilling notes as twenty shilling coins. If altogether she has KSh 3400. Find the number of fifty shilling notes and twenty shilling coins. Step 1: Define variables. Let N_50 be the number of fifty shilling notes. Let N_20 be the number of twenty shilling coins. Step 2: Formulate equations based on the given information. From "She has three times as many fifty shilling notes as twenty shilling coins": N_50 = 3N_20 (Equation 1) From "If altogether she has KSh 3400": 50N_50 + 20N_20 = 3400 (Equation 2) Step 3: Substitute Equation 1 into Equation 2. Substitute N_50 = 3N_20 into 50N_50 + 20N_20 = 3400: 50(3N_20) + 20N_20 = 3400 150N_20 + 20N_20 = 3400 170N_20 = 3400 Step 4: Solve for N_20. N_20 = (3400)/(170) N_20 = 20 So, Mary has 20 twenty shilling coins. Step 5: Substitute the value of N_20 back into Equation 1 to find N_50. N_50 = 3 × 20 N_50 = 60 So, Mary has 60 fifty shilling notes. The number of fifty shilling notes is 60 and the number of twenty shilling coins is 20. Send me the next one 📸