This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to the questions. QUESTION ONE 1) Express: (x+1)/(3) - (2x+4)/(5) as a single fraction in its simplest form. Step 1: Find the least common multiple (LCM) of the denominators 3 and 5, which is 15. Step 2: Rewrite each fraction with the common denominator. (5(x+1))/(15) - (3(2x+4))/(15) Step 3: Combine the fractions over the common denominator. (5(x+1) - 3(2x+4))/(15) Step 4: Expand the terms in the numerator. (5x+5 - (6x+12))/(15) (5x+5 - 6x-12)/(15) Step 5: Simplify the numerator by combining like terms. ((5x-6x) + (5-12))/(15) (-x-7)/(15) The single fraction in its simplest form is (-x-7)/(15). 2) Calculate the value of 0.45 + 0.005. Step 1: Align the decimal points and add the numbers. 0.450 + 0.005 ----- 0.455 The value is 0.455. 3) Make y the subject: p = (2y-c)/(m) Step 1: Multiply both sides of the equation by m. pm = 2y-c Step 2: Add c to both sides of the equation. pm+c = 2y Step 3: Divide both sides by 2. y = (pm+c)/(2) Making y the subject gives y = (pm+c)/(2). 4) The ratio of female to male student at NACE is 5:4. If the number of male students is 1000. Find the total number of students at NACE. Step 1: Set up the ratio of female (F) to male (M) students. (F)/(M) = (5)/(4) Step 2: Substitute the given number of male students, M=1000. (F)/(1000) = (5)/(4) Step 3: Solve for the number of female students, F. F = (5)/(4) × 1000 F = 5 × 250 F = 1250 Step 4: Calculate the total number of students. Total students = F + M = 1250 + 1000 = 2250 The total number of students at NACE is 2250. 5) The diagram below shows the position of town P and town Q. a) Find the bearing of town P from Q. Step 1: Identify the North line at Q. Step 2: Bearings are measured clockwise from the North line. Step 3: The angle given in the diagram from the North line at Q to the line segment QP is 70^. The bearing of town P from Q is 070^. b) Find the bearing of town Q from P. Step 1: Draw a North line at town P, parallel to the North line at town Q. Step 2: The angle between the North line at Q and the line QP is 70^. Step 3: Due to parallel North lines, the alternate interior angle formed by the line PQ and the North line at P (if extended to the left) would be 70^. Step 4: The angle between the line PQ and the South line at P (which is 180^ from North) is also 70^ (alternate interior angles with the angle NQP). Step 5: The bearing of Q from P is measured clockwise from the North line at P. This angle is 180^ (to the South line) plus the 70^ angle. Bearing = 180^ + 70^ = 250^ The bearing of town Q from P is 250^. QUESTION TWO 1) Find the equation of the straight line passing through the points P (6,1) and Q (0,-3). Step 1: Calculate the gradient (m) of the line using the formula m = (y_2 - y_1)/(x_2 - x_1). Let (x_1, y_1) = (6,1) and (x_2, y_2) = (0,-3). m = (-3 - 1)/(0 - 6) = (-4)/(-6) = (2)/(3) Step 2: Use the point-slope form of a linear equation, y - y_1 = m(x - x_1), with one of the points (e.g., P(6,1)) and the gradient. y - 1 = (2)/(3)(x - 6) Step 3: Multiply both sides by 3 to eliminate the fraction. 3(y - 1) = 2(x - 6) 3y - 3 = 2x - 12 Step 4: Rearrange the equation into the form y = mx+c or Ax+By+C=0. 3y = 2x - 12 + 3 3y = 2x - 9 Or, in y = mx+c form: y = (2)/(3)x - (9)/(3) y = (2)/(3)x - 3 The equation of the straight line is y = (2)/(3)x - 3 or 3y = 2x - 9. That's 3 down. 2 left today — send the next one.