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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QUESTION ONE 3) Make y the subject: p = (2y-c)/(m). Step 2: Add c to both sides. pm + c = 2y Step 3: Divide both sides by 2. y = (pm+c)/(2) The subject is 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. Let F be the number of female students and M be the number of male students. Given ratio F:M = 5:4. Given M = 1000. Step 1: Set up the ratio equation. (F)/(M) = (5)/(4) Step 2: Substitute the number of male students. (F)/(1000) = (5)/(4) Step 3: Solve for 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. The diagram shows town Q with a North line (N) and a line segment QP. The angle between the North line at Q and the line segment QP is 70^. a) Find the bearing of town P from Q. Bearing is measured clockwise from the North line. From the diagram, the angle from the North line at Q to the line QP is 70^. The bearing of town P from Q is 070^. b) Find the bearing of town Q from P. Step 1: The bearing of P from Q is 070^. Step 2: To find the bearing of Q from P, we add 180^ to the bearing of P from Q, since 070^ < 180^. Bearing of Q from P = 070^ + 180^ = 250^. The bearing of town Q from P is 250^. 3 done, 2 left today. You're making progress.