Mathematics Questions and Answers

Let the average speed for the first 4 km = x km/h.

Hence, the last 5 km, speed = (x + 2) km/h

Recall: \(Time = \frac{Distance}{Speed}\)

Total time = 1 hour.

\(\therefore \frac{4}{x} + \frac{5}{x + 2} = 1\)

\(\frac{4(x + 2) + 5x}{x(x + 2)} = 1\)

\(9x + 8 = x^{2} + 2x\)

\(x^{2} + 2x - 9x - 8 = 0 \implies x^{2} - 7x - 8 = 0\)

\(x^{2} - 8x + x - 8 = 0\)

\(x(x - 8) + 1(x - 8) = 0\)

\(x = -1; x = 8\)

Since speed cannot be negative, x = 8km/h.

The canal's width = 10cm = 100mm (given) The speed of water = 1000mm 10mm = 1cm 1000mm = 100cm The adjacent sea must give speed x width = 1000 x 100 = 100,000

\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\)

= \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\)

= \(3x - 2 > \frac{1}{4} - x\)

= \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\)

\(x > \frac{9}{16}\)

Given the curve \(y = x^{2} - 6x + 5\)

At minimum or maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)

\(\frac{\mathrm d y}{\mathrm d x} = 2x - 6\)

\(2x - 6 = 0 \implies x = 3\)

Since 3 > 0, it is a minimum point.

When x = 3, \(y = 3^{2} - 6(3) + 5 = -4\)

Hence, the turning point has coordinates (3, -4).

Force to the east = 5 units

force to the North - east = 4 units.

Resultant of the two forces is the square root

52 + 42 = 41 and plus the sum of its resistance

5 x 4\(\sqrt{2}\) = 20\(\sqrt{2}\)

= \(\sqrt{41 + 20 \sqrt{2}}\) units