Question Library

incident ray = XY incidence angle i = 15\( ^{\circ} \) rotation angle \theta = 40º angle that the reflected ray make with AY = i + 2\theta = 15 + [2 * 40] = 15 + 80 = 95º

The image produced by a convex mirror is always virtual, and located behind the mirror. When the object is far away from the mirror the image is upright and located at the focal point.

The relative atomic mass will be [(20 * 1) + (3 * 22)] / (4) (86)/(4) = (43)/(2) = 21.5

If the pressure is increased, the position of equilibrium. moves in the direction of the fewest moles. of gas However, because there is an equal number of particles on both sides, change in pressure will have no effect on the system.

Alkanes undergo a substitution reaction with halogens in the presence of light. For instance, in ultraviolet light , methane reacts with halogen molecules such as chlorine and bromine.

\(\frac{1}{\sqrt{2} + \sqrt{5}}\) \(\frac{1}{\sqrt{2} + \sqrt{5}} \times \frac{(\sqrt{2} - \sqrt{5})}{(\sqrt{2} - \sqrt{5})}\) = \(\frac{\sqrt{2} - \sqrt{5}}{2 - 5}\) = \(\frac{\sqrt{2} - \sqrt{5}}{-3}\) = \(\frac{1}{3} (\sqrt{5} - \sqrt{2})\)

3 - 2 \(\div\) (\(\frac{4}{5}\)) + \(\frac{1}{2}\) 3 - (2 x \(\frac{5}{4}\)) + \(\frac{1}{2}\) = 3 - \(\frac{10}{4}\) + \(\frac{1}{2}\) = 3 - \(\frac{5}{2}\) + \(\frac{1}{2}\) = \(\frac{6 - 5 + 1}{2}\) = \(\frac{2}{2}\) = 1

7 + 2 + 1 = 10 \(\frac{1}{10}\) x 560.70 = N56.07

7 years ago, Father(x - 7) years old, Son (y - 7) years x - 7 = 3(y - 7) x - 7 = 3y - 21 3y - x = -7 + 21 = 14 3y - x = 14 ... (1) In six years time, x + 6 = 2(y + 6) x + 6 = 2y + 12 2y + 12 = x + 6 12 - 6 = x - 2y 6 = x - 2y ... (2)

6x - 5 - x2 = (-1)(-x2 - 5 + 6x) = x2 - 6x + 5 = (x - 5)(x - 1) -(x - 1) = 1 - x = (x - 5)(1 - x)

bx\(^2\) + cx + a = 0 a = b; b = c; c = a x = -b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2a}\) x = -c \(\pm\) \(\frac{\sqrt{c^2 - 4ab}}{2b}\)

The graphical method of solving the equation x3 + 3x2 + 4x - 28 = 0 is by drawing the graphs of the curves y = x2 + 3x + 4 and y = \(\frac{28}{x}\)`.

2log2 x - x log2 (1 + y) = 3 log2 \(\frac{x^2}{(1 + y)^x}\) = 3 = \(\frac{x^2}{(1 + y)^x}\) = 23 = 8 = x2 = 8(1 + y)x

x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x (\(\frac{y^{\frac{9}{7}}}{x^{\frac{45}{8}}}\))\(\frac{1}{9}\) = \(\frac{y^{\alpha}}{y^{\beta}}\) x\(\frac{3}{8}\) x y\(\frac{-6}{7}\) x y\(\frac{1}{7}\) = x\(\alpha\) = x\(\frac{3}{8}\) + \(\frac{5}{8}\) + y\(\frac{6}{7}\) + \(\frac{1}{7}\) = x\(\alpha\)y\(\beta\) x1y\(\frac{-5}{7}\) = x\(\alpha\)y\(\beta\) \(\alpha\) = 1, \(\beta\) = \(\frac{5}{7}\)

Two lines are said to be parallel if the slope of the two lines are equal. The equation : \(3y + 2x + 7 = 0\) \(3y = -2x - 7\) \(y = \frac{-2}{3} x - \frac{7}{3}\) \(\frac{\mathrm d y}{\mathrm d x} = - \frac{2}{3}\) All the options have the same slope except 15y + 10x - 13 = 0.