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To evaluate the expression \( \log_{\sqrt{2}} 4 + \log_{\frac{1}{2}} 16 - \log_{4} 32 \), we can simplify each term separately. 1. Calculate \( \log_{\sqrt{2}} 4 \) \(\log_{\sqrt{2}} 4 = \frac{\log_2 4}{\log_2 \sqrt{2}}\) We know that \( \log_2 4 = 2 \) (since \( 2^2 = 4 \)) and \( \log_2 \sqrt{2} = \log_2 2^{1/2} = \frac{1}{2} \). Thus, \(log_{\sqrt{2}} 4 = \frac{2}{\frac{1}{2}} = 4\) 2. Calculate \( \log_{\frac{1}{2}} 16 \) \(\log_{\frac{1}{2}} 16 = \frac{\log_2 16}{\log_2 \frac{1}{2}}\) We know that \( \log_2 16 = 4 \) (since \( 2^4 = 16 \)) and \( \log_2 \frac{1}{2} = \log_2 2^{-1} = -1 \). Thus, \(\log_{\frac{1}{2}} 16 = \frac{4}{-1} = -4\) 3. Calculate \( \log_{4} 32 \) \(\log_{4} 32 = \frac{\log_2 32}{\log_2 4}\) We know that \( \log_2 32 = 5 \) (since \( 2^5 = 32 \)) and \( \log_2 4 = 2 \). Thus, \(\log_{4} 32 = \frac{5}{2}\) Now, we can substitute these results back into the original expression: \(\log_{\sqrt{2}} 4 + \log_{\frac{1}{2}} 16 - \log_{4} 32 = 4 - 4 - \frac{5}{2}\) Calculating this gives: \(4 - 4 - \frac{5}{2} = 0 - \frac{5}{2} = -\frac{5}{2}\) Thus, the final result is: \(-\frac{5}{2}\) = - 2.5

L ∝ 1/W L = K/W K = WL = 1.5 * 20 30

Area of Trapezium = 1/2(sum of parralel sides) * ht 21 = 1/2(5 + 9)h 42 = 14h h = 42/14 h = 3cm

The locus is the "semicircle with diameter XY" on the given side. By Thales’ theorem, ∠XPY = 90^\circ means P lies on the circle with diameter XY. “One side only” of line XY restricts it to a semicircle.

y = 3 sin(-4x) dy/dx = 3 x - 4 cos (-4x) = -12 cos (-4x)

Let the 4th angle be = x ∴ x + 45 + 90 + 135 = 360 x + 270 = 360 x = 360 - 270 x = 90 ∴ smallest angle = 45o 45o = N28.00 1o = 28.00/45 135o = (28.00/45) * (135/1) = N84.00

12 + 20 + x + 21 + x-1 +28 = 120 2x + 80 = 120 2x = 120 - 80 2x = 40 x = 20

\theta = 30 x 2(∠ at center twice ∠ at circumference) ∴ \theta = 60^\circ Length of arc = \(\frac{\theta}{360}\times 2\pi r\) = \(\frac{60}{360}\times 2 \times \frac{22}{7}\times \frac{21}{1}\) = 22cm

γ + S = 180(Allied angles) S = 180 - γ α + P = 180(Allied angles) p = 180 - α β = P + S β = 180 - α + 180 - γ β = 360 - α - γ β + α + γ = 360

\theta = 45 x 2 (∠ at center twice ∠ at \theta) \theta = 90^\circ Area of sector = \(\frac{\theta}{360}\) x \pir\(^2\) = \(\frac{90}{360}\) x \(\frac{22}{7}\) x 7 x 7 = \(\frac{77}{2}\)cm\(^2\) Area of triangle = \(\frac{1}{2}\) ab sin \theta = \(\frac{1}{2}\) x 7 x 7 sin 90 = \(\frac{1}{2}\) x 7 x 7 x 1 = \(\frac{49}{2}\) cm\(^{2}\) Area of shaded region = \(\frac{77}{2}\) - \(\frac{49}{2}\) = \(\frac{28}{2}\) = 14cm\(^2\)

Midpoint of PQ \(\frac{x_2 + x_1}{2};\frac{y_2 + y_1}{2}\) =\(\frac{-1+(-3)}{2};\frac{6-4}{2}\) =\(\frac{-4}{2};\frac{2}{2}\) = -2, 1 Midpoint of QR = \(\frac{x_2 + x_1}{2};\frac{y_2 + y_1}{2}\) \(\frac{-3+1}{2};\frac{-4-4}{2}\) \(\frac{-2}{2};\frac{-8}{2}\) -1, -4 Midpoint of PQ and QR = (-2,1) and (-1,-4)

Standard deviation = \(\frac{\sum(x - \overline{x})^2}{\text{n}}\) \(\overline{x}\) = \(\frac{ 3 + 6 + 4 + x + 7}{5}\) = 5 = 20 + x = 25 x = 25 - 20 = 5. x x - \(\overline{x}\) (x - \(\overline{x}\))\(^2\) 3 3 - 5 -2 = 4 4 4 - 5 -1 = 1 5 5 - 5 0 = 0 6 6 - 5 1 = 1 7 7 - 5 2 = 4 \(\sum\) (x - \(\overline{x}\))\(^2\) = 10 Standard deviation = \(\frac{\sum(x - \overline{x})^2}{\text{n}}\) = \(\frac{\sqrt{10}}{5}\) = \(\sqrt{2}\)

Represent notebook with x Represent pen with y ∴ 3x+y=35 ---- (i) 2x+2y=30 ---- (ii) Multiply eqn (i) by 2 and eqn (ii) by 1 6x+2y=70 ---- (iii) 2x+2y=30 ---- (iv) Subtract eqn (iv) from eqn (iii) 4x = 40 X = 10 Substitute x = 10 in eqn (i) 3x+y=35 3(10)+y=35 Y = 5 ∴ The cost of pen is N5.00

< QPS = 75^\circ < QRS = 180^\circ - 75^\circ = 105^\circ (opposite angles of a cyclic quadrilateral)

The sum of angles in a triangle is 180o 2xo + (2x -21)o + ∠ZYP = 180o Also ∠ZYP = 180o - (3x+14)o ∴2xo + (2x-21)o + 180o - (3x+14)o = 180 x = 35o