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Area of the bigger semi circle PRO = \(\frac{\pir^2}{2}\) = \(\frac{22}{7}\) x \(\frac{49}{2}\) = 77cm\(^2\) 2nd Area of the smaller circle DYO = \(\frac{22}{7}\) x \(\frac{3.5^2}{2}\) = 19.25cm\(^2\) Total Area = 77 + 19.25 = 96.25 ≈ 96.00cm\(^2\)

B occurs at the point y = 0. When y = 0, we have 4x + 3(0) = 6 4x = 6 x = 1\(\frac{1}{2}\) B = \((1\frac{1}{2}, 0)\)

Tn = a + (n - 1)d; AP; a = 2d; Tn = 2d + (n - 1)d T5 = 2d + (5 - 1)d = 2d + 4d = 6d

a = 1st term ; ar\(^4\) = 5th term ∴ ar\(^4\) = a(2)\(^4\) = 16a; 16a = 45 + a, a = 3; 3(2)\(^4\) = 48

Angular distance = 40o + 20o = 60o using θ/360 x 2πr for distance between p and Q = 60/360 x 2πRcos4o = 1/3πRcos4o

by using trial and error Y = sinx since \(\pi\) in radian = 180º

Sin\(\theta\) = -0.9063; \(\theta\) = sin-1(0.9063) \(\theta\) = -64.99 = -65^\circ \(\theta\) = 180^\circ - (-65); 180 + 65 = 245^\circ

x/25 = tan 30o x = 25/√3 = 25√3/3

x/25 = tan 30o x = 39 x 1/√3 = 13√3m

From the figure, < ATO = 90^\circ < AOT = 180^\circ - 145^\circ = 35^\circ \(\therefore\) < TAO = 180^\circ - (90^\circ + 35^\circ) = 55^\circ (sum of angle in a triangle)

P = {2,1,3,9,1/2}; Q = {1,21/2,3,7} R = {5,4,212} P∪Q∪R = {1/2,1,2,21/2,3,4,5,7,9} = {1/2,1,2,21/2,3,4,5,7,9}

No expanded working has been attached for this question yet. The verified answer is Option B.

Applying the formula 80 = 65 - x + x + 50 - x x = 35

No expanded working has been attached for this question yet. The verified answer is Option A.

x = 5, \(\frac{\sum {fx}}{\sum {f}}\) = 5= \(\frac{\sum {fx}}{30}\) \(\sum {fx}\) = 5 x 30 = 150; 150 - 34 = 116 \(\sum {f}\) = 29 \(\sum {fx}\) = 116 = \(\frac{116}{29}\) = 4