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The roots of the equation are the values of x at the point where the graph crosses the x-axis. And this can be read from the graph to be 1.4 and -1.7.

Given that AC = 16 cm and BD = 12 cm. If O is the center. The diagonals of a rhombus meet at 90º. AO = \(\frac{\text{AC}}{2}\) = 8cm; BO = \(\frac{\text{BD}}{2}\) = 6 cm AB\(^2\) = AO\(^2\) + OB\(^2\) (pythagoras theorem) AB\(^2\) = 64 + 36 = 100; AB = 10 cm Area of △ABO = \(\frac{1}{2}\) \times 8 \times 6 = 24 cm\(^2\) Area of rhombus = 4 \times area of △ABO ⇒ 4 \times 24 = 96 cm\(^2\).

Total surface area = 2\pir\(^2\) + 2\pirh = 2\pir(r + h) ⇒ 2 \times \(\frac{22}{7}\) \times 7 \times (7 + 5) ⇒ 44 \times 12 = 528 cm\(^2\)

Total surface area of cone = \pirl + \pir\(^2\) = \pir(l + r) r = 3 and l = 4 Area = \(\frac{22}{7}\) \times 3 (4 + 3) = \(\frac{22}{7}\) \times 3 \times 7 = 66 cm\(^2\)

Capacity of water tank = (\(\frac{1}{2}\) \times \(\frac{3}{2}\) \times \(\frac{3}{2}\)) m\(^3\) = \(\frac{9}{8}\) m\(^3\) = \(\frac{9}{8}\) \times 1000 litres = 1125 litres. Water tankers capacity = 1500 litres Amount of water remaining = (1500 - 1125) litres = 375 litres

Volume of sphere = \(\frac{4}{3}\)\pir\(^3\) Surface area of sphere = 4\pir\(^2\) Since volume and area = k cm\(^3\) and k cm\(^2\) \(\frac{4}{3}\)\pir\(^3\) = 4\pir\(^2\) \(\frac{4}{3}\)r\(^3\) = 4r\(^2\) 4\pir\(^3\) = 12\pir\(^2\) r\(^3\) - 3r\(^2\) = 0 r\(^2\)(r - 3) = 0 ⇒ r = 3 cm d = 2 x 3 = 6cm.

Triangles OAD is similar to OBC \(\frac{\text{OA}}{\text{AD}}\) = OB But \(\frac{\text{OB}}{\text{BC}}\) = OA + OB Height of the cone is OB Let OA = x; OB = 12 + x \(\frac{\text{x}}{6}\) = \(\frac{(12 + x)}{12}\) Hence \(\frac{\text{x}}{6}\) = \(\frac{(12 + x)}{12}\) ⇒ 12x = 6x + 72 6x = 72 ⇒ x = 12 cm Height of cone is DB = 12 + x = (12 + 12) cm = 24 cm

Angle of sector = 108^\circ; Radius of circle = 3\(\frac{1}{2}\)cm ⇒ \(\frac{7}{2}\) cm Perimeter of sector = 108/360 \times 2\pir + 2r ⇒ \(\frac{108}{360 }\) \times 2 \times \(\frac{22}{7}\) \times \(\frac{7}{2}\) + 2(\(\frac{7}{2}\)) ⇒ 66/10 + 7 cm = 6\(\frac{3}{5}\) + 7 = 13\(\frac{3}{5}\)

∠DCB = ∠CBD = 35º (base angles of an isosceles triangle are equal) ∠ADB = 35º + 35º = 70^\circ (ext. angle of a triangle equals two opp. interior angles) ∠ABD = 90º (angle in a semicircle is a right angle) ∠BAO + ∠ADB + ∠ADB = 180º (sum of angles in a triangle) ∠BAO + 90º + 70º = 180º ∠BAO + 160º = 180º ∠BAO = 180º - 160º ∠BAO = 20º

The sides are in a common ratio ⇒ \(\frac{EC}{AE} = \frac{AB}{CD}\) \(\frac{EC}{AE} = \frac{9}{12}\) EC = \(\frac{8 \times 9}{12}\) = 6cm.

8 appears 4 times (highest frequency) → Mode = 8

Since sin x = cos (90 - x) [complementary angles] .: sin 20 = cos 70 cos 25 = sin 65 . . \(\frac{sin 20}{ cos 70} + \frac{cos 25 }{sin 65}\) =\(\frac{sin 20 }{ sin 20} + \frac{cos 25 }{cos 25}\) = 1 + 1 = 2

Total sum = 20 \times 4 = 80 Remaining sum = 80 – 23 = 57 Remaining observations = 19 Mean = \(\frac{57}{19}\) = 3

a. px - 2qx - 4qy + 2py = x(p - 2q) - 2y(2q - p) = x(p - 2q) + 2y(p - 2q) = (x + 2y)(p - 2q) b. U = {1,2,3,4,5,6,7,8,9,10} P = {1,2,4,6,10} Q = {2,3,6,9} (P ∪ Q) = {1,2,3,4,6,9,10} (P ∪ Q)\(^I\) = Elements in U but not in (P ∪ Q) (P ∪ Q)\(^I\) = {5,7,8} P\(^I\) = {3,5,7,8,9} Q\(^I\) = {1,4,5,7,8,10} P\(^I\) ∩ Q\(^I\) = {5,7,8} Evidently, (P ∪ Q)\(^I\) = P\(^I\) ∩ Q\(^I\)

a. y = c + k(\(\frac{1}{x^2}\)) where c and k are constant b. when x = 1, y = 11 ∴ 11 = c + k(\(\frac{1}{x^2}\)) = > 11 = c + k --- eqn (I) when x = 2, y = 5 ∴ 5 = c + k(\(\frac{1}{x^2}\)) = > 5 = c + k\(\frac{1}{2^2}\) --- eqn (II) Multiply through by 4) 20 = 4c + k - - - - - - - -(III) 11 = c + k --- eqn (I) 20 = 4c + k - - - - - - (III) From eqn(I) k = 11 - c Put k = 11 - c into eqn(III) 20 = 4c + k = 20 = 4c + 11 - c 20 = 3c + 11 3c = 20 - 11 = 9 3c = 9 c = 3. Put c = 3 into eqn(I) 11 = c + k 11 = 3 + k k = 11 - 3 = 8 ∴ c = 3, k = 8. When x = 4, But c = 3, K = 8 y = 3 + \(\frac{8}{4^2}\) y = 3 + \(\frac{1}{2}\) y = \(\frac{7}{2}\).