The number of zeroes of a polynomial is equal to the number of times its graph intersects the x-axis.

From the graph, we can see that the curve intersects the x-axis at 3 distinct points.

Therefore, the number of zeroes of \( p(x) \) is 3.

Given A.P.: \( \dfrac{3}{2}, \dfrac{1}{2}, \dfrac{-1}{2}, \dfrac{-3}{2}, \ldots \)

First term \( a = \dfrac{3}{2} \)

Common difference \( d = \dfrac{1}{2} - \dfrac{3}{2} = -1 \)

Using the formula for nth term: \( a_n = a + (n-1)d \)

\( a_{22} = \dfrac{3}{2} + (22-1)(-1) \)

\( a_{22} = \dfrac{3}{2} + 21(-1) \)

\( a_{22} = \dfrac{3}{2} - 21 \)

\( a_{22} = \dfrac{3}{2} - \dfrac{42}{2} \)

\( a_{22} = \dfrac{-39}{2} \)

To find where the line intersects the x-axis, we set \( y = 0 \):

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

\( 2x = 6 \)

\( x = 3 \)

Therefore, the line intersects the x-axis at \( (3, 0) \).

For one cone:

Radius \( r = 4 \) cm, Slant height \( l = 6 \) cm

Height of one cone \( h = \sqrt{l^2 - r^2} \)

\( h = \sqrt{6^2 - 4^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \) cm

When two identical cones are joined base to base, the total height becomes:

\( 2h = 2 \times 2\sqrt{5} = 4\sqrt{5} \) cm

For the system to be inconsistent, the lines must be parallel but not coincident.

For parallel lines: \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2} \)

Given equations: \( 3x - 7y = 1 \) and \( kx + 14y = 6 \)

\( \dfrac{3}{k} = \dfrac{-7}{14} \)

\( \dfrac{3}{k} = \dfrac{-1}{2} \)

\( k = -6 \)

Now check: \( \dfrac{c_1}{c_2} = \dfrac{1}{6} \) which is not equal to \( \dfrac{3}{k} = \dfrac{-1}{2} \)

Therefore, for \( k = -6 \), the system is inconsistent.

Total number of outcomes when two dice are rolled = \( 6 \times 6 = 36 \)

Favorable outcomes (sum more than 9):

Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes

Sum = 11: (5,6), (6,5) → 2 outcomes

Sum = 12: (6,6) → 1 outcome

Total favorable outcomes = \( 3 + 2 + 1 = 6 \)

Probability = \( \dfrac{6}{36} = \dfrac{1}{6} \)

Let's find the length of diagonal AC:

Point A = \( (2, -2) \), Point C = \( (4, 8) \)

Using distance formula:

\( AC = \sqrt{(4 - 2)^2 + (8 - (-2))^2} \)

\( AC = \sqrt{(2)^2 + (10)^2} \)

\( AC = \sqrt{4 + 100} = \sqrt{104} \)

\( AC = \sqrt{4 \times 26} = 2\sqrt{26} \)

Let's verify with diagonal BD:

Point B = \( (8, 4) \), Point D = \( (-2, 2) \)

\( BD = \sqrt{(-2 - 8)^2 + (2 - 4)^2} \)

\( BD = \sqrt{(-10)^2 + (-2)^2} = \sqrt{100 + 4} = \sqrt{104} = 2\sqrt{26} \)

Therefore, length of diagonal = \( 2\sqrt{26} \)

In right triangle OAP:

\( \angle \text{OAP} = 90^\circ \) (tangent is perpendicular to radius)

\( \angle \text{APO} = 30^\circ \) (given)

OA = 2.5 cm (radius)

Using trigonometric ratio:

\( \cos 30^\circ = \dfrac{\text{OA}}{\text{OP}} \)

\( \dfrac{\sqrt{3}}{2} = \dfrac{2.5}{\text{OP}} \)

\( \text{OP} = \dfrac{2.5 \times 2}{\sqrt{3}} = \dfrac{5}{\sqrt{3}} \) cm

Wait, let me verify with sine ratio:

\( \sin 30^\circ = \dfrac{\text{OA}}{\text{OP}} \)

\( \dfrac{1}{2} = \dfrac{2.5}{\text{OP}} \)

\( \text{OP} = 2.5 \times 2 = 5 \) cm

Therefore, OP = 5 cm

Probability of happening of event = 57% = 0.57

Probability of non-happening of event = 1 - Probability of happening

P(non-happening) = 1 - 0.57 = 0.43

Also, 0.43 = 43%

Therefore, probability of non-happening is 0.43

Length of arc = \( \dfrac{\theta}{360^\circ} \times 2\pi r \)

Given: arc length = \( \dfrac{22}{3} \) cm, radius r = 7 cm

\( \dfrac{22}{3} = \dfrac{\theta}{360^\circ} \times 2 \times \dfrac{22}{7} \times 7 \)

\( \dfrac{22}{3} = \dfrac{\theta}{360^\circ} \times 2 \times 22 \)

\( \dfrac{22}{3} = \dfrac{\theta}{360^\circ} \times 44 \)

\( \dfrac{22}{3} \times \dfrac{360^\circ}{44} = \theta \)

\( \dfrac{1}{3} \times \dfrac{360^\circ}{2} = \theta \)

\( \dfrac{360^\circ}{6} = \theta \)

\( \theta = 60^\circ \)

Therefore, \( \angle \text{AOB} = 60^\circ \)