(i) \(\frac { 13 }{ 64 } =\frac { 13 }{ { 2 }^{ 6 } } \) So,\(\frac { 13 }{ 64 } \) has a terminating decimal expansion
(ii) \(\frac { -71 }{ 125 } =\frac { -71 }{ { 5 }^{ 3 } } \) So,\(\frac { -71 }{ 125 } \) has a terminating decimal expansion.
(iii) \(\frac { 43 }{ 375 } =\frac { 43 }{ { 3 }^{ 1 }\times { 5 }^{ 3 } } \)So,\(\frac { 43 }{ 375 } \) has a non - terminating recurring decimal expansion.
(iv) \(\frac { 31 }{ 400 } =\frac { 31 }{ { 2 }^{ 4 }\times { 5 }^{ 2 } } \) So, \(\frac { 31 }{ 400 } \)has a terminating decimal expansion

The rational numbers between \(-7\over 11\) and \(2\over 11\) is
\(\frac {- 6 }{ 11 } , \frac { -5 }{ 11 } , \frac { -4 }{ 11 },\frac {- 3 }{ 11 },\frac {- 2 }{ 11 },\frac {- 1 }{ 11 } , 0, \frac { 1 }{ 11 } \)

Distance of the chord from the centre
= \(\sqrt { { 15 }^{ 2 }-{ 12 }^{ 2 } } \)
= \(\sqrt { 225-144 } \)
= \(\sqrt { 89 } \)
= 9 cm

B' = {b, c, f, g}
(B')' = U-B'
= { a, b, c, d, e, f, g, h} - {b, c, f, g}
= {a, d, e, h}

{2, 3, 4, 6, 7, 8, 9, 10, 11}

n(A) = 4
n[P(A)] = 2^m = 2⁴
= 16

Number of zeros = 1 (The curve is intersecting at the origin)

We have
\(\frac{x^{3}-4x^{2}+6x}{x}=\frac{x^{3}}{x}-\frac{4x^{2}}{x}+\frac{6x}{x} , x\ne0\)
= x²- 4x + 6

q(y) = 2y - 3
= \(2\left( y-\frac { 2 }{ 3 } \right) \)
\(q\left( \frac { 3 }{ 2 } \right) =2\left( \frac { 3 }{ 2 } -\frac { 3 }{ 2 } \right) \)
= 2 \(\times\) 0 = 0
Hence \(\frac { 3 }{ 2 } \) is the zero of f(z)

A = {b, d, e, g, h}, B = {a, e, c, h}
A – B = {b, d, g}
n(A – B) = 3 .......... (1)
A ∩ B = {e, h}
n(A ∩ B) = 2 , n(A) = 5
n(A) – n(A∩B) = 5 - 2
= 3 ....... (2)
From (1) and (2) we get n(A – B) = n(A) – n(A∩B).

It is not a trapezium

(A – B)′

The geometrical shape of the figure is square

Coefficient of x² is \(\pi \)  and Coefficient of x is -1

A – B = {6, 7, 9}
B – A = {10, 12}
A Δ B = (A – B) ∪ (B – A) = {6, 7, 9} ∪ {10,12}
A Δ B = {6, 7, 9, 10, 12}.

Given A = {1, 2, 6}, B = {2, 3, 4}
A∪B = {1,2,3, 4,6}.

A= {a,b}
Subsets of A are \(\phi \),{a}, {b} and {a, b}.

\(\angle SOA=2x^0\) (Vertically opposite angles are equal)
\(\angle POB+\angle SOA+\angle DOQ=180^0\) (angle of a straight line)
3x° + 2x°+ 4x° = 180°
9x° = 180°
\(x^0=\frac{180}{9}=20^0\)
The value of x = 20°.

Supplement of Right angle (90°) = 180° - 90°
= 90°

Complement of 62°32' = 90° -  62°32'
= 27°28′

A = {1,2,3,4,5,7,9,11}
Since set A contains 8 elements, n(A) = 8.

ASSESSMENT
A= {A, S, E, M, N, T}

The x-coordinate is negative and y – coordinate is positive. So, Point(–7,3) lies in the II quadrant

The x- coordinate is positive and y – coordinate is negative. So, Point(3,–8) lies in the IV quadrant.