(i) \(2\over 3\) = 0.666... = \(0.\overline 6\)

(ii) \(47\over 99\) = 0.4747... = \(0.\overline {47}\)

(iii) \(-{16\over 45}\) = 0.355.... = \(0.3\overline{5}\)

A\(\cup \)B = {1, 3, 5, 7} \(\cup \) {0, 2, 3, 5, 7}
= {0, 1, 2, 3, 5, 7}
(A\(\cup \)B)' = U-(A\(\cup \)B)
= {0, 1, 2, 3, 4, 5, 6, 7} - {0, 1, 2, 3, 5, 7}
= {4, 6}

Number of zeros = 0 (The curve is not intersecting the x-axis)

When y = -1
f(y) = 6y - 3y² + 3
f(-1) = 6(-1) - 3(-1)²+ 3
= -6 - 3+ 3 = -6

From the venn diagram,
A = {5, 10, 15, 20}
B = {10, 20, 30, 40, 50,}
Then A∪B = {5, 10, 15, 20, 30, 40, 50}
A∩B = {10, 20}
n(A) = 4, n(B) = 5, n(A∪B) = 7, n(A∩B) = 2
n(A∪B) = 7 ........... (1)
n(A) + n(B) – n(A∩B) = 4 + 5 - 2 = 7 ......... (2)
From (1) and (2), n(A∪B) = n(A) + n(B) – n(A∩B).

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

\(\angle SOA=2x^0\) (Vertically opposite angels 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 140° = 180⁰- 140°
= 40°

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

The x-coordinate is positive and y – coordinate is positive. So Point (2,5) lies in the I quadrant

Let R be the set of children forgotten their rulers.
Let P be the set of children forgotten their pencil.
n(R) = 20, n(P) = 17, n(R\(\cap \)P) = 5
By using venn-diagram

From the venn -diagram we get
(i) Number of children forgotten only pencil = 12
(ii) Number of children forgotten only ruler 15
(iii) Total number of children in the class 15 + 5 + 12 = 32

Thus, by division method \(\sqrt { 3 } \) = 1.7320508…
It is found that the square root of every positive non perfect square number is an irrational number  \(\sqrt { 2 } ,\sqrt { 3 } ,\sqrt { 5 } ,\sqrt { 6 } ,\sqrt { 7 } \)..... are all irrational numbers.

x³ (x² +x) = x⁵ + x⁴ The degree of the polynomial 5

Draw TE || AB.
\(\angle ABT+\angle ETB=180^0(AB||TE)\)
\(48^0+\angle ETB=180^0\)
\(\angle ETB=180^0-48^0=132^0\)
Similarly \(\angle CDT+\angle DTE=180^0\)
\(24^0+\angle DTE=180^0\)
\(\therefore \angle DTE=180^0-24^0\) = 156⁰
\(\therefore \angle BTE+\angle ETD=132^0+156^0\) = 288°
x = 288°

The distance between the points (-4, 3), (2, -3) is
\(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 2+4 \right) ^{ 2 }+\left( -3-3 \right) ^{ 2 } } \)
\(=\sqrt { \left( { 6 }^{ 2 }+\left( -6 \right) ^{ 2 } \right) } =\sqrt { \left( 36+36 \right) } \)
\(=\sqrt { 36\times 2 } \)
\(=6\sqrt { 2 } \)

Steps for construction:
Step 1: Draw the ΔPQR with the given measures.
Step 2: Construct the perpendicular bisector of (PQ and PR) any two sides and let them meet at S which is the circumcenter.
Step 3: With S as centre and SP = SQ = SR as radius draw the circumcircle to passes through P, Q and R.


Circum radius = 4.3 cm.

Let T be the set of families buy Tamil news Papers
Let E be the set of families buy English news Papers
n(TUE) = 100
n(T) = 65 and n(E) = 55
Let "X" be the number families buy both newspapers.
By using venn-diagram
From the venn-diagrarn we get
65 - x + x + 55 - x = 100
120 - x = 100
120 -100 = x
20 = x
(i) Families buy both newspapers = 20
(ii) Families buy Tamil newspapers only = 65 - x
= 65 - 20
= 45
(iii) Families buy English newspapers only = 55 - x
= 55 - 20
= 35.