Let x = 0.57777 ......... \(\rightarrow\) (1)
10x = 5.77777 ............ \(\rightarrow\) (2)
100x = 57.7777 ........... \(\rightarrow\) (3)

 \(x={52\over 90}={26\over 45}\)
\(\therefore 0.5\overline{7}={26\over 45}\)    

(i) {-3,-2,-1,0,1,4} (ii) {-3,-2,-1,0,1,4} (iii) {-3,-2,0,1,3,4}

It is not a quadrilateral

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

{p, q, r} _____ {w, x, y, z}
Since the element p belongs to {p, q, r} but does not belong to {w, x, y, z}, shows that {p, q, r} \(\nsubseteq \) {w, x, y, z}.

\(\angle AOB =90^0\)
\(\angle BOC +\angle COA=90^0\)
3x⁰+2x⁰ = 90⁰
5x⁰= 90⁰
\(x^0=\frac{90}{5}=18^0\)
The value of x = 18°

The x-coordinate is negative and y – coordinate is negative. So, Point(–1,–3) lies in the III quadrant

Measure BD and DC.
BD = 3.3 cm and DC = 3.3 cm
\(\therefore\) BD=DC=3.3 cm
Measure CE and EA.
CE =4 cm andEA= 4 cm
\(\therefore\) CE=EA=4cm
Distance =\(\sqrt{(x_2+x_1)^2+(y_2-y_1)^2}\)
CG =\(\sqrt{(-3+3)^2+(3-7)^2}\)
\(=\sqrt{0+(-4)^2}=\sqrt{16}=4 \ uints\)
GF =\(\sqrt{(-3+3)^2+(3-1)^2}\)  (The point F is (-3,1) )
\(=\sqrt{0+2^2}=\sqrt{4}=2 \ uints\)
AG =\(\sqrt{(-3-1)^2+(3-0)^2}\)
\(=\sqrt{(-4)^2+(3)^2}=\sqrt{16+9}=\sqrt{25}=5 \ units\)
GD =\(\sqrt{(-3+5)^2+(3-4.5)^2}\)
\(=\sqrt{2^2+(-1.5)^2}=\sqrt{4+2.25}=\sqrt{6.25}=2.5 \ uints\)
BG =\(\sqrt{(-7+3)^2+(2-3)^2}\)
\(=\sqrt{(-4)^2+(-1)^2}=\sqrt{16+1}\)
\(=\sqrt{17}=4.12 \ units\) (Approximately BG = 4)
GE =\(\sqrt{(-3+1)^2+(3-3.5)^2}\)
\(=\sqrt{2^2+(0.5)^2}=\sqrt{4+0.25}\)
\(=\sqrt{4.25}=2.06 \ units\) (Approximately GE = 2)
AG : GD = 5 : 2.5
= 1 : 0.5
=2: 1
BG: GE =4: 2
= 2: 1
CG: GF =4: 2
= 2: 1
AG : GD = BG : GE = CG = GF = 2 : 1

\(\sqrt { 2 } \) =1.414213...
\(\sqrt { 3} \) =1.73205....
\(\frac { 7 }{ 9 } \) =0..7777
\(\frac { 6 }{ 7 } \) =0.857142
\(\frac { 23 }{ 6 } \) =3.8333...
\(\frac { 48 }{ 5 } \) =9.6
(i) Arrange the ten real numbers in the given boxes in ascending order.

(ii) Arrange the same numbers in the boxes given below in descending order.

Let f(x) = 2x⁴- 6x³+3x²+3x-2
g(x) = x² –3x + 2
= x²-2x-x+2
= x(x-2)-1(x-2)
= (x-2)(x-1)
we show that f(x) is exactly divisible by (x–1) and (x–2) using remainder theorem
f(1) = 2(1)⁴-6(1)³+3(1)²+3(1)-2
= 2 – 6 + 3 + 3 – 2 = 0
f(2) = 2(2)⁴-6(2)³+3(2)²+3(2)-2
= 32 – 48 + 12 + 6 – 2 = 0
f(x) is exactly divisible by (x – 1) (x – 2)
i.e., f(x) is exactly divisible by x² –3x + 2
If p(x) is divided by (x -a) with the remainder p(a) = 0 , then (x -a) is a factor of p(x). Remainder Theorem leads to Factor Theorem.

In the given expressin (i), (iii) and (vi) contain only one term.
\(\therefore\) (i), (iii) and (vi) are monomials.
The expression (v) contains two terms. So it is a binomial.
The expression (ii) contains three terms. So it is a trinomial.
The expression (iv) contains four terms. So it is none of the given type. It is a polynomial

AD = 3 units
DA = 3 units (From graph)
Yes, Distance AD = Distance DA
The students are asked to do the activity with different points and verify the result.

If a side is produced in a triangle, the angle formed is called an exterior angle.
Here BC, a side of ΔABC is produced, which lead to the formation of exterior angle ㄥACD.
Also BCD is a straight line ⇒ ㄥBCD is a straight angle.
ㄥBCD = 180⁰
ㄥc + ㄥd = 180⁰  ......(1)
Also in ΔABC ㄥa + ㄥb + ㄥc
From (1) and (2) we get,
ㄥa + ㄥb + ㄥc = ㄥc + ㄥd
∴ ㄥa + ㄥb =ㄥd
Hence, we conclude that the measure of exterior angle formed is equal to the sum of the measure of two opposite interior angles.

To draw the different types of quadrilateral by giving the instructions using their properties.
Example:
(i) It is four sides closed diagram.
(ii) Opposite sides are parallel.
(iii) Four sides are equal in size.
(iv) Diagonals bisect each other at 90°.
(v) Diagonals are not equal.
The diagram represents as a rhombus
Like this for hexagon, parallelogram, Trapezium etc.
Any one must give the clues to draw the diagram.
At last they have to find what it represent.

Distance = \(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+({ y }_{ 2 }-y_{ 1 })^{ 2 } } \)
AB= \(\sqrt { (-2-6)^{ 2 }+(-4+4)^{ 2 } } \)
= \(\sqrt { (-8)^{ 2 }+0 } =\sqrt { 64 } =8\)
BC = \(\sqrt { (2+2)^{ 2 }+(10+4)^{ 2 } } \)
= \(\sqrt { (4)^{ 2 }+(14)^{ 2 } } =\sqrt { 16+196 } =\sqrt { 212 } \)
AC = \(\sqrt { (2-6)^{ 2 }+(10+4)^{ 2 } } \)
= \(\sqrt { (-4)^{ 2 }+(14)^{ 2 } } =\sqrt { 1+-196 } =\sqrt { 212 } \)
BC = AC =\(\sqrt { 212 } \) (Two sides are equal)
∴  ABC is an isosceles triangle.

Distance AB =\(\sqrt { (2+2)^{ 2 }+(-3+8)^{ 2 } } \)
= \(\sqrt { (4)^{ 2 }+(5)^{ 2 } } =\sqrt { 16+25 } =\sqrt { 41 } \)
BC =\(\sqrt { (6-2)^{ 2 }+(2+3)^{ 2 } } \)
= \(\sqrt { (4)^{ 2 }+(5)^{ 2 } } =\sqrt { 16+25 } =\sqrt { 41 } \)
AC = \(\sqrt { (6+2)^{ 2 }+(2+8)^{ 2 } } \)
= \(\sqrt { (8)^{ 2 }+(10)^{ 2 } } =\sqrt { 64+100 } =\sqrt { 164 } \)
=  \(\sqrt { 4\times 41 } =2\sqrt { 41 } \)
AB + BC = AC ⇒ \(\sqrt { 41 } +\sqrt { 41 } =2\sqrt { 41 } \)
∴ The given three points are collinear.

i) Draw a line and mark a point A on it.
ii) Mark a point B such that AB = 9.3 cm.
iii) Mark a point C on this line such that BC = 1 unit
iv) Find the midpoint of AC by drawing perpendicular bisector of AC and let it be O
v) With O as center and OC = OA as radius, draw a semicircle
vi) Draw a line BD, which is perpendicular to AB at B.
vii) Now BD = \(\sqrt { 9.3 } \)  which can be marked in the number line as the value of  BE = BD =\(\sqrt { 9.3 } \)