6 has the maximum frequency 9. Therefore 6 is the mode.

In the figure\(\angle \)POQ = 100⁰
\(\angle \)PQR = 30⁰
\(\angle \)PQR = \(\cfrac { 1 }{ 2 } \)\(\angle \)PQR = \(\cfrac { 1 }{ 2 } \)\(\times\)100 = 50⁰
In\(\triangle \)OPQ = \(\angle \)OQP =\(\angle \)POQ = 180⁰
2\(\angle \)OPQ = 180⁰
2\(\angle \) OPQ = 80⁰
\(\therefore \) \(\angle \)OPQ = 40⁰ 
In \(\triangle \)PRQ,
\(\angle \)R+\(\angle \)P+\(\angle \)Q = 180⁰
50⁰ + (40 + x) + 30⁰ = 180⁰
(40 + x)⁰ = 180⁰ - 80 = 100⁰
x⁰ = 100⁰ - 40⁰ = 60⁰
\(\therefore \) \(\angle \)RPO = x = 60⁰

We know that (a+b)³ ≡ a³+3a²b+3ab²+b³
(2a+3b)³ = (2a)³+3(2a)²(3b)+3(2a)(3b)²+(3b)³
= 8a³+36a²b+54ab²+27b³

(2a-3b+4c)²= (2a+(-3b)+4c)² = (2a)²+(-3b)²+(4c)²+2(2a)(-3b)+2(-3b)(4c)+2(4c)(2a)
= 4a²+9b²+16c²+(-12ab)+(-24bc)+16ca
= 4a²+9b²+16c²-12ab-24bc+16ca

(5x + 4y)(5x −4y)    [we have (a+b)(a-b) = a²-b²]
(5x+4y)(5x-4y) = (5x)²-(4y)²    put [a = 2a, b = 4y]
= 25x²-16y²

(4000000)³ ÷ (0.00002)⁴
\( =\left(4.0 \times 10^{6}\right)^{3} \div\left(2.0 \times 10^{-5}\right) 4 \)
\(=(4.0)^{3} \times\left(10^{6}\right)^{3} \div(2.0)^{4} \times\left(10^{-5}\right)^{4} \)
\(=\frac{64.0 \times 10^{18}}{16.0 \times 10^{-20}} \)
\(=4 \times 10^{18} \times 10^{+20} \)
\(=4.0 \times 10^{38} \)

\( \left(2.75 \times 10^{7}\right)+ \left(1.23 \times 10^{8}\right) \\ =\left(2.75 \times 10^{-1}+1.23\right) \times 10^{8} \\ =(0.275+1.23) \times 10^{8} \\ =1.505 \times 10^{8} \)

(i) \(\frac{1}{\sqrt{50}}=\frac{1}{\sqrt{50}}\times\frac{\sqrt{50}}{\sqrt{50}}=\frac{\sqrt{50}}{\sqrt{50}}=\frac{\sqrt{5\times5\times2}}{5\times5\times2}=\frac{ ̶5̶\sqrt{2}}{ ̶5̶\times5\times2}=\frac{\sqrt{2}}{10}\)
(ii) \(\frac{5}{3\sqrt{5}}=\frac{5}{3\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{ ̶5̶\sqrt{5}}{3\times ̶5̶}=\frac{\sqrt{5}}{3}\)
(iii) \(\frac{\sqrt{75}}{\sqrt{18}}=\frac{\sqrt{3\times5\times5}}{\sqrt{3\times2\times3}}=\frac{5\sqrt{3}\times\sqrt{2}}{3\sqrt{2}\times\sqrt{2}}=\frac{5\sqrt{6}}{3\times2}=\frac{5\sqrt{6}}{6}\)
(iv) \(\frac{3\sqrt{5}}{\sqrt{6}}=\frac{3\sqrt{5}\times\sqrt{6}}{\sqrt{6}\times\sqrt{6}}=\frac{ ̶3̶\sqrt{30}}{ ̶6̶}=\frac{\sqrt{30}}{2}\)

Given A = {0, 2, 4,6, 8} , B = {2, 3, 5, 7} and C = {5, 6, 7, 8}
First, we find \(B\cap C\) = {5, 7} ,\(A\cup (B\cap C)\) = {0, 2, 4, 5, 6, 7, 8} .......... (1)
Next, \(A\cup B\) = {0,2,3,4,5,6,7, 8},  \(A\cup C\) = {0, 2, 4, 5, 6, 7, 8}
Then, \((A\cup B)\cap (A\cup C)\) = {0, 2, 4, 5, 6, 7, 8} .............(2)
From (1) and (2), it is verified that \((A\cup \left( B\cap C \right) =(A\cup B)\cap (A\cup C)\).

Now,  \((B\cap C)=\left\{ \frac { 1 }{ 4 } ,2,\frac { 5 }{ 2 } \right\} \)
\(A\cap (B\cap C)=\left\{ \frac { 1 }{ 4 } ,2 \right\} \)    .......(1)
Then, \(A\cap B=\left\{ 0,\frac { 1 }{ 4 } ,\frac { 3 }{ 4 } ,2 \right\} \)
\((A\cap B)\cap C=\left\{ \frac { 1 }{ 4 } ,2 \right\} \)        .......(2)
From (1) and (2), it is verified that
\((A\cap B)\cap C=A\cap (B\cap C)\).

A = {-2,0,1,3,5}, B = {-1,0,2,5,6}
C = {-1,2,5,6,7}
BUC = {-1,0,2,5,6,7}
A-(BUC) = {-2,1,3}.............(1)
(A-B) = {-2,1,3}
(A-C) = {-2,0,1,3}
(A-B)∩(A-C) = {-2,1,3}............(2)
From (1) and (2), it is verified that
A-(BUC) = (A-B)⋂(A-C)

Let X represent the set of seven values x₁, x₂, x₃, x₄, x_5, x_6, x₇ 
Then \(\bar X={{\sum_{i=1}^7}\over 7}=30\) or \({\sum_{i=1}^7}x_i=210\)
If each value is divided by 3, then the mean of new set of values is
\({{\sum_{i=1}^7}{x_i\over3}\over7}={\left({x_1\over3}+{x_2\over3}+{x_3\over3}+{x_4\over3}+{x_5\over3}+{x_6\over3}+{x_7\over3} \right)\over7}\)
\({{\sum_{i=1}^7}x_i\over21}={210\over 21}=10\)
Aliter
If Y is the set of values obtained by dividing each value of X by 3.
Then, \(\bar Y= \frac {\bar x}{3} = \frac {30}{3}=10\)

\(\bar{X}=\frac{\sum_{i=1}^{8} x_{i}}{n}=\frac{21+30+22+16+24+28+18+17}{8}=\frac{176}{8}=22\)
Deviation of an entry x_i from the arithmetic mean \(\bar X\) is \(x_i-\bar X\), i = 1, 2,........8
Sum of the deviations = (21-22)+(30-22)+(22-22)+(16-22)+(24-22)+(28-22)+(18-22)+(17-22)
= 16–16 = 0. or equivalently, \(\sum_{i=1}^{8}\left(x_{i}-\bar{X}\right)=0\)
Hence, we conclude that sum of the deviations from the Arithmetic Mean is zero.

Let Assumed mean A =170 

Mean \(\bar X = A+{Σfd\over Σf}\)
\(=170+\left(-780\over32\right)\)
Therefore \(\bar X\) = 170 - 24.375
= 145.625

Let p(x) = 8x⁴- 2x² + 6x-7
Standard form = 8x⁴+ 0x³- 2x²+ 6x - 7
Co-efficients are 8 0 - 2 6 -7
Q(x) = 2x+1,
To find zero of 2x + 1, put 2x + 1 = 0; 2x = -1, x = \(-\frac{1}{2}\)
Synthetic division

Quotient = \(\frac{1}{2}\)[8x³-4x²+6] = 4x³-2x²+3
Quotient 4x³ - 2x² + 3 is compared with the given quotient 4x³-px²-qx+3
Co-efficients of x² is p = -2
Co-efficients of x is q = 0
Remainder is - 10

\((y-\frac{1}{y})^3=27\)
y³-\(\frac{1}{y^3}\) = \((y-\frac{1}{y})^3+3y+\frac{1}{y}(y-\frac{1}{y})\)
∵ x³-y³ ≡ (x-y)³+3xy(x-y)
= 27+3\((y-\frac{1}{y})\)
\(∵ (y-\frac{1}{y})^3=27; y-\frac{1}{y}=\sqrt[3]{27}=3\)
= 27+3\(\times\)3 = 27+9 = 36

(i) (x + 5)(x + 6)(x + 7)
Co-efficient of x² = a + b + c = 5 + 6 + 7 = 18
Co-efficient of x²= ab + bc + ca = (5\(\times\)6)+(6\(\times\)7)+(7\(\times\)5)
= 30 + 42 + 35 = 107
Constant term = abc = 5\(\times\)6\(\times\)7
Co-efficient of constant term = 210
(ii) (2x + 3)(2x - 5)(2x - 6)
ஃ Co-efficient of x²= 4(a+b+c) = 4(3+(-5)+(-6))
= 4\(\times\)(-8) = -32
Co-efficient of x = 2(ab+bc+ca)
= 2[3x(-5)+(-5)(-6)+(-6)(3)]
= 2[-15+30-18] = 2x(-3) = -6
Constant term = abc = 3x(-5)x(-6) = 90

(i) The world population is nearly
7000,000,000 = 7.0 \(\times\) 10⁹
(ii) One light year means the distance
9460528400000000 km = 9.4605284 \(\times\) 10¹⁵ km
(iii) Mass of an electron is
0.000 000 000 000 000 000 000 000 000 00091093822 kg = 9.1093822 \(\times\) 10^-31 kg

\(\frac { \sqrt { 7 } -2 }{ \sqrt { 7 } +2 } =a\sqrt{7}+b\)
L.H.S = \(\frac{\sqrt{7}-2\times\sqrt{7}-2}{\sqrt{7}+2\times\sqrt{7}-2}=\frac{(\sqrt{7}-2)^2}{\sqrt{7}^2-2^2}=\frac{\sqrt{7}^2-2\sqrt{7}\times2+2^2}{7-4}\)
=\(\frac{7-4\sqrt{7}+4}{3}=\frac{11-4\sqrt{7}}{3}=\frac{11}{3}-\frac{4\sqrt{7}}{3}\)
\(\frac{-4\sqrt{7}}{3}+\frac{11}{3}=a\sqrt{7}+b\)
ஃa √7=\(\frac{-4\sqrt{7}}{3}\)
a = \(\frac{-4}{3}\)
b = \(\frac{11}{3}\)

U = {x: −4 ≤ x ≤ 4, x \(\in \)Z}, U = {-4,-3,-2,-1,0,1,2,3,4}
A = {x:-4 B = {x :−2 ≤ x ≤ 3,x \(\in \) Z}, B = {-2,-1,0, 1,2,3}
(AUB)' = A'⋂B'
AUB = {-3, -2, -1,0, 1,2, 3}
(AUB)' = {-4,4} ...(1)
A' = {-4, 3, 4}; B' = {-4, -3, 4}
A'∩B' = {-4,4} ...(2)
From (1) and (2) it is verified that
(AUB)' = A'⋂B'

ஃ A-(B∩C)=(A-B)U(A-C)
Hence it is proved.

PQRS is a cyclic quadrilateral
Given ㄥPSR = 70°
ㄥPSR+ㄥPQR = 180° (state reason________)
70° +ㄥPQR = 180°
ㄥPQR = 180°− 70°
ㄥPQR = 110°
In ΔPQR we have, ㄥPQR+ㄥPRQ+ㄥQPR = 180° (state reason_________)
110° +ㄥPRQ + 40° = 180°
ㄥPRQ = 180°− 150°
ㄥPRQ = 30°

(i)
  
\({ x }^{ 0 }=\cfrac { 1 }{ 2 } \angle BOC=\cfrac { 1 }{ 2 } \times \left( { 30 }^{ 0 }+{ 60 }^{ 0 } \right) =\cfrac { 1 }{ 2 } \times { 90 }^{ 0 }={ 45 }^{ 0 }\)
(ii)

\(\angle QPR=\cfrac { 1 }{ 2 } \angle QOR=\cfrac { 1 }{ 2 } \times { 80 }^{ 0 }={ 40 }^{ 0 }\)
In \(\triangle QPR\)
\(\angle R+\angle P+\angle Q={ 180 }^{ 0 }\)
\(\angle Q\) = 180^o-120^o= 60^o
\(\angle Q\) = x^o+ 50^o = 60^o
x^o = 60^o-50^o = 10^o
(iii)

\(\angle \) MPN = 90° (Angle subtended by the diameter is 90°)
\(\angle OMP+\angle OPM={ 180 }^{ 0 }-{ 110 }^{ 0 }={ 70 }^{ 0 }\)
\(\angle MPO=\cfrac { 70 }{ 2 } =35^{ 0 }\)
But \(\angle MPN={ 90 }^{ 0 }\)
\(\therefore\) \(\angle \)OPN = 90^o - 35^o = 55^o
(iv)

Central angle is twice that of angle sub tended on the circumference
\(\angle \)YOZ = 120^o\(\times\)2 = 40^o
\(\therefore\) x+\(\angle\)YOZ = 360^o
x^o = 360^o- 240^o = 120^o
(v)

\(\angle\)BOC = 360^o- 240^o = 120^o
\(\therefore\) \(\angle\)BAC = x^o=\(\cfrac { { 120 }^{ 0 } }{ 2 } \) = 60^o

In \(\triangle \)ABC, AB = 6 cm, LB= 110°, AC = 9 cm

Construction :
Step 1: Draw \(\triangle \)ABC with AB = 6 cm, ∠B =110°, AC = 9 cm
Step 2: Draw perpendicular bisectors of any two sides (BC and AB) to find the mid points of BC and AB.
Step 3: Construct medians AD and CE. Let them meet at G.
Step 4: G is the centroid of the given \(\triangle \)ABC.

In \(\triangle \)ABC,
AB = 4 cm, AC = 3 cm, \(\angle\)A =  90°

Construction :
Step 1: Draw \(\triangle \)ABC with AB = 4 cm, AC = 3 cm, \(\angle\)A = 90°
Step 2: Draw perpendicular bisectors of any two sides (AB and AC) to find the mid points of AB and AC.
Step 3: Draw the medians CD and BE. Let them meet at G.
Step 4: G is the centroid of the given triangle.