Given that side (a) = 10 cm
volume of the cube = a³
= 10 × 10 × 10
= 1000 cm³

write 19⁰59' = 19⁰54' + 5'
From the table we have, cos19⁰54' = 0.9403 

By Pythagoras theorem,
\(AB=\sqrt { { BC }^{ 2 }{ AC }^{ 2 } } \)
= \(\sqrt { { \left( 25 \right) }^{ 2 }-{ 7 }^{ 2 } } \)
= \(\sqrt { 625-49 } =\sqrt { 576 } \) = 24
The six trignometric ratios are
\(sin\theta =\frac { opposite\ side }{ hypotenuse } =\frac { 7 }{ 25 } \)
\(tan\theta =\frac { oppositeside }{ adjacent\ side } =\frac { 7 }{ 24 } \)
\(sec\theta =\frac { hypotenuse }{ adjacent\ side } =\frac { 25 }{ 24 } \)
\(cos\theta =\frac { adjacentside }{ hypotenuse } =\frac { 24 }{ 25 } \)
\(cosec\theta =\frac { hypotenuse }{ oppositeside } =\frac { 25 }{ 7 } \)
\(cot \theta\ \frac { adjacentside }{ oppositeside } =\frac { 24 }{ 7 } \)

31, 33, 63, 33, 28, 29, 33, 27, 27, 34, 35, 28, 32
Writing in ascending order we get
27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 36
Number of values = 13 which is an odd number
\(\therefore\) Median =\(\left( \cfrac{13+1}{2} \right) \) Value
= \(\left(\cfrac{14}{2}\right)^{th}\) Value = 7^th value
= 32

(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) 3\(\sqrt{7}\) + 5\(\sqrt{7}\) = (3 + 5)\(\sqrt{7}\) = 8\(\sqrt{7}\). The answer is irrational.
(ii) 7\(\sqrt{5}\) - 4\(\sqrt{5}\) = (7 - 4)\(\sqrt{5}\) = 3\(\sqrt{5}\). The answer is irrational.

\(\frac { 5\sqrt { 3 } +\sqrt { 2 } }{ \sqrt { 3 } +\sqrt { 2 } } =\frac{(5\sqrt{3}+\sqrt{2})\times(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})\times(\sqrt{3}-\sqrt{2})}\)

=\(\frac{5\sqrt{3}\times\sqrt{3}+\sqrt{2}\times\sqrt{3}-5\sqrt{3}\times\sqrt{2}-\sqrt{2}^2}{\sqrt{3}^2-\sqrt{2}^2}\)
=\(\frac{5\times3+\sqrt{6}-5\sqrt{6}-2}{3-2}=\frac{13-4\sqrt{6}}{1}=13-4\sqrt{6}\)

Let S, C and B represent sets of families who owned Scooter, Car and Bicycle respectively.
Given, n(U)=600 n(S) =\(\frac{3}{5}\) \(\times\) 600 = 360
n(C) = \(\frac{1}{3}\) \(\times\) 600 = 200 n(B)=\(\frac{1}{4}\) \(\times\) 600 = 150
n(S⋂C⋂B)=\(\frac{2}{15}\) \(\times\) 600 = 80
From Venn diagram,
(i) The number of families owned atleast two types of vehicles = 40 + 6 + 10 + 80 = 136
(ii) The number of families owned no vehicle
= 600 – (owned atleast one vehicle)
= 600- (230 + 40 + 74 + 6 + 54 + 10 + 80)
= 600 - 494 = 106

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

Let the breadth of the rectangle be "b"
Length of the rectangle = x² + 4
Area of the rectangle = x⁴ + 9x² + 20

Length x Breadth = x⁴ + 9x² + 20
(x⁴ + 4)x b = x⁴ + 9x² + 20
b = \(\frac{x^4+9x^2+20}{x^2+4}\)
= x²+ 5
Breadth of the rectangle = x²+ 5

(i) In ΔAPB and ΔCQD we have
ㄥAPB = ㄥCQD (90^o each)
AB = CD (opposite sides of parallelogram ABCD)
ㄥABP = ㄥCDQ (AB II CD and AD is a transversal)
Using ASA congruency we have,
ㄥAPB ≅ ㄥCQD
(ii) Since ΔAPB ≅ ㄥCQD
∴ Their corresponding parts are equal.
∴ AP = CQ.

Let M be the set of students opting for Mathematics.
Let S be the set of students opting for Science.
n (M U S) = 40, n (M) = 26, n (S) = 24
n (M U S) = n (M) + n (S) - n (M n S)
40 = 26 + 24 - n (M n S)
n (M n S) = 26 + 24 - 40
= 50 - 40 = 10
∴ Number of students opted for Mathematics and Science = 10

p(x) = x²⁰¹⁸ + 2018
When it is divided by x - 1,
p( 1) = 1²⁰¹⁸+ 2018
= 1 + 2018
= 2019
The Remainder : 2019

Let x = -21.213777 ...... \(\rightarrow\) (1)
Here period of decimal is 4,multiply equation (1) by 10000
10000 x = -21213, 7.777 .............. (2)
(2) - (1)
9999 x = 212116.5640
\( x =\frac{-212116.5640}{9999} \)
\( x =\frac{-2121165640}{99990000} \)
\( =\frac{-190924}{9000}=\frac{-47731}{2250} \)

2x⁴ + 4x² - 3x + 7 - (3x³ - x² + 2x + 1)
= 2x⁴ + 4x² - 3x + 7 - 3x³ + x² - 2x - 1
= 2x⁴ - 3x³ + 4x² + x2 - 3x - 2x + 7 - 1
= 2x⁴ - 3x³ + 5x² - 5x + 6
2x⁴ - 3x³ + 5x² - 5x + 6 must be subtracted to get 3x³ - x² + 2x + 1.

The degree of the polynomial is 4

Using the distance formula, we have
\(AB=\sqrt { \left( 6-3 \right) ^{ 2 }+\left( 4-1 \right) ^{ 2 } } =\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
\(BC=\sqrt { \left( 8-6 \right) ^{ 2 }+\left( 6-4 \right) ^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 } \)
\(AC=\sqrt { \left( 8-3 \right) ^{ 2 }+\left( 6-1 \right) ^{ 2 } } =\sqrt { 25+25 } =\sqrt { 50 } =5\sqrt { 2 } \)
\(AB+BC=3\sqrt { 2 } +3\sqrt { 2 } =5\sqrt { 2 } =AC\)
Therefore the points lie on a straight line.

\(\angle 1:\angle 2=11:7\)
\(\angle 1+\angle 2=180^0\) (Angle of a straight line)
Divide 180° in the ratio 11 : 7
Total ratio 11 + 7 = 18
\(\angle 1=\frac{11}{18}\times180^0=110^0\)
\(\angle 2=\frac{7}{18}\times180^0=70^0\)
\(\angle 3=\angle 1=110^0\)(Vertically opposite angles)
\(\angle 4=\angle 2=70^0\)(Vertically opposite angles)
\(\angle 5=\angle 1=110^0\)(Corresponding angles)
\(\angle 6=\angle 2=70^0\)(Corresponding angles)
\(\angle 7=\angle 5=110^0\)(Vertically opposite angles)
\(\angle 8=\angle 6=70^0\)(Vertically  opposite angles )
\(\therefore\) ∠1 = 110°, ∠2 = 70°, ∠3 = 110°, ∠4 = 70°, ∠5 = 110°, ∠6 = 70°, ∠ 7 = 110° ∠8 = 70°.