Let ‘a’ be the side of the cube.
Given that, total surface area = 864 cm²
6a²= 864
a² = \(\frac{864}{6}\)
a² = 144
Therefore, side (a) = 12 cm
Now, volume of the cube = a³
= 123 = 12 ×12 ×12 = 1728 cm³

Here, the dimensions of a sweet box are Length (l) = 22cm, breadth (b) = 18cm,
height (h) = 10 cm.
Volume of a sweet box = l × b × h
= 22 × 18 × 10 cm³
The dimensions of a carton are
Length (l) = 1m= 100 cm, breadth (b) = 88 cm,
height (h) = 63 cm.

Volume of the carton = l × b × h
= 100 × 88 × 63 cm³
The number of sweet boxes packed =\(\frac{volume \ of \ the \ carton}{volume\ of \ a \ sweet \ box}\)
= \(\frac{100 \times 88\times63}{22\times18\times10}\)
= 140 boxes

Let E be the event that it will rain tomorrow. Then E′ is the event that it will not rain tomorrow.
Since P(E) = 0.91, we have P(E′) = 1−0.91 (how?)
= 0.09
Therefore, the probability that it will not rain tomorrow
= 0.09

(i) sin 38⁰36' + tan 12⁰12'
sin 38⁰36' = 0.6239
tan12⁰12' = 0.2162
sin 38⁰36' + tan 12⁰12' = 0.8401
(ii) tan 60⁰25^' - cos 49⁰20^'
tan 60⁰25' = 1.7603 + 0.0012 = 1.7615
cos 49⁰20^' = 0.6521 - 0.0004 = 0.6517
tan 60⁰25' - cos 49⁰20' = 1.1098

write 64⁰34' = 64⁰30' + 4'
From the table we have, sin64⁰30' = 0.9026 

Given linear equations are
kx + 2y = 3 ......(1)
2x - 3y = 1  .......(2)
\(\left[ \begin{matrix} { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0 \\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0 \end{matrix} \right] \)
Here a₁ = k, b₁ = 2, a₂ = 2, b₂ = −3 ;
For unique solution we take \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \); therefore \(\frac { k }{ 2 } \neq \frac { 2 }{ -3 } \);\(k\neq \frac { 4 }{ -3 } \), that is \(k\neq -\frac { 4 }{ 3 } \)

(i) tan 7°tan 23°tan 60°tan 67°tan 83°
=  tan 7⁰ tan 83⁰ tan 23⁰ tan 67⁰ tan 60⁰ (Grouping complementary angles)
=  tan 7⁰ tan(90⁰ - 7⁰)tan 23⁰ tan(90⁰ - 23⁰)tan 60⁰
=  (tan7⁰.cot7⁰)(tan 23⁰. cot 23⁰)tan 60⁰
=  (1)\(\times\) (1)\(\times\) tan 60⁰
=  tan 60⁰ = \(\sqrt { 3 } \) 
(ii) \(\frac { cos35° }{ sin55° } +\frac { sin12° }{ cos78° } -\frac { cos18° }{ sin72° } \)
\(=\frac { cos\left( 90°-55° \right) }{ sin55° } +\frac { sin\left( 90°-78° \right) }{ cos78° } -\frac { cos\left( 90°-72° \right) }{ sin72° } \) \(\left[ \begin{matrix} { \text Since} \\cos35°=cos\left( 90°-55° \right) \\ sin12°=sin(90°-78°) \\ cos18°=cos\left( 90°-72° \right) \end{matrix} \right] \)
= \(\frac { sin55° }{ sin55° } +\frac { cos78° }{ cos78° } -\frac { sin72° }{ sin72° } \)
= 1+1 - 1 = 1

Let a = 28 m, b = 15 m and c = 41 m
Then, s = \(\frac{a+b+c}{2}=\frac{28+15+41}{2}=\frac{84}{2}\) = 42m
Area of triangular field =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 42(42-28)(42-15)(42-41) } \)
=\(\sqrt { 42\times 14​​\times 17\times 1 } \)
=\(\sqrt { 2\times 3\times 7\times 7\times 2\times 3\times 3\times 3\times 1 } \)
\(=2 \times 3 \times 7 \times 3\)
= 126 m²
Given the cost of levelling is Rs. 20 per m².
The total cost of levelling the field = 20 \(\times\)126 = Rs. 2520.

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
\(sin\theta =\frac { opposite\ side }{ hypotenuse } =\frac { 7 }{ 25 } \)
\(tan\theta =\frac { oppositeside }{ adjacent\ side } =\frac { 7 }{ 24 } \)
\(sec\theta =\frac { hypotenuse }{ adjacent\ side } =\frac { 25 }{ 24 } \)
\(cos\theta =\frac { adjacentside }{ hypotenuse } =\frac { 24 }{ 25 } \)
\(cosec\theta =\frac { hypotenuse }{ oppositeside } =\frac { 25 }{ 7 } \)
\(cot \theta\ \frac { adjacentside }{ oppositeside } =\frac { 24 }{ 7 } \)

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)

Total of the data = 11 + 15 + 17 + x + 1 + 19 + x -2 + 3
= 2x + 64
Mean of the data = \(\frac{2 x+64}{7}\)
= 14 (given)
\(\frac{2 x+64}{7}=\frac{14}{1}\)
2x + 64 = 98
2x = 98 - 64
2x = 34
\(\therefore x=\frac{34}{2}=17\)
The data = 11, 15, 17, 17+1, 19, 17- 2, 3
= 3, 11, 15, 15, 17, 18, 19
15 occurs two times
\(\therefore\) The Mode of data = 15

Let, p(x) = x³+13x²+ 32x + 20
Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0
Hence, (x -1) is not a factor.
Sum of coefficients of even powers and constant term = 13 + 20 = 33
Sum of coefficients of odd powers = 1 + 32 = 33
Hence, (x +1) is a factor of p (x)
Now we use synthetic division to find the other factors

Let A, B and C represent sets of students who speak Tamil, English and French respectively.
Given, n(AUBUC) = 100, n(A) = 85, n(B) = 40, n(C) = 20
n(A⋂B) = 32, n(B⋂C) = 13, n(A⋂C) = 10
We know that,
n(A U B U C) = n(A) + n(B) + n(C) - n(A⋂B) - n(B⋂C) - n(A⋂C) + n(A⋂B⋂C)
100 = 85 + 40 +20+ -32 -13-10 + n(A⋂B⋂C)
Then, n(A⋂B⋂C) = 100 - 90 = 10
Therefore, 10 students speak all the three languages

Given, A = {b,e,f,g} and B = {c,e,g,h}
(i) \(A\cup B\) = {b,c,e,f,g,h}     .....(1)
\(B\cup A\) = {b,c,e,f,g,h}        ......(2)
From (1) and (2) we have \(A\cup B\) = \(B\cup A\)
It is verified that union of sets is commutative.
(ii) \(A\cap B\) = {e,g}  .....(3)
\(B\cap A\) = {e,g}   .....(4)
From (3) and (4) we get, \(A\cap B\) = \(B\cap A\)
It is verified that intersection of sets is commutative.

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.

\(\therefore\) The Quotient : x²+7x–3
The Remainder : 0

Let A be the set of people owned car A
Let B be the set of people owned car B
n( A) = 400, n(B) = 200, n(A \(\cap\) B) = 50
n(A \(\cup\) B) = 500 .........(1)
n(A) + n(B) - n(A \(\cap\) B) = 400 + 200 - 50
= 600 - 50
= 550  ...........(2)
From (1) and (2) we get
\(n(A\cup B)\ne n(A) +n(B)-n(A\cap B)\)
\(\therefore\) The given data is incorrect.

The given polynomial is in standard form or  \(-1+4x-\frac { 6 }{ 5 } { x }^{ 2 }{ +7x }^{ 3 }\)

Let A(–4, –3), B(3, 1), C(3, 6), D(–4, 2) be the four vertices of any quadrilateral ABCD. Using the distance formula.
Let \(d=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(AB=\sqrt { \left( 3+4 \right) ^{ 2 }+\left( 1+3 \right) ^{ 2 } } =\sqrt { 49+16 } =\sqrt { 65 } \)
\(BC=\sqrt { \left( 3-3 \right) ^{ 2 }+\left( 6-1 \right) ^{ 2 } } =\sqrt { 0+25 } =\sqrt { 25 } =5\)
\(CD=\sqrt { \left( -4-3 \right) ^{ 2 }+\left( 2-6 \right) ^{ 2 } } =\sqrt { 49+16 } =\sqrt { 65 } \)
\(AD=\sqrt { \left( -4+4 \right) ^{ 2 }+\left( 2+3 \right) ^{ 2 } } =\sqrt { \left( { 0 }^{ 2 } \right) +\left( { 5 }^{ 2 } \right) } =\sqrt { 25 } =5\)
\(AB=CD=\sqrt { 65 } \) and BC = AD = 5
Here, the opposite sides are equal. Hence ABCD is a parallelogram.

In \(\triangle PQR\ and \triangle XYZ\)
PR = YZ (Given)
QR = XY (Given)
\(\angle QPR=\angle XYZ\) (Given)
By SAS congruency
\(\triangle PQR\cong\triangle XYZ\)

In \(\triangle ABD\ and\ \triangle CDB\)
AB = CD (Given)
AD = BC (Given)
BD is common
By SSS congruency
\(\triangle ABD\cong\triangle CDB\)

Through T draw TE|| AB.
\(\therefore \angle BAT+\angle ATE=180^0(AB||TE)\)
\(140^0+\angle ATE=180^0\)
\(\angle ATE=180^0-140^0=40^0\)
Similarly \(\angle ETC+\angle TCD=180^0(TE||CD)\)
\(\angle ETC+150^0=180^0\)
\(\angle ETC=180^0-150^0=30^0\)
\(x=\angle ATE+\angle ETC\)
= 40° + 30° = 70°
x = 70°