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

Total number of centum scorers = 1184
Therefore n = 1184
(i) Let E₁ be the event of getting a centum scorer in Mathematics.
Therefore n(E₁) = 233, That is, r₁ = 233
\(P({ E })_{ 1 }=\frac { { r }_{ 2 } }{ n } =\frac { 233 }{ 1184 } \)
(ii) Let E₂ be the event of getting a centum scorer in Science.
Therefore n(E₂ ) = 106, That is, r₂ = 106
\(P({ E })_{ 2 }=\frac { { r }_{ 2 } }{ n } =\frac {106 }{ 1184 } \)
P(E^'₂) = 1− P(E₂)
\(=1-\frac { 106 }{ 1184 } \)
\(=\frac { 1078 }{ 1184 } \)

Sample space S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting a number greater than 4
E = {5, 6}
\(P(E)=\frac { Number\ of\ favourable\ outcomes }{ Total\ number\ of\ outcomes } \)
\(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =0.333...\)

The side of the cube (a) = 5 cm
Total Surface Area = 6a² = 6(5²) = 150 sq. cm
Lateral Surface Area = 4a² = 4(5²) = 100 sq. cm

(i) We know that cosec A =  sec(90⁰ A)
sec(90⁰ - A) sec(34⁰)
90⁰ - A =34⁰
We get A = 90° − 34°
A = 56⁰
(ii) We know that tan B = cot(90⁰ - B)
cot(90⁰ - B) = cot 47⁰
90⁰ - B = 47⁰
We get B = 90° − 47°
B = 43⁰

Given x – 2y = 7 …(1)
2x + 3y = 7 …(2)
When x = 5, y = −1 we get
From (1) x – 2y = 5 – 2(−1) = 5 + 2 = 7 which is RHS of (1)
From (2) 2x + 3y = 2(5) + 3(−1) = 10−3 = 7 which is RHS of (2)
Thus the values x = 5, y = −1 satisfy both (1) and (2) simultaneously. Therefore (5,−1) is a solution of the given equations.

In the given right angled triangle, note that for the given angle \(\theta \), PR is the ‘opposite’ side and PQ is the ‘adjacent’ side.
\(sin\theta =\frac { opposite\ side }{ hypotenuse } =\frac { PR }{ QR } =\frac { 35 }{ 37 } \)
\(cos\theta =\frac { adjacent\ side }{ hypotenuse } =\frac { PQ }{ QR } =\frac { 12 }{ 37 } \)
\(tan\theta =\frac { opposite\ side }{ adjacent\ side } =\frac { PR }{ PQ } =\frac { 35 }{ 12 } \)
It is enough to leave the ratios as fractions. In case, if you want to simplify each ratio neatly in a terminating decimal form, you may opt for it, but that is not obligatory.

\(A\cap B\) = {5,7}
\(B\cap A\)= {5,7}
From (3) and (4) we get, \(A\cap B=B\cap C\)
It is verifield that insersection of sets is commutative.

Given, A = {2,5,6,7} and B = {3,5,7,8}
\(A\cup B\) = {2,3,5,6,7,8}  ...........(1)
\(B\cup A\) = {2,3,5,6,7,8} ..............(2)

a^x = \(\underline { { a }^{ x } } \)
a^x+y = \(\underline { { a }^{ x } } \). ay
\(\therefore\) a^x+y+z = \(\underline { { a }^{ x } } \). a^y. a^z    

Let p (x) = x³ + 6x² - 7x - 60
By factor theorem (x+4) is a factor of p(-4)  = 0
p(-4) = (-4)³ + 6(-4)² -7(-4) -60 = -64 + 96 + 28 - 60 = 0
Therefore,(x+4) is a factor of x³ + 6x² + (-7x) -60

(0.00000004)³=\(\left( 4.0\times 10^{ -8 } \right) ^{ 3 }=\left( 4.0 \right) ^{ 3 }\times \left( { 10 }^{ -8 } \right) ^{ 3 }\)
=  \(64 \times 10^{-24}=6.4 \times 10 \times 10^{-24}=6.4 \times 10^{-23}\)

\(3\sqrt { 162 } \times 7\sqrt { 50 } \times 6\sqrt { 98 } \) = \(\left( 3\times 9\sqrt { 2 } \times 7\times 5\sqrt { 2 } \times 6\times 7\sqrt { 2 } \right) \)
= \(3 \times 7 \times 6 \times 9 \times 5 \times 7 \times \sqrt{2} \times \sqrt{2} \times \sqrt{2}\)  = 79380\(\sqrt { 2 } \)

The data given is11,15,17, x+1, 19, x-2, 3
\(\bar{x}=4\)
\(\bar{x}=\cfrac{\Sigma x}{n}=\cfrac{11+15+17+x+1+19+x-2+3}{7}\)
=\(\cfrac{66+2x-2}{7}=\cfrac{64+2x}{7}\)
\(\cfrac{64+2x}{7}=14\)
64 + 2x = 98
2x =  98 - 64 = 34
\(x=\cfrac{34}{2}=17\)
The data 11, 15, 17, x+1, 19, x-2, 3
= 11, 15,17, 18, 19, 15, 3

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)

\(\bar { x } =20.2\) 
\( \bar{x}=\cfrac{\Sigma fx}{\Sigma f}\)
\(=\cfrac { 10\times 6+15\times 8+20p+25\times 10+30\times 6 }{ 6+8+p+10+6 } \)
\(20.2=\cfrac { 60+120+20p+250+180 }{ 30+p } \)
(30+p)20.2 = 610+20 p
606+20.2p = 610+20 p
20.2 p-20 p = 610-606 = 4
0.2 p = 4
\(\Rightarrow \) \(p=\cfrac { 4\times 10 }{ 0.2\times 10 } =\cfrac { 40 }{ 2 } =20\)

(i) \( \sqrt{3} =3^{\frac{1}{2}} =3^{\frac{6}{12}} =\sqrt[12]{3^{6}} =\sqrt[12]{729} \)
(ii) \( \sqrt[4]{3} =3^{\frac{1}{4}} =3^{\frac{3}{12}} =\sqrt[12]{3^{3}} =\sqrt[12]{27} \)
(iii) \( \sqrt[3]{3} =3^{\frac{1}{3}} =3^{\frac{4}{12}} =\sqrt[12]{3^{4}} =\sqrt[12]{81} \)
The last row has surds of same order.

(i) 
(ii) 
(iii) 
(iv) 

In rectangle ABCD, P is the mid-point of AB.
Q is the mid-point of BC. R is the mid-point of CD
S is the mid-point of DA. AC is the diagonal
Now in ΔABC,
PQ = \(\frac { 1 }{ 2 } \)AC and PQ || AC        .......(1)
Similarly in ΔACD,
SR = \(\frac { 1 }{ 2 } \) AC and SR || AC        .......(2)
From (1) and (2) we get,
PQ = SR and PQ II SR
Similarly by joining BD, we have
PS = QR and PS II QR
i.e. Both pairs of opposite sides of quadrilateral PQRS are equal and parallel.
∴ PQRS is a parallelogram.

(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.

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.

(3, 5)

A (- 6, 4), B (- 3, -3) and C (2, 2)

AB = \(\sqrt { (5+7)^{ 2 }+(10+3)^{ 2 } } \)
= \(\sqrt { (12)^{ 2 }+(13)^{ 2 } } =\sqrt { 144+169 } =\sqrt { 313 } \)
BC = \(\sqrt { (15-5)^{ 2 }+(8-10)^{ 2 } } \)
= \(\sqrt { 10^{ 2 }+(-2)^{ 2 } } =\sqrt { 100+4 } =\sqrt { 104 } \)
CD = \(\sqrt { (3-15)^{ 2 }+(-5-8)^{ 2 } } \)
= \(\sqrt { (-12)^{ 2 }+(-13)^{ 2 } } =\sqrt { 144+169 } =\sqrt { 313 } \)
AD =\(\sqrt { (3+7)^{ 2 }+(-5+3)^{ 2 } } \)
= \(\sqrt { (10)^{ 2 }+(-2)^{ 2 } } =\sqrt { 100+4 } =\sqrt { 104 } \)
AB = CD = \(\sqrt { 313 } \) and BC = AB = \(\sqrt { 104 } \) (Opposite sides are equal)
∴  ABCD is a parallelogram.

0.86 = \({8\over 10}+{6\over 100}={80+6\over 100}\)
        = \(86\over 100\) (or) \(43\over 50\)

p(x) = 6x² - 7x + 2 ; q(x) = 6x³ - 7x + 15
p(x) + q(x) = 6x² - 7x + 2 +6x³ - 7x + 15
= 6x³ + 6x²-7x + 2+ 15
= 6x³+6x²–14x+17
The degree of the polynomial is 3

In the given diagram
AB = CD (Given)
BD is common.
\(\angle ABD=\angle BDC\) (alternate angles)
(Since AC and CD are parallel and BD is the transversal)
By SAS congruency
\(\therefore\triangle ABD\cong\triangle CDB\)

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.