(4m² + 3n) − (3m+ 9n² ) − (3m² − 6n² ) + (5m−n)
  = 4m² + 3n − 3m − 9n² − 3m² + 6n² + 5m − n
  = (4m² − 3m² ) + (3n − n) + (−3m+ 5m) + (−9n² + 6n²)
  = m² + 2n + 2m − 3n²
Hence, the degree of the expression is 2.

This can be written as (4x² + 3xy + 9y² ) + (2x² − 9xy + 6y²)
Let us group the like terms, thus we have
(4x² + 2x² )+ (3xy − 9xy)+ (9y² + 6y² ) = x² (4 + 2)+ xy (3 − 9) + y² (9 + 6)
  = 6x² − 6xy +15y²
Thus, the degree of the expression is 2.

The terms of the given expression are 12xyz, 3x³ y²z, z⁸
Degree of each of the terms : 3, 6, 8
Terms with highest degree : z⁸.
Therefore, degree of the expression is 8.

In −3p³q², the sum of powers of p and q is 5 (that is, 3 + 2). Thus, the degree of the expression is 5.

In the above pattern of polygons, in the first shape (x = 1), we get a closed shape called a triangle. Similarly the second shape (x = 2 ) gives a four sided polygon and the third shape (x = 3) is a five sided polygon and continuing in the same way two more shapes are formed. If the number of match sticks required to form each of the shapes is taken as y, then the values of x and y are tabulated as given below.

Observe the table and express y in terms of x as below:
when x = 1, y = 3 = 1 + 2
when x = 2 , y = 4 = 2 + 2
when x = 3, y = 5 = 3 + 2
when x = 4 , y = 6 = 4 + 2
when x = 5, y = 7 = 5 + 2
Hence, each of the values of y which we get from the table is 2 more than x. That is
y = x + 2 .
Therefore,
6^th shape (x = 6) will have y = 8 = 6 + 2 (8 match sticks)
7^th shape (x = 7) will have y = 9 = 7 + 2 (9 match sticks)
8^th shape (x = 8) will have y = 10 = 8 + 2 (10 match sticks)
We clearly see that the next three shapes will require 8, 9 and 10 matchsticks.

6¹³ x 48¹³ ÷ 12¹³ = 6¹³ x (48¹³ ÷ 12¹³) [BIDMAS]
= 6¹³x\(\left( \frac { 48 }{ 12 } \right) ^{ 13 }\)    [since \(\frac { { a }^{ m } }{ { b }^{ m } } =\left( \frac { a }{ b } \right) ^{ m }\)]
= 6¹³ x 4¹³
= (6 x 4)¹³            [Since, a^m x b^m = (a x b)^m]
= (24)¹³

4³ x 2³ x 5³ = (4 x 2 x 5)³ = 40³   [Rule extended for 3 numbers]

(2⁶)² x (2⁴)⁷ = 2^{6 x 2} x 2^{4 x 7}    [since (a^m)ⁿ = a^mxn]
= 2¹² x 2²⁸
=2^{12 + 28} = 2⁴⁰    [since a^m x aⁿ = a^m+n]

25 x 32 x 625 x 64 = (5 x 5) x (2 x 2 x 2 x 2 x 2) x (5 x 5 x 5 x 5) x (2 x 2 x 2 x 2 x 2 x 2)
= 5² x 2⁵ x 5⁴ x 2⁶
= (5² x 5⁴ ) x (2⁵ x 2⁶ )    [grouping exponential numbers with the same base]
= 5²⁺⁴ x 2⁵⁺⁶ = 5⁶ x 2¹¹

Exterior angle = sum of two interior opposite angles
128⁰ = x + x
128 = 2x
\(\frac { 128 }{ 2 } =\frac { 2x }{ 2 } \) [on both sides, divide by 2]
x = 64⁰

Let, ㄥNMO = y
Exterior angle = sum of two interior opposite angles
y = 62⁰ + 31⁰
= 93⁰ 

Given angles are 65⁰ and 35⁰.
Let the third angle be x
65⁰ + 35⁰ + x = 180⁰
100⁰ + x = 180°
x = 180⁰ – 100⁰
x = 80⁰

Let ㄥA = x
We know that, ㄥA + ㄥB + ㄥC = 180⁰ (angle sum property)
x + 44⁰ + 31⁰ = 180⁰
x + 75⁰ = 180⁰
x = 180⁰ – 75⁰
x = 105⁰

13² = 13 x13 = 169

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁹

Dividing by 3, we get

729 = 3 x 3 x 3 x 3 x 3 x 3 = 3⁶
Also, 729 = 9 × 9 × 9 = 9³

Here, l = 9 m, b = 7 m
Area of the Room = l x b
= 9 x 7
= 63 m²
L = l + 2w = 8 + 2(3) = 8 + 6 = 14 m
B = b + 2w = 5 + 2(3) = 5 + 6 = 11m
Area of the room including verandah = L × B
= 14 x 11
= 154 m²
The area of the verandah = Area of the room including verandah − Area of the room
= 154 − 63
= 91m²
The cost of cementing the floor for 1sq.m = Rs.15
Therefore, the cost of cementing the floor of the verandah = 91 x 15 = Rs.1365.

Here, L = 10 m B = 8 m
Area of the floor = L x B
= 10 x 8
= 80 m²
Area of the carpet = l x b
= 7 x 5
= 35 m²
Therefore, the total area of the floor not covered by the carpet = 80 − 35
= 45 m²

The radius of the inner circle r = 21m
The path is around the inner circle.
Therefore, the radius of the outer circle, R = 21 + 14 = 35 m
The area of the circular path = π(R² − r² ) sq.units
= \(\frac { 22 }{ 7 } \)x(35² - 21²)
= \(\frac { 22 }{ 7 } \) x (35 x 35) - (21 x 21)
= \(\frac { 22 }{ 7 } \) x (1225 - 441)
= \(\frac { 22 }{ 7 } \) x 784
= 22 x 112 = 2464 m²

To find the perimeter of the quadrant disc, we need to find the circumference of quadrant shape.
Given that radius (r) = 14 cm.
We know that the circumference of circle = 2πr units.
So, the circumference of the quadrant arc = \(\frac { 1 }{ 4 } \) x 2πr
 =\(\frac { \pi r }{ 2 } \)
 =\(\frac { 22 }{ 7 } \times \frac { 14 }{ 2 } \)
 = 22 cm
Given, the radius of the circle = 14 cm
Thus, perimeter of required quadrant shaped disc = 14 +14 + 22
 = 50 cm.

The radius of the poultry farm is = 420 m
The length of the barbed wire for fencing the poultry farm is equal to the circumference of the circle.
We know that the circumference of the circle = 2πr units
= 2 x \(\frac { 22 }{ 7 } \) x 420
= 2 x 22 x 60
The length of the barbed wire to fence the poultry farm = 2640 m
The cost of fencing the poultry farm at the rate of Rs.12 per metre = 2640 x 12
= Rs.31,680
Given that he has Rs.30,000 with him.
The excess amount required = Rs.31,680 − Rs.30,000 = Rs.1,680.

Here the distance travelled by the tip of the seconds hand of a clock in 1 minute is the circumference of the circle and the length of the seconds hand is the radius of the circle. So, r = 56 mm
Circumference of the circle, C = 2πr units
= 2 x \(\frac { 22 }{ 7 } \) x 56
= 2 x 22 x 8
= 352 mm
Therefore, distance travelled by the tip of the seconds hand of a clock in 1 minute is 352 mm.

Radius of circular disc (r) = 14 cm
Circumference of the disc = 2πr units
= 2 x \(\frac { 22 }{ 7 } \) x 14
= 88 cm

Compare 0.05 and 0.09
 By using the steps given above, the integral parts and tenths places are equal. By comparing the hundredth place, we get 5 < 9. Therefore, 0.05 < 0.09.
 So far we discussed about the comparison of two decimal numbers. Extending this, we can arrange the given decimal numbers in ascending or descending order.

(i) The place value of 7 in 0.37 is Hundredth.
(ii) The place value of 7 in 2.73 is Tenth.
(iii) The place value of 7 in 28.271 is Hundredth.