This can be written as (3x³ − 2x² − 7x + 6)− (x³ − x² + x + 3)
When there is a –ve sign before the brackets, it can be removed by changing the sign of every term inside the bracket.
(3x³ − 2x² − 7x + 6)−(x³ − x² + x + 3) = 3x³ − 2x² − 7x + 6 − x³ + x² −x−3
= (3x³ − x³ )+ (−2x² + x² )+ (−7x − x)+ (6 − 3)
= x³ (3 −1)+ x² (−2 +1)+ x (−7 −1)+ (6 − 3)
= 2x³ − x² − 8x + 3
Hence, the degree of the expression is 3.

In −4xy²z³, the sum of powers of x, y and z is 6 (that is, 1 + 2 + 3). Thus, the degree of the expression is 6.

In x⁵, the exponent is 5. Thus, the degree of the expression is 5.

\(\frac { { 6 }^{ 4 } }{ { 6 }^{ 0 } } \) = 6^4-0 = 6⁴ (or) \(\frac { { 6 }^{ 4 } }{ { 6 }^{ 0 } } \)=\(\frac { { 6 }^{ 4 } }{ 1 } \) = 6⁴  [since 6⁰ = 1]

\(\frac { { 10 }^{ 8 } }{ { 10 }^{ 6 } } \) = 10^8-6  = 10²

Exterior angle = sum of two interior opposite angles
6y + 2 = 26⁰ + 36⁰
6y + 2 = 62⁰
Subtract 2 from both sides
6y = 62 − 2
6y = 60⁰
\(\frac { 6y }{ 6 } =\frac { 60 }{ 6 } \) [on both sides, divide by 6]
y = 10⁰
ㄥUWY = 6y + 2 = 6(10) + 2 = 62⁰ .
We can conclude that ㄥXWV = ㄥUWY, because they are vertically opposite angles as well as exterior angles.

Exterior angle = sum of two interior opposite angles
135⁰ = z + 40⁰
Subtract 40⁰ on both sides
135⁰ − 40⁰ = z + 40⁰ − 40⁰
z = 95⁰
 

Let, ㄥSRQ = x
Exterior angle = sum of two interior opposite angles
x = 38⁰ + 44⁰ = 82⁰

Given, ㄥSUT = 70⁰
ㄥUST = ㄥSTU = x [Angles opposite to equal sides]
ㄥSUT + ㄥUST + ㄥSTU = 180⁰
70⁰ + x + x = 180⁰
70⁰ + 2x = 180⁰
2x = 180⁰ – 70⁰
2x = 110⁰
x = \(\frac { { 110 }^{ 0 } }{ 2 } \) = 55°

(i) Given angles 80⁰, 70⁰, 50⁰
Sum of the angles = 80⁰ + 70⁰ + 50⁰ = 200⁰ ≠ 180⁰
The given angles cannot form a triangle.
(ii) Given angles 56⁰, 64⁰, 60⁰
Sum of the angles = 56⁰+ 64⁰+ 60⁰ = 180⁰
The given angles can form a triangle.

243 = 3 x 3 x 3 x 3 x 3 = 3⁵

1000 = 10 x10 x10 = 10³

From the dimensions of the ground we have,
L = 30 m; B = 19 m; l = 27 m; b = 16 m
Area of the kho kho ground = L × B
= 30 x 19
= 570 m²
Area of the play field = l x b
= 27 x 16
= 432 m²
Area of the lobby = Area of Kho-Kho ground − Area of the play field
= 570 − 432
= 148 m²

Here L = 23 cm B = 11 cm
Area of the chart = L x B
= 23 x 11
= 253 cm²
l = L − 2w = 23 − 2(3) = 23 − 6 = 17 cm
b = B − 2w = 11− 2(3) = 11− 6 = 5 cm
Area of the picture 17 x 5 = 85 cm²
Therefore, the area of the margin = 253 − 85
  = 168 cm².

The radius of the outer circle, R = 10 m
The radius of the inner circle, r = 3 m
The area of the circular path =Area of outer circle −Area of inner circle
= πR² − πr²
= π(R² − r² ) sq.units
= \(\frac { 22 }{ 7 } \) x (10² - 3²)
= \(\frac { 22 }{ 7 } \) x (10 x 10)-(3 x 3))
= \(\frac { 22 }{ 7 } \)x (100 - 9)
= \(\frac { 22 }{ 7 } \) x 91
= 286 m²

In this shape, we have to calculate the circumference of semicircle on each side of the rectangle. The outer boundary of this figure is made up of semicircles of two different sizes. Diameters of each of the semicircles are 7 cm and 14 cm. We know that the circumference of the circle = πd units.
Circumference of the semicircular part = \(\frac { 1 }{ 2 } \)πd units
Hence, the circumference of the semicircle having diameter 7 cm is,
= \(\frac { 1 }{ 2 } \) x \(\frac { 22 }{ 7 } \) x 7 = 11 cm
Circumference of the pair of semicircular parts (II and IV) = 2 x 11 = 22 cm
Similarly, circumference of the semicircle having diameter 14 cm is,
= \(\frac { 1 }{ 2 } \) x \(\frac { 22 }{ 7 } \) x 14 = 22 cm
Circumference of the pair of semicircular parts (I and III) = 2 x 22 = 44 cm.
Perimeter of the given shape = 22 + 44 = 66 cm.

The distance covered in one rotation
=  the circumference of the circle
= 2πr units
= 2 x \(\frac { 22 }{ 7 } \) x 77
= 2 x 22 x 11
= 484 cm
The distance covered in one rotation = 484 cm
The distance covered in 35 rotations = 484 x 35
= 16940 cm

Circumference of the circle, C = 2πr units
The circumference of the given circle = 132 m
\(\frac { C }{ 2\pi } =r\)
r = \(\frac { 132 }{ 2\times \frac { 22 }{ 7 } } \)
= \(\frac { 132 }{ 2 } \times \frac { 7 }{ 22 } \)
= 21 m
d = 2r
= 2 x 21
= 42 m

Given, d = 6 cm, d = 2r = 6 cm, r = 3 cm
Circumference of a circle = 2πr units
 = 2π × 3
 = 18.84
The circumference is 18.84 cm.

The whole number parts of the three decimal numbers are equal.
The digits at tenths place are also equal.
The digits at hundredths place are 0, 1 and 5. Here 0 < 1 < 5
Therefore, the ascending order is 4.90, 4.91, 4.95.

Compare 99.56 and 99.65
Here the whole number parts of the given two numbers are equal.
Comparing the digits at tenths place, we get 5 < 6.
Therefore, 99.56 < 99.65

40 + 6 + \(\frac { 7 }{ 10 } +\frac { 2 }{ 100 } +\frac { 6 }{ 1000 } \) = 46.726

537 m = \(\frac { 537 }{ 1000 } \)km = 0.537 km

3 + \(\frac { 5 }{ 10 } +\frac { 3 }{ 100 } +\frac { 4 }{ 1000 } \) = 3.534