Additive inverse of \(\frac { 2 }{ 3 } \) is \(-\frac { 2 }{ 3 } \)

Let I be x. Then according to the question,
\(\frac { x-8 }{ 12 } \)=11
=> x - 8 = 12 x 11
=> x - 8 = 132
=> x = 132 + 8
=> x = 140
∴There are 11 players in a full team of cricket.
Hence, I am the number 140.

We have [(3²)³ x 3⁶] \(\div\) 3⁶
= [3^2x3 X 3⁶] \(\div\) 3⁶              \([\because (a^m)^n =a^{mn}]\)
= [3⁶ X 3⁶]\(\div\) 3⁶
= [3^{6 + 6}] \(\div\) 3^{6                           \([\because a^m\times a^n =a^{m+n}]\)}
= 3¹² \(\div\) 3⁶
= 3^{12 - 6                                       \([\because a^m\div a^n =a^{m-n}]\)}
= 3⁶
Thus, [(3²)³ X 3⁶] \(\div\) 3⁶ = 3⁶.

Steps of Construction:
Step I: Firstly, we draw a rough sketch of triangle with given measures marked on it.

Step I: Draw a line segment AB of length 4.5 cm,
Step II: Draw an angle of 90° on point Band produce it to Y.
Step III: With B as centre, draw an arc of 4.5 cm which intersects ray BY to C
Step IV: Join CA.

Thus, ΔABC is the required right angled isosceles triangle.

\(\therefore\) Radius of the circular pipe = 4 cm [given]
\(\therefore\) Length of the cord wrapped around the circular pipe = Circumference of the circular pipe
                                = 2\(\pi\)r = 2 x 3.14 x 4= 25.12 cm
\(\therefore\) Length of the cord wrapped around the square box = Perimeter of the square box
                               = 4 x Side = 4 x 4 = 16 cm
Since Pragya wrapped the square box by the cord whose length is equal to the circumference of pipe and 25.12 cm > 16 cm.
Hence, (25.12 - 16) cm i.e. 9.12 cm cord is left with Pragya.

Given, radius of circular card sheet = 14 cm
\(\therefore\) Area of circular card sheet = \(\pi\) x (Radius)²
                                          = \(\frac{22}{7}\times (14)^2=\frac{22}{7}\times 14\times 14\) = 616 cm²
Area of circle of radius 3.5 cm = \(\frac{22}{7}\times\) (3.5)²
                    = \(\frac{22}{7}\times 3.5\times 3.5=\frac{269.5}{7}\) = 38.5 cm²
\(\therefore\)  Area of two circles = 2 x 38.5 = 77 cm²
Now, length of rectangle, L= 3 cm
and breadth of rectangle, b = 1cm
\(\therefore\) Area of the rectangle = Length X Breadth = 3 x 1 = 3 cm²
Now, area of remaining sheet = Area of circular card sheet - (Area of two circles + Area of rectangle)
                                           = 616 - (77 + 3) = 616 - 80 = 536 cm²
Hence, the area of remaining card sheet is 536 cm².

Given, radius of a circular sheet (outer circle), R = 4 cm and radius of removed circular sheet (inner circle), r = 3 cm.
We know that, Area of a circle = \(\pi\) x (Radius)2
\(\therefore\) Area of circular sheet of radius 4 cm = \(\pi\) x (4)²
                                                         = 3.14 x 16 = 50.24 cm²
and area of circular sheet of radius 3 cm = \(\pi\) x (3)²
                                                           = 3.14 x 9 = 28.26 cm²
\(\therefore\) Area of remaining sheet= Area of circular sheet of radius 4 cm - Area of circular sheet of radius 3 cm
                                      = 50.24 - 28.26 = 21.98 cm²
Hence, the area of the remaining sheet is 21.98 cm².

Given rational numbers are \(\frac { 1 }{ 3 } \) and  \(\frac { 1 }{ 4 } \).
LCM of 3 and 4 is 12,
\(\frac { 1\times 4 }{ 3\times 4 } =\frac { 4 }{ 12 } \) and \(\frac { 1\times 3 }{ 4\times 3 } =\frac { 3 }{ 12 } \)
Now, \(\frac { 4\times 10 }{ 12\times 10 } =\frac { 40 }{ 120 } \) and \(\frac { 3\times 10 }{ 12\times 10 } =\frac { 30 }{ 120 } \)
\(\frac { 30 }{ 120 } <\frac { 31 }{ 120 } <\frac { 32 }{ 120 } <\frac { 33 }{ 120 } <\frac { 34 }{ 120 } <\frac { 35 }{ 120 } <.......<\frac { 40 }{ 120 } \)
Hence, 5 rational numbers between \(\frac { 1 }{ 3 } \) and \(\frac { 1 }{ 4 } \) are
\(\frac { 31 }{ 120 } ,\frac { 32 }{ 120 } ,\frac { 33 }{ 120 } ,\frac { 34 }{ 120 } ,\frac { 35 }{ 120 } .\)

-13

We have HCF of 30 and 45 = 15
\(\therefore\) \(\frac{-30}{45}=\frac{(-30)\div15}{45\div15}=\frac{-2}{3}\)

Since, the mass of Uranus = 86,800,000,000,000,000,000,000,000 kg
= 8.68 X 1025 kg [Standard form]

\(\therefore\)  Required sum = 3m - 4n - 3mn - 3

(2²⁰ \(\div\) 2¹⁵) x 2³ = (2^20-15)x2³
=2⁵ x2³ =2⁵⁺³ =2⁸

\(\therefore\)   Area of a rectangle = Length x Width
\(\Rightarrow\)   600 = 30 x Width   \(\Rightarrow\) Width = \(\frac{600}{30}=\frac{60}{3}\) = 20 m.

Given rational number are \(\frac { 2 }{ 3 } \) and \(\frac { 4 }{ 7 } \) . Now, to add \(\frac { 2 }{ 3 } \) and \(\frac { 4 }{ 7 } \) . Now to add \(\frac { 2 }{ 3 } \) and \(\frac { 4 }{ 7 } \).
Firstly, find LCM of 3 and 7, which is 21.
\(\therefore \) \(\frac { 2 }{ 3 } =\frac { 2 }{ 3 } \times \frac { 4 }{ 7 } =\frac { 14 }{ 21 } \) and \(\frac { 4 }{ 7 } =\frac { 4 }{ 7 } \times \frac { 3 }{ 3 } =\frac { 12 }{ 21 } \) 
Thus, \(\frac { 2 }{ 3 } +\frac { 4 }{ 7 } =\frac { 14 }{ 21 } +\frac { 12 }{ 21 } =\frac { 14+12 }{ 21 } =\frac { 26 }{ 21 } \).

We have, 729 = 3 x 3 x 3 x 3 x 3 x 3 = 3⁶

Here, base= 3 and exponent = 6, since 3 repeated 6 times.

In the given parallelogram ABCD, DC = Base = 8 cm and AE = Height = 6cm
\(\therefore\) Area of the parallelogram = DC x AE = 8 X 6 cm² = 48 cm².

Given, side = 2.5 m, Area of a square = (Side)² \(\Rightarrow\) Area of square = (2.5)² = 2.5 x 2.5 = 6.25 m²
So, area of square is 6.25 m².

Here, Length of the rectangular park = 70 m
Breadth of the rectangular park = 45 m
The cross paths are shown by EFGH and PQRS in the figure.
Now, PQ =3 m and PS = 45 m
EH = 3 m and EF = 70 m
KL = 3 m and KN = 3 m
Now, area of the path = [Area of rectangle PQRS] + [Area of rectangle EFGH]- [Area of square KLMN]
= [PS x PQ] + [EF x EH] - [KL x KN]
= [45 x 3] m² + [70 x 3] m² - [3 x 3] m²
= 135 m² + 210 m² - 9 m² = 336 m²
Now, cost of constructing the path = Rs. 110 x 336 = Rs. 36960.
Note: While finding the area of cross roads, the area of the middle square (here, KLMN) is taken twice
which is to be subtracted once.

Let ABCD be the rectangular park of sides 40 m and 25 m, and the shaded region represents the path 2.5 m wide.
Now, PQ = (40 + 2.5 + 2.5) m = 45 m
PS = (25 + 2.5 + 2.5) m = 30 m
\(\therefore\)Area of rectangle ABCD = I x b = 40 m x 25 m
= 1000 m2
Area of rectangle PQRS = 45 m x 30 m = 1350 m²
So, Area of the path = [Area of rectangle PQRS] - [Area of rectangle ABCD]
= 1350 m² - 1000 m² = 350 m²

A = 2 + 4x + 8x²
B = -3-5x + x²
C = 1 + 3x-7x²
A+B+C=?
A + B + C = (2 + 4x + 8x²) + (- 3 - 5x + x²)+ (1 + 3x-7x²)
= 2 + 4x + 8x² - 3 - 5x + x² + 1 + 3x -7x²
= x²(8 + 1 - 7) + x( 4 - 5 + 3) + 2 - 3 + 1
=2x²+2x+0
=2x²+2x
= 2x(x + 1).

(a) \(\frac{12^4\times 9^3\times 4}{6^3\times 8^2\times 27} =\frac{(3\times 2^2)^4\times (3^2)^3\times 2^2}{(2\times 3)^3\times (2^3)^2\times 3^3}\)
\(=\frac{3^4\times 2^8\times 3^6\times 2^2}{2^3\times3^3\times2^6\times3^3}\)
\(=\frac{2^{8+2}\times 3^{6+4}}{2^{6+3}\times 3^{3+3}}\)
\(=\frac{2^{10}\times 3^{10}}{2^9\times 3^6}=2^{10-9}\times 3^{10-6}\)
= 2 x 3⁴ = 2 x 81 = 162.
(b) 2³ x a³ x 5a⁴
= 8 x a³ x 5 x a⁴
= 8 x 5 x a³ x a⁴
= 40 x a³ + 4
= 40 x a⁷
= 40a⁷.

Speed of light = 3 x 10⁸ m/s
and time = 8 min. = 8 x 60 = 480 seconds
∴ Distance = Speed X Time
= 3 x 10⁸ x 480 m
= 3 x 10⁸ x 48 x 101m
= 144 x 10⁹m
\(={144\over100}\times10^9\times100m\)
=1.44 X 10⁹⁺² m
= 1·44 X 10¹¹m

\((a)\left(-{4\over 3}\right)\times\left(-{4\over 3}\right)\times\left(-{4\over 3}\right)\times\left(-{4\over 3}\right)\times\left(-{4\over 3}\right)\)
\(={(-4)\times(-4)\times(-4)\times(-4)\times(-4)\over 3\times3\times3\times3\times3}\)
\(={(-4)^5\over (3)^5}=\left(-4\over 3\right)^5\)
\((b)\left(-{8\over 3}\right)\times\left(-{8\over 3}\right)\times\left(-{8\over 3}\right)\times\left(-{8\over 3}\right)\times\left(-{8\over 3}\right)\)
\(={(-8)\times(-8)\times(-8)\times(-8)\times(-8)\over 3\times3\times3\times3\times3}\)
\(={(-8)^5\over (3)^5}=\left(-8\over 3\right)^5\)

(a) (6^-1 - 8^-1)^-1 + (2^-1 - 3^-1)^-1
\(=(\frac{1}{6}-\frac{1}{8})^{-1}+(\frac{1}{2}-\frac{1}{3})^{-1}\)
\(=(\frac{4-3}{24})^{-1}+(\frac{3-2}{6})^{-1}\)
\((\frac{1}{24})^{-1}+(\frac{1}{6})^{-1}\)
= 24 + 6 = 30
(b) \({ \left\{ { 6 }^{ -1 }+{ \left( \frac { 3 }{ 2 } \right) }^{ -1 } \right\} }^{ -1 }\)
\(=(\frac{1}{6}+\frac{2}{3})^{-1}\)
\(=(\frac{1+4}{6})^{-1}=(\frac{5}{6})^{-1}\)
\(=\frac{6}{5}\)

Sequence is \(\frac { -3 }{ 5 } ,\frac { 7 }{ -10 } ,\frac { -5 }{ 6 } \\ \)
L.C.M.of 5, 10 and 6=30
\(\Rightarrow \frac { -3 }{ 5 } ,\frac { 7 }{ -10 } ,\frac { -5 }{ 6 } \)
\(\Rightarrow -\frac { 3\times 6 }{ 5\times 6 } ,\frac { 7\times 3 }{ -10\times 3 } ,-\frac { 5\times 5 }{ 6\times 5 } \)
\(\Rightarrow -\frac { 18 }{ 30 } ,\frac { 21 }{ 30 } ,\frac { 25 }{ 30 } \)
Since \(-\frac { 25 }{ 30 } <\frac { 7 }{ -10 } <\frac { -3 }{ 5 } \)
Hence sequence in ascending order is
\(\frac { -5 }{ 6 } <\frac { 7 }{ -10 } <\frac { -3 }{ 5 } \).