- 3x³y² +2x²y³
             -3x²y³-5y^{4
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- 3x³y² +x²y³-5y^{4
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Sum =- 3x³y² +x²y³-5y⁴
Now,                 x⁴ +x³y² + x²y³ + y⁴
                            - 3x³y² - x²y³ - 5y⁴
                            (+)     (+)        (+)       
                       __________________ 
Difference =     x⁴ +4x³y²+2x²y³+ 6y⁴
                       ___________________

Let \(\angle EFC=x\)
Then \(\angle AFD=x\)
It is given that CD intersects line AB at F.
Therefore,\(\angle CFB=\angle AFD\) (vertically opposite angles)
So,x=50⁰
But \(\angle EFA=\angle AFD\) which gives \(\angle EFA=50^0\)
Now, \(\angle CFB+\angle EFA+\angle EFC=180^0\) [as AB is a straight line]
\(50^0+50^0+\angle EFC=180^0\)
\(\angle EFC=180^0-100^0\)
Thus,\(\angle EFC=80^0\)

We have, \(\frac { { 3 }^{ 5 }\times { 10 }^{ 5 }\times 25 }{ { 5 }^{ 7 }\times { 6 }^{ 5 } } =\frac { { 3 }^{ 5 }\times { \left( 2\times 5 \right) }^{ 5 }\times { 5 }^{ 2 } }{ { 5 }^{ 7 }\times { 2 }^{ 5 }\times { 3 }^{ 5 } } \)
[ \(\because\)10 = 2x5, 25 = 5x5=5², 6=2x3]
\(=\frac { { 3 }^{ 5 }\times { 2 }^{ 5 }\times { 5 }^{ 5 }\times { 5 }^{ 2 } }{ { 5 }^{ 7 }\times { 2 }^{ 5 }\times { 3 }^{ 5 } } \)    [ \(\because\)(ab)^m = a^m x b^m]
\(=\frac { { 3 }^{ 5-5 }\times { 2 }^{ 5-5 }\times { 5 }^{ 5+2 } }{ { 5 }^{ 7 } } \)  [ \(\because\)a^m\(\div\)aⁿ=a^m-n and a^m x aⁿ = a^m+n]
=3⁰ x 2⁰ x5^7-7 =3⁰ x2⁰ x5⁰
=1x1x1=1                [ \(\because\)a⁰ = 1]

Area of lawn excluding the area of the flower bed = Area of the whole land - Area of the flower bed
                    = \(50-\frac{88}{7}=\frac{350-88}{7}=\frac{262}{7}\) = 37.43 m²

On putting m = 2 in \(\frac { 5m }{ 2 } \) -4, we get
\(\frac { 5m }{ 2 } \) - 4= \(\frac { 5\times 2 }{ 2 } -4=\frac { 10 }{ 2 } \)- 4=5 - 4 = 1

ab² + 2a²b + 4

We have, (-2)³ x (-10)³
= (-2) x (-2) x (-2) x (-10) x (-10) x (-10)
= (-8) x (-1000) = 8000    [ \(\because\)(-1) odd number =-1]
Hence, the value of (-2)³ x (-10)³ is 8000.

We have, 5x²y - 5x² + 3yx² - 3y² +x² - y² + 8xy² - 3y²
= 5x²y + 3yx² - 5x² + x² - x² - 3y² -y² - 3y²  + 8xy   [rearranging terms]
= x²y (5+3) + x² (-5 + 1) +y² (-3 -1 -3) + 8xy²
= 8x²y - 4x² - 7y² + 8xy

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, l||m and t is a transversal.
We know that, the sum of pair of interior angles on the same side of a transversal is supplementary.
\(\therefore\) \(\angle\)z+60° = 180°  \(\Rightarrow\) \(\angle\)z = 180°-60°=120°

\(\angle 2=40°,\angle 3=140°,\angle 4=40°\)

Complement angle of 57° = 90° - 57° = 33°

\(\angle \) AOB and \(\angle \)BOC are adjacent angles because \(\angle \)AOB and \(\angle \)BOC have a common vertex and a common arm OB. They do not have common interior point, as shown in adjoining figure.

In the given figure \(\angle \) 1 and \(\angle \)2 are adjacent angles because they have a common vertex and a common arm but no common interior point.

\(\frac { 9 }{ 32 } \)

-10x³+5x²+15x-4

Given, length of a rectangular piece of land, I = 500 m and breadth of a rectangular piece of land, b = 300 m
Cost of 1m² land = Rs 10000
\(\therefore\) Cost of 150000 m² land = Rs 10000 x 150000
                                       = Rs 1500000000
Hence, the cost of 150000 m² land at the rate of 10000 per m² is Rs 1500000000.

Since, QP||RS, PR is a transversal line.
\(\therefore \quad \angle QPR=\angle PRS=65°\) [alternate angles]
\(\angle a=65°\) \(\left[ \angle \quad PRS=a \right] \)
\(\angle PQS=\angle b=70°\) [corresponding angles]
Hence, \(\angle \)a and \(\angle \)b are 65° and 70°, respectively.

\(\frac { { 5 }^{ -2 }\times { 3 }^{ -3 }\times (125)^{ 2/3 } }{ (27)^{ -2/3 }\times (32)^{ -1/5 } } \)
∵ 125 = (5)³ = 5 x 5 x 5
So, (125)^2/3 =(5)^{3 x 2/3} =5² and 27=(3)³
∴ (27)^-2/3 = {(3)³}^-2/3 =\({ (3 })^{ 3\times \frac { -2 }{ 3 } }\) =(3)²
32 = 2 x 2 x 2 x 2 x 2 = (2)⁵
So, (32)^-1/5 = {(2)⁵}^-1/5 = \({ (2) }^{ 5\times \frac { (-1) }{ 5 } }\)=(2)^-1
Now \(\frac { { 5 }^{ -2 }\times { 3 }^{ -3 }\times { 5 }^{ 2 } }{ (3)^{ -2 }\times (2)^{ -1 } } \)     \(\left[ \because a^{ -m }=\frac { 1 }{ { a }^{ m } } \right] \)
= 5^-2 x 3^-3 x 3² x 2¹ x 5²
=5^-2+2 x 3^-3+2 x 2¹
a^m x aⁿ =a^m+n
=5⁰ x 3^-1 x 2¹ = 1 x \(\frac { 1 }{ 3 } \times 2=\frac { 2 }{ 3 } \).

60°

Substituting a = 3 and b = 2 in
(a) a + b, we get
a + b = 3 + 2 = 5
(b) 7a - 4b,  we get
7a - 4b = 7 x 3 - 4 x 2
= 21 - 8 = 13
(c) a² + 2ab + b², we get
a² + 2ab + b² = 3² + 2 x 3 x 2 + 2²
= 9 + 2 x 6 + 4
= 9 + 12 + 4 = 25.
(d) a3 - b3, we get
a³ - b³ = 3³ - 2³ = 3 x 3 x 3 - 2 x 2 x 2
= 27 -8 = 19.

(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.