X⁴ - 4X³y³ + 2y⁴

Given, line segment AB = 6 cm. If perpendicular bisector constructeg on AB. Then, the length of equal parts will be \(\frac{6}{2}=3\) cm.
∴ Perpendicular bisector divide the line in two equal parts.

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

Given, base of a parallelogram = 5 cm
and height of a parallelogram = 4.8 cm
\(\therefore\) Area of a parallelogram = Base x Height
                                     = 5 x 4.8 = 24 cm².

100 m.

\(\frac { -3 }{ 65 } \)

114

1

\(\therefore\) Breadth of outer figure = 12 cm
Breadth of inner figure = 12 - 4 - 4 = 4 cm
Length of outer figure = 18 cm
Length of inner figure = 18 - 4 = 14 cm
So, area of inner rectangular figure
            = 14 x 4 = 56 cm².

Here, if we add an expression to x² + xy + Y², then we get sum 2x² + 3xy. So if, we subtract x2 + xy+ y² from 2x² + 3xy, then we will get the expression.
\(\therefore\) Required expression  = (2x² + 3xy) - (x² +xy +y²)
\(2x^2+3xy\\\ \ x^2+xy+y^2\\-\ \ -\ \ \ \ \ -\\\\\_\_\_\_\_\_\_\_\_\_\_\_\\\ \ x^2+2xy-y^2\\\_\_\_\_\_\_\_\_\_\_\_\_\)
Hence, x²+2xy-y² should be added to x²+xy+y² to obtain 2x²+3xy.

Given, \(\Delta\)ABC is right angled at A, AB = 5 cm, BC = 13 cm and AC = 12 cm.
On taking AC as base and AB as height, we get
Area of \(\Delta\)ABC = \(\frac{1}{2}\times\) Base x Height
                      = \(\frac{1}{2}\times\) 12 x 5 = \(\frac{1}{2}\times\) 60 = 30 cm².
Also, AD is perpendicular to BC.
On taking BC as base and AD as height, we get
Area of \(\Delta\)ABC = \(\frac{1}{2}\times\) BC x AD
\(\Rightarrow\)   \(\frac{1}{2}\times\) BC x AD = 30
\(\Rightarrow\)   \(\frac{1}{2}\times\) 13 x AD = 30
\(\therefore\)   AD = \(\frac{30\times 2}{13}=\frac{60}{13}=4\frac{8}{13}\) cm.
Hence, the length of AD is \(4\frac{8}{13}\) cm.

\(\frac { 1 }{ 3 } \)

\(4\frac { 4 }{ 5 } \div \frac { 24 }{ 5 } +8\frac { 1 }{ 4 } =\frac { (5\times 4)+4 }{ 5 } \div \frac { 24 }{ 5 } +\frac { (8\times 4)+1 }{ 4 } \)
\(\frac { 24 }{ 5 } \div \frac { 24 }{ 5 } +\frac { 33 }{ 4 } \)
Since , the reciprocal  of  \(\frac { 24 }{ 5 } \) is \(\frac { 5 }{ 24 } \) 
\(\\ \therefore \quad \frac { 24 }{ 5 } \times \frac { 5 }{ 24 } +\frac { 33 }{ 4 } =1+\frac { 33 }{ 4 } =\frac { 37 }{ 4 } \)

\(\frac { 3 }{ 4 } \times \left( \frac { -2 }{ 3 }  \right) =\frac { 3\times (-2) }{ 4\times (3) } =\frac { -6 }{ 12 } =\)\(-\frac { 1 }{ 2 } \)

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).

Area of one tile (parallelogram shape)
= base x height
= 3 x 4.5
= 13.5 cm²
Area of such 2760 tiles = 2760 x 13.5
= 37,260 cm²
= 3.726 m²
Cost of polishing = 3.726 x 20
= Rs. 74.52.