Method 1:
\(\frac { -18 }{ -42 } =\frac { -18\div (-2) }{ -42\div (-2) } =\frac { 9\div 3 }{ 21\div 3 } =\frac { 3 }{ 7 } \) (dividing by –2 and 3 successively)
Method 2:
The HCF of 18 and 42 is 6 (Find it). Thus, we can get its standard form by dividing it by 6.
\(\frac { -18 }{ -42 } =\frac { -18\times (-1) }{ -42\times (-1) } =\frac { 18 }{ 42 } =\frac { 18\div 6 }{ 42\div 6 } =\frac { 3 }{ 7 } \)

Comparing (y-5)³ with (a-b)³,we get a = y,b = 5
(a-b)³ = a³- 3a²b + 3ab + 3ab²-b²
(y - 5)³ = (y)³ + 3(y)²(5) + 3(y)(5)² - (-5)³
= (y)³+ 3y²(5) + 3(y)(25)-125
(y - 5)³ = y³-15y² + 75y-125

Guna fixed a single door of 3 feet wide.
Radius of this single door = 3 feet.
Nathan fixed a double door each of 1 1/2 feet wide.
Radius of each of this double door
\(=\frac{3}{2} \text { feet. } \)
The area required for the single door
\(=\frac{\theta}{360} \times \pi r^{2} \)
\(=\frac{120}{360} \times 3.14 \times 3 \times 3 \)
= 9.42 m²    ............(i)
The area required for the double door
\(=2 \times \frac{\theta}{360} \times \pi r^{2} \)
\(=2 \times \frac{120}{360} \times 3.14 \times \frac{3}{2} \times \frac{3}{2} \)
= 4.71 m²       ........................(ii)
From (i) and (ii), the double door requires minimum area.

Now, from Fig PQ = PR
⇒\(\angle\)Q = \(\angle\)R (angles opposite to equal sides are equal)
⇒ \(\angle\)x = \(\angle\)y
⇒\(\angle\)x+ \(\angle\)y + 50^o = 180^o (angle sum property in ΔPQR)
⇒ 2\(\angle\)x = 130^o
⇒\(\angle\)x = 65^o
⇒ \(\angle\)y = 65^o

\(\frac { 1 }{ 8 } \left( \frac { 1 }{ 6 } -\frac { 1 }{ 4 } \right) =\frac { 1 }{ 8 } -\left[ \frac { (1\times 2)-(1\times 3) }{ 12 } \right] \)
\(=\frac { 1 }{ 8 } -\left( \frac { 2-3 }{ 12 } \right) \)
\(=\frac { 1 }{ 8 } -\left( -\frac { 1 }{ 12 } \right) \)
\(=\frac { 1 }{ 8 } +\frac { 1 }{ 12 } =\frac { (1\times 3)+(1\times 2) }{ 24 } \)
\(=\frac { 3+2 }{ 24 } =\frac { 5 }{ 24 } \)

Capacity of the small water jug = 3\(\frac{4}{5}\)litres.
Capacity of the big jug = 2\(\frac{2}{3}\) times the small one.
\(2\frac { 2 }{ 3 } \times 3\frac { 4 }{ 5 } =\frac { 8 }{ 3 } \times \frac { 19 }{ 5 } =\frac { 152 }{ 15 } \)
= \(10\frac { 2 }{ 5 } \) litres
Capacity of the large jug = \(10\frac { 2 }{ 5 } \) litres

The rational number for the point marked on the number line is  -3\(\frac{2}{3}\) = \(\frac{-11}{3}\)

Triangular Prism

The given figure is a combination of a semicircle and a triangle.
diameter = 6 cm
radius = 3 cm
base = 6 cm
height = 9 cm
The shaded area of the figure = Area of the semicircle + Area of the triangle
\(=\frac{1}{2} \pi \mathrm{r}^{2}+\frac{1}{2} \mathrm{bh} \)
\(=\frac{1}{2} \times 3.14 \times 3 \times 3+\frac{1}{2} \times 6 \times 9 \)
= 14.13 + 27 = 41.13 cm²

From this figure, perimeter
= 10 m + 7 m + 10 m + L
\(\begin{equation} =27 \mathrm{~m}+\frac{\theta}{360} \times 2 \pi \mathrm{r} \end{equation}\)
\(=27+\frac{180}{360} \) \(\times 2 \times \frac{22}{7} \times \frac{7}{2}\)
= 27 + 11 = 38 m
Area of the shaded part - Area of the rectangle - Area of the semicircle
=\(​​(l\times b)-\frac { 1 }{ 2 } \times \pi { r }^{ 2 }\)
= \((10\times 7)-\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \)
= 50.75 m²

Length of the arc of the sector l = 44 cm
Perimeter of the sector P = 64cm
l+ 2r = 64cm
44 + 2r = 64
2r = 64-44
2r = 20
r = \(\frac{20}{2}\) = 10cm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 44\times 10 }{ 2 } \) cm² = 22 x 10 cm² = 220 cm²
Area of the sector = 220 cm²

(x-1)²-(x-2)²={(x-1+(x-2)}{(x-1)-(x-2)}
= (2x-3)-(x-1-x+2)
= (2x-3)X1=2x-3

Radius of the big semicircle = 14 cm
∴ Area of big semicircle = \(\frac { 1 }{ 2 } \)πr² sq. units
= \(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 14\times 14\)
= 308 cm²
Radius of small semi circles = 7cm
Area of 2 small semi circles = \(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 7\times 7\)
= 154 cm²
∴ Required area = 308 + 154 cm²
= 462 cm²

Now, x² + 8x + 16
This can be written as x² + 8x + 42
Comparing this with a² + 2ab + b² = (a + b)² we get a = x; b = 4
(x²) + 2(x)(4) + (4)² = (x + 4)²
x² + 8x + 16 = (x + 4)²

Radius of a circular quadrant, r = 3.5 cm and side of a square, a = 3.5 cm.
The given figure is formed by the joining of 4 quadrants of a circle with each side of a square. The boundary of the given figure consists of 4 arcs and 4 radii.
(i) Perimeter of the given combined shape
= 4 x length of the arcs of the quadrant of a circle + 4 x radius
\(=\left( 4\times \frac { 1 }{ 4 } \times 2\pi r \right) +4r\)
\(=\left( 4\times \frac { 1 }{ 4 } \times 2\times 3.5 \right) +(1\times 3.5)\)
= 22 + 14 = 36 cm (approximately)
(ii) Area of the given combined shape
= area of the square + 4 x area of the quadrants of the circle
\({ a }^{ 2 }=\left( 4\times \frac { 1 }{ 4 } \times \pi { r }^{ 2 } \right) \)
\(=(3.5\times 3.5)+\left( \frac { 22 }{ 7 } \times 3.5\times 3.5 \right) \)
A = 12.25 + 38.5 = 50.75 cm² (approximately)