\(\frac { -6 }{ -11 } \times \left( -4 \right) =\frac { 6 }{ 11 } \times \frac { \left( -4 \right) }{ 1 } =\frac { 6\times (-4) }{ 11\times 1 } =\frac { -24 }{ 11 } \)

Method 1:
\(\frac { 48 }{ -84 } =\frac { 48\div (-2) }{ -84\div (-2) } =\frac { -24\div 2 }{ 42\div 2 } =\frac { -12\div 3 }{ 21\div 3 } =\frac { -4 }{ 7 } \) (dividing by –2, 2 and 3 successively)
Method 2:
The HCF of 48 and 84 is 12 (Find it). Thus, we can get its standard form by dividing it by  -12.
\(=\frac { 48\div (-12) }{ -84\div (-12) } =\frac { -4 }{ 7 } \)  

\(3.0=\frac { 30 }{ 10 } =\frac { 3 }{ 1 } \)

= 6x² − 8xy + 15xy − 20y²
= 6x² + 7xy− 20y²(simplify the like terms

Radius, r = 35 cm and n = 5
Length of each of the arcs, l = \(\frac{1}{n}\times 2\pi r\) units
\(=\frac{1}{5}\times2\times \pi\times35\)
l = 14 π cm

Length of the wire = \(\frac{4}{5}m\)
\(=\frac { 4\times 100 }{ 5 } cm=80cm\)
Number of equal pieces made from it = 80 \(\div\) 8 = 10 cm
∴ Length of each small pieces = 10 cm

\(\frac { -3 }{ 5 } =\frac { -3\times 2 }{ 5\times 2 } =\frac { -6 }{ 10 } \)
\(\frac { -3 }{ 5 } =\frac { -3\times 3 }{ 5\times 3 } =\frac { -9 }{ 15 } \)
\(\frac { -3 }{ 5 } =\frac { -3\times 4 }{ 5\times 4 } =\frac { -12 }{ 20 } \)
\(\frac { -3 }{ 5 } =\frac { -3\times 5 }{ 5\times 5 } =\frac { -15 }{ 25 } \)
Four equivalent rational numbers of \(\frac { -3 }{ 5 } \)are \(\frac { -6 }{ 10 } ,\frac { -9 }{ 15 } ,\frac { -12 }{ 20 } ,\frac { -15 }{ 25 } \)

Square Pyramid

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 = 50 mm
Radius r = 14mm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 50\times 14 }{ 2 } \) mm² = 50 x 7 mm² = 350 mm²
Area of the sector = 350mm²

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

Given:
MA = 4 cm, AT = 3.6 cm,
TH = 4.5 cm, MH = 5 cm and ∠A = 85°


Steps:
1. Draw a line segment MA = 4 cm.
2. Make ∠A = 85°.
3. With A as centre, draw an arc of radius 3.6 cm. Let it cut the ray AX at T.
4. With M and T as centres, draw arcs of radii 5 cm and 4.5 cm respectively and let them cut at H.
5. Join MH and TH.
6. MATH is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral MATH = \(\frac12\) × d × (h₁+ h₂) sq.units
= \(\frac12\) x 5.1 x (3.9 + 2.8)
= 2.55 x 6.7 = 17.09 cm²

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)