\(\frac { 30 }{ 12 } \)

We have, 2¹⁰ or 10²
\(\therefore\) 2¹⁰ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024
and 10² = 10 x 10 = 100
\(\because\) 1024 > 100 \(\Rightarrow\) 2¹⁰ > 10²
Hence, 2¹⁰ is greater than 10².

Expression 5 - 3t is a binomial because it contains 2 unlike terms 5 and -3t

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.

∴ Mixed fraction =Quotient \(\frac{Reminder}{Divisor}=2 \frac{2}{3}\)

The required expression
= (2x³-3x² + 4x + 5) - (2x³ + 7x² - 2x + 7)
= 2x³ - 3x² + 4x + 5 - 2x³ - 7x² + 2x - 7
= (2x³ - 2x³) + (-3x²-7x²) + (4x + 2x)+ (5-7)
= (2 - 2)x³ + (-3 - 7)x² + (4 + 2)x + (5 - 7)
= (O)x³ + (-10)x² + (6)x + (-2)
=-10x²+6x-2
Thus, the expression (2x³ - 3x² + 4x + 5) is greater than ( 2x³ + 7x² - 2x + 7) by the expression (-10x² + 6x - 2).

Area of parallelogram ABCD = AB x AE
= 8 x 4 cm² = 32 cm²
Let altitude corresponding to AD be h. Then,
h x AD = 32
or h x 6 = 32
or h = \(\frac { 32 }{ 6 } \) = \(\frac { 16 }{ 3 } \)
Thus, altitude corresponding to AD is \(\frac { 16 }{ 3 } \)cm

In order to construct the ΔABC, we follow the following steps:
(a) Draw \(\angle XBY\) of measure 60°.
(b) From ray BX, cut off line segment BC of length 6 cm.

(c) From ray BY, cut off line segment BA of length 5 cm.
(d) Join AC to obtain the required triangle ABC, where \(\angle B\) = 60°, AB = 5 cm and BC = 6 cm.

\(\frac { 5 }{ 4 } +\left( \frac { -11 }{ 4 } \right) =\frac { 5+(-11) }{ 4 } =\frac { -6 }{ 4 } =\frac { -3 }{ 2 } \) [ on dividing numerator and denominator by 2 ] 

Given, 2ⁿ⁺²- 2ⁿ = c x 2ⁿ
⇒ 2ⁿ⁺² = 2ⁿ x 2², 2ⁿ⁺¹ = 2ⁿ x 2¹
So, 2ⁿ x 2² - 2ⁿ x 2¹ + 2ⁿ = c x 2ⁿ
Taking ~ common from both the side, we get
2ⁿ (2²-2¹)= c x 2ⁿ
⇒ (2²-2¹+1)c
∴ c = 4-2+1   [∵ 2²= 2 x 2=4, 2¹=2]
=2 +1 =3

In Fig. (i), there are 3 cubes when observed from the view shown by the arrow mark. In Fig. (ii), there are 3 cubes when observed from the view shown by the arrow mark. In Fig. (iii), there are 4 cubes when observed from the view shown by the arrow mark.

Three boys earned = Rs 235.50
∴ Average amount earned = \(\frac{235.50}{3}=\frac{23550}{300}=\frac{2355}{30}\)=Rs 78.50
Hence, each boy earned Rs 78.50

We have,2(a² +ab)+3-ab = 2a² +2ab+3-ab
= 2a² + 2ab - ab + 3   [rearranging terms]
= 2a² + ab(2 -1) + 3 = 2a² + ab + 3
On putting a = 5 and b = - 3, we get
The value of 2a² + ab + 3 = 2(5)²  5 x (-3) + 3
= 2 x 5 x 5 + 5 x(-3)+3 = 50 - 15 +3 = 38
Hence, the simplified expression is 2a² + ab + 3 and its required value is 38

Given, diameter of the circular table top = 1.6 m
Then, radius of the circular table top = \(\frac{1.6}{2}\) = 0.8 m
\(\therefore\)  Area of the circular table top = \(\pi\)r² = 3.14 x (0.8)2
                                              = 3.14 x 0.8 x 0.8 = 2.0096 m²
Now, cost of polishing 1 sq m = Rs 15
\(\therefore\) Cost of polishing 2.0096 m² = Rs (15 x 2.0096) = Rs 30.144
                                             = Rs 30.14 (approx.)
Hence, the cost of polishing the circular tabletop at the rate of Rs 15 per sq m² is Rs 30.14 (approx.)

Number of lines of symmetry for the given figures arc as follows:

\(\frac { 8 }{ 3 } \)

-2b

Line of symmetry = 4
Order of symmetry = 8 (angle of rotation = 45°)

96 eggs

(i) Net for triangular prism

(ii) Net for tetrahedron,

(iii) Net for cuboid

\(\therefore\) Side of square pizza = 45 cm
\(\therefore\) Area of a square pizza = (Side)²
                                     = (45)² = 2025 cm²
Diameter of circular pizza = 50 cm
Radius = \(\frac{50}{2}\) = 25 cm.
\(\therefore\) Area of a circle = \(\pi r^2\)
\(\therefore\) Area of the circular pizza = \(\pi r^2\)
                                         = \(\frac{22}{7}\times 25\times 25=\frac{22}{7}\times 625\)
                                         = \(\frac{13750}{7}\) = 1964.28 cm²
Price of 1 cm square pizaa  = \(\frac{2025}{150}\) = Rs 135
Price of 1 cm circular pizaa  = \(\frac{1964.29}{160}\) = Rs 12.27
Hence, the circular pizza is a better deal.

Given, PQ = 5 cm, mㄥPQR = 105° and m ㄥQRP = 40°
In ΔPQR, by angle sum property, we have
ㄥPQR + ㄥQRP + ㄥRPQ = 180°
⇒ 105° + 40° + ㄥRPQ = 180°
⇒ ㄥRPQ = 180° -145° = 35°
Thus, we have PQ = 5 cm, ⇒ ⇒P = 35° and ⇒Q = 105°
Now, to draw ΔPQR, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch of traingle with measures marked on it.

Step II Draw a line segment PQ = 5 cm.

Step III At point P, draw a ray PX making an angle of 350 with PQ i.e., ㄥQPX = 35⁰.

Step IV At point Q, draw a ray QY making an angle of 105° with PQ i.e. ㄥPQY = 105°.

Step V Extend the ray PX and QY. Let rays PX and QY intersect at R.

Thus, ΔPQR is the required triangle.

\(\frac{-7}{11}=\frac{-7\times-2}{11\times-2}=\frac{14}{-22}\)

Let ABCD be the quadrilateral in which AC is a diagonal, such that AC = 40 cm

Suppose DE \(\bot \) AC and BF \(\bot \) AC, such that
DE = 8.4 cm and BF = 6.6 cm.
Now, area of the quadrilateral ABCD
= [Area of \(\Delta\)ABC] + [Area of \(\Delta\)ADC]
=\(\left[ \frac { 1 }{ 2 } \times AC\times BF \right] \)+\(\left[ \frac { 1 }{ 2 } \times AC\times DE \right] \)
=\(\left[ \frac { 1 }{ 2 } \times 40cm\times 66cm\right] \)+\(\left[ \frac { 1 }{ 2 } \times 40cm\times 8.4cm\right] \)
=\(\left[ \frac { 1 }{ 2 } \times 40\times \frac { 66 }{ 10 } { cm }^{ 2 } \right] \)+​​​​​​​​​​​​​​\(\left[ \frac { 1 }{ 2 } \times 40\times \frac { 84 }{ 10 } { cm }^{ 2 } \right] \)
= [2 x 66 cm²] + [2 x 84 cm²]
= 132 cm² + 168 cm² = 300 cm²
Thus, the area of the quadrilateral ABCD = 300 cm².

A = 2a - 3b
B=-3a+4b
C = -a + b
A+B+C=?
A + B + C = (2a-3b) + (-3a + 4b) + (-a + b)
= 2a-3b-3a + 4b-a + b
= a(2 - 3 -1) + b(- 3 + 4 + 1)
= a(2 - 4) + b(1 + 1)
=-2a+2b
A+B-C=?
A + B - C = (2a- 3b) + (- 3a + 4b) - (- a + b)
= 2a-3b-3a + 4b + a-b
= a(2-3 + 1) + b(-3 + 4-1)
= a(0) + b(-4 + 4)
= a(0) + b(0)
= 0.

\(⇒\ \left(1\over 5\right)^{5+19}=\left(1\over 5\right)^{8x}\)
[∵ a^m x aⁿ = a^m+n]
\(⇒\ \left(1\over 5\right)^{24}=\left(1\over 5\right)^{8x}\)
When bases are equal, then by equating their exponents, we get
8x = 24
\(∴\ x={24\over 8}=3\)

Given pattern is
\(-\frac { 1 }{ 3 } ,\frac { 2 }{ 6 } ,\frac { 3 }{ 9 },-\frac { 4 }{ 12 } ...\)
Here, \(-\frac { 1 }{ 3 } =\frac { (-1)\times 1 }{ 3\times 1 } \)
\(-\frac { 2 }{ 6 } =\frac { (-1)\times 2 }{ 3\times 2 } \)
\(-\frac { 3 }{ 9 } =\frac { (-1)\times 3 }{ 3\times 3 } \)
and \(-\frac { 4 }{ 12 } =\frac { (-1)\times 4 }{ 3\times 4 } \)
Hence, next four numbers are
\(\frac { (-1)\times 5 }{ 3\times 5 } =-\frac { 5 }{ 15 } \)
\(\frac { (-1)\times 6 }{ 3\times 6 } =-\frac { 6 }{ 18 } \)
\(\frac { (-1)\times 7 }{ 3\times 7 } =-\frac { 7 }{ 21 } \)
\(\frac { (-1)\times 8 }{ 3\times 8 } =-\frac { 8 }{ 24 } \).