Now, PS is the bisector of ∠P
∴ ∠APS = ∠TPS = 35^o
From ΔAPS,
∠A+∠APS+∠ASP=180^o (angle sum property in Δ APS)
∠A + 35 + 100^o = 180^o
⇒∠A = 180^o−135^o = 45^o
From ΔTPS,
∠T + ∠TPS + ∠TSP = 180^o (angle sum property in Δ TPS)
⇒∠T + 35^o + 80^o = 180^o
⇒∠T = 180^o − 115 = 65^o

\(\left( \frac { 4 }{ 3 } -\left( \frac { -3 }{ 2 } \right) \right) +\left( \frac { -5 }{ 3 } \div \frac { 30 }{ 12 } \right) +\left( \frac { -12 }{ 9 } \times \frac { -27 }{ 16 } \right) =\left( \frac { 4 }{ 3 } +\frac { 3 }{ 2 } \right) +\left( \frac { -5 }{ 3 } \times \frac { 12 }{ 30 } \right) +\left( \frac { -12 }{ 9 } \times \frac { -27 }{ 16 } \right) \)
= \(\left( \frac { 8 }{ 9 } +\frac { 9 }{ 6 } \right) +\left( \frac { -1 }{ 1 } \times \frac { 4 }{ 6 } \right) +\left( \frac { -3 }{ 1 } \times \frac { -3 }{ 4 } \right) \)
= \(\left( \frac { 17 }{ 6 } \right) +\left( \frac { -4 }{ 6 } \right) +\left( \frac { 9 }{ 4 } \right) \)
= \(\left( \frac { 17-4 }{ 6 } \right) +\frac { 9 }{ 4 } =\frac { 13 }{ 6 } +\frac { 9 }{ 4 } \)
= \(\frac { 26+27 }{ 12 } =\frac { 53 }{ 12 } \)

Let the other number be x
Given, \(\frac { 3 }{ 7 } x=\frac { -2 }{ 3 } \)
Multiplying by the reciprocal of \(\frac { 3 }{ 7 } \) that is, \(\frac { 7 }{ 3 } \)
\(\Rightarrow \frac { 7 }{ 3 } \times \frac { 3 }{ 7 } \times x=\frac { 7 }{ 3 } \times \frac { -2 }{ 3 } \)
\(\Rightarrow x=\frac { -14 }{ 9 } \)

\(\frac { -5 }{ 8 } \times 7=\frac { -5 }{ 8 } \times \frac { 7 }{ 1 } =\frac { -5\times 7 }{ 8\times 1 } =\frac { -35 }{ 8 } \)

LCM of 9, 3, 12 = 36
\(\frac { -5 }{ 9 } +\frac { -4 }{ 3 } +\frac { 6 }{ 12 } =\frac { -5 }{ 9 } \times \frac { 4 }{ 4 } +\frac { -4 }{ 3 } \times \frac { 12 }{ 12 } +\frac { 7 }{ 12 } \times \frac { 3 }{ 3 } \)
= \(\frac { -20 }{ 36 } +\frac { -48 }{ 36 } +\frac { 21 }{ 36 } =\frac { -20-48+21 }{ 36 } \)
= \(\frac { -47 }{ 36 } \)

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

Now,(103)³ = (100 + 3)³
Comparing this with (a+  b)³,we get a = 100,b = 3
(ab)³ = a³ + 3a²b + 3ab² + b³ replacing a, b values,
(100+3)³ = (100)³ +3(100)²(3) + 3(100)(3)² + (3)³
= 1000000 + 3(10000)(3) + 3(100)(9) + 27
= 1000000 + 9000 + 2700 + 27
(103)³ = 1092727

We know, 998 can be expressed as (1000 − 2)
Now (a - b)a² - 2ab + b²
(1000 - 2)² = (1000)22(1000(2) + (2)²
= 1000000 - 4000 + 4
(998)² = 996004

Given TA = TP
∴ ∠P = ∠A (if angles then sides)

Proof:

Proof:

Now, \(\frac { PQ }{ PR } =\frac { 20 }{ 15 } =\frac { 4 }{ 3 } \)
\(\frac { PR }{ PS } =\frac { 15 }{ 11.25 } =\frac { 4 }{ 3 } \)
Also, \(\frac { QR }{ RS } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)
We find \(\frac { PQ }{ PR } =\frac { PR }{ PS } =\frac { QR }{ RS } \)
That is, their corresponding sides are proportional.
∴ By SSS Similarity, ΔPQR ～ ΔPRS

Proof

The resultant expression
=3xy × (2x+3y+50)
=3xy(2x)+3xy(3y)+3xy(50)
=6x²y+9xy²+150xy

The area of the square carpet, A = (side × side) sq. units.
= 3x2×3x2
= 3×3×x²×x²
A = 9x⁴ sq.m

(i) Here, we have 2 types of car with 5 different colors and 3 models in each color.

Total number of different way to buy a new car by Madhan = 2 x 5 x 3 = 30. (You can also find the number of ways from the figure shown above).
(ii) If the white colour is not available in Type 2, then..

For Type 1, we have 5 colors and 3 models and hence there are 5 x 3 = 15 choices.
For Type 2, we have only 4 colors and 3 models and hence there are 4 x 3 = 12 choices.
Therefore, the total number of different ways 15 + 12 = 27 ways.
The above example illustrates both the addition and multiplication principles.

Assuming that the question Q₁ is answered True, questions Q₂ and Q₃ can be answered as TT, TF, FT, and FF in 4 ways.
Similarly, assuming that the question Q₂ is answered False, Q₂ and Q₃ can also be answered as TT, TF, FT, and FF in 4 ways.
Thus, as each question has only two options (True or False), the number of ways of answering these 3 questions in a slip test is 2 x 2 x 2 = 8 possible ways

There are 6 different outcomes, namely 1, 2, 3, 4, 5 and 6. All the 6 outcomes cannot occur simultaneously. Hence, there are 6 different possible outcomes in rolling a die.

Proof:

Also, their corresponding sides are proportional
⇒ \(\frac { BI }{ FA } =\frac { BG }{ FT } \Rightarrow \frac { 10 }{ FA } =\frac { 5 }{ 8 } \)
⇒ FA = \(\frac { 10\times 8 }{ 5 } =\frac { 80 }{ 5 } \) = 16 cm

Given that ΔPEN ～ ΔPAD,
∴ \(\frac { PE }{ PA } =\frac { EN }{ AD } \Rightarrow \frac { 4 }{ 7 } =\frac { 6 }{ x } \Rightarrow x=\frac { 42 }{ 4 } \)
Also, \(\frac { PE }{ PA } =\frac { PN }{ PD } =\frac { y }{ y+5 } \)
i.e. \(\frac { 4 }{ 7 } =\frac { y }{ y+5 } \) ⇒ 4y + 20 =7y ⇒ 7y - 4y = 20
3y = 20 ⇒ \(y=\frac { 20 }{ 3 } \)cm

Now, from Fig in ΔABC \(\angle\)A = x (vertically opposite angles)
Similarly \(\angle\)B = \(\angle\)C = \(\angle\)x (Why?)
⇒ \(\angle\)A + \(\angle\)B + \(\angle\)C = 180^o (angle sum property in ΔABC)
⇒ 3x = 180^o
⇒ x = 60^o
⇒ y = 180^o − 60^o = 120^o

Now, from Fig \(\angle\)140^o+ \(\angle\)z = \(\angle\)180^o (linear pair)
⇒ \(\angle\)z = 180^o −140^o = 40^o
Also \(\angle\)x+ \(\angle\)z= \(\angle\)70 + \(\angle\)z (exterior angle property)
⇒\(\angle\)x=70^o
Also \(\angle\)z+ \(\angle\)y + 70^o = 180^o (angle sum property in ΔABC)
⇒ 40^o + \(\angle\)y + 70^o = 180^o
⇒ \(\angle\)y = 180^o − 110^o = 70^o

Area of the dining table of Kamalesh = πr² sq. units
\(=\frac{22}{7}\times70\times70\)
A = 15400 sq.cm (approximately.)

Area of the circular quadrant dining table of Tharun
\(=\frac{1}{4}\pi r^2=\frac{1}{4}\times \frac{22}{7}\times140 \times140\)
A = 15400 sq.cm (approximately.)
We find that, the area of the dining tables of both of them have the same area.

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 arc ,\(l=\frac { { \theta }^{ 0 } }{ { 360 }^{ 0 } } \times 2\pi r\)
\(=\frac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times 2\times \pi \times 21\)
\(=\frac { 1 }{ 3 } \times 2\times \pi \times 21\)
\(l=14\pi cm(or)\)
\(=14\times \frac { 22 }{ 7 } \)
= 44 cm (approximately)