(i) -275°C
T =  -275°C
T^oC = (T + 273)K
= (-275 + 273)K = -2K
-275°C =  -2 Kelvin
(ii) 45°C
T = 45°C
T^oC = (T + 273)K
(45 + 273)K = 318K
45°C = 318 Kelvin
(iii) -400°C
T = -400°C
T^oC = (T + 273)K
= (-400 + 273) K = -127 K
-400°C = -127 Kelvin
(iv) -273°C
T = -273°C
T^oC = (T + 273)K
= (-273 + 273) K = 0K.
-273°C = 0 Kelvin

(i) 13x + 12y - 5 from 27x + 5y - 43
27x + 5y-43 - (13x + 12y- 5)
= 27x + 5y - 43 - 13x - 12y + 5
= (27 - 13) x + (5 - 12) y + (- 43) + 5
= 14x + (-7)y + (-38)
= 14x - 7y - 38
(ii) 3p + 5 from p - 2q + 7
= P - 2q + 7 + (- 3p -5)
= (1-3)p-  2q + 2
= -2p - 2q  + 2
(iii) m + n from 3m - 7n
Additive inverse of (m + n) is -m -n,
= 3m -7n - (m + n) = 3m -7n + (-m -n)
= (3m - m) + (-7n - n)
= (3 - 1)m + (-7 -1) n
= 2m - 8n
(iv) 2y + Z from 6z - 5y
Additive inverse of 2y + z = -2y - z.
6z - 5y-(2y + z)
= ( 6z -5y) +(-2y - z)
= (62 - z) + (-5y - 2y)
= (6 - 1)z + (-5 -2)y
= 5z + (-7)y = 5z - 7y

(i) 

∴. 8 + (-12) = -4
(ii) 

∴ (-3) + (-5) = -8
(iii) (-100) + (-10) = -100-10 = -110
(iv) 20 + (-72) = 20 - 72 = -52
(v) 82 + (-75) = 82 - 75 = 7
(vi) -48 + (-15) = -48 - 15 =- 63
(vii) -225 + (-63) = 225 - 63 = -288

3 x 5 = 15
(-3) x (-5) = +15
Two pairs :3,5 and -3, -5.

-70 + 20 = x - 10
-50 = x - 10
x = -50 + 10 = -40
= -40

To reach 4th step of water level = 1 jump
To reach 20th step of water level \(=\frac{20}{4}=5 \text { jumps }\)
The dog wants to take 5 jumps to reach the water level.

Let x be an integer
\(\frac{x}{7}=-3\)
X = -3 x 7 = -21
The inleger is -21.

Loss of calory in 30 days = 4800
∴ Loss of calory in 1 day \(=\frac{4800}{30}\) = 160 calories
∴ 160 calories lost per day

(5a - 3b + 2c) + Expression = a - 4b - 2c 
Expression = (a - 4b - 2c)- (5a- 3b + 2c)
= (a - 4b - 2c) + (-5a - 3b +2c)
= (a - 5a) + (-4b + 3b) + (-2c - 2c)
= (1 - 5)a + (-4 + 3)b + (-2 - 2)c
= -4a - b - 4c

The two real - life situations where we use parallel lines are
(i) Railway tracks
(ii) Edges of a door frame

(i) 3a = 126°
a = \(\frac{126°}{3} \) = 42°
a = 42°
(ii) 4a + 13^o + 135° = 180°
4a + 148° = 180°
4a = 180° - 148°
4a = 32°
a = \(\frac{32°}{4}\)
a = 8°
(iii)  8a + 29° = 45°
8a = 45° - 29^o
8a = 16°
a = 2°
(iv) 6a = 90°
a = \(\frac{32°}{4}\) 
a = -15°

From all the figures, all the pair of angles are alternate interior (or) exterior angle.
i) y = 28^o
ii) y = 58^o
iii) y = 123^o
iv) y = 108^o

Given : h = 50 cm, n = 200 cm,
b = 150 cm
Area of trapezium shaped table 
\(=\frac{1}{2} h \times(a+b) \text { sq.units } \)
\(=\frac{1}{2} \times 50(200+150) \)
\(=\frac{1}{2} \times 50(350) \)
= 8750 sq.cm
The cost of the glass used to cover the table per 10 sq.cm =  Rs. 6
The cost of the glass used to cover the table for 8750 sq.cm \(=\frac{8750}{10} \times 6\) = Rs. 5250

Given: Height of parallelogram \(=\frac{1}{4} of\ its\ base \)
h \(=\frac{1}{4} b \)
Area of the parallelogram = 676
b x h = 676
\(b \times \frac{1}{4} b=676\)
b x b = 676 x 4
b x b = 26 x 26 x 2 x 2
b = 26 x 2
Base (b) = 52 cm
\(h=\frac{1}{4} \times b \)
\(b=\frac{1}{4} \times 52 \)
h = 13

The cost of a dozen soaps = Rs. 396
ie., The cost of 12 soaps = Rs. 396
The cost of 1 soap = 33
The cost of 35 soaps = 33 x 35 = Rs. 1155

-47 -(-12) = -47 + 12 = -35
A student got -35.
No, he is not correct

Kala's bodytemperature when woke up = 102^o F
After 2 hours then taking medicine her
temperature lowered by (-) = -2^oF
Then her temperature = 102^o - 2^o
= 100^oF
Her body temperature was 100^oF

From the figure
∠FEH = ∠FEG + ∠GEH
120° =  34° + ∠GEH
120° - 34° = ∠GEH 
∠GEH = 86°

(i) 8x + 3x = (8 + 3) x = 11x
(ii) 7mn + 5mn = (7 + 5)mm = 12mm
(iii) -9y + 11y + 2y = (-9 + 11 + 2) y = (2 + 2) y = 4y

To cover 100 km quantity of petrol required = 2 litres
\(\therefore\) To cover 1km petrol required = 2litres
To cover 1km petrol required = \(\cfrac { 2 }{ 100 } =\cfrac { 1 }{ 50 } litres\)
\(\therefore\) To cover 250km petrol required = \(\cfrac { 1 }{ 50 } \times 250=5litres\)
5 litres of petrol required to cover 250 km

Given : Area = 100 sq.cm, d₁ = 8 cm, d₂ = ?
Area of rhombus = 100 sq.cm
\(\frac{1}{2} \times\left(d_{1} \times d_{2}\right)=100 \)
\(\frac{1}{2} \times\left(8 \times d_{2}\right)=100 \)
8 x d₂ = 100 x 2
\(\mathrm{d}_{2}=\frac{100 \times 2}{8} \Rightarrow \mathrm{d}_{2}=25 \mathrm{~cm}\)

Given base b = 14 cm ; Height h = 9 em
Area of the rhombus = b x h sq. units
= 14 x 9
Area of rhombus = 126 sq.cm

Five pairs are :
(0,2) Reason:0 + 2 = 2
(-2,4) Reason:-2 + 4 = 2
(-1,3) Reason:-1 + 3 = 2
(1; 1) Reason:1 + 1 = 2.
(-3,5) Reason -3 + 5 = 2

(i) Both Team A and B scored 10 marks. Quiz is draw.
(ii) Yes, we can add integers in any orders since addition is commutative

Pencils received : (+)
Number of pencils Kabilan has = 10
Number of pencils father gove = 6
= 16
Pencils given : (-)
Number of pencils gave to Senthil = 2
Number of pencils gave to Karthick = 3
Number of penciis gave to Sister = 8
= 13
Number of pencils left = Received - Given
= 16-13
Number of pencils left = 3