Let us form table of values for each line and then fix the ordered pairs to be Plotted.
Graph of y = 7 - x

Points to be plotted : (-2, 9), (-1, 8), (0, 7), (1, 6), (2, 5)
Graph of y = x - 3

Points to be plotted (-2, -5), (-1, -4), (0, -3), (1, -2), (2, -1) 

Let the vertices of the ABC be A(x₁, y₁), B(x₂, y₂ ) and C(x₃, y₃) and the given mid-points of the sides AB, BC and CA are (5, 1), (3, −5) and (−5, −1) respectively. Therefore
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =5 \Rightarrow\)x₁ + x₂ = 10 ...(1)
\(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =3 \Rightarrow\)x₂ + x₃ = 6 ...(2)
\(\frac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } =-5 \Rightarrow\)x₃ + x₁ = –10 ...(3)
Adding (1), (2) and (3)
2x₁ + 2x₂ + 2x₃ = 6
x₁ + x₂ + x₃ = 3 ...(4)
(4) − (2) \(\Rightarrow\) x₁ = 3 − 6 = −3 
(4) − (3) \(\Rightarrow\) x₂ = 3 +10 = 13
(4) − (1) \(\Rightarrow\) x₃ = 3 −10 = −7
\(\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =1\)\(\Rightarrow\)y₁ + y₂ = 2 …(5)
\(\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } =-5\)\(\Rightarrow\)y₂ + y₃ = –10 …(6)
\(\frac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } =1\)\(\Rightarrow\)y₃ + y₁ = –2 …(7)
Adding (5), (6) and (7),
2y₁ + 2y₂ + 2y₃ = −10
y₁ + y₂ + y₃ = −5 ...(8)
(8) − (6) \(\Rightarrow\) y₁ = −5 +10 = 5
(8) − (7) \(\Rightarrow\) y₂ = −5 + 2 = −3
(8) − (5) \(\Rightarrow\) y₃ = −5 − 2 = −7
Therefore the vertices of the triangles are A(−3, 5), B(13, −3) and C(−7, −7).

Volume of a cuboid = 1800 cm³
l = 15 cm h = 12 cm b = ?
l \(\times\) b \(\times\) h = V
15 \(\times\) b \(\times\) 12 = 1800

Bredth = 10 cm

p(A) = 0.72
P(A') = 1 - 0.72 = 0.28

Total n(S) = 10,000
Defective n(A) = 25
Number of defective laptop is 25
Number of good laptop is = 10000 - 25 = 9975
Let A be the event of choosing a good laptop. n(A) = 9975
\(\mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{9975}{10000}\)
P(A) = 0.9975
Probability of getting a good laptop is 0.9975

(i) side ofa cube = 8 m
TSA of the cube = 6a² = 6 \(\times\) 64 = 384 m²
LSA of the cube = 4a² = 4 \(\times\) 64 = 256 m²
(ii) side a = 21 cm
TSA= 6a² = 6 \(\times\) 21 \(\times\) 21 = 2646 cm².
LSA = 4a² = 4 \(\times\) 21 \(\times\) 21 = 1764 cm².
(iii) side a = 7.5 cm
TSA = 6a² = 6 \(\times\) 7.5 \(\times\) 7.5 cm² = 337.5 cm²
LSA = 4a² = 4 \(\times\) 7.5 \(\times\) 7.5 cm² = 225 cm².

Let the time taken by the larger pipe be x hours
and Set the time taken by the smaller pipe be y hours.
\(\cfrac { 1 }{ x } +\cfrac { 1 }{ y } =\cfrac { 1 }{ 24 } \)
In 1.hour the larger pipe can fill it = \(\cfrac { 1 }{ x } \)
In 1hour the smaller pipe can fill it  = \(\cfrac { 1 }{ y } \)
\(\cfrac { 8 }{ x } +\cfrac { 18 }{ y } =\cfrac { 1 }{ 2 } \)
Put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
\(a+b\ =\cfrac { 1 }{ 24 } \)
24a + 24b = 1
24a + 24b - 1 = 0    ...(1)
8a + 18b = \(\cfrac { 1 }{ 2 } \)
16a+36b = 1
16a + 36b - 1 = 0    ..(2)
For cross multiplication method, we write the co-efficients as
\(\cfrac { a }{ -24+36 } =\cfrac { b }{ -16+24 } =\cfrac { 1 }{ 864-384 } \)
\(\cfrac { a }{ 12 } =\cfrac { b }{ 8 } =\cfrac { 1 }{ 480 } \)
\(\cfrac { a }{ 12 } =\cfrac { 1 }{ 480 } \ \)  \( \cfrac { b }{ 8 } =\cfrac { 1 }{ 480 } \\ \)

\(\therefore\) x = 40, y = 60

Let the number of 2 rupee coins be x
Let the number of 5 rupee coins be y
x+y = 80
2x+ 5y = 220
x +y- 80 = 0
2x + 5y - 220 = 0
For cross multiplication method, we write the co-efficients as

\(\cfrac { x }{ -220 } =\cfrac { y }{ -160+220 } =\cfrac { 1 }{ 5-2 } \)
\(\cfrac { x }{ 180 } =\cfrac { y }{ 60 } =\cfrac { 1 }{ 3 } \)
\(\therefore \ \cfrac { x }{ 180 } =\cfrac { 1 }{ 3 } \ \ \cfrac { y }{ 60 } =\cfrac { 1 }{ 3 } \)

x = 60 y = 20
\(\therefore\) No. of 2 rupee coins = 60
No. of 5 rupee coins = 20

x₁ y₁    x₂ y₂   m : n 
A (4, -3), B (9, 7), 3 : 2
By section formula \(P\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) =P(x,y) \)

\(P(x,y)=\left( \frac { 3(9)+2(4) }{ 3+2 } ,\frac { 3(7)+2(-3) }{ 3+2 } \right) \)
\(=\left( \frac { 27+8 }{ 5 } ,\frac { 21-6 }{ 5 } \right) =\left( \frac { 35 }{ 5 } ,\frac { 15 }{ 5 } \right) =(7,3)\)

If \(2cos\theta =\sqrt { 3 } \)
\(cos\theta =\cfrac { \sqrt { 3 } }{ 2 } \)
\(x=\sqrt { { 2 }^{ 2 }-\sqrt { { 3 }^{ 2 } } } =\sqrt { 4-3 } =\sqrt { 1 } =1\)
\(\therefore sin\theta =\cfrac { 1 }{ 2 } \)
\(cos\theta =\cfrac { \sqrt { 3 } }{ 2 } \)
\(tan\theta =\cfrac { 1 }{ \sqrt { 3 } } \)
\(coseec\theta =2\)
\(sec\theta =\cfrac { 2 }{ \sqrt { 3 } } \)
\(cot\theta =\sqrt { 3 } \)

(i) \(sinB=\cfrac { 12 }{ 13 } \)
(ii) \(secB=\cfrac { 1 }{ cosB } \)
=  \(\cfrac { 1 }{ 5/3 } =\cfrac { 13 }{ 5 } \)
(iii) \(cotB=\cfrac { 1 }{ tanB } =\cfrac { 1 }{ 12/5 } =\cfrac { 5 }{ 2 } \)
(iv) 

(v) \(\tan C=\cfrac { 12 }{ 16 } =\cfrac { 3 }{ 4 } \)
(vi)  \(cosecC=\cfrac { 1 }{ sinC } =\cfrac { 1 }{ 12/20 } =\cfrac { 20 }{ 12 } =\cfrac { 5 }{ 3 } \)

By the pythagoras theorem,
\(AD=\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \)
= \(\sqrt { 144 } =12\)
AC =\(\sqrt { { 12 }^{ 2 }+{ 16 }^{ 2 } } \)
= \(\sqrt { { 144 }+{ 256 } } \)
= \(\sqrt { 400 } =20\) 

\(M(x,y)=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\((-4,2)=\left( \frac { -3+{ x }_{ 2 } }{ 2 } ,\frac { 7+{ y }_{ 2 } }{ 2 } \right) \left( { x }_{ 1 }{ y }_{ 1 } \right) \)
\(\frac { -3+{ x }_{ 2 } }{ 2 } =-4\quad \frac { 7+{ y }_{ 2 } }{ 2 } =2\)
-3 + x₂ = -8
7 +y₂ = 4
x₂ = -8 +3
y₂ = 4 -7 = -3
x₂ = -5
The other end is (-5, -3)

Cube
Side (a) = 12cm
Cuboid
length (l) = 18cm
breadth (b) = 16cm
height (h) = ?
Here, Volume of the Cuboid =Volume of the Cube
l × b × h = a³
18 ×16 × h = 12 ×12 ×12
h = \(\frac{12\times12\times12}{18\times16}\)
h = 6 cm
Therefore, the height of the cuboid is 6 cm

Let E be the event that it will rain tomorrow. Then E′ is the event that it will not rain tomorrow.
Since P(E) = 0.91, we have P(E′) = 1−0.91 (how?)
= 0.09
Therefore, the probability that it will not rain tomorrow
= 0.09

Total number of staff members = 42
The relative frequencies:
Car users \(=\frac { 7 }{ 42 } =\frac { 1 }{ 6 } \)
Two-wheeler users \(=\frac { 20 }{ 42 } =\frac { 10 }{ 21 } \)
Cycle users \(=\frac { 15 }{ 42 } =\frac { 5 }{ 14 } \)

The side of the cube (a) = 5 cm
Total Surface Area = 6a² = 6(5²) = 150 sq. cm
Lateral Surface Area = 4a² = 4(5²) = 100 sq. cm

(i) sin 38⁰36' + tan 12⁰12'
sin 38⁰36' = 0.6239
tan12⁰12' = 0.2162
sin 38⁰36' + tan 12⁰12' = 0.8401
(ii) tan 60⁰25^' - cos 49⁰20^'
tan 60⁰25' = 1.7603 + 0.0012 = 1.7615
cos 49⁰20^' = 0.6521 - 0.0004 = 0.6517
tan 60⁰25' - cos 49⁰20' = 1.1098

write 70⁰13' = 70⁰12' + 1'
From the table we have, tan70⁰12' = 2.7776 

(i) \(\frac { sin\ 49° }{ cos\ 41° } \) 
sin 49⁰ = sin(90⁰ - 41⁰) = cos 41⁰,  since 49⁰ + 41⁰ = 90⁰ (complementary),
Hence on substituting sin 49^o = cos41^o we get, \( \frac { cos\ 41° }{ cos\ 41° } \)= 1
(ii) \(\frac { sec\ 63° }{ cosec\ 27° } \)
sec63^o = sec (90^o- 27^o) = cosec27^o, here, 63^o and 27^o are complementary angles
we have \(\frac { sec\ 63° }{ cosec\ 27° } =\frac { cosec\ 27° }{ cosec\ 27° } =1\)

Given the sides are in the ratio 9:10:11, let the sides be 9k, 10k, 11k
The perimeter of the triangular park = 300 m
9k +10k +11k = 300m
30k = 300
k = 10m
Therefore, the sides are a = 90 m, b = 100 m, c = 110 m
s=\(\frac{1+b+c}{2}=\frac{90+100+110}{2}=\frac{300}{2}\)= 150 m
Hence, Area of triangular park = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 150\times (150-90)(150​​-100)(150-110) } \)
= \(\sqrt { 150\times 60​​\times 50\times 40 } \)
= \(\sqrt { 3\times 50\times 20​​\times 3\times 50\times 2\times 20 } \)
= 50 \(\times\) 20 \(\times\) 3\(\sqrt{2}\)
= 3000 \(\times\)1.414 = 4242 m²

(i) 8x- 3y = 12      ...(1)
5x-2y = 7      ..(2)
8x- 3y-12 = 0
5x-2y- 7 = 0
For cross multiplication method, we write the co-efficients as

\(\cfrac { x }{ (-3)(-7)(-2)(-12) } =\cfrac { y }{ (-12)(5)-(-7)(8) } =\cfrac { 1 }{ (8)(-2)-(5)(-3) } \)
\(\cfrac { x }{ 21-24 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\cfrac { x }{ -3 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\therefore \ \cfrac { x }{ 3 } =\cfrac { 1 }{ -1 } \ \cfrac { y }{ -4 } =\cfrac { 1 }{ -1 } \)
x = 3 , y = 4
\(\therefore\)  Solutions: x = 3; y = 4
(ii) 6x+ 7y-11 = 0
5x+ 2y-13 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { x }{ -91-(-22) } =\cfrac { y }{ -55-(-68) } =\cfrac { 1 }{ 12-35 } \)
\(\cfrac { x }{ -91+22 } =\cfrac { y }{ -55+78 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { 1 }{ -23 } \quad \cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)

x = 3, y= -1
\(\therefore\)  x = 3; y= -1
(iii) 

\(\cfrac { 2 }{ x } +\cfrac { 3 }{ y } -5=0\)
\(\cfrac { 3 }{ x } -\cfrac { 1 }{ y } +9=0\)
In (1), (2) Put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
(1) \(\Rightarrow\) 2a + 3b - 5 = 0
(2) \(\Rightarrow\) 3a - b + 9 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { a }{ (3)(9)-(-1)(-5) } =\cfrac { b }{ (-5)(3)-(9)(2) } =\cfrac { 1 }{ (2)(-1)-(3)(3) } \)
\(\cfrac { a }{ 27-5 } =\cfrac { b }{ -15-18 } =\cfrac { 1 }{ -2-9 } \)
\(\cfrac { a }{ 22 } =\cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)
\(\therefore \ \cfrac { a }{ 22 } =\cfrac { 1 }{ -11 } \cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)

a = -2    b = 3
\(a=\cfrac { 1 }{ x } =-2\quad b=\cfrac { 1 }{ y } =3\)
\(\therefore \ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)
solution \(\ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)

The centroid G(x, y) of a triangle whose vertices are (x₁, y₁), (x₂ , y₂ ) and (x₃ , y₃) is given by

G(x,y)=G\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 }3 }{ } \right) \)
We have (x₁, y₁) = (6, −1); (x₂ , y₂ ) = (8, 3); (x₃ , y₃) = (10, −5)
The centroid of the triangle
G(x, y) =G\(\left( \frac { 6+8+10 }{ 3 } ,\frac { -1+3-5 }{ 3 } \right) \)
= G \((\frac{24}{3},\frac{-3}{3})\)= G(8, -1)

Given x – 2y = 7 …(1)
2x + 3y = 7 …(2)
When x = 5, y = −1 we get
From (1) x – 2y = 5 – 2(−1) = 5 + 2 = 7 which is RHS of (1)
From (2) 2x + 3y = 2(5) + 3(−1) = 10−3 = 7 which is RHS of (2)
Thus the values x = 5, y = −1 satisfy both (1) and (2) simultaneously. Therefore (5,−1) is a solution of the given equations.