Volume of the cylindrical tank = Volume of the rectangle tank
πr²h = 28 x 16 x 11 m³

\(h=\frac { 16\times 11 }{ 88 } =2m\)

PQ II RS


Also

\(\angle \therefore \Delta POQ\sim \Delta SOR\) (AAA similarity criterion)

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 75 }{ 5 } \)=15
d'=\(\frac { x-\bar { x } }{ c } =\frac { x-A }{ c } \)
A is assumed mean c is common factor.
Here A= 15, C = 5
σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x c
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
If 3 is added to each value, we get 8, 13, 18,23, 28 as new values.

∴ σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
S.D. doesn't change when a number is added or subtracted to the values.

p²x² + (P² -Comparing this with ax' + bx + c = 0, we have
a=p²
b=p²-q²
c =-q²
D = b²-4ac
= (P²-q²)-4xp²x-q²
= (P²-q²)²+ 4p² q²
= (P²+q²)²>0
So, the given equation has real roots given by
\(\alpha =\frac { -b-\sqrt { D } }{ 2a } =\frac { -({ p }^{ 2 }-{ q }^{ 2 })+({ p }^{ 2 }+{ q }^{ 2 }) }{ { 2p }^{ 2 } } \)
\(=\frac { { q }^{ 2 } }{ { p }^{ 2 } } \)
\(\beta =\frac { -b-\sqrt { D } }{ 2a } =\frac { -({ p }^{ 2 }-{ q }^{ 2 })+({ p }^{ 2 }+{ q }^{ 2 }) }{ { 2p }^{ 2 } } \)
=-1

We have 6 = 2¹ x 3¹ and
20 = 2 x 2 x 5 = 2² x 5¹
You can find HCF (6, 20) = 2 and LCM (6, 20) = 2 x 2 x 3 x 5 = 60.
As done in your earlier classes. Note that HCF (6, 20) = 21 = product of the smallest power of each common prime factor in the numbers.
LCM (6, 20) = 2² x 3¹ x 5¹ = 60.
= Product of the greatest power of each prime factor, involved in the numbers.

Let A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) be the given paints. One way at showing that ABCD is a square is to use the property that all its sides should be equal and both its diagonals should be equal.
Now,
AB = \(\sqrt { (1-4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
BC = \(\\ \sqrt { (4+1)^{ 2 }+(2+1)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
CD = \(\sqrt { (-1+4)^{ 2 }+(-1-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
DA =\(\sqrt { (1+4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
AC = \(\sqrt { (1+1)^{ 2 }+(7+1)^{ 2 } } =\sqrt { 4+64 } =\sqrt { 68 } \)
BD = \(\\ \sqrt { (4+4)^{ 2 }+(2-4)^{ 2 } } =\sqrt { 64+4 } =\sqrt { 68 } \)
Since, AB = BC = CD = DA and AC = BD, all the four sides at the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is a square.

\(\frac { { x }^{ 2 }-16 }{ x+4 } \div \frac { x-4 }{ x+4 } =\frac { \left( x+4 \right) \left( x-4 \right) }{ \left( x+4 \right) } \times \left( \frac { x+4 }{ x-4 } \right) \) = x + 4

Range R = 13.67
Largest value L = 70.08
Range R = L - S
13.67 = 70.08-S
S = 70.08 - 13.67 = 56.41
Therefore, the smallest value is 56.41

 f(x) = \(\frac { 2x+1 }{ x-9 } \)
If x - 9 = 0,then
x = 9

The domain is all values of x that make the expression defined.
\(i.e., (-\infty, 9) \cup(9, \infty) \)
\(i.e., (x / x \neq 9) \Rightarrow R-\{9\}\)

Let the x intercept be ‘a’ and y intercept be ‘– a’.
The equation of the line in intercept form is \(\frac { x }{ a } +\frac { y }{ b } =1\)
gives \(\frac { x }{ a } +\frac { y }{ -a } =1\) (Here b = – a)
Therefore, x − y = a ...(1)
Since (1) passes through (5, 7)
Therefore, 5 - 7 = a gives a = − 2
Thus the required equation of the straight line is x − y = − 2 ; or x − y + 2 = 0

Let r be the radius of the hemisphere.
Given that, base area = \(\pi\)r² = 1386 sq. m
T.S.A. = 3 \(\pi\)r² sq.m
= 3 x 1386 = 4158
Therefore, T.S.A. of the hemispherical solid is 4158 m².

Let BD =  x cm, then DC = (6 – x)cm
AD is bisector of\(\angle\) A
Therefore by Angle Bisector Theorem
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac { 10 }{ 14 } =\frac { x }{ 6-x } \quad \frac { 5 }{ 7 } =\frac { x }{ 6-x } \)
So, 12x = 30 we get, \(x=\frac { 30 }{ 12 } =2.5\)
Therefore, BD = 2.5 cm, DC = 6−x = 6−2.5 = 3.5 cm

2, 5, 10, 17,....,
= 1² + 1, 2² + 1, 3² + 1, 4² + 1 ...
Here every term is obtained by adding 1 to its square
The general term a_n = n² + 1

\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } } \)=\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta } } \) [multiply numerator and denominator by the conjugate of 1 - cos\(\theta \)]
=\(\sqrt { \frac { (1+cos\theta { ) }^{ 2 } }{ (1-cos\theta { ) }^{ 2 } } } \) =\(\frac { 1+cos\theta }{ \sqrt { si{ n }^{ 2 }\theta } } \) [since sin²\(\theta \) + cos²\(\theta \) = 1]
=\(\frac { 1+cos\theta }{ sin\theta } =cosec\theta +cot\theta \)

x² + x + 7 = 0
Let y=x2+x+7
Step 1:

Step 2:
Points to be plotted: (-4, 19), (-3, 13), (-2, 9), (-1, 7), (0, 7), (1, 9), (2, 13), (3, 19), (4, 27)
Step 3:
Draw the parabola and mark the co-ordinates of the parabola which intersect with the x-axis.

Step 4:
The roots of the equation are the points of intersection of the parabola with the x axis. Here the parabola does not intersect the x axis at any point.
So, we conclude that there is no real roots for the given quadratic equation.

Draw the parabola using the points (-5, 44), (-4, 30), (-3, 18), (-2, 8), (-1, 10), (0, -6), (1, -10), (2, -12), (3, -12), (4, -10)

To solve the equation X² - 5x - 14 = 0, subtract X² - 5x - 14 = 0 from y = X² - 5x - 6.
 is a straight line parallel to x axis.
The co-ordinates of the points of intersection of the line and the parabola forms the solution set for the
equation X² - 5x - 14 = 0.
∴ Solution {-2, 7}

Given radius r = 3.6 cm
Length of the tangents PA = PB = 6.2cm

Construction:
Steps
(1) with centre at o, drawn a circle of radius 3.6 cm.
(2) Draw a line OP = 7.2 cm,
(3) Draw a perpendicular bisector of OP, which cuts OP at M.
(4) With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 6.2 cm.

Construction:
Steps (1) Draw QR = 6.5 cm.
Steps (2) Draw \(\angle RQE={ 60 }^{ 0 }\)
Steps (3) Draw \(\angle FQE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector XY to ER which intersects QF at O and ER at G.
Steps (5)  With O as center and OQ as radius drawn a circle
Steps (6) XY intersects QR at G. On XY, from G marked an arc at M, such that GM = 4.5 cm
Steps (7) Drawn AB through M which is parallel to QR
Steps (8)  AB meets the circle at P and S
Steps (9) Joined QP and RP Then \(\triangle\)PQR is the required triangle.
Steps (10) Here \(\triangle\)SQR is also another required triangle.

Let AB be the ladder and CA be the wall with . the window at A.
Also, BC = 2.5 m and CA = 6 m
From Pythagoras theorem
AB² =  BC²+ CA²
= (2.5)² + (6)²
= 42.25
AB = 6.5
Thus, length at the ladder is 6.5 m.

We are given that sin 3A = cos (A - 26°) ...(1)
Since sin 3A = cos(90° - 3A) we can write (1) as
cos (90° - 3A) = cos (A - 26°)
Since 90° - 3A and A - 26° are both acute angles
90° - 3A = A - 26°
which gives A = 29°

Volume of a cylinder \(=\pi { r }^{ 2 }h\)
Volume of a cone \(=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
Volume of a sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
Their ratio V₁ : V₂ : V₃

\(h:\frac { h }{ 3 } :\frac { 4r }{ 3 } \)
3h: h: 4r
3h : h : 2(2r)    (where 2r = h)
∴  V₁ : V₂ : V₃ = 3: 1 : 2

CV =\(\frac { \sigma }{ \bar { x } } \) x 100 ⇒ \(\bar { x } =\frac { \sigma }{ CV } \) x 100
\(\bar { x } =\frac { 15.6 }{ 6.9 } \) x 100  = 22.6

The number of rose plants in the 1st, 2nd, 3rd, . . . rows are
23,21, 19, ... 5
It forms an A.P.
Let the number of rows in the flower bed be n.
Then a = 23, d = 21 - 23 = -2/a = 5.
As, a_n = a + (n - 1)d i.e. t_n = a + (n - 1)d
We have 5 = 23 + (n - 1)(-2)
i.e. -18 = (n - 1)(-2)
n = 10
ஃ There are 10 rows in the flower bed.

R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

Since the given points are collinear, the area a the triangle formed by them must be 0, i.e.,
= \(\frac { 1 }{ 2 } \) [2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0
= \(\frac { 1 }{ 2 } \) [-4k] = 0
k = 0 
Area of ΔABC
= \(\frac { 1 }{ 2 } \)[2(0 + 3) + 4(-3 - 3) + 6(3 - 0)] = 0

In right triangle HIJ,
sin11° =\(\frac { IJ }{ HJ } \)
0.1908 = \(\frac { IJ }{ 180 } \); IJ = 34.34 km
Distance  of J to the North of H = 34.34km.

King spade, Queen spade, Jack spade are removed
∴ total number of cards = 52 - 3 = 49.
(i) Probability (diamond)= \(\frac { 13 }{ 49 } \)
(ii) Probability (queen) \(\frac { 4-1 }{ 49 } =\frac { 3 }{ 49 } \)
(iii) Probability (spade) = \(\frac { 13-3 }{ 49 } =\frac { 10 }{ 49 } \)
(iv) Probability (heart bearing number 5) = \(\frac { 13-5 }{ 49 } =\frac { 8 }{ 49 } \)

Given lines are 5x - 6y - 2 and 3x + 2y = 10.
Let us solve the equations to get the point of intersection

\(x=\frac{32}{14}=\frac{16}{7} \)
Substituting \(x=\frac{16}{7} \text { in }(2) \)
\(3\left(\frac{16}{7}\right)+2 y =10 \)
\(2 y =10-\frac{48}{7}=\frac{22}{7} \)
\(y =\frac{11}{7} \)
The point of intersection is \(\left(\frac{16}{7}, \frac{11}{7}\right)\)
Slope of the line 4x - 7y + 13 = 0 is
\(-\frac{a}{b}=\frac{-4}{-7}=\frac{4}{7}\)
Slope of the perpendicular line is \(-\frac{7}{4}\)
Now, equation of the line passing through \(\left(\frac{16}{7}, \frac{11}{7}\right)\) and having slope \(m=-\frac{7}{4}\) is
\(y-y_{1} =m\left(x-x_{1}\right) \)
\(y-\frac{11}{7} =\frac{-7}{4}\left(x-\frac{16}{7}\right) \)
\(\frac{7 y-11}{7} =\frac{-7 x}{4}+4 \)
28y - 44 = -49x + 112
49x + 28y - 156 = 0

We have  \(0.\overline { 123 } \) = 0.123123123....
= 0.123 + 0.000123 + 0.000000123 +....
\(\frac{t_{2}}{t_{1}}=\frac{0.000123}{0.123}=\frac{0.123}{123}=0.001\)
It is a G.P. with a = 0.123 and r = 0.001
Sum of infinity \(=\frac{a}{1-r} \)
\(=\frac{0.123}{1-0.001}=\frac{0.123}{1-\frac{1}{1000}} \)
\(=\frac{0.123}{\frac{999}{1000}}=\frac{123}{999} \)
\(\therefore 0 . \overline{123}=\frac{41}{333}\)

Hollow Hemisphere 
Internal diameter = 6 cm
Internal radius 'r' = 3 cm
External diameter = 10 cm
External radius 'R' = 5 cm
\(\left.\begin{array}{l} \text { Volume of hemisphere (or) } \\ \text {Volume of material used } \end{array}\right\}=\frac{2}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \text { cu. units }\)
\(=\frac{2}{3} \pi\left(5^{3}-3^{3}\right) \)
\(=\frac{2}{3} \pi(125-27)=\frac{196 \pi}{3} \mathrm{~cm}^{3} \)
Cylinder
Diameter = 14 cm
radius = 7 cm
height = h
Volume of cylinder \(=\pi r^{2} h\ cu. units \)
\(=\pi(7)^{2} h \)
\(=49 \pi h \mathrm{~cm}^{3} \)
Given that hollow hemisphere is melted and cast into a solid cylinder
Volume of cylinder = volume of hollow hemisphere
\(49 \pi h =\frac{196 \pi}{3} \)
\(h =\frac{196}{3 \times 49}=\frac{4}{3}=1.33 \)
Height of the cylinder = 1.33 cm.

A, B, C ⊆ N
f : A ⟶ B be defined by f(x) = 2x + 1
g : B ⟶ C be defined by g(x) = x²
f o g = f [g (x)] = f(x²) = 2x² + 1
g o f = g [f (x)] = g(2x - 1) = (2x + 1)²
= 4x² + 4x + 1
'x can take any real value and can produce any real value. Thus, the domain and range of f o g and g o f is R (Set of real numbers).