Clearly bucket forms frustum of a cone such that the radii of its circular ends are r₁ = 28 cm, r₂ = 7 cm, h = 45 cm
Capacity of the bucket = volume of the frustum
\(\Rightarrow \frac { 1 }{ 3 } \times \pi h[{ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 }]\)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 45[{ 28 }^{ 2 }+{ 7 }^{ 2 }+28\times 7)]\)
= 22 x 15 x (28 x 4 + 7 + 28)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)

PQ II RS


Also

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

Let assumed mean A = 30
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 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
To convert the values for 100, all the values will be multiplied by 2. Therefore the new values are 40, 50, 60, 70, 80.
Let A = 60, C = 10

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 10 
=\(\sqrt { 2 } \) x 10
= 10\(\sqrt { 2 } \)
S.D. also be multiplied by 2. It is also true for the division also.

9x²-9(a+b)x+(2a²+5ab+2b²)=0
Comparing this with ax² + bx + c = O.
a =9
b = -9(a + b)
c = (2a² + 5ab + 2b²)
∴ ∆=B²-4AC
⇒ 81(a+b)²-36(2a²+5ab+2b²)
⇒ 9a² + 9b² - 18ab
⇒ 9(a - b)2> 0
∴ the roots are real and given by
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 12a+6b }{ 18 } =\frac { 2a+b }{ 3 } \)
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 6a+12b }{ 18 } =\frac { a+2b }{ 3 } \)

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.

By joining B to D, you will get too triangles ABD and BCD.
Now, the area of ΔABD
=\(\frac { 1 }{ 2 } \)[ -5(-5 - 5) + (-4)(5 -7) + 4(7 + 5)]
=\(\frac { 1 }{ 2 } \)(50 + 8 + 48)
=\(\frac { 106 }{ 2 } \) = 53 square units.
Also, the area of ΔBCD
=\(\frac { 1 }{ 2 } \) = [-4(-6 - 5) - 1(5 + 5) + 4(-5 + 6)]
=\(\frac { 1 }{ 2 } \)(44 - 10 + 4)
= 19 square units.
So, the area of quadrilateral ABCD
= 53 + 19 = 72 square units.

\(\left[ \begin{matrix} x & y-z & z+3 \end{matrix} \right] +\left[ \begin{matrix} y & 4 & 3 \end{matrix} \right] =\left[ \begin{matrix} 4 & 8 & 16 \end{matrix} \right] \)
x + y = 4  ...(1)
y - z + 4 = 8 ..(2)
From (3), we get z =10
From (2), we get y -10 + 4 = 8
From (2), we get y = 14
From (1) we get  x + 14 = 4
x = -10
x = -10, y = 14, z = 10

From (1) and (2) we have
\(\frac{A Q}{A B} =\frac{A R}{A D} \)
\(\frac{A B}{A Q} =\frac{A D}{A R} \)
\(\frac{A Q+Q B}{A Q} =\frac{A R+R D}{A R} \)
\(1+\frac{Q B}{A Q} =1+\frac{R D}{A R} \)
\(\Rightarrow \frac{Q B}{A Q} =\frac{D R}{A R} \)

f(k) - 2k - 1
f o f(k) = 5
f(f(k)) = f(2k - 1) = 5
⇒ 2(2k-1) -1 = 5
4k - 2 -1 = 5 ⇒ 4k = 8
k = 2

For comparing two data, first we have to find their coefficient of variations
Mean \(\bar { { x }_{ 1 } } \) = 155 cm, variance σ₁² = 72.25 cm²
Therefore standard deviation σ₁ = 8.5
Coefficient of variation C.V₁ = \(\frac { { \sigma }_{ 1 } }{ \bar { { x }_{ 1 } } } \) x 100%
C.V₁ = \(\frac { 8.5 }{ 15.5 } \) x 100% = 5.48% (for heights)
Mean \(\bar { { x }_{ 2 } } \) = 155 cm, variance σ₂² = 72.25 kg²
Standard deviation σ₂ = 5.3 kg
Coefficient of variation CV₂ = \(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100%
C.V₂ = \(\frac { 5.3 }{ 46.50 } \) x 100% = 11.40% (for weights)
C.V₁ = 5.48% = and CV₂ = 11.40%
Since C.V₂ > C.V₁,  the weight of the students is more varying than the height.

Given slope 'm' = 0
tan θ = 0 = tan 0⁰
θ = 0⁰

First term a = 5; common difference d = 6.
A.P.is given by., a + d, a + 2d, a + 3d,......
In this case 5, 5 + 6, 5 + 2(6),5 + 3 (6),...
5, 11, 17 ,23,...
The required A.P. is 5, 11,17,23,....

Given that, radius r = 7 cm
Now, total surface area of the cone = \(\pi\)r(l + r)sq. units
T.S.A = 704 cm²
704 = \(\frac{22}{7}\times7(l+7)\)
32 = l + 7 implies l = 25 cm
Therefore, slant height of the cone is 25 cm.

1 + \(\frac { co{ t }^{ 2 }\theta }{ 1+cosec\theta +1 } \) = 1+ \(\frac { cose{ c }^{ 2 }\theta -1 }{ cosec\theta +1 } \) [since cosec²-1 = cot²\(\theta \)
= 1+\(\frac { (cosec\theta +1)(cosec\theta -1) }{ cosec\theta +1 } \) 
1 +( cosec\(\theta \)-1) = cosec\(\theta \)

\(\Delta ACD\sim \Delta ABC\)
So,
\(\cfrac { AC }{ AB } =\cfrac { AD }{ AC } \)
AC²  = AB ·AD
Similarly \(\Delta BCD\sim \Delta BAC\)
So,
\(\cfrac { BC }{ BA } =\cfrac { BD }{ BC } \)
BC²  = BA·BD
From (1) and (2)
\(\cfrac { { BC }^{ 2 } }{ AC^{ 2 } } =\cfrac { BA.BD }{ AB.AD } =\cfrac { BD }{ AD } \)

No. of balls required \(=\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \frac { 4 }{ 3 } \pi \times { \left( \frac { r }{ 4 } \right) }^{ 3 } } \)
\(={ r }^{ 3 }\times \frac { 4 }{ r } \times \frac { 4 }{ r } \times \frac { 4 }{ r } =64\)

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

-2, 2, -2, 2, -2
t₂ - t₁ = 2-(-2) = 4
t₃ - t₂ = -2 -2 = -4
t₄ - t₃ = 2 - (-2) = 4
It is not an A.P.

F(5) = 3x² - 10
= 3(5)²- 10 = 75 - 10 = 65

We know that diagonals at a parallelogram bisect each other.
So, the coordinates of the mid-point at AC = coordinates of the mid-point at BD.
i.e., \(\left[ \frac { 6+9 }{ 2 } ,\frac { 1+4 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 2+3 }{ 2 } \right] \)
\(\left[ \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 5 }{ 2 } \right] \)
\(\frac { 15 }{ 2 } =\frac { 8+P }{ 2 } \)
P = 7

LHS (AB)C
AB = \({ \left[ \begin{matrix} 1 & -1 & 2 \end{matrix} \right] }_{ 1\times 3 }{ \left[ \begin{matrix} 1 & -1 \\ 2 & 1 \\ 1 & 3 \end{matrix} \right] }_{ 3\times 2 }=\left[ \begin{matrix} 1-2+2 & -1-1+6 \end{matrix} \right] =\left[ \begin{matrix} 1 & 4 \end{matrix} \right] \)
(AB)C = \({ \left[ \begin{matrix} 1 & 4 \end{matrix} \right] }_{ 1\times 2 }\times { \left[ \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right] }_{ 2\times 2 }=\left[ \begin{matrix} 1+8 & 2-4 \end{matrix} \right] =\left[ \begin{matrix} 9 & -2 \end{matrix} \right] \) ....(1)
RHS = A(BC)
BC = \({ \left[ \begin{matrix} 1 & -1 \\ 2 & 1 \\ 1 & 3 \end{matrix} \right] }_{ 3\times 2 }\times { \left[ \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right] }_{ 2\times 2 }=\left[ \begin{matrix} 1-2 & 2+1 \\ 2+2 & 4-1 \\ 1+6 & 2-3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 3 \\ 4 & 3 \\ 7 & -1 \end{matrix} \right] \)
A(BC) = \({ \left[ \begin{matrix} 1 & -1 & 2 \end{matrix} \right] }_{ 1\times 3 }{ \left[ \begin{matrix} -1 & 3 \\ 4 & 3 \\ 7 & -1 \end{matrix} \right] }_{ 3\times 2 }\)
A(BC) = \(\left[ \begin{matrix} -1-4+14 & 3-3-2 \end{matrix} \right] =\left[ \begin{matrix} 9 & -2 \end{matrix} \right] \) ....(2)
From (1) and (2), (AB)C = A(BC).

Total number of balls = blue balls + red balls
n(S) = 12 + x
(i) The probability that it will be a red ball:
Let A be the event of selecting a red ball
n(A) = x
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)} \)
\(\mathrm{P}(\mathrm{A})=\frac{x}{12+x} \)
(ii) If 8 more red balls are put in the bag then the number of red balls = x + 8
Total number of balls = 12 + x + 8
Probability of drawing a red ball
\(\mathrm{P}(\mathrm{B}) =\frac{n(B)}{n(S)} \)
\(=\frac{x+8}{12+x+8}=\frac{x+8}{x+20} \)
If the probability of drawing a red ball will be twice that of the probability (i), then
\(\frac{x+8}{x+20}=2 \times\left[\frac{x}{12+x}\right]\)
(x + 8)(12 + x) = 2x (x + 20)
12x + 96 + x² + 8x = 2x² + 40x
2x²  - x² + 40x - 12x - 8x - 96 = 0
x² + 20x - 96 = 0
(x - 4) (x + 24) = 0
x = 4 and x = - 24 (not possible)
By applying the value of x in P(A)
we get P(A) \(=\frac{4}{12+4}=\frac{4}{16}\)
\(P(A)=\frac{1}{4}\)
x = 4

Let the amount he saved in the first year be x
Then x + (x + 100) + (x + 200) + ... 10 terms
= 16500
x+ x + 100 + x + 200 + ... 10 terms = 15500
(x + x + ... 10 terms) + (100 + 200 + ... 9 terms)
= 16500
\(10 x+\frac{9}{2}[2(100)+8(100)]=16500\)
10x = 16500 - 4500
10x = 12000
\(x=\frac{12000}{10}=1200\)
His 1^st year saving = Rs 1200

Given that intercepts are in the ratio 2 : 5
\(\frac{a}{b} =\frac{2}{5} \)
\(a =\frac{2 b}{5} \)
Equation of the line in Intercepts form is \(\frac{x}{a}+\frac{y}{b}=1\)
\(\frac{x}{\left(\frac{2 b}{5}\right)}+\frac{y}{b}=1 \)
\(\frac{5 x}{2 b}+\frac{y}{b}=1 \)
5x + 2y = 2b
This passes through ( 1, - 4)
5(1) + 2(-4) = 2b
\(5-8=2 b \Rightarrow b=-\frac{3}{2} \)
\(a =\frac{2 b}{5}=\frac{2\left(-\frac{3}{2}\right)}{5}=-\frac{3}{5} \)
Equation of a straight line is
\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{\left(-\frac{3}{5}\right)}+\frac{y}{\left(-\frac{3}{2}\right)}=1\)
\(\frac{5 x}{-3}+\frac{2 y}{-3}=1\)
5x + 2y + 3 = 0.

Let h and r be the height and radius of the cone and cylinder.
Let l be the slant height of the cone.
Given that, h = 2.4 cm and d = 1.4 cm ; r = 0.7 cm
Here, total surface area of the remaining solid} C.S.A. of the cylinder + C.S.A. of the cone + area of the bottom
= 2\(\pi\)rh + \(\pi\)rl + \(\pi\)r² sq.units
Now, \(\\ \\ l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { 0.49+5.76 } =\sqrt { 6.25 } =2.5cm\)
Area of the remaining solid = 2\(\pi\)rh + \(\pi\)rl + \(\pi\)r² sq.units
= \(\pi\)r(2h + l + r)
\(\frac { 22 }{ 7 } \times 0.7\times [(2\times 2.4)+2.5+0.7]\)
Therefore, total surface area of the remaining solid is 17.6 m²

LHS
\(\left( \frac { 1+ta{ n }^{ 2 }A }{ 1+co{ t }^{ 2 }A } \right) =\frac { 1+ta{ n }^{ 2 } }{ 1+\frac { 1 }{ ta{ n }^{ 2 }A } } \)
=\(\frac { 1+ta{ n }^{ 2 }A }{ \frac { ta{ n }^{ 2 }A+1 }{ ta{ n }^{ 2 }A } } =ta{ n }^{ 2 }A...(1)\)
RHS
\({ \left( \frac { 1-tanA }{ 1-cotA } \right) }^{ 2 }{ \left( \frac { 1-tanA }{ 1-\frac { 1 }{ tanA } } \right) }^{ 2 }\) 
=\({ \left( \frac { 1-tanA }{ \frac { tanA-1 }{ tanA } } \right) }^{ 2 }\) = (-tan A)² = tan²A        ...(2)
From (1) and (2),\(\left( \frac { 1+ta{ n }^{ 2 }A }{ 1+co{ t }^{ 2 }A } \right) ={ \left( \frac { 1-tan }{ 1-cotA } \right) }^{ 2 }\)

\(f(x)= \begin{cases}x+2 & \text { if } x>1 \\ 2 & \text { if }-1 \leq x \leq 1 \\ x-1 & \text { if }-3<x<-1\end{cases}\)
i) f(3) = 3 + 2 = 5
ii) f(0) = 2
iii) f(-1.5) = -1.5 - 1 = -2.5
iv) f(2) + f( -2) = ( 2 + 2 ) + ( -2 -1)
= 4 - 3 = 1

(2x-3)(x+2)=0
2x² - 3x + 4x - 6 = 0
2x² + 1x-6 = 0
Let y = 2x² +X - 6= 0

Step 2:
The points to be plotted: (-4,22), (-3, 9), (-2, 0), (-1, -5), (0, -6), (1, -3), (2,4), (3,15), (4, 30)
Step 3:
Draw. the parabola and mark the co-ordinates of the intersecting point of the parabola with the x-axis.

Step 4:
The points of intersection of the parabola with the x-axis are (-2, 0) and (1.5,0).
Since the parabola intersects the x-axis at two points, the equation has real and unequal roots
∴ Solution {-2, 1.5}

Step 1: Draw the graph of y = 2x² by preparing the table of values as below

Step 2 : To solve 2x² - x - 6 = 0, subtract 2x² - x - 6 = 0 from y = 2x²

The equation y = x + 6 represents a straight line. Draw the graph of y = x + 6 by forming table of values as below

Step 3 : Mark the points of intersection of the curve y = 2x² and the line y = x + 6. That is, (–1.5, 4.5) and (2,8)
Step 4 : The x coordinates of the respective points forms the solution set {–1.5,2} for 2x² - x - 6 = 0

Given, radius = 4 cm
Construction
Step 1 : With O as the centre, draw a circle of radius 4 cm.
Step 2 : Take a point L on the circle. Through L draw any chord LM.
Step 3 : Take a point M distinct from L and N on the circle, so that L, M and N are in anti clockwise direction. Join LN and NM.
Step 4 : Through L draw a tangent TT' such that \(\angle\)TLM =\(\angle\)MNL
Step 5 : TT' is the required tangent.

Construction:
Step (1) Draw \(\bar { AB } =5.5cm\)
Step (2) Draw \(\angle BAE={ 25 }^{ 0 }\)
Step (3) Draw \(\angle FAE={ 90 }^{ 0 }\)
Step (4) Drawn the perpendicular bisector XY to AB which intersects AF at O and AB at G.
Step (5) With O as center and OA as radius drawn a circle
Step (6) XY intersects AB at G. On XY from G marked an arc at M such that GM = 4 cm
Step (7) Drawn PQ through M which is parallel to AB.
Step (8) PQ meets the circle at C and S.
Step (9) Joined AC and BC. Now \(\triangle\)ABC is the required triangle