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)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)

Given XY IIA C

So,

\(\therefore \Delta ABC\sim \Delta XbY\) (AAA similarity criterion)
So, \(\cfrac { ar(ABC) }{ ar(XBY) } =\left( \cfrac { AB }{ XB } \right) ^{ 2 }\)         ...(1)
ar(ABC) = 2ar(XBY)
\(\cfrac { ar(ABC) }{ ar(ABC) } =\cfrac { 2 }{ 1 } \)           ...(2)
From (1) and (2),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

PQ II RS


Also

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

Let assumed mean A = 30
C = 5

x	d' = \(\frac { x-30 }{ 5 } \)	d'2
20	-2	4
25	-1	1
30	0	0
35	1	1
40	2	4
	Σd' = 0	Σd'2 = 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 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

x	d' = \(\frac { x-60 }{ 10 } \)	d'2
40	-2	4
50	-1	1
60	0	0
70	1	1
80	2	4

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.

An arrow diagram
 

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

It is not a function as the vertical line PQ cuts the graph at two points

4 x 4

in right triangle ABC [see fig.(b)]
tan\(\theta \) =\(\frac { 8 }{ 3 } \)
= tan^-1(2.66)
\(\theta \) = \(69.4°\)(since tan \(69.4°\)=2.6604)
\(\angle \)BAC = \(69.4°\)

√2t² - 3t + 3√2 = 0
a = √2, b = -3, c = 3√2
∆ = b²-4ac
= (-3)² - 4 x √2 x 3√2
= 9 - 24
= -15 < 0.
∴ The roots are not real.

\(\sqrt { 1+\frac { 1 }{ { x }^{ 6 } } +\frac { 2 }{ { x }^{ 3 } } } =\sqrt { { 1 }^{ 2 }+2.1.\frac { 1 }{ { x }^{ 3 } } +{ \left( \frac { 1 }{ { x }^{ 3 } } \right) }^{ 2 } } \)
\(=\sqrt { { \left( 1+\frac { 1 }{ { x }^{ 3 } } \right) }^{ 2 } } \left| 1+\frac { 1 }{ { x }^{ 3 } } \right| \)

 43.5, 13.6, 18.9,38.4,61.4,29.8
Largest value L= 61.4
Smallest value S = 13.6
R = L - S
= 61.4 - 13.6 = 4
Co-efficient of range = \(\frac { L-S }{ L+S } \)
= \(\frac { 47.8 }{ 75 } =0.64\)
Range = 47.8; co-efficient of range = 0.64.

Given points (2, 3) and (- 7, - 1)
Equation of the line passing through (x₁ , y₁) and (x₂, y₂) is
\( \frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y-3}{-1-3}=\frac{x-2}{-7-2} \)
9 (y - 3) - 4 (x - 2)
9y - 27 = 4 x - 8
4 x - 9y +19 = 0

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

Given lines 4x + 5y − 13 = 0 .....(1)
x − 8y + 9 = 0 .....(2)
To find the point of intersection, solve equation (1) and (2)

\(\frac { x }{ 45-104 } =\frac { y }{ -13-36 } =\frac { 1 }{ -32-5 } \)
\(\frac { x }{ -59 } =\frac { y }{ -49 } =\frac { 1 }{ -37 } \)
\(x=\frac { 59 }{ 37 } \),\(y=\frac { 49 }{ 37 } \)
Therefore, the point of intersection (x, y) = \(\left( \frac { 59 }{ 37 } ,\frac { 49 }{ 37 } \right) \)
The equation of line parallel to Y axis is x = c.
It passes through (x, y) = \(\left( \frac { 59 }{ 37 } ,\frac { 49 }{ 37 } \right) \). Therefore, c = \(\frac { 59 }{ 37 } \)
The equation of the line x = \(\frac { 59 }{ 37 } \) gives 37x - 59 = 0

Given \(\triangle\)LMN is a right angled triangle with \(\angle\)L = 90^o
By Pythagoras theore
Since PL ∥ OQ,PL= r = OP
we have  NR=NP=NL−PL
[ NR and NP are tangents]
= (6 - r)cm
MR = MQ = ML- LQ = (8 - r) cm
[MR and MQ are tangents]
NM = NR+ RM
= (6 - r + 8 - r)cm
= (14 - 2r) cm
Now NM² = NL² + LM² [By pythagoras Theorem]
(14-2r)2 =  82+62
(14-2r)2 = 64+36

\( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} =\sec \theta+\tan \theta \)
\(\mathbf{L H S} =\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} \)
\(=\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}\)
[Multiplying the Numerator and denominator by \(\sqrt{1-\sin \theta}\)]
\( =\sqrt{\frac{1^{2}-\sin ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \)
\(=\sqrt{\frac{\cos ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right] \)
\(=\frac{\cos \theta}{1-\sin \theta} \)
\(=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta} \)
[Multiplying Numerator and denominator by \(1+\sin \theta\)]
\( =\frac{\cos \theta(1+\sin \theta)}{1^{2}-\sin ^{2} \theta}=\frac{\cos \theta(1+\sin \theta)}{\cos ^{2} \theta} \)
\({\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]} \)
\(=\frac{1+\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \)
\(=\sec \theta+\tan \theta=\text { RHS }\)

Given points (3, - 1), (a, 3) and (1, - 3)
Let the points be A (3, - 1), B (a, 3) and C (1,- 3)
Slope of AB = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{-1-3}{3-a}=\frac{-4}{3-a} \)
Slope of BC = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{3+3}{a-1}=\frac{6}{a-1} \)
Since, the points A, B and C are collinear.
Slope of AB = Slope of BC
\(\frac{-4}{3-a}=\frac{6}{a-1}\)
-2 (a - 1) = 3(3 - a)
-2a + 2 = 9 - 3a
3a - 2a = 9 - 2
a = 7

f(x) = 3 x -2, g(x) = 2x + k
f o g = f(g(x)) = f(2x + k) = 3(2x + k) - 2 = 6x + k - 2
Thus, f o g(x) = 6x + 3k - 2
g o f(x) = g(3x - 2) = 2(3x - 2) + k
Thus, g o f(x) = 6x - 4 + k
Given that f o g = g o f
Therefore, 6x + 3k-2 = 6z - 4 + k
6x - 6x + 3k - k = -4 + 2 ⇒ -1

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

Let the distance be 'd'
Speed = 500 km/hr
Time = 't' hours
Distance = Time x Speed
d(t) = 500 t

(i) Set builder form of R = ((x, y) | y = x - 2, x \(\in \) P, y \(\in \) Q}
(ii) Roster form R = {(5 , 3),(6 , 4)(7 , 5)}
(iii) Domain of R = {5, 6, 7} and range of R = {3, 4, 5}

A x B = {(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A x B}. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A x B}. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.

Let the number of small spheres obtained be n.
Let r be the radius of each small sphere and R be the radius of metallic sphere.
Here, R = 16 cm, r = 2 cm
Now, n\(\times\)(Volume of a small sphere) = Volume of big metallic sphere
\(n\left( \frac { 4 }{ 3 } { \pi r }^{ 2 } \right) =\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
\(n\left( \frac { 4 }{ 3 } \pi \times { 2 }^{ 3 } \right) =\frac { 4 }{ 3 } \pi \times { 16 }^{ 3 }\)
8n = 4096 gives n = 512
Therefore, there will be 512 small spheres.