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

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

OA· OB = OC . OD (Given)

so, \(\cfrac { OA }{ OC } =\cfrac { OD }{ OB } \)
Also we have 
(vertically opposite angles) ...(2)
From (1) and (2)
\(\Delta AOD\sim \Delta COB\)( SAS similarity criterion)
So

(corresponding angles of similar triangles)

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.

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.

x²-x-1 = 0, Here a = 1, b = -1, c = -1
∆ = b²-4ac = (-1)²-4 x 1 x -1
= 1 + 4
= 5 > 0
∴ The roots are real and unequal 

Mean \(\bar x\) = 15
Coefficient of variation (C.V) = 48
Coefficient of variation C.V \(=\frac{\sigma}{x} \times 100 \%\)
\(48 =\frac{\sigma}{15} \times 100 \% \)
\(\frac{48 \times 15}{100} =\sigma \)
\(\sigma=\frac{36}{5}=7.2 \)
Standard deviation \(\sigma=7.2 \)

Let AB = x be the distance between foot of the house and the observer at the median of the road.
DB = 2x is the width of the road.
Height of the house BC = \(4 \sqrt{3} m\)
From the right triangle \(\triangle\) ABC
\( \therefore \tan 30^{\circ} =\frac{B C}{A B} \)
\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x} \)
\(x =4 \sqrt{3} \times \sqrt{3}=4 \times 3=12 \mathrm{~m} \)
\(\text { Width of the road } =2 \times x=2 \times 12=24 \mathrm{~m}\)
Width of the road = 24 m.

The 17^th term can be obtained by subtracting the sum of first 16 terms from the sum of first 17 terms
\({ S }_{ 17 }=\frac { 5\times \left( 17 \right) ^{ 2 } }{ 2 } +\frac { 3\times 17 }{ 2 } =\frac { 1445 }{ 2 } +\frac { 51 }{ 2 } =748\)
\({ s }_{ 16 }=\frac { 5\times \left( 16 \right) ^{ 2 } }{ 2 } +\frac { 3\times 16 }{ 2 } =\frac { 1280 }{ 2 } +\frac { 48 }{ 2 } =664\)
Now, t₁₇ = S₁₇ - S₁₆ = 748 - 664 = 84

Required Ratio = C.S.A. of the sphere: C.S.A. of the cylinder : C.S.A. of the cone
\(4\pi { r }^{ 2 }:2\pi rh:\pi rl,\ (l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { 2r }^{ 2 } } =\sqrt { 2r } units)\)
\(=4:2:\sqrt { 2 } =2\sqrt { 2 } :\sqrt { 2 } :1\).

f(x) = 2x + 3, g(x) = 1 - 2x, h(x) = 3x
Now, (f o g)(x) = f(g(x)) = f(1 - 2x) = 2(1 - 2x) + 3 = 5 - 4x
Then, (f o g) o h(x) = (f o g)(3x) = 5 - 4(3x) = 5 - 12x..(1)
(g o h)(x) = g(h(x)) = g(3x) = 1 - 2(3x) = 1 - 6x
So, f o (g o h)(x) = f(1 - 6x) = 2(1 - 6x) + 3 = 5 - 12x...(2)
From (1) and (2), we get (f o g) oh = f o (g o h)

f: N ⟶ N
f(x) = 2x - 1
N = {1,2,3,4,5, ...}
When x = 1, f(1) = 2( 1) - 1 = 1
When x = 2, f(2) = 2(2) - 1 = 3
When x = 3, f(3) = 2(3) - 1 = 5
When x = 4, f(4) = 2(4) - 1 = 7
When x = 5, f(5) = 2(5) - 1 = 9

This function maps every element from the domain to element that is twice minus one the original. 2x - 1 is always an odd number when x\(\in\)N.
Clearly, each element from the domain is mapped to different element in the co-domain. So, the function is one-to-one. On the other hand, there are no elements in the domain that would map to even numbers. So, the function is not onto. 
Hence f is one one but not onto.

In Is \(\triangle\)ABC ~ \(\triangle\)PQR
\(\frac { PQ }{ AB } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } ;\frac { QR }{ BC } =\frac { 4 }{ 10 } =\frac { 2 }{ 5 } \)
Since \(\frac { 1 }{ 2 } \neq \frac { 2 }{ 5 } ,\frac { PQ }{ AB } \neq \frac { QR }{ BC } \)
The corresponding sides are not proportional.
Therefore \(\triangle\)ABC  is not similar to \(\triangle\)PQR.

\(\Delta ACD\sim \Delta ABC\)
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)
\(\cfrac { { BC }^{ 2 } }{ AC^{ 2 } } =\cfrac { BA.BD }{ AB.AD } =\cfrac { BD }{ AD } \)

cot 85° + cos 75°
= cot(90° - 5°) + cos(90° - 15°)
= tan 5° + sin 15°

Radius of the cylinder be r
Height of the cylinder be h
Total surface area of the article
= CSA of cylinder + CSA of 2 hemispheres
= 2ㅠrh + 2πr² = 2πr(h + 2r)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times (10+2\times 3.5)\)
= 22 x 17 = 374 cm²

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

1,1,1,2,2,2,3,3,3
t₂ - t₁ = 1-1 = 0
t₃ - t₂ = 1-1 = 0
t₄ - t₃ = 2-1 = 1
Here t₂ - t₁ ≠ t₃ - t₂
ஃ It is not an A.P.

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

The area of the triangle formed by the given points is equal to
= \(\frac { 1 }{ 2 } \) [-1.5 (-2 - 4) + 6 (4 - 3) + (-3) (3 + 2)]
= \(\frac { 1 }{ 2 } \) [9 + 6 - 15] = 0
We can have a triangle at area 0 square units? What does this mean?
If the area of a triangle is 0 square units, then its vertices will be collinear.

Product of face values 6: { (1, 6), (2, 3), (6, 1), (3,2)}
Difference of face value 5: {(1, 6), (6, 1)}
P(product 6) = \(\frac { 4 }{ 6\times 6 } =\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)
p( difference 5) = \(\frac { 2 }{ 2\times 6 } =\frac { 1 }{ 18 } \)

Let CF be the height of the man; AD be the height of the window; BC is the distance between the observer and the house.
From the right triangle \(\triangle\)CBD
\( \tan 45^{\circ} =\frac{D B}{B C} \)
\(1 =\frac{D B}{5} \)
DB = 5m     ...(1)
From the right triangle CBA
\( \tan 60^{\circ} =\frac{A B}{C B} \)
\(\sqrt{3} =\frac{A D+D B}{5} \)
\(5 \sqrt{3} =\mathrm{AD}+5 \quad[\because \text { from }(1) D B=5 \mathrm{~m}] \)
\(\mathrm{AD} =5 \sqrt{3}-5=5(\sqrt{3}-1) \)
\(\mathrm{AD} =5(1.732-1) \)
\( {[\text {Given } \sqrt{3}=1.732] }\)
= 5 x 0.732 = 3.660
Height of the window = 3.66 m

(i) Surface area of a sphere Radius of sphere = r
Surface area = 4r² sq. units
(ii) Curved surface area of cylinder 
Radius of cylinder = r
Height of cylinder = r + r = 2r
Curved surface area \(=2 \pi r h\ sq. units \)
\(=2 \pi r(2 \mathrm{r}) \)
\(=4 \pi r^{2} \text { sq. units } \)
(iii) Ratio of the areas \(=\frac{\text { Surface area of sphere }}{\text { CSA of cylinder }} \)
\(=\frac{4 \pi r^{2}}{4 \pi r^{2}}=\frac{1}{1} \)
Ratio = 1 : 1.

(a) Slope of AB = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { -2+2 }{ 6-1 } =0\)
(b) Slope of CD \(\frac { 1-1 }{ 2-5 } =\frac { 0 }{ -3 } =0\)

Let f (x) = 12 (x⁴ - x³)
= 12x³ (x - 1)
g(x) = 8(x⁴- 3x³ + 2x²)
= 8x² (x² - 3x + 2)
= 8x² (x - 1) (x - 2)
Given LCM = 24 x³ (x - 1) (x - 2)
We know f(x) x g (x) = LCM x GCD
\(\therefore \mathrm{GCD} =\frac{f(x) \times \mathrm{g}(\mathrm{x})}{L C M} \)
\(= \frac{12 x^{3}(x-1) \times 8 x^{2}(x-1)(x-2)}{24 x^{3}(x-1)(x-2)} \)
GCD = 4x² (x - 1)

The points are P(-1, 5, 3), Q(6, -2) , R(-3, 4)
Area of Δ PQR = \(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
= \(\frac{1}{2}\) { (3 + 24 - 9) - (18 + 6 - 6) }
= \(\frac{1}{2}\) { 18 - 18 } = 0
Therefore, the given points are collinear.

Since the remainders are 4, 5 respectively the required number is the HCF of the number 445 - 4 = 441, 572 - 5 = 567.
567  = 441  x 1 + 126
441 = 126 x 3 + 63 
126 = 63 x 2 + 0
Therefore HCF of 441, 567 = 63 and so the required number is 63

x²-9=0
Let y=x²-9
Step 1:

Step 2:
The points to be plotted: (-4,7), (-3, 0), (-2, -5), (-1, -8), (0, -9), (1, -8), (2, -5), (3, 0), (4, 7)
(v) Real and equal roots
Step 3: 
Draw the parabola and mark the co-ordinates of the parabola which intersect the x-axis.

Step 4:
The roots of the equation are the co-ordinates of the intersecting points (-3, 0) and (3, 0) of the parabola with the x-axis which are -3 and 3 respectively.
Step 5:
Since there are two points of intersection with the x axis, the quadratic equation has real and unequal roots.
∴ Solution{-3, 3}

y=x²+3x-4

Draw the parabola using the points (-4, 0), (-3, -4), (-2, -6), (-1, -6), (0, -4), (1, 0), (2, 6), (3, 14), (4,24).

To solve: X² + 3x - 4 = 0 subtract X² + 3x - 4 = 0 from y = X² + 3x - 4

The points of intersection of the parabola with the x axis are the points (-4, 0) and (1, 0), whose x - co-ordinates (-4, 1) is the solution, set for the equation X2 + 3x - 4 = 0.

Construction:
(1) With O as the centre, draw a circle of radius 4.5 cm.
(2) Taken a point L on the circle through L drawn as chord LM
(3) Taken a point M distinct from L and N on the circle so that L, M, N are anti-clock wise direction. Joined LN and NM.
(4) Through 'L' drawn a tangent TT' such that \(\angle T L M=\angle M N L\)
(5) TT' is the required tangent.

Given, diameter (d) = 6 cm, we find radius \((r)=\cfrac { 6 }{ 2 } =3cm\)

Construction
Step 1: With centre at O, draw a circle of radius 3 cm.
Step 2: Draw a line OP of length 8 cm.
Step 3: Draw a perpendicular bisector of OP, which cuts OP at M.
Step 4: With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
Step5: Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 7.4 cm.
Verification : In the right angle triangle OAP,PA² = OP² - OA² = 64 -9 = 55
\(PA=\sqrt { 55= } 7.4\ cm\) (approximately) .