\(\sqrt { 6x+7 } -(2x-7)=0\)
\(\sqrt { 6x+7 } -(2x-7)\)
Squaring both sides
6x+7=(2x-7)²
\(\sqrt { 6x+7 } -(2x-7)\)
\(\Rightarrow 6x+7=4x^{ 2 }-28x+49\\ \Rightarrow 4x^{ 2 }-34x+42=0\\ \Rightarrow 2x^{ 2 }-14x-3x+21=0\\ \Rightarrow 2x(x-7)-3(x-7)=0\\ \Rightarrow (x-7)(2x-3)=0\\ \Rightarrow x=7\quad or\quad x=\frac { 3 }{ 2 } \)

Tangents drawn from an exterior point to a circle are equal. 
AD = AF = x cm (say)
BD = BE = y cm (say)
CE = CF = z cm (say) 
Now AB = 8 cm
⇒  AD + BD = 8 cm
⇒  x + y = 8    .........(i)
BC = 10cm
⇒ BE + CE = 10 cm
⇒ y + z = 10     .......(ii)
CA = 12 cm
⇒ CF + AF = 12 cm
⇒  z+x = 12   .....(iii)
On adding (i), (ii) and (iii), we have
2(x + y + x) = 30
x + y + z = 15     .....(iv)
Subtracting (ii) from (iv), we have x = 15 - 10 = 5  
Subtracting (iii) from (iv), we have y = 15 - 12 = 3
Subtracting (i) from (iv), we have z = 15 - 8 = 7
ஃ AD = x cm = 5 cm,  BE = y cm = 3 and CF = z cm = 7 cm

Here, a is the first term, d is the common difference and \({ a }_{ n }\) is the \({ n }^{ th }\) term of an A.P.
Here, \(a=-18.9,d=2.5,{ a }_{ n }=3.6,n=......\)
​​​​​​We know that  \({ a }_{ n }=a+(n-1)d\)
\(\Rightarrow \) \(3.6 = -18.9 + (n-1)2.5\)
\(\Rightarrow \) \(3.6 = -18.9 + 2.5n - 2.5\)
\(\Rightarrow \) \(3.6 = -21.4 + 2.5n\)
\(\Rightarrow \) \(2.5n = 3.6 + 21.4\)
\(\Rightarrow \ n=\frac { 25 }{ 2.5 } =10\)   Hence, \(n = 10\)

16, 14 - Creativity involves in the use of skills and imagination to produce something new and innovative.
Vision - ability to think about future.

Total number of ways to draw ball from bag = 24
Number of ways to draw a white ball = x
\(\therefore \) Probability of drawing a white ball = \(\frac { x }{ 24 } \) 
A.T.Q., \(\frac { x }{ 24 } \)=y ....(i)
When 12 more white balls are added to the bag,
Number of balls in the bag = 24 + 12 = 36 Number of white balls in the bag = (x + 12)
\(\therefore \)Probability of drawing a white ball
=\(\frac { x+12 }{ 36 } \)
A.T.Q., \(\\ \frac { x+12 }{ 36 } =\frac { 5 }{ 3 } \times x\)
\(\Rightarrow \frac { x+12 }{ 36 } =\frac { 5 }{ 3 } \times \frac { x }{ 24 } \quad \{ using\quad (i)\} \)
\(\therefore \quad x+12=\frac { 36\times 5x }{ 72 } \)
\(\Rightarrow x+12=\frac { 35x }{ 2 } \)
\(\Rightarrow \) 2x+24=5x\(\Rightarrow \)3x=24\(\Rightarrow \)x=8

-n(2n+1)

Given: Two circles C₁ and C₂ touch each other at P where C₁ be inside of C₂. T be a point on the tangent at P, where TQ and TR are the tangents, segments to the circles. 
To prove: TQ = TR 
Proof: In circle C₁, TP and T Q are two tangents from an external point T. Then they are equal in length. 
I.e.,  TP = TQ ...(i)
Similarly, in circle C₂, TP = TR  .....(ii)
From (i) and (ii), we get 
TQ = TR Proved.

\(\angle \)TOQ + \(\angle \)TPQ = 180^o
\(\angle \)TOQ = 110^o
Also \(\angle \)TOQ = 2\(\angle \)TRQ
[angle subtended by an arc at centre of the circle is twice the angle subtended by it in alternate segment]
⇒ 110^o = 2 \(\angle \)TRQ ⇒ \(\angle \)TRQ = 55^o

Here, a = 14, d = 11 - 14 = -3
Let  a_n = -1 
 a + (n - 1)d = -1
14 + (n - 1)(-3) = -1
(n - 1)(-3) = -1 - 14
n - 1 = \(\frac{-15}{-3}\) = 5
n = 6
6th term of the AP is -1.

2x²+x+4=0
Now adding square of (\(\frac {1}{2}\) coefficient x) on both sides, we get
\(\Rightarrow \quad { x }^{ 2 }+\frac { 1 }{ 2 } x+{ \left( \frac { 1 }{ 4 } \right) }^{ 2 }=-2+{ \left( \frac { 1 }{ 4 } \right) }^{ 2 }\quad \Rightarrow \quad { \left( x+\frac { 1 }{ 4 } \right) }^{ 2 }=-2+\frac { 1 }{ 16 } \)
\(\Rightarrow \quad { \left( x+\frac { 1 }{ 4 } \right) }^{ 2 }=\frac { -32+1 }{ 16 } \quad \Rightarrow \quad { \left( x+\frac { 1 }{ 4 } \right) }^{ 2 }=\frac { -31 }{ 16 } \)
\(\Rightarrow \quad { \left( x+\frac { 1 }{ 4 } \right) }^{ 2 }={ \left( \frac { \sqrt { -31 } }{ 16 } \right) }^{ 2 }\)
Which is impossible as square root of negative integers is not possible.
Hence, the roots do not exist.

(2x-1)(x-3)=(x+5)(x-1)
2x²-6x-x+3=x²-x+5x-5
2x²-6x-x+3-x²+x-5x+5=0
x²-11x+8=0
Which is of the form ax²+bx+c=0. Hence the given equation is a quadratic equation.

Clearly, \(\angle QAB=60° and QBA=60°\)
[\(\because \Delta \)ABQ is equilateral]
So, \(\angle PAQ=\angle PAB+\angle QAB=60°+60°=120°\)
Similarly,  \(\angle PBQ=120°\)
\([\because PAB=PBA,\quad as\quad \angle PA=PB]\)   .....(i)
Now, in \(\Delta PAQ\quad and\quad \Delta PBQ,\)
Pa = PB    [tangents from external point]
\(\Rightarrow\) AQ = BQ   [\(\Delta\)ABQ equilateral]
\(\Rightarrow\) \(\angle\)PAQ =\(\angle\)PBQ   [each 120\(°\), shown above]
So, \(\Delta PAQ\cong \Delta PBQ\)  [by SAS congruence rule]
\(\Rightarrow\)\(\angle\)APQ = \(\angle\)BPQ   [by CPCT] ....(ii)
Let QP intersect AB at M.
Now, in \(\Delta\)PAM and \(\Delta\)PBM,
\(\angle\)APM = \(\angle\)BPM   [from Eq. (ii)]
\(\Rightarrow\)  PA = PB
\(\Rightarrow\)  PM = PM
So, \(\Delta\)PAM \(\cong \) \(\Delta\)PBM [[by SAS congruence rule]
\(\Rightarrow\)  AM = BM
and \(\angle\)AMP =  \(\angle\) BMP  [by CPCT]...(iii)
But  \(\angle\) AMP + \(\angle\)BMP =180\(°\)
\(\Rightarrow\) \(\angle\)AMP + \(\angle\)AMP =180\(°\)
\(\Rightarrow\) 2 \(\angle\)AMP = 180\(°\)
\(\Rightarrow\) \(\angle\) AMP = 90 \(°\)
From Eqs. (iii) and (iv), we get that QP bisects AB at right angles.

Let BC = h m be the height of the pedestal and CD = 1.6m be the length of the statue, which is standing on the pedestal.
Again, let point A be a fixed point on the ground such that the angles of elevation of the top of the statue and bottom of the statue (i.e. top of the pedestal) are
\(\angle D A B=60^{\circ} \text { and } \angle C A B=45^{\circ}\)
Also, let AB = x m.
In right angled \(\Delta\)ABD, \(\tan 60^{\circ}=\frac{P}{B}=\frac{B D}{A B}\)
\(\begin{array}{lll} \Rightarrow & \sqrt{3}=\frac{B C+C D}{x} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right] \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & \sqrt{3}=\frac{h+1.6}{x} & \\ \end{array}\)
\(\begin{array}{lll} \Rightarrow & h=\sqrt{3} x-1.6 \end{array}\) ...(i)
In right angled \(\Delta\)CBA, tan 45° \(=\frac{B C}{A B}\)
\(\begin{array}{ll} \Rightarrow & 1=\frac{h}{x} \quad\left[\because \tan 45^{\circ}=1\right] \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & x=h \end{array}\)
On putting x = h in Eq. (i), we get
\(h=\sqrt{3} h-1.6 \Rightarrow h(\sqrt{3}-1)=1.6\)
\(\Rightarrow \quad h=\frac{1.6}{(\sqrt{3}-1)} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)  [rationalising] 
\(\begin{aligned} & =\frac{1.6(\sqrt{3}+1)}{(\sqrt{3})^2-(1)^2}\left[\because(a+b)(a-b)=a^2-b^2\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1.6}{2}(\sqrt{3}+1)=0.8(\sqrt{3}+1) \mathrm{m} \end{aligned}\)
Hence, the height of the pedestal is \(0.8(\sqrt{3}+1) \mathrm{m}\).

Let PQ and PR be two tangents drawn from an external point P to a circle with centre O.

To prove \(\angle\)QOR = 180° – \(\angle\)QPR
or \(\angle\)QOR + \(\angle\)QPR = 180°
Proof In \(\Delta\)OQP and \(\Delta\)ORP,
PQ = PR   [\(\because\) tangents drawn from an external point are equal in length]
OQ = OR      [radii of circle]
OP  = OP   [common sides]
\(\therefore\)  \(\Delta\)OQP \(\cong\) \(\Delta\)ORP   [by SSS congruence rule]
Then, \(\angle\)QPO = \(\angle\)RPO    [by CPCT]
and \(\angle\)POQ = \(\angle\)POR    [by CPCT]
\(\left.\begin{array}{ll} \Rightarrow & \angle Q P R=2 \angle O P Q \\ \text { and } & \angle Q O R=2 \angle P O Q \end{array}\right\}\)    ...(i)
Now, in right angled \(\Delta\)OQP, \(\angle\)QPO + \(\angle\)QOP = 90°
\(\Rightarrow\)  \(\angle\)QOP = 90° - \(\angle\)QPO
\(\Rightarrow\)  2\(\angle\)QOP = 180°- 2 \(\angle\)QPO
[multiplying both sides by 2]
\(\Rightarrow\) \(\angle\)QOR = 180° - \(\angle\)QPR      [from Eq. (i)]
\(\Rightarrow\)   \(\angle\)QOR + \(\angle\)QPR = 180°      Hence proved.

Let the number of green marbles out of 24 marbles in a jar be x.
Probability of getting green marbles = \(\frac { 2 }{ 3 } \)
\(\frac { x }{ 24 } =\frac { 2 }{ 3 } \)
3x = 48
x = 16
 Number of green marbles = 16
Hence, the number of blue marbles = 24 - 16 
=8

Let A be the centre of the balloon (spherical) whose radius is r.
Let AB be the height of the ballon i.e., h units, such that
ㄥDPC=ፀ, ㄥAPB-ф
Since ΔPDA and ΔPCA are congruent, therefore
ㄥAPC=ㄥAPD=ፀ/2
In rt. ㄥed ΔPCA,
\(\frac { AC }{ AP } =sin\frac { \theta }{ 2 } \)
⇒ AP=\(AC\frac { 1 }{ sin\frac { \theta }{ 2 } } =r.cosec\frac { \theta }{ 2 } \) or
Consider rt . ㄥed ΔPBA, we have
\(\frac { AB }{ AP } \)=sin ф
AB=AP.sinф
h=r.cosec\( \frac { \theta }{ 2 } \).sinф
h=r sinф.cosec\( \frac { \theta }{ 2 } \).

Total number of points on the circle = 8
(i) Let E₁ = Event of getting arrow at number 8
\(\therefore\) Number of outcomes favourable to E₁ = 1
Probability that arrow cones at number 8,
\(P\left(E_1\right)=\frac{1}{8}\)
(ii) Let E₂ = Event of getting arrow at an odd number
Here, odd numbers are 1, 3, 5 and 7.
\(\therefore\) Number of outcomes favourable to E₂ = 4
Probability that arrow comes at an odd number,
\(P\left(E_2\right)=\frac{4}{8}=\frac{1}{2}\)
(iii) Let E₃ = Event of getting arrow at a number greater than 2, i.e. at 3, 4, 5, 6, 7 or 8
\(\therefore\) Number of outcomes favourable to E₃ = 6
Probability that arrow comes at a number greater than 2,
\(P\left(E_3\right)=\frac{6}{8}=\frac{3}{4}\)
(iv) P(a number less than 9) = \(\frac{8}{8}\)=1