Given: l and m are the tangent to a circle such that l || m, intersecting at A and B respectively.
To prove: AB is a diameter of the circle.
Proof:
A tangent at any point of a circle is perpendicular to the radius through the point of contact.
{tex}\therefore{/tex} {tex}\angle X A O = 90 ^ { \circ }{/tex}
and {tex}\angle Y B O = 90 ^ { \circ }{/tex}
Since {tex}\angle X A O + \angle Y B O = 180 ^ { \circ }{/tex} 
An angle on the same side of the transversal is 180°.
Hence the line AB passes through the centre and is the diameter of the circle.

We have to  find the greatest number that divides 445, 572 and 699 and leaves remainders of 4, 5 and 6 respectively. This means when the number divides 445, 572 and 699 leaves remainders 4, 5 and 6 is that
445 - 4 = 441, 572 - 5 = 567 and 699 - 6 = 693 are completely divisible by the  required number.For the highest number which divides the above numbers can be calculated by HCF .
Therefore, the required number  is the  H.C.F. of 441, 567 and 693 Respectively.
First, consider 441 and 567.
By applying Euclid’s division lemma, we get
567 = 441 {tex}\times{/tex} 1 + 126
441 = 126 {tex}\times{/tex} 3 + 63
126 = 63 {tex}\times{/tex} 2 + 0.
Therefore, H.C.F. of 441 and 567 = 63
Now, consider 63 and 693
again we have to  apply Euclid’s division lemma, we get
693 = 63 {tex}\times{/tex} 11 + 0.
Therefore, H.C.F. of 441, 567 and 693  is  63
Hence, the required number is 63. 63 is the highest number which divides 445,572 and 699 will leave 4,5 and 6 as remainder respectively.

We have,
{tex}8 x ^ { 2 } - 22 x - 21 = 0{/tex}
{tex}\Rightarrow{/tex} 8x² - 28x + 6x - 21 = 0
{tex}\Rightarrow{/tex} 4x(2x - 7) + 3 (2x - 7) = 0
{tex}\Rightarrow{/tex} (2x - 7) (4x + 3) = 0 {tex}\Rightarrow{/tex} 2x - 7 = 0 or, 4x + 3 = 0 {tex}\Rightarrow{/tex} {tex}x = \frac { 7 } { 2 } \text { or } x = - \frac { 3 } { 4 }{/tex}
Thus, {tex}x = \frac { 7 } { 2 } \text { and } x = - \frac { 3 } { 4 }{/tex} are two roots of the equation 8x² - 22x - 21 = 0

Let three consecutive numbers be x, (x + 1) and (x + 2)
Let x = 6q + r 0 {tex}\leq r < 6{/tex}
{tex}\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}
{tex}\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}
if x = 6q then which is divisible by 6
{tex}\text { if } x = 6 q + 1{/tex}
{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}
{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}
{tex}= 6 ( 3 q + 1 ) \cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}
which is divisible by 6
if x = 6q + 2
{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}
{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}
{tex}= 6 ( 2 q + 1 ) \cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}
Which is divisible by 6
{tex}\text { if } x = 6 q + 3{/tex}
{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}
{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}
which is divisible by 6
{tex}\text { if } x = 6 q + 4{/tex}
{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}
{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}
which is divisible by 6
if x = 6q + 5
{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}
{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}
which is divisible by 6
{tex}\therefore {/tex} the product of any three natural numbers is divisible by 6.

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Let the common ratio term of income be x and expenditure be y.
So, the income of first person is Rs.9x and the income of second person is Rs.7x.
And  the expenditures of first and second person is 4y and 3y respectively.
Then, Saving of first person =9x - 4y
and  saving of second person = 7x - 3y
As per given condition
 9x - 4y = 200
 {tex}\Rightarrow{/tex} 9x - 4y - 200=0    ... (i)
and, 7x - 3y = 200 
{tex}\Rightarrow{/tex} 7x - 3y - 200 =0   ..... (ii)
Solving equation (i) and (ii) by cross-multiplication, we have 
{tex}\frac { x } { 800 - 600 } = \frac { - y } { - 1800 + 1400 } = \frac { 1 } { - 27 + 28 }{/tex}
{tex}\frac { x } { 200 } = \frac { - y } { - 400 } = \frac { 1 } { 1 }{/tex}
{tex}\Rightarrow{/tex} x =200 and y =400
So, the solution of equations is x = 200 and y = 400.
Thus, monthly income of first person = Rs.9x = Rs.(9 {tex}\times{/tex} 200)= Rs.1800
and, monthly income of second person = Rs.7x = Rs.(7{tex}\times{/tex} 200)= Rs. 1400

According to the question,
{tex}(x - 5)(x - 6) = \frac{{25}}{{{{\left( {24} \right)}^2}}}{/tex}
{tex}\Rightarrow x(x - 6) - 5(x - 6) = \frac{{25}}{{{{(24)}^2}}}{/tex}
{tex}\Rightarrow {x^2} - 6x - 5x + 30 - \frac{{25}}{{{{(24)}^2}}} = 0{/tex}
{tex}\Rightarrow {x^2} - 11x + 30 - \frac{{25}}{{{{(24)}^2}}} = 0{/tex}
{tex}\Rightarrow {x^2} - 11x + \frac{{30 \times {{24}^2} - 25}}{{{{(24)}^2}}} = 0{/tex}
{tex} \Rightarrow {x^2} - 11x + \frac{{30 \times 576 - 25}}{{{{(24)}^2}}} = 0{/tex}
{tex} \Rightarrow {x^2} - 11x + \frac{{17280 - 25}}{{{{(24)}^2}}} = 0{/tex}
{tex}\Rightarrow {x^2} - \frac{{264x}}{{24}} + \frac{{145}}{{24}} \times \frac{{119}}{{24}} = 0{/tex}
{tex}\Rightarrow {x^2} - \left( {\frac{{145}}{{24}} + \frac{{119}}{{24}}} \right)x + \frac{{145}}{{24}} \times \frac{{119}}{{24}} = 0{/tex}
{tex}\Rightarrow {x^2} - \frac{{145}}{{24}}x - \frac{{119}}{{24}}x + \frac{{145}}{{24}} \times \frac{{119}}{{24}} = 0{/tex}
{tex}\Rightarrow x\left( {x - \frac{{145}}{{24}}} \right) - \frac{{119}}{{24}}\left( {x - \frac{{145}}{{24}}} \right) = 0{/tex}
{tex}\Rightarrow \left( {x - \frac{{145}}{{24}}} \right)\left( {x - \frac{{119}}{{24}}} \right) = 0{/tex}
{tex}\Rightarrow x - \frac{{145}}{{24}} = 0{/tex} or {tex}x - \frac{{119}}{{24}} = 0{/tex}
{tex}\Rightarrow x = \frac{{145}}{{24}}{/tex} or {tex}x = \frac{{119}}{{24}}{/tex}

Let α=5

Given α+β=0

5+β=0,β=-5

So the polynomial f(x)=(x-5)[x-(-5)]

=(x-5)(x+5)

=x²-25

Let a be the first term and d be the common difference of the given A.P.
Then, 8^th term = a₈ = a + 7d
and 2^nd term = a₂ = a + d
According to given information,
{tex}a _ { 8 } = \frac { a _ { 2 } } { 2 }{/tex}
{tex}\Rightarrow{/tex} 2a₈ = a₂
{tex}\Rightarrow{/tex} 2(a + 7d) = a + d
{tex}\Rightarrow{/tex} a + 13d = 0...(i)
Now, 11^th term = a₁₁ = a + 10d and 4^th term = a₄ = a + 3d
According to the given information,
{tex}a _ { 11 } - \frac { a _ { 4 } } { 3 } = 1{/tex}
{tex}\Rightarrow{/tex} 3a₁₁ - a₄ = 3
{tex}\Rightarrow{/tex} 3(a + 10d) - (a + 3d) = 3
{tex}\Rightarrow{/tex} 3a + 30d - a - 3d = 3
{tex}\Rightarrow{/tex} 2a + 27d = 3...(ii)
Multiplying equation (i) by 2, we get
2a + 26d = 0...(iii)
Subtracting (iii) from (ii), we get
d = 3
{tex}\Rightarrow{/tex} a + 13(3) = 0
{tex}\Rightarrow{/tex} a = -39
Now, 15^th term = a_{​​​​​​15} = a + 14d = -39 + 14(3) = -39 + 42 = 3

Hence, 15^th term is 3 .