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Mêēt Ghøträ 6 years, 4 months ago
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Shravan Kumar 6 years, 4 months ago
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????? ?????? 6 years, 4 months ago
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Sadique Khan 6 years, 4 months ago
Posted by Pranam Shetty 6 years, 4 months ago
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Yogita Ingle 6 years, 4 months ago
(x+a)+(x-b) = x (x - b) + a (x -b)
= x2 - bx + ax - ab
= x2 - (b - a)x - ab
Posted by Nayan Nanjappa 6 years, 4 months ago
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Charan Pinnpureddy 6 years, 4 months ago
Posted by The Insane Guy 6 years, 4 months ago
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Sneha Gowda 6 years, 4 months ago
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Posted by Abhishek Sahariya 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
Let the present age of Nisha be x years.
Then, as per given condition
Asha's age (in years) is 2 more than the square of her daughter Nisha's age.
So, Asha's age = x2 + 2
Difference in their ages = x2 + 2 - x
And, Asha's age when Nisha's age will be (x2 + 2) years = ( x2 + 2) + (x2 + 2 - x)
If Asha's age is one year less than 10 times the present age of Nisha = 10x - 1
So, As per given condition
(x2 + 2 - x) + (x2 + 2) = 10x - 1
{tex}\Rightarrow{/tex} 2x2 - x + 4 = 10x - 1
{tex}\Rightarrow{/tex} 2x2 - 11x + 5 = 0
{tex}\Rightarrow{/tex} 2x2 - 10x - x + 5 = 0
{tex}\Rightarrow{/tex} 2x(x - 5) - 1(x - 5) = 0
{tex}\Rightarrow{/tex} (x - 5)(2x - 1) = 0
{tex}\Rightarrow{/tex} x - 5 = 0 or 2x - 1 = 0
{tex}\Rightarrow{/tex} x = 5 or {tex}x = \frac{1}{2}{/tex}
{tex}\Rightarrow{/tex} x = 5 [as age can't be a fraction]
{tex}\Rightarrow{/tex} x2 + 2 = 52 + 2 = 27
Hence, Nisha's age is 5 years and Asha's age is 27 years.
Posted by Ivin Renjith 6 years, 4 months ago
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I ❤ My India??{ Diya }? 6 years, 4 months ago
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Yogita Ingle 6 years, 4 months ago
For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such that
a = bq + r , where 0≤r<b
Explanation: Thus, for any pair of two positive integers a and b; the relation
a = bq + r , where 0≤r<b
will be true where q is some integer.
Manjeet Yadav 6 years, 4 months ago
Posted by Madhavanunni Km 6 years, 4 months ago
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Aadi Ahluwalia 6 years, 4 months ago
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Sia ? 6 years, 3 months ago
Let the present age of one friend be x years.
Also, sum of ages of both friends = 20 years
hence age of 2nd friend will be (20 - x) years.
4 years ago, age of 1st friend = (x - 4 ) years.
age of 2nd friend= (20-x)- 4 = (16-x) years.
According to the question;
(x - 4 )( 1 6 - x ) = 48
{tex}\Rightarrow{/tex} x2 - 20x + 112 = 0
Let D be the discriminant of this quadratic. Then,
D ={tex}b^2-4ac{/tex} = 400 - 448 = -48 < 0. (here, a=1 b=-20, c=112)
So, above equation does not have real roots. Hence, the given situation is not possible.
Posted by Subhash Chandra Sneh Lata 6 years, 4 months ago
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Khushi ?? 6 years, 4 months ago
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Sia ? 6 years, 3 months ago
Given,
tan A = n tan B
{tex} \Rightarrow{/tex} tanB = {tex}\frac{1}{n}{/tex}tan A
{tex}\Rightarrow{/tex} cotB = {tex}\frac { n } { \tan A }{/tex}..........(1)
Also given,
sin A = m sin B
{tex}\Rightarrow{/tex} sin B = {tex}\frac{1}{m}{/tex}sin A
{tex}\Rightarrow{/tex} cosec B = {tex}\frac { m } { \sin A }{/tex}.....(2)
We know that, cosec2B - cot2B = 1, hence from (1) & (2) :-
{tex} \quad \frac { m ^ { 2 } } { \sin ^ { 2 } A } - \frac { n ^ { 2 } } { \tan ^ { 2 } A } = 1{/tex}
{tex}\Rightarrow \quad \frac { m ^ { 2 } } { \sin ^ { 2 } A } - \frac { n ^ { 2 } \cos ^ { 2 } A } { \sin ^ { 2 } A } = 1{/tex}
{tex}\Rightarrow \quad \frac { m ^ { 2 } - n ^ { 2 } \cos ^ { 2 } A } { \sin ^ { 2 } A } = 1{/tex}
{tex}\Rightarrow{/tex} m2 - n2cos2A = sin2A
{tex}\Rightarrow{/tex} m2 - n2cos2A = 1 - cos2A
{tex}\Rightarrow{/tex} m2 - 1 = n2cos2A - cos2A
{tex}\Rightarrow{/tex} m2 - 1 = (n2 - 1) cos2A
{tex}\Rightarrow \quad \frac { m ^ { 2 } - 1 } { n ^ { 2 } - 1 } ={/tex} cos2A
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