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Posted by Preet Kotiya 6 years, 10 months ago
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Kartik Tomar 6 years, 10 months ago
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Posted by Geetanand Yadav/ Rao Sahab??????✌??? 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
(a - b)x + (a + b)y = a2 - 2ab - b2 ...(1)
(a + b)(x + y) = a2 + b2 {tex}\Rightarrow{/tex} (a + b)x + (a + b)y = a2 + b2 ....(2)
Subtracting equation (2) from (1), we obtain:
(a - b)x - (a + b)x = (a2 - 2ab - b2) - (a2 + b2)
{tex}\Rightarrow{/tex} (a - b - a - b)x = -2ab - 2b2
{tex}\Rightarrow{/tex} -2bx = -2b(a + b)
{tex}\Rightarrow{/tex} x = a + b
Substituting the value of x in equation (1), we obtain:
(a - b) (a + b) + (a + b)y = a2 - 2ab - b2
{tex}\Rightarrow{/tex} a2 - b2 + (a + b)y = a2 - 2ab - b2 {tex}\Rightarrow{/tex} (a + b)y = -2ab
{tex}\Rightarrow y = \frac{{ - 2ab}}{{a + b}}{/tex}
Posted by Shivani Patel 6 years, 10 months ago
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Posted by Abhay Agrahari 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
{tex}\sqrt { 3 x ^ { 2 } + 6 } = 9{/tex}
3x2 + 6 = 81
3x2 = 81 - 6 = 75
x2 = 25
{tex}\therefore{/tex} x = ± 5
Hence, Positive root = 5
Posted by Prateek Sharma 6 years, 10 months ago
- 2 answers
Gaurav Seth 6 years, 10 months ago
The numbers are :-
45 and 27
Lets find their HCF
BY Euclid's division Lemma :-
a = bq + r
45 = 27 x 1 + 18
27 = 18 x 1 + 9
18 = 9 x 2 + 0
HCF = d = 9
Given:-
d = 27x+45y
9 = 27x + 45y
9 = 27 - 18 x 1
18 = 45 - 27 x 1
Therefore ,
9 = 27 - [ 45 x 27 x 1 ] x 1
= 27 x 2 - 45
9 = 27 x 2 + 45 x [ -1 ]
x = 2
y = -1
Geetanand Yadav/ Rao Sahab??????✌??? 6 years, 10 months ago
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Geetanand Yadav/ Rao Sahab??????✌??? 6 years, 10 months ago
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Geetanand Yadav/ Rao Sahab??????✌??? 6 years, 10 months ago
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Dhirendra Behera 6 years, 10 months ago
Geetanand Yadav/ Rao Sahab??????✌??? 6 years, 10 months ago
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Madhur Dubey ??? 6 years, 10 months ago
Posted by Taneya Gupta 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let {tex} \alpha{/tex} and {tex} \frac { 1 } { \alpha }{/tex} be the zeros of (a2 + 9)x2 + 13x + 6a.
Then, we have
{tex} \alpha \times \frac { 1 } { \alpha } = \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ 1 = {tex} \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ a2 + 9 = 6a
⇒ a2 - 6a + 9 = 0
⇒ a2 - 3a - 3a + 9 = 0
⇒ a(a - 3) - 3(a - 3) = 0
⇒ (a - 3) (a - 3) = 0
⇒ (a - 3)2 = 0
⇒ a - 3 = 0
⇒ a = 3
So, the value of a in given polynomial is 3.
Posted by Ashwani Yadav 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
The number which appears most often in a set of numbers.
Example: in {6, 3, 9, 6, 6, 5, 9, 3} the Mode is 6 (it occurs most often).
Chetna ?✌️ 6 years, 10 months ago
Posted by Ashwani Yadav 6 years, 10 months ago
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Posted by Jeevak 1527 6 years, 10 months ago
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Chetna ?✌️ 6 years, 10 months ago
Posted by Diltej Singh Maan 6 years, 10 months ago
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Posted by Shubhendra Pratap Singh 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
We have,
{tex}\frac{1}{{(x - 1)(x - 2)}} + \frac{1}{{(x - 2)(x - 3)}}{/tex} {tex}+ \frac{1}{{(x - 3)(x - 4)}} = \frac{1}{6}{/tex}
{tex}\Rightarrow{/tex}{tex} (x - 3)(x - 4) + (x - 1)(x - 4) + (x - 1)(x - 2) = {/tex}{tex}\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}
[{tex}\because{/tex} Multiplying both sides by (x -1)(x - 2)(x - 3)(x - 4)]
{tex}\Rightarrow{/tex} {tex}x^2 - 4x - 3x + 12 + x^2 - 4x - x + 4 + x^2 - 2x - x + 2 ={/tex} {tex}\frac{1}{6}{/tex}{tex}[(x - 1)(x - 2)(x - 3)(x - 4)]{/tex}
{tex}\Rightarrow{/tex} {tex}3x^2 - 15x + 18 ={/tex} {tex}\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}3(x^2 - 5x + 6) ={/tex} {tex}\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}18[x^2 - 3x - 2x + 6] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}18[x(x - 3) - 2(x - 3)] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} {tex}18(x - 3)(x - 2) = (x - 1)(x - 2)(x - 3)(x - 4){/tex}
{tex}\Rightarrow{/tex} 18 = (x - 1)(x - 4)
{tex}\Rightarrow{/tex} 18 = x2 - 4x - 1x + 4
{tex}\Rightarrow{/tex} x2 - 5x + 4 - 18 = 0
{tex}\Rightarrow{/tex} x2 - 5x - 14 = 0
In order to factorize x2 - 5x - 14, we have to find two numbers 'a' and 'b' such that.
a + b = - 5 and ab = -14
Clearly, -7 + 2 = -5 and (-7)(2) = -14
{tex}\therefore{/tex} a = -7 and b = 2
Now,
x2 - 5x - 14 = 0
{tex}\Rightarrow{/tex} x2 - 7x + 2x - 14 = 0
{tex}\Rightarrow{/tex} x(x - 7) + 2(x - 7) = 0
{tex}\Rightarrow{/tex} (x - 7)(x + 2) = 0
{tex}\Rightarrow{/tex} x - 7 = 0 or x + 2 = 0
{tex}\Rightarrow{/tex} x = 7 or x = -2
Posted by Vishnu Sharma 6 years, 10 months ago
- 3 answers
Shubham Tiwari 6 years, 10 months ago
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Sia ? 6 years, 4 months ago
We have, {tex} \frac{1}{{\sec A + \tan A}} - \frac{1}{{\cos A}} = \frac{1}{{\cos A}} - \frac{1}{{\sec A - \tan A}}{/tex}
{tex}\Rightarrow \frac{1}{{\sec A + \tan A}} + \frac{1}{{\sec A - \tan A}} = \frac{1}{{\cos A}} + \frac{1}{{\cos A}}{/tex}
LHS {tex}= \frac{1}{{\sec A + \tan A}} + \frac{1}{{\sec A - \tan A}}{/tex}
{tex}= \frac{{\sec A - \tan A + \sec A + \tan A}}{{(\sec A + \tan A)(\sec A - \tan A)}}{/tex}
{tex}= \frac{{2\sec A}}{{{{\sec }^2}A - {{\tan }^2}A}}{/tex} {tex} \left[ {\because (a + b)(a - b) = ({a^2} - {b^2}} \right]{/tex}
{tex}= \frac{{2\sec A}}{1}{/tex} {tex} \left[ {\because {{\sec }^2}A - {{\tan }^2}A = 1} \right]{/tex}
= 2secA
RHS {tex}= \frac{1}{{\cos A}} + \frac{1}{{\cos A}}{/tex}
{tex}= \frac{{1 + 1}}{{\cos A}}{/tex}
{tex}= \frac{2}{{\cos A}}{/tex}
= 2 secA
LHS = RHS
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