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Ask QuestionPosted by Aditya Kumar 7 years, 5 months ago
- 3 answers
Prashant Chaudhary 7 years, 5 months ago
Aditya Kumar 7 years, 5 months ago
Posted by Sanjay Chopra 7 years, 5 months ago
- 1 answers
Posted by Aayushi Agrawal 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
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}
Posted by Aayushi Agrawal 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
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}
Posted by Aditya Choudhery 7 years, 5 months ago
- 4 answers
Posted by Ayush Gupta 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Let n = 4q + 1 (an odd integer)
{tex}\therefore \quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}
{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \quad \text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}
{tex}= 16{q^2} + 8q{/tex}
{tex}= 8 \left( 2 q ^ { 2 } + q \right){/tex}
= 8m, which is divisible by 8.
Posted by Anish Agarwal 7 years, 5 months ago
- 4 answers
Divya Yadav 7 years, 5 months ago
Hyfa Kasim 7 years, 5 months ago
Posted by Vikash Awase 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
The maximum number of zeroes that a polynomial of degree 3 can have is three
Posted by Vikash Awase 7 years, 5 months ago
- 1 answers
Posted by Prachi Bhomrajka 6 years, 4 months ago
- 1 answers
Posted by Vishesh Kundu 7 years, 5 months ago
- 1 answers
Posted by Manish Arora 7 years, 5 months ago
- 4 answers
Sai Kumar Yellale 7 years, 5 months ago
Sonali Aggarwal 7 years, 5 months ago
H2=B2+P2
H stands for hypotenuse, B stands for base and P stands for perpendicular of the right angled triangle.
Priya Kumari 7 years, 5 months ago
Arohi . 7 years, 5 months ago
Posted by Sahil Hussain 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
The given system of equations may be written as
2x + 3y = 7
So, 2x + 3y - 7 = 0
and (p + q) x +(2p -q)y - 21 = 0
The given system of equations is of the form
a1x + b1 y + c1 =0
a2x + b2y + c2=0
where, a1 = 2, b1 = 3, c1 = -7 and a2 =(p+ q), b2= (2p -q), c2= -21
We have, {tex}\frac{a_1}{a_2}=\frac2{p\;+\;q}\;,\frac{b_1}{b_2}=\frac3{2p-\;q}\;\text{ and }\frac{c_1}{c_2}=\frac{-7}{-21}=\frac{\;1}3{/tex}
If {tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }{/tex} then the pair of linear equations has infinitely many solutions.
The given system of equations will have infinite number of solutions, if
{tex}\frac { 2 } { p + q } = \frac { 3 } { 2 p - q } = \frac { 7 } { 21 }{/tex}
{tex}\Rightarrow \quad \frac { 2 } { p + q } = \frac { 3 } { 2 p - q } = \frac { 1 } { 3 }{/tex}
{tex}\Rightarrow \quad \frac { 2 } { p + q } = \frac { 1 } { 3 } \text { and } \frac { 3 } { 2 p - q } = \frac { 1 } { 3 }{/tex}
{tex}\Rightarrow{/tex} p + q = 6 and 2p - q = 9
{tex}\Rightarrow{/tex} (p + q) + (2p - q ) = 6 + 9
{tex}\Rightarrow{/tex} 3p = 15 [On adding]
{tex}\Rightarrow{/tex} p = 5
Putting p = 5 in p + q = 6 or, 2p - q = 9, we get q = 1.
Hence, the given system of equations will have infinitely many solutions, if p = 5 and q = 1.
Posted by Fairy Mehta 7 years, 5 months ago
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Vampire Dukedom 7 years, 5 months ago
Posted by Sahil Prajapati 7 years, 5 months ago
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Posted by Sai Kumar Yellale 7 years, 5 months ago
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Posted by Rajneesh Kumar 5 years, 8 months ago
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Posted by Suraj Jat 7 years, 5 months ago
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Sachin Kumar 7 years, 5 months ago
Posted by Mohit Gehlot 7 years, 5 months ago
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Posted by Pratyush Pal 5 years, 8 months ago
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Posted by Teesha Nagdev 7 years, 5 months ago
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Posted by Xyz Abc 7 years, 5 months ago
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Posted by Mayank Kunwar 7 years, 5 months ago
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Posted by Priyanshu Yadav 7 years, 5 months ago
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Posted by Thakurz Crimenal 7 years, 5 months ago
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Posted by Monu Gupta 7 years, 5 months ago
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Posted by Sudharm Bhargat 7 years, 5 months ago
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Posted by Sudharm Bhargat 7 years, 5 months ago
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Posted by Ravi Chandran 7 years, 5 months ago
- 2 answers
Ravi Chandran 7 years, 5 months ago
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Shruti Gupta 7 years, 5 months ago
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