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Ask QuestionPosted by Arshalan Khan 8 years, 3 months ago
- 3 answers
Posted by Lokesh Kumar 8 years, 3 months ago
- 2 answers
Posted by Amal Jose 8 years, 3 months ago
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Posted by Lakshay Agarwal 8 years, 3 months ago
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Posted by Reena Karim 8 years, 3 months ago
- 1 answers
Rashmi Bajpayee 8 years, 3 months ago
Given: {tex}p\left( x \right) = \sqrt 2 {x^2} + kx + 3{/tex}
If (x - 2) is a factor of p(x), then
p(2) = 0
=> {tex}\sqrt 2 {\left( 2 \right)^2} + k\left( 2 \right) + 3 = 0{/tex}
=> {tex}4\sqrt 2 + 2k + 3 = 0{/tex}
=> {tex}2k = - 4\sqrt 2 - 3{/tex}
=> {tex}k = {{ - \left( {4\sqrt 2 + 3} \right)} \over 2}{/tex}
Posted by Akanksha Nigam 8 years, 3 months ago
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Posted by Prabhjot S 8 years, 3 months ago
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Posted by Ishika Jain 8 years, 3 months ago
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Rashmi Bajpayee 8 years, 3 months ago
x = {tex}3\sqrt 2 + \sqrt 3 {/tex}
Then, {tex}{1 \over x} = {1 \over {3\sqrt 2 + \sqrt 3 }} \times {{3\sqrt 2 - \sqrt 3 } \over {3\sqrt 2 - \sqrt 3 }} = {{3\sqrt 2 - \sqrt 3 } \over {15}}{/tex}
And {tex}x + {1 \over x} = 3\sqrt 2 + \sqrt 3 + {{3\sqrt 2 - \sqrt 3 } \over {15}}{/tex} = {tex}{{45\sqrt 2 + 15\sqrt 3 + 3\sqrt 2 - \sqrt 3 } \over {15}}{/tex} = {tex}{{48\sqrt 2 + 14\sqrt 3 } \over {15}}{/tex}
Now, {tex}{x^3} + {1 \over {{x^3}}} = \left( {x + {1 \over x}} \right)\left( {{x^2} + {1 \over {{x^2}}} - x \times {1 \over x}} \right){/tex}
= {tex}\left( {x + {1 \over x}} \right)\left[ {{{\left( {x + {1 \over x}} \right)}^2} - 2 \times x \times {1 \over x} - 1} \right]{/tex}
= {tex}\left( {x + {1 \over x}} \right)\left[ {{{\left( {x + {1 \over x}} \right)}^2} - 3} \right]{/tex}
= {tex}\left( {{{48\sqrt 2 + 14\sqrt 3 } \over {15}}} \right)\left[ {{{\left( {{{48\sqrt 2 + 14\sqrt 3 } \over {15}}} \right)}^2} - 3} \right]{/tex}
= {tex}\left( {{{48\sqrt 2 + 14\sqrt 3 } \over {15}}} \right)\left[ {{{4608 + 588 + 1344\sqrt 6 } \over {225}} - 3} \right]{/tex}
= {tex}\left( {{{48\sqrt 2 + 14\sqrt 3 } \over {15}}} \right)\left[ {{{5196 + 1344\sqrt 6 - 675} \over {225}}} \right]{/tex}
= {tex}\left( {{{48\sqrt 2 + 14\sqrt 3 } \over {15}}} \right)\left( {{{4521 + 1344\sqrt 6 } \over {225}}} \right){/tex}
= {tex}\left( {{{48\sqrt 2 + 14\sqrt 3 } \over {15}}} \right)\left( {{{1507 + 448\sqrt 6 } \over {75}}} \right){/tex}
= {tex}{{72336\sqrt 2 + 21504\sqrt {12} + 21098\sqrt 3 + 6272\sqrt {18} } \over {1125}}{/tex}
= {tex}{{72336\sqrt 2 + 43008\sqrt 3 + 21098\sqrt 3 + 18816\sqrt 2 } \over {1125}}{/tex}
= {tex}{{91152\sqrt 2 + 64106\sqrt 3 } \over {1125}}{/tex}
Posted by Sahil Pachouri 8 years, 3 months ago
- 1 answers
Posted by Samyak Mehta 8 years, 3 months ago
- 2 answers
Arpit Mishra 8 years, 3 months ago
Posted by Aniket Shil 8 years, 3 months ago
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Hi M Anshu Pathak 8 years, 3 months ago
Soumya Ghoshal 8 years, 3 months ago
In which figure have three side and addition of three angles is 180°,this is called triangle.
Posted by Surendra Saroj 8 years, 3 months ago
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Posted by Vyom Singh 8 years, 3 months ago
- 1 answers
Posted by Zahabiya Talwala 8 years, 3 months ago
- 1 answers
Sridarsan Sah 8 years, 3 months ago
Posted by Sanket Mane 8 years, 3 months ago
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Posted by Naresh Jangir 8 years, 3 months ago
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Posted by Sanket Mane 8 years, 3 months ago
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Sridarsan Sah 8 years, 3 months ago
Posted by Rajat Jain 8 years, 3 months ago
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Soumya Ghoshal 8 years, 3 months ago
Let the side x cm.
Semi-Perimeter ={tex}\frac{{(x+x+x)}}{2}=\frac{{3x}}{2}{/tex}cm
Now , area of the equilateral triangle is
= {tex}\sqrt{{(3x/2)(x/2)(x/2)(x/2)}}{/tex}
={tex}\frac{{√3x^2}}{4}{/tex}cm²
Posted by Srikant Kumar 8 years, 3 months ago
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Posted by Nitin Sharma 8 years, 3 months ago
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Hans Raj 8 years, 3 months ago
52773636
= 5x10000000 + 2x1000000 + 7x100000 x 7x10000 x 3000 + 6x100 + 3x10 + 6
Posted by Amit Singh 8 years, 3 months ago
- 1 answers
Posted by Vanshita Yadav 8 years, 3 months ago
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Posted by Abhishek Keshri 8 years, 3 months ago
- 2 answers
Posted by Patel Heli K 8 years, 3 months ago
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Posted by Harleen Thind 8 years, 3 months ago
- 2 answers
Hans Raj 8 years, 3 months ago
25 - (1 x 1/3) x 3 x 16
= 25 - (1/3 x 3 x 16)
= 25 - 16
= 9
Posted by Sahana Uppal 8 years, 3 months ago
- 1 answers
Dharmendra Kumar 8 years, 3 months ago
8x3+27y3=23x3+33y3=(2x)3+(3y)3 [by using anbn=(ab)n]
Now by using identity
(a+b)3=a3+b3+3ab(a+b)
a3+b3=(a+b)3 - 3ab(a+b) ---------(1)
Now 8x3+27y3=(2x)3+(3y)3
= (2x+3y)3-3(2x)(3y)(2x+3y) [by using equation (1) put a=2x and b=3y]
= (2x+3y)3 - 18xy(2x+3y)
= (12)3 - 18×6×12 [given 2x+3y =12 and xy= 6]
=1728 -1296
= 432
Posted by Purva Mittal 8 years, 3 months ago
- 0 answers
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Rashmi Bajpayee 8 years, 3 months ago
2a4
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