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Ask QuestionPosted by Neeraj Dhiman 6 years, 10 months ago
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Posted by Sneha Singh 6 years, 10 months ago
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Khushboo Giri 6 years, 10 months ago
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- 2 answers
Posted by Yash Balhra 6 years, 10 months ago
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Savrit Mor 6 years, 10 months ago
Posted by Muhammed Shahabas 6 years, 10 months ago
- 3 answers
Ram Kushwah 6 years, 10 months ago
3x+y-1 =0
(2k-1)x+(k-1)y-2k-1 =0
{tex}\begin{array}{l}\frac{\mathrm a1}{\mathrm a2}=\frac{\mathrm b1}{\mathrm b2}\neq\frac{\mathrm c1}{\mathrm c2}\\\frac3{2\mathrm k-1}=\frac1{\mathrm k-1}\\3\mathrm k-3=2\mathrm k-1\\\mathrm k=2\\\mathrm{Now}\;\frac{\mathrm c1}{\mathrm c2}=\frac{-1}{-2\mathrm k-1}=\frac15\neq\frac{\mathrm a1}{\mathrm a2}(=\frac33=1)\end{array}{/tex}
Hence for k=1 the eqns has no solution
Puja Sahoo? 6 years, 10 months ago
Honey ??? 6 years, 10 months ago
Posted by Jasleen Kaur 6 years, 10 months ago
- 4 answers
Ram Kushwah 6 years, 10 months ago
No need to worry about sst,
sst is always week for maths good students
Answer all question write something what you remember
Some time topper in mathsgets minimum marks in sst. so not to worry.
before exam read small and MCQ aur rat lo
Read like stories,write like stories.
All the best
Posted by Nidhi Chaudhary 6 years, 10 months ago
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Aryan Yadav 6 years, 10 months ago
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Shivay Brahmåñ 6 years, 10 months ago
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- 2 answers
Ram Kushwah 6 years, 10 months ago
{tex}\begin{array}{l}(a^2+9)x^2+13x+6a\\\end{array}{/tex}
Supose the zeros are m and 1/m
then
m x 1/m=6a/(a2+9)
a2+9=6a
a2-6a+9=0
(a-3)^2=0
a=3,3
Posted by Yukta Mandavi 6 years, 10 months ago
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Choudhary Anil 6 years, 10 months ago
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Geetanand Yadav 6 years, 10 months ago
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Ram Kushwah 6 years, 10 months ago
(4-1/n)+(4-2/n)+.......
=4n-1/n(1+2+3........n)
{tex}\begin{array}{l}\text{=4n-}\frac1{\mathrm n}\times\frac{\mathrm n(\mathrm n+1)}2\\\text{=4n-}\frac{n+1}2\\\text{=}\frac{8n-n-1}2\\\text{=}\frac{7n-1}2\end{array}{/tex}
Posted by Yashvi Jaiswal 5 years, 8 months ago
- 1 answers
Posted by Harsh Upadhyay 6 years, 10 months ago
- 2 answers
Ram Kushwah 6 years, 10 months ago
LCM of 8,15,21=840
110000/840=130.95
hence the greatest no divisible by 840 or divisible by 8,15,21:
=840*130=109200
Posted by Mustafa Khan 6 years, 10 months ago
- 2 answers
Ritika Bisht 6 years, 10 months ago
Ritika Bisht 6 years, 10 months ago
Posted by Aniesh G 6 years, 10 months ago
- 5 answers
Gaurav Seth 6 years, 10 months ago
54, 51, 48,...........
First term a = 54
Common difference d = 51 - 54 = -3
Let n is the number of terms.
Given, sum = 513
=> (n/2)*{2a + (n - 1)d} = 513
=> n{2a + (n - 1)d} = 513*2
=> n{2*54 + (n - 1)*(-3)} = 1026
=> n{108 - 3n + 3} = 1026
=> n{111 - 3n} = 1026
=> 111n - 3n2 = 1026
=> 3n2 - 111n + 1026 = 0
=> n2 - 37n + 342 = 0 {divide by 3}
=> (n - 18)*(n - 19) = 0
=> n = 18, 19
Hence, number of terms is either 18 or 19
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