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  • 3 answers

Suryesh Kumar 1 year, 10 months ago

D = b² - 4ac , => D = 36 - 12 => D = 24 , Here D is greater than 0 (D >0 ) . Hence the nature of roots is real and unequal .

Sankalp Dinesh 1 year, 10 months ago

Real and distinct roots.

Ritik Modi 1 year, 10 months ago

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  • 3 answers

Gurjot Kaur 1 year, 10 months ago

k=-10

Harsh Yadav 1 year, 10 months ago

K=-10

Ridhima Goyal 1 year, 10 months ago

K = -10
  • 2 answers

Sankalp Dinesh 1 year, 10 months ago

an = 4n + 5--- (1), Putting value of n as 3 in the first equation, I wil get a3 = 4(3) + 5 = 12 + 5 = 17. So, the third term of this A.P. is 17.

Pavankumar Dindalakoppa 1 year, 10 months ago

Please answer
  • 4 answers

Aijaz Ansari 1 year, 10 months ago

21

Harsh Yadav 1 year, 10 months ago

21

Ridhima Goyal 1 year, 10 months ago

21

Harsh Vardhan 1 year, 10 months ago

21
  • 1 answers

Suryesh Kumar 1 year, 10 months ago

1/sin
  • 5 answers

Alka Dohare 1 year, 10 months ago

Nb

Amitosh Singh 1 year, 10 months ago

Dont study

Mohd Kamil 1 year, 10 months ago

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Mohd Kamil 1 year, 10 months ago

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[Cosec theta-cot theta+1÷cosec theta+cot theta-1]=1-cos theta ÷sin theta
  • 1 answers

Ridhima Goyal 1 year, 10 months ago

By practicing mcqs daily with calculation in mind
  • 1 answers

Sankalp Dinesh 1 year, 10 months ago

Sum of observations/Total number of observations i.e. 1+2+3+4+5/5 = 15/5 = 3 is the required Arithmetic mean..
  • 1 answers

Ritik Modi 1 year, 10 months ago

0.491
  • 2 answers

Ridhima Goyal 1 year, 10 months ago

(a+b)² = a² + b² + 2ab

Sankalp Dinesh 1 year, 10 months ago

(a+b)² = a² + 2ab + b².
  • 0 answers
  • 1 answers

Harsh Gehlawat 1 year, 10 months ago

2-x-(2k+2) and zero is -4 then Put x = -4 2-(-4)-(2k+2) 2+4-2k-2 6-2-2k 4-2k Put it equal to zero 4-2k=0 -2k=-4 2k=4 K=4/2 K=2 Hence with a simple solution answer is ready
  • 3 answers

Ashish Kumar 1 year, 10 months ago

yes

Rashmi , Tadiyal 1 year, 10 months ago

Yes

Raman Raj 1 year, 10 months ago

Yes
  • 1 answers

Harsh Gehlawat 1 year, 10 months ago

2
  • 1 answers

Harsh Gehlawat 1 year, 10 months ago

Cos A
  • 1 answers

Ishant Sharma 1 year, 10 months ago

I have difficulty in trigonometry
  • 2 answers

Ashish Kumar 1 year, 10 months ago

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Few rules to keep homework help section safe, clean and informative.
  • Don't post personal information, mobile numbers and other details.
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Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

Saksham Joshi 1 year, 10 months ago

Land
  • 3 answers

Anand Yadav 1 year, 10 months ago

√2 is a irrational number √2b=a { a and b are the co prime } √2b=a { both side are squaring} 2b=a { a is divisible by 2 } {2 is divisible by a } Let a=√2c. { c is an integer } 2b=√2c {squaring of √2c } 2b=2c. { c is divisible by 2} { b is divisible by 2} Where a and b are common factor So contradiction is a is √2 irrational number when wrong assertion

Mehakpreet Sandhu 1 year, 10 months ago

Watch dear sir video on YouTube

Aditya Singh 1 year, 10 months ago

You write direct Latsume to √2 is a rational number Where , √2=p/q =√2p=
  • 1 answers

Miss Anshi 1 year, 10 months ago

Questions is wrong?
  • 1 answers

Sankalp Dinesh 1 year, 10 months ago

You want only steps to solve or you want solution with steps??

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