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ffg
  • 3 answers

Shrish Kashyap 7 years, 5 months ago

hii

Diljani . 7 years, 5 months ago

Hii

Durga Verma 5 years, 8 months ago

hii
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  • 1 answers

Akash Sharma 7 years, 6 months ago

Sin Cos tan cosec sec cot
  • 3 answers

Aman Singh 5 years ago

Cos θ = (Adjacent side)/(Hypotenuse side) = √(1-sin^2 θ) = 1/(sec θ) = sin(90-θ) = sin((π/2)-θ) = sin(90+θ) = sin((π/2)+θ) = (sin 2θ)/(2 cos θ) = √(1+cos 2θ)/2 = (tan θ)/(sin θ) = (cos θ)/(cot θ)

Arin Dev 7 years, 6 months ago

1/sin thita

Disha Sethi 7 years, 6 months ago

1/sec thita,
  • 2 answers

Piyush Pathak 7 years, 6 months ago

Jitna tum chaho usme bs practice chahiye .....

Anmol Kumar 7 years, 6 months ago

pura
  • 2 answers

Varsha Yadav 7 years, 5 months ago

Daily 1 hour

Goku The Saiyan 7 years, 6 months ago

Minimum 2 hours
  • 1 answers

Pragnya Paramita Padhi 7 years, 6 months ago

Use droun study videos
  • 4 answers

Akshay Singh 7 years, 6 months ago

9

Isha Malhotra 7 years, 6 months ago

9

Anushk Tiwari 7 years, 6 months ago

9 is he a small child

Shweta Kushwaha 7 years, 6 months ago

9
  • 1 answers

Sia ? 6 years, 6 months ago

Let a be the first term and d be the common difference of the given A.P.
Clearly, in an A.P. consisting of 11 terms, {tex} \left( \frac { 11 + 1 } { 2 } \right) ^ { t h }{/tex} i.e. 6th term is the middle term.
{tex}\text{ it is given that the middle term =30}{/tex}
So a+5d=30 .....(1)

.{tex}S_{11}=\frac{11}{2}(2a+10d){/tex}
= 11(a+5d)
But a+5d=30 from (1)
Hence S​​​​​​11 = 11 × 30= 330

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  • 1 answers

Sia ? 6 years, 6 months ago

Check Question Papers here : <a href="https://mycbseguide.com/cbse-question-papers.html">https://mycbseguide.com/cbse-question-papers.html</a>

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Mahanta Bhagat 7 years, 6 months ago

Using variables, representing a real no. in the form of an equation.
  • 1 answers

Sia ? 6 years, 6 months ago

since 3 is the root of the equation, x = 3 must satisfy the equation.
Applying  x = 3 in the equation {tex}{x^2} - 5x + 6 = 0{/tex}
gives,  {tex}{\left( 3 \right)^2} - 5 ( 3) + 6 = 0{/tex}
{tex} \Rightarrow {/tex}{tex}9 - 15 + 6 = 0{/tex}
{tex} \Rightarrow {/tex}{tex}15 - 15 = 0{/tex}
{tex} \Rightarrow {/tex}{tex}0 = 0{/tex}
{tex} \Rightarrow {/tex}L.H.S. = R.H.S.
Hence, {tex}{x^2} - 5x + 6 = 0{/tex} is a required equation which has 3 as root.

  • 1 answers

Akash Choudhary 7 years, 6 months ago

The hcf is 8
  • 1 answers

S.Md Sameer 7 years, 6 months ago

x=2, y=3
  • 1 answers

Rajiv Ranjan 7 years, 6 months ago

4
  • 1 answers

Vaishnavi Bongu 7 years, 6 months ago

I don't know
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  • 1 answers

Naaz Khatoon 7 years, 6 months ago

Tmhara question wrong h ya x2 - x -2 hoga tb hi bnaga
  • 1 answers

Rajiv Ranjan 7 years, 6 months ago

Google sir se puch lo
  • 1 answers

Isha Parashar 7 years, 6 months ago

nth term means the last term. It can be written as An

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