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  • 2 answers

Ayush Shukla???? 6 years, 10 months ago

0

Harsh Vardhan Singh 6 years, 10 months ago

Wrong question
  • 1 answers

Ram Kushwah 6 years, 10 months ago

{tex}\begin{array}{l}\left(\frac{1-\mathrm{tanA}}{1-\mathrm{cotA}}\right)^2=\left(\frac{1-\mathrm{tanA}}{1-{\displaystyle\frac1{\mathrm{tanA}}}}\right)^2\\=\left(\mathrm{tanA}\frac{(1-\mathrm{tanA})}{\mathrm{tanA}-1}\right)^2\\=(-\mathrm{tanA})^2=\tan^2\mathrm A\end{array}{/tex}

  • 3 answers

Fatma Naayab 6 years, 10 months ago

triangle theorem n construction both were asked in 4 marks

Fatma Naayab 6 years, 10 months ago

prepare at least triangle theorem

Aman Ahammad K 6 years, 10 months ago

Study surface area and volumes it's important
  • 1 answers
If the number 14n ,for any n , were to end with zero , then it would be divisible by 5 . that is the prime factorization of 14 and would contain the prime 5 it is not possible because 14 n =[7]2n , show the only prime in the factorization of 14 and S7 so the uniqueness of the fundamental theorem of arithmetic guarantees that
  • 1 answers

Ram Kushwah 6 years, 10 months ago

x,y,z=mishra

  • 0 answers
  • 1 answers

Vineeth Reddy 6 years, 10 months ago

Get me answer..
  • 0 answers
  • 2 answers

Ram Kushwah 6 years, 10 months ago

a=5,d=8 l=121

an=a+(n-1)d

121=5+(n-1)x8

116=8n-8

8n=124

n=124/8=15.5

So 121 cannot be term of this AP,So question is not correct(this is not an AP)

pl check question again

Rahul Singh 6 years, 10 months ago

1008
  • 1 answers

Mr Manish Singh 6 years, 10 months ago

9
  • 3 answers

Ankit Kumar 6 years, 10 months ago

Answer is 100

Ankit Kumar 6 years, 10 months ago

Sorry this ans is incorrect

Ankit Kumar 6 years, 10 months ago

140
  • 1 answers

Aarohi Singh 6 years, 10 months ago

an = a + (n-1)d
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  • 0 answers
  • 1 answers

Ayush Shukla???? 6 years, 10 months ago

U can find by mid point formula
  • 1 answers

Ankit Kumar 6 years, 10 months ago

Psq. + 1 upon psq._ 1
  • 2 answers

Deepanshu Mahajan 6 years, 10 months ago

LHS 1/cosA + sinA/cosA =p 1+sinA/cosA =p 1+sinA=pcosA 1+sinA=p√1-sin sqA Squaring both side (1+sinA )sq = p sq. (1 -sin sq.A) (1+sinA) (1+sinA) =p sq. (1-sin sq.A) (1+sinA) (1+sinA)/(1-sinA)(1+sinA) =p sq. {:.a sq.+b sq =(a+b)( a-b)} (1+sinA)/(1-sinA) = p sq. (1+sinA) =p sq.( 1 - sin A) 1 + sinA = p sq. -p sq sinA SinA + p sq.sinA = p sq-1 sin A (1 + p sq.) = p sq-1 Sin A = (p sq. -1)/(p sq+1) CosecA = (p sq.+1) / (p sq.-1) {sinA = 1/ cosecA} H.P

Deepanshu Mahajan 6 years, 10 months ago

SecA+tanA =p
  • 1 answers

Dpsc Singh 6 years, 10 months ago

A+B+C = 180° B+C=180°-A B+C/2=90°-A/2 Cos(B+C/2)=cos(90°-A/2) Cos(B+C/2)=sin(A/2) Hence proved
  • 1 answers

Sia ? 6 years, 4 months ago


Let O be the position of the bird and B be the position of the boy. Let FG be the building and G be the position of the girl.
In {tex}\triangle{/tex}OLB,
{tex}\frac { OL } { B O }= \sin 30 ^ { \circ }{/tex}
{tex}\Rightarrow \quad \frac { O L } { 100 } = \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow{/tex} OL = 50 m
OM = OL - ML
= OL - FG
= 50 - 20 = 30 m
In {tex}\triangle{/tex}OMG
{tex}\frac {OM}{OG}=sin45°=\frac{1}{√2}{/tex}
OG = OM√2 = 30 √2 = 42.3 meter

  • 2 answers

Devanshu Sharma 6 years, 10 months ago

Let the first zero be x And second be y A.T.Q x+y= -3 = -b upon a xy=2=c upon a We have a=1 , b=3 (because -b= -3 so b=3) and c =2 ax square +bx +c=0 X square +3x+2=0

Affu 😊 6 years, 10 months ago

x^2 + 3x + 2 ... Its too easy
  • 2 answers

Fatma Naayab 6 years, 10 months ago

calculate cumulative frequency then write less than upper limit similarly write all from class interval n put the value of cf bcoz in less than value is same as cf in more than type write lower limit of class interval n their value in 1st u can write to last c.f in only 1st then from 2nd just subtract the frequency no. from 1st , from 2nd value subtract 2nd of frequency n so on then u can draw easily

Devanshu Sharma 6 years, 10 months ago

Commulative frequency ko adjust krke less than or more than mein f bnake graph draw ki a jaenga
  • 1 answers

Siddhartha Jain 6 years, 10 months ago

What

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