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  • 2 answers

Riya Trivedi 6 years, 11 months ago

Muskan what is this

Muskan Kumari 6 years, 11 months ago

Please jaldi pado new year se pelly
  • 1 answers

Shubhendra Pratap Singh 6 years, 11 months ago

U have ab =12 Now aby= -(d)÷a 12y= -(24)/1 = -24 y=-2 = x+2 will be factor of f(x) On division u will have quotient x2-7x+12 Factorisation will give zeroes -2 , 3,and 4 Thats ur answer
  • 3 answers

Yogita Ingle 6 years, 11 months ago

x = 2 , y = 3
2x + 3y - 13
= 2(2) + 3(3) - 13
= 4 + 9 - 13
= 13 - 13
= 0

Puja Sahoo 6 years, 11 months ago

Yes

S Sihag Ji 6 years, 11 months ago

0 is answer
  • 1 answers

Sia ? 6 years, 6 months ago

Let a be any positive integer and b = 6
∴ by Euclid’s division lemma
a = bq + r, 0≤ r and q be any integer q ≥ 0
∴ a = 6q + r,
where, r = 0, 1, 2, 3, 4, 5
If a is even then then remainder by division of 6 is 0,2 or 4

Hence r=0,2,or 4

or A is of form 6q,6q+2,6q+4
As, a = 6q = 2(3q), or
a = 6q + 2 = 2(3q + 1), or
a = 6q + 4 = 2(3q + 2).
If these 3 cases  a is an even integer.
but if the remainder is 1,3 or 5 then r=1,3 or 5

or A is of form 6q+1,,6q+3 or,6q+5
Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1, 
Case 2: a = 6q + 3 = 6q + 2 + 1,
= 2(3q + 1) + 1 = 2n + 1, 
Case 3: a = 6q + 5 = 6q + 4 + 1
= 2(3q + 2) + 1 = 2n + 1

This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

  • 1 answers

Shivam The Galaxy 6 years, 11 months ago

Euclid's Division Lemma :- a = bq +r 117 > 65 117 = 65 × 1 + 52 ----> [ 2 ] 65 = 52 x 1 + 13 -----> [1] 52 = 13 x 4 + 0 HCF = 13 13 = 65m + 117n From [ 1] , 13 = 65 - 52 x 1 From [2] , 52 = 117 - 65 x 1 ----> [3] Hence , 13 = 65 - [ 117 - 65 x 1 ] ------> from [3] = 65 x 2 - 117 = 65 x 2 + 117 x [-1 ] m = 2 n = -1
  • 6 answers

S Sihag Ji 6 years, 11 months ago

Thanks bhai

Shivam The Galaxy 6 years, 11 months ago

Equation will be like this 4k^2 -12k+0=0 then taking 4k as common then 4k(k-3)=0 And u will get the answer as mentioned

Shivam The Galaxy 6 years, 11 months ago

Right answer is K=0 and K = 3

S Sihag Ji 6 years, 11 months ago

2k^2-6k-5=0,meri ye equation ban rhi hai , but k ki value nhi aa rhi

You . 6 years, 11 months ago

Answer KYa aa rha h mera to koi square = 3k aa rha h

S Sihag Ji 6 years, 11 months ago

Plz solve this problem
  • 6 answers

Riya Trivedi 6 years, 11 months ago

Thanks Pooja

Puja Sahoo 6 years, 11 months ago

Riya height of the tree will be 50(root 3 - 1).

Riya Trivedi 6 years, 11 months ago

Thanks shivam

Shivam The Galaxy 6 years, 11 months ago

Then tan 30= h/x x=h√3 Now, Tan 45 = h/100-x h = 100-h√3 h = 100/√3+1 h = 100(√3-1)/2 h = 50(√3-1)

Riya Trivedi 6 years, 11 months ago

Shivam maine procss kiya but answer nhi mila aap answer bta do

Shivam The Galaxy 6 years, 11 months ago

Let Height of tree be 'h' and distance of tree from point p and q be x and 100-x respectively
  • 7 answers

Anurag Mishra 6 years, 11 months ago

S2u N tnks

Cutie Pie ? 6 years, 11 months ago

Same to you Miss Yadav

Riya Trivedi 6 years, 11 months ago

Same to u Kiran

Rishabh Jain 6 years, 11 months ago

Same to you and thanks kiran jiiii

Prashant Chaudhary 6 years, 11 months ago

Thanks and same to you Kiran ji ??

Amira Roy❤ 6 years, 11 months ago

Same to you

S Sihag Ji 6 years, 11 months ago

Same to u sis☺?☺
  • 2 answers

Divya Garg 6 years, 11 months ago

First we take LCM and the answer we will obtain is...xsq.+(x+1)sq./(x+1)x= 34/15. Then opening the brackets...2xsq.+2x+1/xsq.+x= 34/15. Now cross multiplication...30xsq.+30x+15= 34xsq.+34x. Now sloving this equation we get the values of x=-5/2 and 3/2

Shivam The Galaxy 6 years, 11 months ago

2x^2 +2x+1= 34x^2+34x/15 Now 4x^2 +4x-15=0 (4x^2-10x)+(6x-15)=0 Taking common 2x(2x-5)+3(2x-5)=0 (2x+3)(2x-5)=0 X= -3/2 and 5/2.
  • 5 answers

Nivedita Marothiya 6 years, 11 months ago

But from my point of view trigonometry too

Shivam The Galaxy 6 years, 11 months ago

Triangles in my opinion

Nivedita Marothiya 6 years, 11 months ago

Which is the most complicated chapter of ncert?

Shivam The Galaxy 6 years, 11 months ago

My pleasure no problem

Amisha Sharma ? 6 years, 11 months ago

Bt kisliye
  • 1 answers

Shivam The Galaxy 6 years, 11 months ago

Let the vertices of the triangle are A B and C and point of intersection of Altitude be D . Then, In triangle ABD and ABC angle A = angle A angle ADB= angle ABC So triangle are similar by AA criteia of similarity Hence, AB/AC = BD /BC a/c=x/b Therefore, ab = cx
  • 3 answers

Amisha Sharma ? 6 years, 11 months ago

Give measurements

Anurag Mishra 6 years, 11 months ago

Where is the figure

Shivam The Galaxy 6 years, 11 months ago

I want the photograph bro
  • 1 answers

Shivam The Galaxy 6 years, 11 months ago

Hey sulakh complete ur question
  • 2 answers

Nivedita Marothiya 6 years, 11 months ago

Sorry but it's not.

Shivam The Galaxy 6 years, 11 months ago

Answered already
  • 1 answers

Sahitha?? Sharma 6 years, 11 months ago

Replace (seco^2-tano^2) in the place of 1 in tano+seco-1
  • 3 answers

Divya Garg 6 years, 11 months ago

Yes it is in AP... The nth term will be 5n-2 n=a+(n-1)d n=3+(n-1)5 n=3+5n-5 n=5n-2 Therefore the nth term for the given AP is 5n-2

Akshita Wadhwa 6 years, 11 months ago

It is in AP(3,8,13,....)

Manoj Kumar 6 years, 11 months ago

The given series is not in AP.
  • 0 answers
  • 2 answers

Nivedita Marothiya 6 years, 11 months ago

Nooo bcz this type of question is given in RD Sharma class 10

Divya Garg 6 years, 11 months ago

I think this question is incomplete...as no value is given to us whereas according to the language of this question the values must be given to us...
  • 2 answers

Ananya P 6 years, 11 months ago

Given - ABIIDC Construction - Draw a line EFIIDC Proof- As EFIIDC (construction) But DCIIAB Therefore, EFIIAB In triangle ADB, DE/EA=OD/OB (BPT).....(1) In triangle DAC, OC/OA=DE/EA (BPT).... (2) From eq. 1 and 2, OC/OA=OD/OB OC/OD=OA/OB Hence,proved

Shivam S 6 years, 11 months ago

Prove that triangles aob and cod are similar. Then you will get the proof
  • 1 answers

Sahitha?? Sharma 6 years, 11 months ago

Speed in upstream is =x-y km/hr Speed in downstream=x+y km/hr Distance/Speed =Time :::(24/x-y)-(24/x+y)= total time
  • 3 answers

Mutyala Gayathri 6 years, 11 months ago

My mam gave me the marks list of every chapter: Real no: 6m Polynomials: 3m Linear eq: 5m Quadratic eq: 5m AP: 7m Triangles: 8m Coordinate: 6m Intraduction to trignometry: 8m Applications: 4m Circles: 3m Const: 4m Areas related: 3m Surface areas: 7m Statistics: 7m Probability: 4m

Sneha Chaudhari 6 years, 11 months ago

it depends if u hv English or social then first go through the chp n then learn one answer if cant remember then learn n write in rough

Mutyala Gayathri 6 years, 11 months ago

By reading
  • 1 answers

Aradhana Singh 6 years, 11 months ago

Let a positive odd integer be ,a ,and , b =6 6q +1, 6q +3 ,6q +5 a= bq + r a = 6q + r a = 6q + 1 (r =1, 3, 5) a = 6q + 3 a = 6q + 5. Proved
  • 1 answers

Akshita Wadhwa 6 years, 11 months ago

Let n be the number of balls. n=Area of big sphere/Area of 1 small sphere. n=1000
  • 5 answers

Anurag Mishra 6 years, 11 months ago

Solve*

Anurag Mishra 6 years, 11 months ago

Aolve ncert

Shraddha Soni 6 years, 11 months ago

Yes

Rishabh Jain 6 years, 11 months ago

Toh upset kyo ho rahi hai aap

Prashant Chaudhary 6 years, 11 months ago

Toh iss Mai rone ki kya baat ???
  • 0 answers
  • 2 answers

H H 6 years, 11 months ago

(P^2-1)(p^2+1)

Aradhana Singh 6 years, 11 months ago

Kya question itna hi hai .
  • 1 answers

Sia ? 6 years, 6 months ago

Given: ABCD is a parallelogram and E is a point on BC. The diagonal BD intersects AE at F.
To prove: {tex}A F \times F B = E F \times F D{/tex}
Proof: Since ABCD is a parallelogram, then its opposite sides must be parallel.
{tex}\therefore{/tex}In {tex}\triangle ADF{/tex} and {tex}\triangle EBF{/tex}
{tex}\angle \mathrm { FDA } = \angle \mathrm { EBF }{/tex} and {tex}\angle F A D = \angle F E B{/tex} [Alternate interior angles]
{tex}\angle A F D = \angle B F E{/tex} [vertically opposite angles]
{tex}{/tex} Therefore,by AAA criteria of similar triangles,we have,
{tex}\triangle \mathrm { ADF } = \triangle \mathrm { EBF }{/tex}
Since  the corresponding sides of similar triangles are proportional. Therefore,we have,
{tex} \frac { A F } { F D } = \frac { E F } { F B }{/tex}
{tex}\Rightarrow A F \times F B = E F \times F D{/tex}

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