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  • 2 answers

Vaidik Goel 4 years, 4 months ago

We will use the formula to find the LCM of 306 and 657. LCM (a, b) = (a × b) / HCF (a, b) Given two numbers 306 and 657 and their HCF is 9. To find : LCM(306,657 ) using the formula, We have LCM (306, 657) = (306 × 657) / HCF (306, 657) = 201042 / 9  = 22338 Thus, LCM of 306 and 657 is 22338

Atharv Srivastava 4 years, 4 months ago

**** up your bur
  • 3 answers

Anil Ojha 4 years, 4 months ago

A=1, B= 2a1, C= -a1-2 ATQ, alpha+beta= alpha*beta =-b/a=c/a => -b=c =-(2a1)=-a-2 =-a1=-2 =a1=2 Hence value of a1= 2 MERITNATION OP

Chitrakshee . 4 years, 4 months ago

please, anyone, tell

A. Sandeep A. Sandeep 4 years, 4 months ago

Ffghj
  • 1 answers

Gauri Singhal 4 years, 4 months ago

Using Euclid's division lemma 652= 375 multiply by 1+227 375= 277 multiply by 1+98 277= 98 multiply by 2+81 98= 81 multiply by 1+17 81= 17 multiply by 5+6 17= 6 multiply by 2+5 6= 5 multiply by 1+1 5= 1 multiply by 5+0 So HCF=1 HOPE THE ANSWER IS CORRECT
  • 2 answers

Ritik Varshney 4 years, 4 months ago

a=bq+r (0<r<b)

Nothing Nothing 4 years, 4 months ago

195 by 38220
  • 1 answers

Nisha Kumari 4 years, 4 months ago

2ײ+k×+3=0
  • 5 answers

Sweta Kumari 4 years, 4 months ago

a = bq+r ,where 0_<r<b

Jaya Maurya 4 years, 4 months ago

a=bq+r (Euclid's equation) Where, 0<=r<b

Sakshi Yadav 4 years, 4 months ago

A=bq+r, 0<r<b.

Prajwal G N 4 years, 4 months ago

a=bq+r where 0<b<r and a is not equal 0

Rudrashrya Thappa 4 years, 4 months ago

a=bq+r where 0<r<b
  • 1 answers

Preeti Dabral 4 years, 4 months ago

LHS
=(1+1/tan²A).(1+1/cot²A)
=(1+cot²A).(1+tan²A)
=cosec²A.sec²A
=1/(sin²A.cos²A)
=1/{(1-cos²A)cos²A}
=1/(cos²A-cos⁴A) = RHS(Proved)

  • 4 answers

Himanshi Sharma 4 years, 4 months ago

-5/3

Alok Kumar Singh 4 years, 4 months ago

K=4/3

Yash .S 4 years, 4 months ago

How to solve please tell

Anshika Sharma 4 years, 4 months ago

K will be ⁴/³
  • 1 answers

Karthik Moyye 4 years, 4 months ago

For any action to be repeated,find the LCM of 40 and 60 ie 120,then convert 120 into hours. 120/60(1 hour=60 minutes) so it is again in two hours. Now add 2 hours to 9am and you will get 11:00am
  • 1 answers

Sia ? 4 years, 4 months ago

Given:

An army contingent of 612 members is to march behind an army band of 48 members.

Here, HCF of 612 and 48 will give the maximum number of columns in which the two groups can march.

So, using Euclid's division algorithm

612=48×12+36

⇒48=36×1+12

→36=12×3+0

∴HCF(612,48)=12

Hence, the maximum no of columns in which they can march is 12 

  • 1 answers

Himanshi Sharma 4 years, 4 months ago

(x+1)(x-5)
  • 3 answers

Himanshi Sharma 4 years, 4 months ago

3x-4y=7_eq1 x=7+4y/3_eq2 Put value of x in eq 1 You get y=0 Put value of y in eq 1 x=7/3

Satwik Banerjee 4 years, 4 months ago

The second equation please

Tejas Bramhe 4 years, 4 months ago

Plz type the whole question
  • 0 answers
  • 1 answers

Preeti Dabral 4 years, 4 months ago

Let p be any positive integer and b = 3. Applying division Lemma with p and b =3 ,

we have
p = 3q + r, where 0 {tex}\leq{/tex} r < 3 and q is some integer

So r=0,1,2

If r=0 , p=3q

If r=1, p=3q+1

If r=2, p=3q+2

Therefore any positive integer  is of form 3q,3q+1,3q+2 for some integer q.

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