Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Sahbaj Ali 6 years, 8 months ago
- 1 answers
Posted by Dazzling Alka 6 years, 8 months ago
- 2 answers
Yogita Ingle 6 years, 8 months ago
Using Euclid’s Division Lemma
a = bq+r , o ≤ r < b
963 = 657×1 + 306
657 = 306×2 + 45
306 = 45×6+36
45 = 36×1+9
36 = 9×4+0
∴ HCF (657, 963) = 9.
Posted by Arpit Mishra 6 years, 8 months ago
- 7 answers
Yogita Ingle 6 years, 8 months ago
let us assume that √7 be rational.
then it must in the form of p / q [q ≠ 0] [p and q are co-prime]
√7 = p / q
=> √7 x q = p
squaring on both sides
=> 7q2= p2 ------> (1)
p2 is divisible by 7
p is divisible by 7
p = 7c [c is a positive integer] [squaring on both sides ]
p2 = 49 c2 --------- > (2)
substitute p2 in equ (1) we get
7q2 = 49 c2
q2 = 7c2
=> q is divisible by 7
thus q and p have a common factor 7.
there is a contradiction as our assumption p & q are co prime but it has a common factor.
so that √7 is an irrational.
Thanks Ki Baarish ........???? 6 years, 8 months ago
Thanks Ki Baarish ........???? 6 years, 8 months ago
Posted by Prakhar Gupta 6 years, 8 months ago
- 0 answers
Posted by Ahacker Paswan 6 years, 9 months ago
- 1 answers
Posted by Sabitha D 6 years, 9 months ago
- 0 answers
Posted by Anil Sahu 6 years, 9 months ago
- 0 answers
Posted by Ananya Yadav 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
6(ax + by) = 3a + 2b
6ax + 6by = 3a + 2b.........(i)
6(bx - ay) = 3b - 2a
6bx - 6ay = 3b - 2a.........(ii)
Multiplying (i) by a and (2) by b,
So, 6a2x + 6aby = 3a2 + 2ab .......... (iii)
And 6b2x - 6aby = 3b2 - 2ab ......... (iv)
Add (iii) and (iv), we get
6a2x + 6aby + 6b2x - 6aby = 3a2 + 2ab + 3b2 - 2ab
⇒ 6a2x + 6b2x = 3a2 + 3b2
⇒6 (a2x + b2x) = 3(a2 + b2)
⇒{tex}x = \frac { 3 \left( a ^ { 2 } + b ^ { 2 } \right) } { 6 \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 3 } { 6 } = \frac { 1 } { 2 }{/tex}
Substituting {tex}x = \frac 12{/tex} in (i),we get
{tex}6 a \times \frac { 1 } { 2 } + 6 b y = 3 a + 2 b{/tex}
⇒ 3a + 6by = 3a + 2b
⇒ 6by =3a + 2b -3a
⇒ 6by = 2b
⇒ {tex}y = \frac { 2 b } { 6 b } = \frac { 1 } { 3 }{/tex}
Hence, the solution is {tex}x = \frac { 1 } { 2 } , y = \frac { 1 } { 3 }{/tex}
Posted by Aryan Singh 6 years, 9 months ago
- 1 answers
Posted by Danish Saingh 6 years, 9 months ago
- 1 answers
Tinku Munda 6 years, 9 months ago
Posted by Akanksha Kumari 6 years, 9 months ago
- 0 answers
Posted by Vinayak Tyagi 6 years, 9 months ago
- 1 answers
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
Posted by Rohan Sankhla 6 years, 9 months ago
- 0 answers
Posted by Pooja Arora 6 years, 9 months ago
- 1 answers
Tinku Munda 6 years, 9 months ago
Posted by Avanish Singh 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
We have 4n where n = 1, 2, 3, 4.......
if n = 1 then 4n = 41 = 4
if n = 2 then 4n = 42 = 16 and so on
If a number ends with zero then it is divisible by 5.
Here, 4 and 16 are not divisible by 5.
Therefore, 4n can never end with zero.
Posted by Poonam Chhonker 6 years, 9 months ago
- 0 answers
Posted by Ukani Ellis 6 years, 9 months ago
- 1 answers
Posted by Rohan Sinha 6 years, 9 months ago
- 1 answers
Yogita Ingle 6 years, 9 months ago
Let us assume that √3 is a rational number
That is, we can find integers a and b (≠ 0) such that √3 = (a/b)
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
√3b = a
⇒ 3b2=a2 (Squaring on both sides) → (1)
Therefore, a2 is divisible by 3
Hence a is also divisible by 3.
So, we can write a = 3c for some integer c.
Equation (1) becomes,
3b2 =(3c)2
⇒ 3b2 = 9c2
∴ b2 = 3c2
This means that b2 is divisible by 3, and so b is also divisible by 3.
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are co-prime.
This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.
Posted by Sunita Banchod 6 years, 9 months ago
- 5 answers
Posted by Khushal Badaghaiya 6 years, 9 months ago
- 4 answers
Yogita Ingle 6 years, 9 months ago
- It has two equal sides.
- It has two equal angles, that is, the base angles.
- When the third angle is 90 degree, it is called a right isosceles triangle.
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
Posted by Last Bencher 6 years, 9 months ago
- 1 answers
Posted by .... ? 6 years, 9 months ago
- 4 answers
Harsh Mishra 6 years, 9 months ago
Aapne lagta Hai, Set kee Definition Naheen padhi Hai achche se.......'cause.......Sets are usually symbolized by uppercase, italicized, boldface letters such as A, B, S, or Z. Each object or number in a set is called a member or element of the set.Examples include the set of all computers in the world, the set of all apples on a tree, and the set of all irrational numbers between 0 and 1......... toh Aapne Dekh hee liya hoga Definition se, kee Set ek Particular Thing ka hota hai, Agar differ hai, Toh woh Anoother set Hoga........
Harsh Mishra 6 years, 9 months ago
First thing A set Doesn't contains differ Elements.....and Second thing is that...... Agar kar raha hoga, Then ........It'll be another set not the subset of the same.......
.... ? 6 years, 9 months ago
Posted by Arzoo Ansari 6 years, 9 months ago
- 3 answers
Posted by Vedant Yeole 6 years, 9 months ago
- 2 answers
Posted by Priyansh Bhardwaj 6 years, 9 months ago
- 1 answers
Posted by Sen Raj 6 years, 9 months ago
- 3 answers
?D? ??Wish Me At 12Pm...?? 6 years, 9 months ago
Posted by Moksh Khandelwal 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
{tex}\frac{xy}{x + y}{/tex} = {tex}\frac{1}{5}{/tex} {tex}\Rightarrow{/tex} {tex}\frac{x + y}{xy}{/tex} = {tex}\frac{5}{1}{/tex}
{tex}\Rightarrow{/tex}{tex}\frac{1}{y}{/tex}+ {tex}\frac{1}{x}{/tex} {tex}= 5{/tex}....(i)
and {tex}\frac{xy}{x - y}{/tex} = {tex}\frac{1}{7}{/tex} {tex}\Rightarrow{/tex} {tex}\frac{x - y}{xy}{/tex} {tex}= 7{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{1}{y}{/tex} - {tex}\frac{1}{x}{/tex} = 7...(ii)
Now solve equation (i) and (ii) by assuming {tex}\frac{1}{y}{/tex} {tex}= a{/tex} and {tex}\frac{1}{x}{/tex} {tex}= b{/tex}
{tex}\therefore{/tex} eq.(i) and (ii) becomes
{tex}\Rightarrow{/tex}{tex}b = -1{/tex}.....(iii)
Putting the value of {tex} b = -1{/tex} from eq. (iii) in equation (i), we get
{tex}a - 1 = 5{/tex} {tex}\Rightarrow{/tex} {tex}a = 6{/tex}
Now, {tex}\frac{1}{y}{/tex} {tex}= 6{/tex} {tex}\Rightarrow{/tex} {tex}y ={/tex} {tex}\frac{1}{6}{/tex}
and {tex}\frac{1}{x}{/tex} = -1 {tex}\Rightarrow{/tex} {tex}x = -1{/tex}
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Vinod Kumar 6 years, 8 months ago
2Thank You