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  • 2 answers

Ayush Singh 1 year, 8 months ago

√2 = p/q, where 'p' and 'q' are integers, q ≠ 0 and p, q have no common factors (except 1).

Prakhar Sharma 1 year, 8 months ago

Let us assume on the contrary that 2 ​ is a rational number. Then, there exist positive integers a and b such that 2 ​ = b a ​ where, a and b, are co-prime i.e. their HCF is 1 ⇒( 2 ​ ) 2 =( b a ​ ) 2 ⇒2= b 2 a 2 ​ ⇒2b 2 =a 2 ⇒2∣a 2 [∵2∣2b 2 and 2b 2 =a 2 ] ⇒2∣a...(i) ⇒a=2c for some integer c ⇒a 2 =4c 2 ⇒2b 2 =4c 2 [∵2b 2 =a 2 ] ⇒b 2 =2c 2 ⇒2∣b 2 [∵2∣2c 2 ] ⇒2∣b...(ii) From (i) and (ii), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means that our supposition is wrong. Hence, 2 ​ is an irrational number.
  • 5 answers

Chinmayi P P 1 year, 8 months ago

1

Rahul Singh Bisht 1 year, 8 months ago

1

Ayush Singh 1 year, 8 months ago

😌0

Soham Singh 1 year, 8 months ago

1

Omya Aryan 1 year, 8 months ago

1
  • 1 answers

Omya Aryan 1 year, 8 months ago

Construction is not in syllabus
  • 0 answers
  • 3 answers

Anjali Nagar 1 year, 8 months ago

Which have more than 2 factor

Ashutosh Sharma 1 year, 8 months ago

Factor more than 2

Omya Aryan 1 year, 8 months ago

The number which has more than 2 factors is factors is known as composite number. OR numbers other than prime number is known as composite number.
  • 0 answers
  • 0 answers
  • 3 answers

Alok Kumar 1 year, 8 months ago

X²+7x+6 X²+6x+x+6 X(x+6)+1(x+6) (X+1)(x+6) Zeros of the polynomial X+1=0 X=-1 X+6=o X=-6 Sum of zeros Alpha + beta =-b\a -6+(-1)=-7\1 7 =7 Alpha ×beta =c\a -6×(-1). =6\1 6. =6 PROVED

Omya Aryan 1 year, 8 months ago

Sum of zeros = -b/a = -7……Product of zeros = c/a = 6

Omya Aryan 1 year, 8 months ago

x²+7x+6……x²+6x+1x+6……x(x+6)+1(x+6)……(x+1)(x+6)……x=-1 or x=-6
  • 2 answers

Preeti Dabral 1 year, 8 months ago

Prime Factors of 3233 are: 53, 61.

Rajeev Rajpoot 1 year, 8 months ago

What is Reflection of light
  • 1 answers

Preeti Dabral 1 year, 8 months ago

f(x) = 4x2 + 4x - 3
{tex}\Rightarrow \quad f \left( \frac { 1 } { 2 } \right) = 4 \left( \frac { 1 } { 4 } \right) + 4 \left( \frac { 1 } { 2 } \right) - 3{/tex}
= 1 + 2 - 3 = 0
and {tex}f \left( - \frac { 3 } { 2 } \right) = 4 \left( \frac { 9 } { 4 } \right) + 4 \left( - \frac { 3 } { 2 } \right) - 3{/tex}
= 9 - 6 - 3 = 0
{tex}\therefore \frac { 1 } { 2 } , - \frac { 3 } { 2 }{/tex}are zeroes of polynomial 4x2 + 4x - 3.
Sum of zeroes = {tex}\frac { 1 } { 2 } - \frac { 3 } { 2 } = - 1 = \frac { - 4 } { 4 }{/tex}
{tex}= - \frac { \text { Coefficient of } x } { \text { Coefficient of } x ^ { 2 } }{/tex}
Product of zeroes = {tex}\left( \frac { 1 } { 2 } \right) \left( - \frac { 3 } { 2 } \right) = \frac { - 3 } { 4 }{/tex}
{tex}= \frac { \text { Constant term } } { \text { Coefficient of } x ^ { 2 } }{/tex}
Verified.

  • 1 answers

Avinash Tripathi 1 year, 9 months ago

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  • 2 answers

Anvesha Sahay 1 year, 9 months ago

All chapters are important we can't choose any particular chapter to do practice of it and get passed easily.

Nishant Singh 1 year, 9 months ago

Statics
  • 1 answers

Akhilesh Shankhdhar 1 year, 9 months ago

Prove टू अपऑन अंडर रूट 2 इज irrational
  • 5 answers

Raunak Yadav 1 year, 8 months ago

√1/2

Deepak Sharma 1 year, 9 months ago

½ is the value of sin30°

Anushka Sharma 1 year, 9 months ago

1/2 is the value of sin30°

Varsha Chauhan 1 year, 9 months ago

0.5

Hrishikesh Debnath 1 year, 9 months ago

One and half
  • 2 answers

Khursheed Alam 1 year, 9 months ago

Ajns

Shreyash Kumar Jaiswal 1 year, 9 months ago

1/2 * 1/2 + √3/2 * √3/2 1/4 +3/4 4/4 1
  • 2 answers

Sarika Joshi 1 year, 9 months ago

1/√2 + 0 is 1/√2

Shreyash Kumar Jaiswal 1 year, 9 months ago

1/2 + 1 = 1/2+2/2 = 3/2
  • 2 answers

Paritosh Rajput 1 year, 9 months ago

Prove

Navkirat Kaur 1 year, 9 months ago

Let √2 is rational √2=a/b Where a and b are integers and b is not equal to 0 Rational is not equal to Irrational Our assumption is wrong √2 is irrational
  • 2 answers

Preeti Dabral 1 year, 9 months ago

17×5×11×3×2+2×11=17×5×3×22+22

=22(17×5×3+1)
=22(255+1)=2×11×256 -----(i)


Equation (i) is divisible by 2, 11 and 256, so it has more than 2 prime factors.

∴ (17×5×11×3×2+2×11) is a composite number.

Payal Kumari Kushwaha 1 year, 8 months ago

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

Few rules to keep homework help section safe, clean and informative.
  • Don't post personal information, mobile numbers and other details.
  • Don't use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.
  • Be nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.
  • Ask specific question which are clear and concise.

Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.
  • 1 answers

Preeti Dabral 1 year, 9 months ago

787

  • 1 answers

Preeti Dabral 1 year, 9 months ago

Height of the cylinder (h) = 14 cm,
Base diameter = 7 cm
{tex}\Rightarrow{/tex} Radius of the base of the cylinder (r) = 3.5 cm
Volume of the cylinder = {tex}\pi r ^ { 2 } h{/tex}
{tex}\frac { 22 } { 7 } \times 3.5 \times 3.5 \times 14{/tex}
= 22 {tex}\times{/tex}3.5 {tex}\times{/tex}3.5{tex}\times{/tex}14
= 539 cm3
Radius of the conical holes (r1) = 2.1 cm,
Height of the conical holes (h1) = 4 cm,
volume of the conical hole {tex}= \frac { 1 } { 3 } \pi r _ { 1 } ^ { 2 } h _ { 1 }{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times 2.1 \times 2.1 \times 4{/tex}
= 18.48 cm3
Volume of the two conical hole = 2 {tex}\times{/tex}18.48
= 36.96 cm3
Volume of the remaining solid = Volume of the cylinder - Volume of two conical hole
= 539 - 36.96
= 502.04 cm3

  • 2 answers

K_Hushii 🍁 1 year, 9 months ago

3median =mode + 2mean

Omya Aryan 1 year, 9 months ago

3 Median = Mode + 2 Mean
  • 2 answers

Surviver 🪐 1 year, 9 months ago

Given , a6+a9 = 136 a12+a15 = 136 a12+a15 = 256 Solu- Dys

Dushyant Shukla 9 months, 4 weeks ago

Chapter 3ka question 5
  • 2 answers

Vidyut Kumaran 1 year, 9 months ago

We have to prove that √5 is an irrational number It can be proved using the contradiction method Assuming √5 as a rational number, i.e., can be written in the form a/b where a and b are integers with no common factors other than 1 and b is not equal to zero. √5/1 = a/b √5b = a By squaring on both sides 5b2 = a2 b2 = a2/5 .... (1) It means that 5 divides a2. It means that it also divides a a/5 = c a = 5c By squaring on both sides a2 = 25c2 Substituting the value of a2 in equation (1) 5b2 = 25c2 b2 = 5c2 Prove that √5 is irrational. b2/5 = c2 As b2 is divisible by 5, b is also divisible by 5 a and b have a common factor as 5 It contradicts the fact that a and b are coprime This has arisen due to the incorrect assumption as √5 is a rational number. Therefore, √5 is irrational. Prove that √5 is irrational. It is proved that √5 is irrational.

Bhavnoorpreet Kaur 1 year, 9 months ago

Let us assume that 5^ is an rational number. So we have to find co-prime in the form of p/q where q is not equals to zero. Such that 2^ = p/q So b 2^=a 2b^2 =a^ According according to 1.8 divisible by 2 squarring on both sides =2^b^2 =a^2. Substituting the values we get 2c=a Substituting for a we get 2b^2=4c^2 B^2=2c^2 So this means that 2 / b^2 square and also divide by b therefore A and B have at least common factor but this contradict ic the no common factors rather than one this contradiction has been that due to our wrong assumption. So we concluded that it is irrational

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