No products in the cart.

Ask questions which are clear, concise and easy to understand.

Ask Question
  • 1 answers

Manav Sharma 2 weeks, 3 days ago

To find the equations of the medians through points A and B, we first find the midpoints of the sides opposite to these vertices: 1. Midpoint of BC (opposite to A): \((-0.5, 2, 0)\) 2. Midpoint of AC (opposite to B): \((-0.5, 1.5, -0.5)\) Now, let's find the equations of the medians: ### Median through A: Equation: \(x + 1.5y - 2.5 = 0\) ### Median through B: Equation: \(2x + 3y - 2z - 4 = 0\) Now, let's solve these equations simultaneously to find the centroid's coordinates. By solving the equations of the medians simultaneously, we find the centroid's coordinates as (1.5, 5/4, 11/8).
  • 1 answers

Manav Sharma 2 weeks, 3 days ago

The integration of \( \frac{e^x}{(1 + x^2)^{3/2}} \times \frac{x}{\sqrt{1 + x^2}} \) is \( -\frac{e^x}{\sqrt{1 + x^2}} + C \), where \( C \) is the constant of integration.
  • 1 answers

Manav Sharma 2 weeks, 3 days ago

To find the number of onto (surjective) functions from set \( A \) to set \( B \), where \( |A| = 4 \) and \( |B| = 3 \), we can use the formula: \[ \text{Number of onto functions} = |B|^{|A|} - \binom{|B|}{1} \times (|B| - 1)^{|A|} + \binom{|B|}{2} \times (|B| - 2)^{|A|} - \binom{|B|}{3} \times (|B| - 3)^{|A|} \] In this case, \( |A| = 4 \) and \( |B| = 3 \), so we have: \[ \text{Number of onto functions} = 3^4 - \binom{3}{1} \times 2^4 + \binom{3}{2} \times 1^4 - \binom{3}{3} \times 0^4 \] \[ = 81 - 3 \times 16 + 3 \times 1 - 1 \times 0 \] \[ = 81 - 48 + 3 - 0 \] \[ = 36 \] So, there are \( 36 \) possible onto functions from set \( A \) to set \( B \).
  • 1 answers

Manav Sharma 3 weeks, 1 day ago

To find the number of distinct integers between 100 and 1000, we subtract 1 from 1000 (to exclude 1000 itself), then subtract 99 (to account for the numbers from 100 to 999), and add 1 (to include 100). So, the number of distinct integers between 100 and 1000 is \(1000 - 100 - 1 + 1 = 900\). Therefore, there are 900 distinct integers that can be arranged between 100 and 1000.
  • 1 answers

Manav Sharma 2 weeks, 3 days ago

Let's check the properties of the relation \( R \) defined in \( \mathbb{R} \) (the set of real numbers) by \( R = \{(a, b) : a \times b \text{ is an irrational number}\} \): 1. **Reflexive:** A relation \( R \) is reflexive if for every element \( a \) in the set \( A \), the pair \( (a, a) \) belongs to \( R \). In this case, for any real number \( a \), \( a \times a = a^2 \) is always a real number, not necessarily irrational. So, \( R \) is not reflexive. 2. **Symmetric:** A relation \( R \) is symmetric if for every pair \( (a, b) \) in \( R \), the pair \( (b, a) \) also belongs to \( R \). If \( a \times b \) is irrational, it doesn't imply that \( b \times a \) is irrational. For example, \( \sqrt{2} \times \sqrt{3} \) is irrational, but \( \sqrt{3} \times \sqrt{2} \) is also irrational. So, \( R \) is symmetric. 3. **Transitive:** A relation \( R \) is transitive if for every pair \( (a, b) \) and \( (b, c) \) in \( R \), the pair \( (a, c) \) also belongs to \( R \). If \( a \times b \) and \( b \times c \) are irrational, it doesn't necessarily imply that \( a \times c \) is irrational. For example, \( \sqrt{2} \times \sqrt{3} \) and \( \sqrt{3} \times \sqrt{2} \) are both irrational, but \( \sqrt{2} \times \sqrt{2} = 2 \) is rational. So, \( R \) is not transitive. In summary: - \( R \) is not reflexive. - \( R \) is symmetric. - \( R \) is not transitive.
  • 1 answers

Manav Sharma 3 weeks, 1 day ago

To prove this identity, let's denote: - \( \theta_1 = \tan^{-1}\left(\frac{2}{11}\right) \) - \( \theta_2 = \tan^{-1}\left(\frac{7}{24}\right) \) We want to prove that: \[ \tan^{-1}\left(\frac{2}{11}\right) + \tan^{-1}\left(\frac{7}{24}\right) = \tan^{-1}\left(\frac{1}{2}\right) \] We'll use the tangent addition formula: \[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta} \] Let's apply this formula: \[ \tan(\theta_1 + \theta_2) = \frac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11} \cdot \frac{7}{24}} \] \[ = \frac{\frac{48 + 77}{264}}{1 - \frac{14}{264}} \] \[ = \frac{\frac{125}{264}}{\frac{250 - 14}{264}} \] \[ = \frac{\frac{125}{264}}{\frac{236}{264}} \] \[ = \frac{125}{236} \] Now, let's find \( \tan^{-1}\left(\frac{125}{236}\right) \): \[ \tan^{-1}\left(\frac{125}{236}\right) = \tan^{-1}\left(\frac{1}{2}\right) \] Therefore, we've proven the identity: \[ \tan^{-1}\left(\frac{2}{11}\right) + \tan^{-1}\left(\frac{7}{24}\right) = \tan^{-1}\left(\frac{1}{2}\right) \]
  • 2 answers

Prem Raghav 1 month ago

X belongs to [-3,7] answer

Ashish Kr Harsh 1 month 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.
  • 0 answers
  • 1 answers

Subham Sarangi 3 weeks, 6 days ago

So answer is xy=c
  • 1 answers

Manav Sharma 3 weeks, 1 day ago

To find the integral of \( \sin^3(x) \cos\left(\frac{x}{2}\right) \), we can use a substitution. Let: \[ u = \sin(x) \] \[ du = \cos(x) dx \] Now, we can rewrite the integral as: \[ \int u^3 \cos\left(\frac{x}{2}\right) du \] Since \( du = \cos(x) dx \), we need to express \( \cos(x) \) in terms of \( u \). From the trigonometric identity \( \sin^2(x) + \cos^2(x) = 1 \), we have: \[ \cos^2(x) = 1 - \sin^2(x) \] \[ \cos(x) = \sqrt{1 - u^2} \] Now, substitute \( \cos(x) = \sqrt{1 - u^2} \) into the integral: \[ \int u^3 \sqrt{1 - u^2} du \] This integral can be solved using trigonometric substitution. Let: \[ u = \sin(\theta) \] \[ du = \cos(\theta) d\theta \] Now, rewrite the integral in terms of \( \theta \): \[ \int \sin^3(\theta) \sqrt{1 - \sin^2(\theta)} \cos(\theta) d\theta \] \[ \int \sin^3(\theta) \sqrt{\cos^2(\theta)} \cos(\theta) d\theta \] \[ \int \sin^3(\theta) \cos^2(\theta) d\theta \] \[ \int \sin^3(\theta) (1 - \sin^2(\theta)) d\theta \] \[ \int (\sin^3(\theta) - \sin^5(\theta)) d\theta \] This integral can be solved using standard trigonometric integral formulas. After integrating, don't forget to revert back to the original variable \( x \) using the original substitution.
  • 1 answers

Alka Gupta 1 month, 1 week ago

Integration means inverse process of differentiation and integrand is a function you want to integrate
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.
  • 3 answers

Shubham Verma 1 month, 4 weeks ago

1

Tanishq Pratap 1 month, 4 weeks 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.

Achyuth R 2 months ago

1
  • 1 answers

Nikhil Kumar 1 month, 4 weeks ago

0.6
  • 1 answers

Vɪʝɑɣ Gangwar 2 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

Shanika Sharma 2 months, 1 week ago

Let √5 be a rational number. then it must be in form of p/q where, q≠0 ( p and q are co-prime) √5=p/q √5×q=p Squaring both side 5q²=p² --------------(1) p² is divisible by 5. So, p is divisible by 5. p=5c Squaring both side p²=25c2 -----------(2) Put p² in eqn.(1)
  • 3 answers
2.5
11/2

Account Deleted 2 months, 1 week ago

5.5.........
  • 0 answers
  • 1 answers

Prachi Kumari 2 months, 2 weeks ago

π/2

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App