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Manav Sharma 1 month, 1 week 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).
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Manav Sharma 1 month, 1 week 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.
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Manav Sharma 1 month, 1 week 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 \( B \).
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Manav Sharma 1 month, 2 weeks 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.
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Manav Sharma 1 month, 1 week 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.
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Manav Sharma 1 month, 2 weeks 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: \[ \tan^{-1}\left(\frac{2}{11}\right) + \tan^{-1}\left(\frac{7}{24}\right) = \tan^{-1}\left(\frac{1}{2}\right) \]
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Prem Raghav 1 month, 3 weeks ago

X belongs to [-3,7] answer

Ashish Kr Harsh 1 month, 3 weeks ago

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Manav Sharma 1 month, 2 weeks 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.
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Smile 😇 2 months ago

Integration means inverse process of differentiation and integrand is a function you want to integrate
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Shubham Verma 2 months, 3 weeks ago

1

Tanishq Pratap 2 months, 3 weeks ago

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Achyuth R 2 months, 3 weeks ago

1
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Vɪʝɑɣ Gangwar 2 months, 4 weeks ago

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