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# Tutor profile: Vaidehi T.

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Vaidehi T.
Mathematics Expert at an Engineering college
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## Questions

### Subject:Trigonometry

TutorMe
Question:

If \$\$cos{x} = 4\$\$, then find the value of \$\$ cos{2x}.\$\$

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Vaidehi T.

As we know that, \$\$ cos{2x} = 2cos^2{x} - 1.\$\$ Plug the value of \$\$cos{x}\$\$, \$\$\therefore cos{2x} = 2(4)^2 - 1\$\$ \$\$= 32 - 1\$\$ \$\$= 31.\$\$

### Subject:Algebra

TutorMe
Question:

Find the equation of the line that passes through two points \$\$(2, 1),\, (3,4)\$\$.

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Vaidehi T.

When we are given two points and we have to find the equation of the line passing through those points, we can use two point formula. Let's say \$\$A(x_1, y_1)\$\$ and \$\$B(x_2, y_2)\$\$ are two points then the formula is defined as \$\$y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1} (x - x_1) \$\$. For the given data, we can take \$\$x_1 = 2,\, x_2 = 3,\, y_1 = 1,\, y_2 = 4.\$\$ Plugging all these values in the formula to find the equation of a line: \$\$y - 1 = \dfrac{4 - 1}{2 - 1} (x - 2) \$\$ \$\$\therefore y - 1 = 3(x-2)\$\$ \$\$\therefore y - 1 = 3x - 6\$\$ \$\$ \therefore y = 3x - 5.\$\$

### Subject:Differential Equations

TutorMe
Question:

Solve the following first order differential equation: \$\$\dfrac{dy}{dx} = x^2 y^2.\$\$

Inactive
Vaidehi T.

In order to solve this first order differential equation, we can separate variables like this \$\$ \dfrac{dy}{y^2} = x^2\, dx.\$\$ Now, integrating both the sides, we get \$\$\dfrac{y^{-2+1}}{-2+1} = \dfrac{x^{2+1}}{2+1} + C\$\$ \$\$ \therefore \dfrac{-1}{y} = \dfrac{x^3}{3} + C\$\$ \$\$ \therefore \dfrac{x^3}{3} + \dfrac{1}{y} + C = 0.\$\$

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