# Tutor profile: Regina M.

## Questions

### Subject: Pre-Calculus

Simplify the following equation: $$\frac{x^3x^-5y^4}{y^-5y^2x^8}$$

Begin by moving all negative exponents from the numerator to the denominator or vice-versa: $$\frac{x^3y^5y^4}{x^5y^2x^8}$$ Then, combine any like terms in the numerator and denominator. We can add the y exponents in the numerator together and similarly the x exponents together in the denominator: $$\frac{x^3y^9}{x^{13}y^2}$$ Expand the terms as necessary to help with visualization: $$\frac{x^3y^2y^7}{x^3x^{10}y^2}$$ Cancel the same terms from the numerator and the denominator: $$\frac{y^7}{x^{10}}$$

### Subject: Calculus

Find the area under the curve $$y =x^{2}$$ from [0,6].

$$ \int_0^6 x^{2}\mathrm{d}x $$ $$\frac{x^{3}}{2} |_0^6$$ $$\frac{6^{3}}{2} - \frac{0^{3}}{2}$$ $$\frac{216}{2}$$ $$108$$

### Subject: Algebra

Consider the following equations: $$2x + y = 5$$ $$y - x = 2$$ Are you able to solve for x and y? Why? Solve for x and y by adding and subtracting the equations from one another and then solving accordingly.

Yes, you are able to solve for x and y because there are two unknown variables and two equations given. If you subtract the second equation from the first: $$(2x + y) - (y - x) = 5 - 2$$ $$3x = 3$$ $$x = 3$$ Plug 3 into the second equation: $$y - x = 2$$ $$y - 3 = 2$$ $$y = 5$$

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