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Elis C.
Tutor at New York Institute of Technology
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Pre-Calculus
TutorMe
Question:

Find the roots for the quadratic equation: $$x^{2} + 3x + 2 = 0$$

Elis C.

$$a = 1, b = 3, c = 2$$ $$x_1 = \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$$ = $$\frac{-3 + \sqrt{3^{2} - 4*1*2}}{2*1}$$ = $$\frac{-3 + \sqrt{9 - 8}}{2}$$ = $$\frac{-3 + \sqrt{1}}{2}$$ = $$\frac{-3 + 1}{2}$$ = $$\frac{-2}{2}$$ = -1 $$x_2 = \frac{-b - \sqrt{b^{2} - 4ac}}{2a}$$ = $$\frac{-3 - \sqrt{3^{2} - 4*1*2}}{2*1}$$ = $$\frac{-3 - \sqrt{9 - 8}}{2}$$ = $$\frac{-3 - \sqrt{1}}{2}$$ = $$\frac{-3 - 1}{2}$$ = $$\frac{-4}{2}$$ = -2

Calculus
TutorMe
Question:

Find the integral: $$\int_\ (\frac{x-3}{x+9}){d}x$$

Elis C.

$$\int_\ (\frac{x-3}{x+9}){d}x$$ = $$\int_\ (\frac{x + 9 - 12}{x+9}){d}x$$ = $$\int_\ [(\frac{x + 9 }{x+9}) - (\frac{12 }{x+9})]{d}x$$ = $$\int_\ (\frac{x + 9 }{x+9}){d}x$$ - $$\int_\ (\frac{12 }{x+9}){d}x$$ = $$\int_\ (1){d}x$$ - $$12* \int_\ (\frac{1}{x+9}){d}(x+9)$$ = $$x - 12ln|x + 9| + c$$

Algebra
TutorMe
Question:

$$[10 (x + 3) - 4] + [4 (x-6) + 10] = 26$$ Find x.

Elis C.

$$(10x + 30 - 4) + (4x - 24 + 10) = 26$$ $$(10x + 26) + (4x - 14) = 26$$ $$10x + 26 + 4x - 14 = 26$$ $$10x + 4x + 26 - 14 = 26$$ $$14x + 12 = 26$$ $$14x + 12 - 12 = 26 - 12$$ $$14x + 0 = 14$$ $$14x = 14$$ $$\frac{14x}{14}$$ = $$\frac{14}{14}$$ $$1x = 1$$ $$x = 1$$

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