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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.
Answer:

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

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

Algebra
TutorMe
Question:

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

Elis C.
Answer:

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

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