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Tutor profile: Oladeji O.

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Oladeji O.
Former math teacher w/ 17 yrs tutoring exp
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Questions

Subject: Pre-Calculus

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Question:

Find lim (x->3) (x^2 - 9)/(x-3)

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Oladeji O.
Answer:

Direct substitution in this case yields 0/0, which is undefined. We therefore try a different approach. To solve this problem, we first observe that x^2 - 9 is a difference of squares, and can be factored as (x-3)(x+3), so we rewrite the problem as lim (x->3) [(x-3)(x+3)]/(x-3). We have a common factor of (x-3) in the numerator and denominator. This cancels out, leaving lim (x->3) (x+3)/1 = lim (x->3) (x+3). Since we no longer have a denominator, we can now use direct substitution because the possibility of dividing by zero has been eliminated: lim (x->3) (x+3) = 3+3 = 6.

Subject: Calculus

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Question:

Differentiate with respect to x. Assume y is a function of x. 3y^2 - 16x^3 + x^2 - 4 = 0

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Oladeji O.
Answer:

According to the chain rule, to find the derivative of terms involving y, we differentiate the given terms, and multiply the result by the derivative of y with respect to x. We also differentiate other terms with respect to x: 6yy' - 48x^2 + 2x = 0 Next, we divide both sides by the Greatest common factor of 2: 3yy' - 24x^2 + x = 0 To isolate the term that contains y as a derivative of x, we first gather other terms that are strictly defined in terms of x to the other side of the equation: 3yy' = 24x^2 - x Finally, we divide both sides by 3y: y' = (24x^2 - x)/3y

Subject: Algebra

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Question:

Solve for x: x^2 - 10x - 75 = 0

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Oladeji O.
Answer:

We factor this quadratic equation by seeking two factors of our constant term, -75, whose sum is the linear coefficient, -10. The only two numbers that work are -15 and 5. We now factor by writing these numbers in two pairs of parentheses, each following a linear term of x to get a product of binomials: (x - 15)(x + 5) = 0. Since the only way two numbers could multiply to a product of zero is if one of them is zero, we have two cases. Either x - 15 = 0 or x + 5 =0. We solve both cases and have: x - 15 = 0. Add 15 to both sides: x = 15, and x + 5 = 0 Subtract 5 from both sides: x = -5. The solutions are 15 and -5. Check: x = 15: 15^2 - 10*15 -75 = 225 - 150 - 75 = 75 -75 = 0. x = -5: (-5)^2 -10*(-5) - 75 = 25 + 50 - 75 = 75 - 75 = 0.

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