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Max S.
Yale senior Economics major, experienced tutor
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Pre-Calculus
TutorMe
Question:

Apply the law of cosines to solve for length z in triangle XYZ where: y = 6.5, x = 9.4, Z = 131 degrees

Max S.
Answer:

The Law of Cosines states: c^2 = a^2 + b^2 − 2ab cos(C) We'll use x = a, y = b, z = c here. z^2 = x^2 + y^2 - 2xy cos(Z) z^2 = (9.4^2) + (6.5^2) + 2(9.4)(6.5) cos(131 degrees) z^2 = 88.36 + 42.25 − 122.2 × (−0.656) z^2 = 130.61 + 80.17 z^2 = 210.78 z = sqrt(210.78) z = 14.5

Calculus
TutorMe
Question:

How can I use the chain rule to fully differentiate y = ((2x + 1)^3 + 5)^2? .

Max S.
Answer:

Pardon the formatting here, but we have to break down the question systematically to allow ourselves to arrive at the answer. We see that this expression is built up of smaller embedded expressions, that is: y = v^2, where v = u^3 + 5, and u = 2x + 1 This sets us up for the use of the chain rule, where we "unwrap" the derivatives of the individual components. dy/dx = (dy/dv) * (dv/du) * (du/dx) = (2v) * (3u^2) * (2) now we replace v and u with their "x" counterparts: dy/dx = (2(u^3 + 5)) * (3(2x+1)^2) * (2) = (2(2x+1)^3 + 5) * (3(2x+1)^2) * 2 = 12((2x + 1)^3 + 5)(2x + 1)^2 hooray! (this is much easier when spoken out compared to when writing it out!)

Algebra
TutorMe
Question:

How can I understand the Rational Roots Theorem to solve for the zeroes in a complex polynomial equation, where mere factoring doesn't do the trick?

Max S.
Answer:

The Rational Roots Theorem is one of the most important tools in all of algebra, pertaining to solving polynominal equation. I call it the "P over Q" equation, because all you really need to do is set up division problem: taking all factors of "p" (the constant term) divided by all factors of "q" (the leading coefficient). We then go through the options iteratively, using synthetic division to determine the rational roots of P(x), which allows us to systematically "factor" the equation, that is, break down the degree of the polynominal until it becomes solvable using basic factoring techniques or through the quadratic equation.

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