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# Tutor profile: Charlie T.

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Charlie T.
Student at Vanderbilt University
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## Questions

### Subject:Music Theory

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Question:

Each of these three progressions are in a major key and contain an issue. Identify the issues and fix the progressions by changing the placement of chords, changing a chord's quality, or removing a chord altogether. Try to remain as faithful to the original progressions as possible. a) I - III - IV - V7 - I b) I - I6 - IV - ii6 - V - viidim6 - I c) I - vi - IV6 - IV - Ger+6 - ii - V - vi

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Charlie T.

a) Unless it is marked as a modally borrowed flat III chord, the iii chord is always minor in a major chord progression. Because it is not marked as flat III, the second chord in this progression does not work. It should be written as: I - iii - IV - V7 - I. b) This is a retrogression: a V chord has the strongest pull towards the I of any diatonic chord of the major scale. Moving from a V to a viidim6 would thus weaken the cadence to the I chord. A corrected version of this progression would reverse the order of the V and viidim6 chords, or remove the viidim6 chord entirely: I - I6 - IV - ii6 - viidim6 - V - I or I - I6 - IV - ii6 - V - I. c) An augmented 6th chord always pulls toward the dominant, so resolving somewhere other than to V does not make sense. This progression could be properly rewritten as: I - vi - IV6 - IV - Ger+6 - V - vi or I - vi - IV6 - IV - ii - Ger+6 - V - vi or I - vi - IV6 - IV - ii - V - I.

### Subject:Calculus

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Question:

Find the derivative of the function f(x) = (lnx - x^2) * (e^x).

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Charlie T.

The product rule states that a(x) * b(x) = a'(x) * b(x) + a(x) * b'(x), where a(x) and b(x) are two differentiable functions. Applying this rule here, we can consider f(x) = a(x) * b(x) where a(x) = lnx - x^2 and b(x) = e^x. So f'(x) = a'(x) * b(x) + a(x) * b'(x). By the difference rule, we know that a(x) = lnx - x^2 becomes a'(x) = d/dx(lnx) - d/dx(x^2). The derivative of lnx is 1/x, and the derivative of any function of the form x^n is n * x^(n - 1). So a'(x) = 1/x - 2x^1 = 1/x - 2x. The derivative of e^x is e^x, so b'(x) = e^x. So f'(x) = (1/x - 2x) * (e^x) + (lnx - x^2) * (e^x). By the distributive property, this is equivalent to: f'(x) = (e^x) * (lnx - x^2 - 2x + 1/x)

### Subject:Algebra

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Question:

Find the values of x on the domain [-5, 5] that satisfy the following equation: x^4 - 5x^3 = 24x^2

Inactive
Charlie T.

To solve for x, first place all expressions involving x on one side of the equation. This can be done by subtracting 24x^2 from both sides: x^4 - 5x^3 - 24x^2 = 0. Now, factor out x^2: (x^2 * x^2) - (x^2 * 5x) - (x^2 * 24) = 0 --> (x^2) * (x^2 - 5x - 24) = 0. The expression x^2 - 5x - 24 can be factored by considering the expression (x + a) * (x + b), where a * b = -24 and ax + bx = -5x --> a + b = -5. The possible values (a, b) that satisfy a * b = -24 are (-1, 24), (-2, 12), (-3, 8), (-4, 6), (-6, 4), (-8, 3), (-12, 2), and (-24, 1). Since (-8, 3) is the only pair that satisfies the equation a + b = -5, x^2 - 5x - 24 can be rewritten as (x - 8) * (x + 3). Note: although pairs of values where b is negative and a is positive would also work, we don't need to worry about them. The pair (3, -8) will result in the same factoring expression, because multiplication is communicative. Now we have the equation (x^2) * (x - 8) * (x + 3) = 0 --> x^2 = 0, x - 8 = 0, x + 3 = 0. Solving for x results in x = -3, 0, 8. But x=8 lies outside of the domain [-5, 5], so our final answer is: x= -3, 0

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