# Tutor profile: Nicolas B.

## Questions

### Subject: Music Theory

Spell root-position diatonic seventh chords on each scale degree of the D major scale. Provide a roman numeral analysis for each chord.

D, F#, A, C# Roman numeral analysis: IM7 (major seventh chord) E, G, B, D Roman numeral analysis: ii7 (minor seventh chord) F#, A, C#, E Roman numeral analysis: iiim7 (minor seventh chord) G, B, D, F# Roman numeral analysis: IVM7 (major seventh chord) A, C#, E, G Roman numeral analysis: V7 (Dominant seventh chord or Mm7 chord) B, D, F#, A Roman numeral analysis: vi7 (minor seventh chord) C#, E, G, B Roman numeral analysis: viiø7 (half-diminished seventh chord) Remember that for chords with a minor third (minor triads and diminished triads), we use lowercase letters for the roman numeral analysis. Only chords with a major third are notated with capital letters in their roman numeral analysis. In minor seventh chords, only a lowercase roman numeral and a 7 is necessary to notate the analysis, so ii7 is correct as opposed to iim7. In major seventh chords, we need to use a M7 as opposed to a 7 to distinguish between Mm7 chords (7) and MM7 chords (M7).

### Subject: Calculus

A rancher is building a rectangular pasture for her horses out of fencing. She has $$1000$$ yards of fencing to build the pasture and wants does not want to have fencing left over. Find the dimensions of the pasture that would maximize its enclosed area.

This problem deals with optimization, which requires us to find the local maximum of a function by finding critical points. Since we need to find the maximum area of the pasture, we need to create a formula for its area. To write the pasture's area as a function of its length and width, we can let $$A$$ represent the area of the pasture, and $$l$$ and $$w$$ will represent the length and width of the rectangle. Therefore $$A=l \times w$$. Do not forget that the area of the rancher's pasture is constrained by the amount of fencing she has available, which is $$1000$$ yards of fencing. This means that the perimeter of the rectangular pasture must be $$1000$$ yards, since the rancher does not want to have any unused fencing. If we let $$P$$ represent the perimeter of the rectangle, then $$P=2l+2w$$. Since we know that the perimeter, $$P$$, in this case is $$1000$$ yards, then $$1000=2l+2w$$. The two formulas we have now created are as follows: $$A=l \times w$$ and $$1000=2l+2w$$. Since we need to find the maximum area of a rectangle with a perimeter of 1000 yards, we need to find a local maximum in our area function, $$A=l \times w$$, by taking the derivative with respect to one variable. Since there are two variables in this equation, we need rewrite one variable in terms of the other. This is where we can use the perimeter formula $$1000=2l+2w$$. We can choose the variable we want to rewrite, so for this example I will rewrite $$l$$ in terms of $$w$$. 1) Since $$1000=2l+2w$$, we can use algebra to isolate $$l$$ on one side of the equation. $$1000=2l+2w$$ Subtract $$2w$$ from both sides: $$1000-2w=2l$$ Divide both sides by $$2$$: $$500-w=l$$ 2) Substitute $$500-w$$ for $$l$$ in the area function. $$A=(500-w) \times w$$ Multiply each term in $$500-w$$ by $$w$$ using the distributive property: $$A=w^2-500w$$ Now we have written the area $$A$$ of this pasture in terms of one variable, its width $$w$$. 3) Take the derivative of function $$A$$ with respect to $$w$$ and find critical points that would suggest a local minimum or maximum. Use the power rule to find the derivative of function $$A$$: $$\frac{dA}{dw}=2w-500$$ Find a critical value of $$w$$, which is a value of $$w$$ where the derivative of function $$A$$ is equal to $$0$$. Set $$\frac{dA}{dw}$$ equal to $$0$$ in $$\frac{dA}{dw}=2w-500$$ and solve for $$w$$ using algebra. $$2w-500=0$$ $$2w=500$$ $$w=250$$ 4) We now know that the function $$A=w^2+500w$$ has a critical value at $$w=250$$, which indicates that a local minimum or maximum could exist at this value of $$w$$. Use a number line test for the function $$\frac{dA}{dw}=2w-500$$ to see that it changes from positive to negative at $$w=250$$. This means that the function $$A=w^2+500w$$ has a local maximum at $$w=250$$, so the width of our area-optimized rectangular pasture is $$250$$ yards. 5) Finally solve for the value of $$l$$ using $$l=500-w$$, since we found that $$w=250$$ $$l=500-250$$ $$l=250$$ We have now found that the dimensions, the length and width, of the rectangular pasture that would yield the maximum enclosed area is a length of $$250$$ yards and a width of $$250$$ yards.

### Subject: Algebra

Solve the following system of equations using the substitution method. $$4x+9y=47$$ $$x-y=2$$

To use the substitution method for solving systems of equations, we need to first use one of the equations to solve for one variable. In this case, it would be easiest to use $$x-y=2$$ to solve for $$x$$ as follows. 1) $$x-y=2$$ 2) Isolate $$x$$ on one side of the equation: $$x=y+2$$ Now we can use $$y+2$$ as another way to write $$x$$ in the other equation, $$4x+9y=47$$. This allows us to solve for a numerical value for the other variable $$y$$. 1) $$4(y+2)+9y=47$$ 2) Remember to multiply both terms in $$y+2$$ by $$4$$ using the distributive property: $$4y+8+9y=47$$ 3) Combine like terms: $$13y+8=47$$ 4) Isolate the $$y$$ term on one side of the equation by subtracting $$8$$ from each side: $$13y=39$$ 5) Divide each side by $$13$$: $$y=3$$ Now, we have a numerical value for one variable. Remember that we solved for $$x$$ in terms of $$y$$. We found that $$x=y+2$$. Since we now know that $$y=3$$ to solve this system of equations, we can substitute this number in place of $$y$$ in $$x=y+2$$ to find a numerical solution for $$x$$. 1) $$x=y+2$$, and $$y=3$$. 2) Therefore, $$x=3+2$$, so $$x=5$$. Using the substitution method, we have found that the solution for the given system of equations is $$(5, 3)$$, since $$x=5$$ and $$y=3$$.

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