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# Tutor profile: Aaron N.

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Aaron N.
Tutor for more than three years
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## Questions

### Subject:Indonesian

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Question:

List three types of noun phrases in Bahasa Indonesia and give an example of each!

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Aaron N.

1. Noun + Noun Lemari (cupboard/closet) Es (ice) Lemari + es = lemari es (a refrigerator) 2. Noun + Adjective Rumah (house) baru (new) Rumah + baru = rumah baru (a new house) 3. Noun + Verb Mobil (car) balap (race) Mobil + balap = mobil balap (a race car)

### Subject:Basic Math

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Question:

This is just a fun question to test our logical reasoning. A magic square is a square that is filled with numbers starting from 1 such that the sum of each row, column, and diagonal is equal. An example of a normal $$3\times3$$ magic square would be: \begin{bmatrix} 4 & 9 & 2 \\ 3 & 5 & 7\\ 8 & 1 & 6 \end{bmatrix}We can see that the magic sum equals 15. Given this information, can you figure out a general formula to determine the magic sum of any $$n\times n$$ square?

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Aaron N.

Since it is an $$n\times n$$ square and we start with $$1$$, we can safely assume that we are filling the square up until $$n^{2}$$. Next, we need to see if this problem has certain patterns that we can turn into a general formula. A $$3\times 3$$ square has $$3$$ rows/columns. We can see that an $$n\times n$$ square would have $$n$$ rows/columns. If we try and find the total sum from $$1$$ until $$9$$, we get the result of $$45$$. Since the magic sum of the $$3\times 3$$ square is $$15$$, we can see that: $$\frac{45}{3} = 15$$ This means that the magic sum is determined by the total sum of all the numbers divided by the number of rows/columns : $$M=\frac{\sum_{i=1}^{n^{2}} i}{n}$$ Using the arithmetic series formula, we can simplify the formula into: $$M=\frac{n^{2}(n^{2}+1)}{n}$$ $$M=n(n^{2}+1)$$ Q.E.D.

### Subject:Algebra

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Question:

One of the most well-known algebraic identities is $$(a+b)^{2}=a^{2}+2ab +b^{2}$$. Given that $$a\neq b$$, prove that $$(a+b)^{2}=\frac{a^{3}+a^{2}b-ab^{2}-b^{3} }{a-b}$$ !

Inactive
Aaron N.

The problem may seem complicated, but we can do a step-by-step solution to simply it: 1. Group numerators with similar properties $$(a+b)^{2} = \frac{a^{3}-b^{3} }{a-b} + \frac{a^{2}b-ab^{2}}{a-b}$$ 2. Use the identity $$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$$ $$(a+b)^{2} = \frac{(a-b)(a^{2}+ab+b^{2})}{(a-b)} + \frac{a^{2}b-ab^{2}}{a-b}$$ 3. Use the distributive property in reverse to make $$a^{2}b-ab^{2}=ab(a-b)$$ $$(a+b)^{2} = \frac{(a-b)(a^{2}+ab+b^{2})}{(a-b)} + \frac{ab(a-b)}{(a-b)}$$ 4. From here, we can cancel $$(a-b)$$ from each numerator and denominator $$(a+b)^{2} = (a^{2}+ab+b^{2}) + ab$$ 5. Since all terms are in additions, we can use the commutative property to combine the same term, which brings us back to our first identity $$(a+b)^{2} = a^{2}+2ab+b^{2}$$ Q.E.D.

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