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Neil Gerson A.
I've been tutoring for two months now, but I will assure anyone that it's gonna turn out well.
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Basic Chemistry
TutorMe
Question:

Balance the following chemical reaction: $$ Cu \; + \; HNO_3 \; \longrightarrow \; Cu(NO_3)_2 \; + \; NO \; + \; H_2O$$

Neil Gerson A.
Answer:

$$ 3Cu \; + \; 8HNO_3 \; \longrightarrow \; 3Cu(NO_3)_2 \; + \; 2NO \; + \; 4H_2O$$

C++ Programming
TutorMe
Question:

Write a C++ code that recursively computes the nth power of a.

Neil Gerson A.
Answer:

function power (int a, int n) { // In this code, let the problem look like a^n = ? if (n == 1) { return a; } else { return a * power(a, n - 1); } }

Algebra
TutorMe
Question:

How many distinct solutions (not necessarily real) exist for the equation? $(x-1)(x^2-1)(x^3-1)\ldots(x^10-1)=0$1$.

Neil Gerson A.
Answer:

Recall that if we have two numbers or terms that when multiplied together results to 0, its either one of them is 0, or both of them are 0, i.e. if $AB = 0$, then $A = 0$ or $B = 0$. Applying this concept to the equation, we can let $A$ and $B$ (take note we can arbitrarily partition the equation in two parts) to be $A = (x - 1)(x^2 - 1)$ and the remaining parts of the equation to be $B$, giving us either $(x - 1)(x^2 - 1) = 0$ or $(x^3 - 1)(x^4 - 1) \ldots (x^10 - 1) = 0$. Consider $A$ to be 0, as stated in the concept above. Take note that it can again be divided into two terms, $(x - 1)$ and $(x^2 - 1)$. Applying the concept again, we have either $(x - 1) = 0$ or $(x^2 - 1) = 0$. Solving the first one, we have $x = 1$. The second one gives us $x = \pm 1$. Applying the same concept repetitively on $B$, we have two distinct solutions for $B$, which is $x = 1$ and $x = -1$. Thus, taking the solutions of $A$ and $B$, we have two distinct solutions for the equation.

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