# Tutor profile: Johnathan A.

## Questions

### Subject: Geometry

Is the triangle with the given sides below an acute triangle, an obtuse triangle, or a right triangle? 2, 4, 7

-For this problem, we need to know the conditions for the sides of an acute triangle, an obtuse triangle, and a right triangle. -First, we label the three numbers given, which represent the three sides of the triangle, with the variables 'a', 'b', and 'c'. Since 'c' is always the longest side of the triangle, 'c' must be 7 (c = 7). Sides 'a' and 'b' can be any of the two other sides. So for this problem, let us have 'a' be 2 (a = 2), and 'b' be 4 (b = 4). -For sides a, b, and c, we can have the following three situations. -An acute triangle must satisfy the following: a^2 + b^2 > c^2 ('^2' means squared, or to the second power). -An obtuse triangle will have the following inequality: a^2 + b^2 < c^2. -Lastly, a right triangle must satisfy this equation, which is also known as the Pythagorean Theorem: a^2 + b^2 = c^2. -Now, we must test our values for a, b, and c to determine which situation matches with these values. - If one tests each of the three cases, one would see that the three sides of the triangle, a,b, and c, make an obtuse triangle, as shown below: obtuse triangle: a^2 + b^2 < c^2 a = 2, b = 4, and c = 7 (from beginning) 2^2 + 4^2 < 7^2 (replace (substitute) values a, b, and c in inequality with known values) 4 + 16 < 49 (simplify) 20 < 49 (this is true) -Since the values a, b, and c for the sides of this triangle satisfy this inequality that represents an obtuse triangle, this means that these three sides make up an obtuse triangle.

### Subject: Pre-Algebra

Jake and Rachael are both selling candy bars for their own school fundraisers. Jake sells his candy bars for $3 per candy bar, where he had already earned $15 before this day of the fundraiser. Rachael started off with no money, but sells her candy bars for $6 per candy bar. When do Jake and Rachael have the same amount of money?

-First, we need to identify the variable in the problem. Variables represent quantities in a math problem, and since this problem discusses how many chocolate bars are sold, this must be our variable. We will call the number of chocolate bars sold 'x'. -We now need to write the expressions for the amount of money that each student receives. Jake sells his candy bars for $3 each, where he has already earned $15. This means that we have the expression 3x + 15 (each often means multiply, which is what we are doing to 3 and x here!). Using the same idea, Rachael sells candy bars for $6 each. Thus, we have the expression for what Rachael earns, which is 6x. -Due to the fact that Jake and Rachael are charging different amounts of money for their candy bars, we know that at some point, both Jake and Rachael will have the same amount of money. In this case, Rachael will eventually catch up with Jake due to having each candy bar she sells be more expensive. In order to determine at what number of candy bars sold will Jake and Rachael have the same amount of money, we equate (make equal) the two expressions above, and solve for 'x'. - 3x + 15 = 6x (Jake) (Rachael) - 3x + 15 = 6x -3x -3x - 15 = 3x - 5 = x Therefore, Jake and Rachael will have earned the same amount of money after they have both sold 5 candy bars each.

### Subject: Algebra

Robert buys medium pizzas (p) and liter sodas (s) for their school math club. Each medium pizza they purchase is $12 and each liter soda is $2. Their purchase is a total of $40. Also, the number of medium pizzas is 1 more than the number of liter sodas. How many medium pizzas (p) and liter sodas (s) did Robert purchase?

-This is a typical system of equations problem, where a system is a collection of two or more equations. With any word problem, we need to identify what is being asked. Here, they are asking how many of each item is being purchased (in this case, medium pizzas and liter sodas). -Since we know that the variable representing medium pizzas is 'p', and that each medium pizza is $12, we have the expression 12p (the word 'each' in math often means multiply!). We use a similar approach for the liter sodas. The variable representing liter sodas is 's', and each liter soda costs $2. This means that we have the expression 2s. -At this point, we write our equations. -We combine the expressions from above (12p and 2s) by adding them (since we are buying both pizza and sodas), and making their sum equal to the total spent, which was $40. Therefore, we have our first equation: Equation 1: 12p + 2s = 40 -With our second equation, it will relate the how much of one food item was bought compared to another. For this problem, we know from the beginning that the number of medium pizzas purchased has to be 1 more than the number of liter sodas purchased. When we see the word 'more', that tells us that addition will occur. Thus, we have the number of pizzas (p) being equal to 1 more than the number of sodas (s + 1). Our second equation is: p = s + 1 -Now, we solve our system of equations to determine how many medium pizzas and liter sodas we need to satisfy the system. We use the substitution method for solving this system. - Step 1: Substitute (replace) p in first equation with second equation. 12p + 2s = 40, p = s + 1 12(s + 1) + 2s = 40 (notice that we now have 1 variable, making it easy to solve for s!) - Step 2: Solve for variable s. 12(s + 1) + 2s = 40 12s + 12 + 2s = 40 (distribute) 14s + 12 = 40 (combine like terms) 14s + 12 = 40 (solve for 's') -12 -12 14s = 28 s = 2 (This means that Robert bought 2 sodas) -Step 3: Solve for variable p (we now replace 's' in the other equation with what we found in the previous step). p = s + 1 p = 2 + 1 (since s = 2 from previous step) p = 3 (This means that Robert bought 3 medium pizzas) -Step 4: Write solution. In order to have satisfied this system, Robert had to have bought 3 medium pizzas and 2 sodas for his math club.