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Tutor profile: Camille P.

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Camille P.
UCLA graduate tutor
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Questions

Subject: Trigonometry

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Question:

Given sin x = 5/13, find tan x.

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Camille P.
Answer:

Firstly sin x = opposite/hypotenuse. Using this for the problem above, we then know that opposite = 5 and hypotenuse = 13. With this we can then draw out a right triangle to have a better idea of the image and, looking at this, we would be able to see that we can use the Pythagorean theorem to solve for the length of the adjacent/third side of the triangle. The Pythagorean theorem tells us that a^2 + b^2 = c^2 (with a and b being the lengths of the adjacent and opposite sides of the triangle and c being the hypotenuse length). Plugging in our known values into this equation then gives us a^2 + 5^2 = 13^2. We can simplify this to be a^2 + 25 = 169. We can then solve for a by first subtracting 25 from both sides to isolate a^2, giving us a^2 = 144. After this, we can then solve for a by taking the square root of both sides of the equal sign, leaving us with a = 12. With this we have the lengths of all three sides of the right triangle and can find the value of tan x, which is known to be tan x = opposite/adjacent. Plugging in the known values, we then get tan x = 5/12.

Subject: Basic Math

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Question:

54x8=

Inactive
Camille P.
Answer:

First, we stack the numbers vertically, with each digit lined up with the another below it. Then the first step is to multiply the rightmost digit of the bottom number to the top digits. If we write 8 in the bottom, we multiply the 8 first by 4, which give us 32. We then write 2 on the answer line, right below the number 8 and carry the 3 on top of the 5. We then multiply 8 by the second number from the right, 5, which gives us 40. But, we must add the 3 that we carried over, giving us 43. As there is no more number to carry over to, we simply write 43 in the answer right next to the 2, giving us 432.

Subject: Algebra

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Question:

Given x+y=4 and 3x-y=8, what are the values of x and y?

Inactive
Camille P.
Answer:

With this system of equations problem, we can use either the process of elimination or substitution. For elimination, we can line up the two equations vertically and add the two equations to each other. This will leave us with 4x=12 (+y and -y will cancel out). To solve for x, we can then divide both sides (4x and 12) by 4 to get x=3. We can then substitute x=3 into either equation to solve for the y variable. For substitution, we can use one of the equations to solve for/isolate one variable. For example, using the first equation, x+y=4, we can solve for/isolate x to get x=4-y (as we subtract 4 from both sides). We can then use this new equation (x=4-y) to substitute 4-y in the second equation for x. This will give us 3(4-y)-y=8. This then leaves us with only one variable in the equation, allowing us to solve for y by simplifying. We will then multiply the 3 into the parenthesis to finally get, 12-3y-y=8. Simplifying this will give 12-4y=8. To solve for y, we must isolate the variable, which we can do by subtracting 12 from both sides. This will give us -4y=-4. We can solve for y by dividing both sides by -4, leaving us with y=1. Again, we can plug this y value into either equation to solve for the value of the x variable.

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