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Tutor profile: Junghune N.

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Junghune N.
M.S. in Material Engineering for Math, Physics, and Chemistry Tutoring
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Questions

Subject: Pre-Calculus

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Question:

You are investing for your child's college education by depositing $25,000 into an account that compounds interest monthly at 3% interest. How long do you have to wait for the balance to reach $45,000?

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Junghune N.
Answer:

The compound interest formula is the following: $(B = P\bigg(1 + \frac{r}{n} \bigg)^{nt}$) where $$B$$ is the balance, $$P$$ is the principal or the starting balance, $$n$$ is the number of times interest is compounded in a year, $$r$$ is the percent interest in decimal form, $$t$$ is the number of years elapsed. We plug in 45,000 for $$B$$, 25,000 for $$P$$, 12 for $$n$$, and .03 for $$r$$ to get: $(45000 = 25000\bigg(1 + \frac{.03}{12} \bigg)^{12t}$) $(\frac{45000}{25000} = \bigg(1 + \frac{.03}{12} \bigg)^{12t}$) $(1.8 = (1.0025)^{12t}$) To deal with the fact that t is in the exponent, let's take the $$log$$ of both sides. $(log(1.8) = log((1.0025)^{12t})$) There is a property of logarithms such that: $(log (x)^y = y \thinspace log(x)$) We can use this property to move $$12t$$ down from the exponent as shown below: $(log (1.8) = 12t \thinspace log(1.0025)$) Rearrange the terms to directly solve for t: $(\frac{1}{12}\frac{log(1.8)}{log(1.0025)} = t$) $$ t = 19.62 \thinspace years$$

Subject: Trigonometry

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Question:

You designed a mischievous robot to pelt bypassers with a dodgeball at an initial speed of 40 ft/second. Due to your excitement, you rush to do a test throw at an unknown angle, $$\theta$$ and measured that the dodgeball landed 50 feet away. The engineer that helped you design the robot tells you that the distance traveled by the dodgeball is given by the following equation: $(\frac{1}{12} v_{o}^2 \sin(2\theta)= r$) Determine the throwing angle of the dodgeball.

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Junghune N.
Answer:

Since we are solving for the launch angle, $$\theta$$, let's rearrange the given equation: $(\frac{1}{12} v_{o}^2 \sin(2\theta)= r$) $(\sin(2\theta)= 12 \frac{r}{v_{o}^2}$) We plug in 40 for $$v_{o}$$ and 50 for $$r$$ to get: $(\sin(2\theta) = \bigg(12\cdot \frac{50}{40^2}\bigg)$) $(\sin(2\theta) = \frac{60}{160} = \frac{3}{8}$) We can use the following trigonometric identity: $(\sin(2\theta) = 2\sin(\theta)\cos(\theta)$) to rewrite the expression for $$\theta$$ as: $(2\sin(\theta)\cos(\theta)= \frac{3}{8}$) $(2\sin(\theta)\cos(\theta) - \frac{3}{8} = 0$) We can directly graph the above expression to find the zeroes. The zeroes are located at: $$\theta = 11.01^{\circ}$$ and $$78.99^{\circ} $$. Thus, the throwing angles of the dodgeball are: $$\theta = 11.01^{\circ}$$ and $$78.99^{\circ} $$.

Subject: Algebra

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Question:

You make picture frames for a side hustle. Earlier in the day, a client requested you to make a frame for a portrait that was brought back from their travels abroad. After preparing all the materials, you drove to your workshop to build the frame, but forgot to bring the exact specifications with you. Because you're really lazy and don't feel like driving back to your home, you figure that you can just do the math to figure out the dimensions. You recall that you need 15 feet of wood to build the client's frame and that its width was 3 feet longer than its height. What are the dimensions of the frame that you need to build?

Inactive
Junghune N.
Answer:

We have 2 important pieces of information here: 1.) You need 15 feet of wood to build the frame. 2.) the frame's width is 3 feet longer than its height. Information piece #1 tells you that the perimeter of the frame is 15 feet. Thus, if we take $$h$$ as the frame's height and $$w$$ as its width, then: $(2w + 2h = 15$) Information piece #2 tells you that the frame's width is 3 feet longer than its height. Thus: $(w = 3 + h$) We now have a system of equations: $(2w + 2h = 15$) $(w = 3 + h$) We can plug in $$3 + h$$ for $$w$$ in the first equation to solve for the frame's height: $(2w + 2h = 15$) $(2(3 + h) + 2h = 15$) $(6 + 2h + 2h = 15$) $(4h = 9$) $(h = \frac{9}{4}feet = 2.25 feet$) Now that we know that the frame's height is 2.25 feet tall, we can plug that back into the 2nd equation of the system: $(w = 3 + h$) $(w = 3 + 2.25$) $(w = 5.25 feet$) Thus, the dimensions of your picture frame are 2.25 feet tall and 5.25 feet wide. Let's check our work: $(5.25 = 3 + 2.25$) Check. $(2*5.25 + 2*2.25 = 10.5 + 4.5 = 15$) Check.

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