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# Tutor profile: Ian V.

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Ian V.
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## Questions

### Subject:Pre-Calculus

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Question:

Find the real solution(s) for: $$2\sqrt{2x} = x-6$$

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Ian V.

Since there is a square root, we can try squaring both sides of the equation: $$(2\sqrt{2x})^2 = (x-6)^2$$ On the left side, everything is multiplied together, so we can use the distributive property of exponents. On the right side, we must expand into a polynomial: $$2^2(\sqrt{2x})^2 = (x-6)(x-6)$$ Simplify: $$4(2x) = x^2-6x-6x+36$$ $$8x = x^2-12x+36$$ Subtract $$8x$$ from each side: $$0 = x^2-20x+36$$ Factor the polynomial and solve: $$0 = (x-18)(x-2)$$ $$x=18, x=2$$ We can check these solutions in the original equation to make sure: $$x=18$$: $$2\sqrt{2(18)} = (18)-6$$ $$2\sqrt{36} = 12$$ $$\pm12=12$$, so $$x=12$$ is a valid solution. $$x=2$$: $$2\sqrt{2(2)} = 2-6$$ $$2\sqrt{4} = -4$$ $$\pm4 = -4$$, so $$x=2$$ is also a valid solution.

### Subject:Trigonometry

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Question:

A lone tree casts a shadow on a flat plain. In the afternoon, the shadow is 24 meters long when the sun is 40 degrees above the horizon. A: How tall is the tree? B: How long will the shadow be soon before dusk, when the sun is 22 degrees above the horizon?

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Ian V.

PART A: A right triangle can be imagined, with the length of the shadow, $$l$$ and height of the tree, $$T$$ forming a right angle at the base of the tree, and an angle of 40 degrees between the hypotenuse and $$l$$. Using TOA (from SOH CAH TOA), $$tan(\theta)=\frac{opposite}{adjacent}$$ $$tan(40) = \frac{T}{24}$$ Solving for T, $$T=24tan(40)$$ Using a calculator, $$T = 20.1$$ meters PART B: The height of the tree, $$T$$, is the same as in part A. Now, we are solving for the length of the shadow, $$l$$, with the angle between $$T$$ and $$l$$ at 22 degrees. Using the same formula, $$tan(22) = \frac{T}{l}$$ Multiply both sides by $$l$$ and divide by $$tan(22)$$, $$l=\frac{T}{tan(22)}$$ Substitute the expression for $$T$$, $$l = 24\frac{tan(40)}{tan(22)}$$ We can plug this into a calculator to get the length of the shadow. $$l=49.8$$ meters

### Subject:Algebra

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Question:

A colony of bacteria currently has a population of 500. The population doubles every hour. Using the following equation, $$P(t) = Ae^{rt}$$ where $$P(t)$$ is the population of the colony at time $$t$$; $$A$$ is the population at $$t=0$$; and $$r$$ is a constant related to growth rate How many hours until the colony reaches a population of one million?

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Ian V.

We want to find $$t$$ when $$P$$ is 1,000,000 and $$A$$ is 500, but $$r$$ is an unknown. Since we know that the population doubles every hour, we can set $$t=1$$ and $$P(1)=2A$$ and simplify: $$P(t) = Ae^{rt}$$ $$(2A) = Ae^{(1)r}$$ The $$A$$s cancel out, and $$2 = e^{r}$$ We can simplify this by taking the natural log of each side: $$ln(2) = ln(e^{r})$$ Using properties of logarithms, we get: $$r=ln(2)$$ Now, we can use the initial equation and plug in the current population, final population, and rate constant, and simplify: $$1,000,000=500e^{tln(2)}$$ $$\frac{1,000,000}{500}=\big({e^{ln(2)}}\big)^t$$ $$2000=2^{t}$$ We need to take the logarithm of each side, and $$log_{2}$$ will simplify the process: $$log_{2}(2000)=log_{2}(2^{t})$$ Using properties of logs, pull the $$t$$ in front and simplify $$t=log_{2}(2000)$$ Using a calculator, $$t=10.97$$ hours

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