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Zach N.

Student at Kansas State University

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Trigonometry

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Question:

Triangle ABC is a right triangle. Line segment AB is the hypotenuse and angle C is the right angle. The line segment AB is equal to 6 and line segment BC is equal to $$ \pi $$. Solve for the sine of angle A.

Zach N.

Answer:

Since we are trying to find sine, we take the opposite side length divided by the side length of the hypotenuse. The length of the opposite side to angle A, BC, is $$ \pi $$ and the hypotenuse, AB, is 6, so the sine of angle A would be $$ \frac{\pi}{6} $$. Looking at our unit circle, we know that $$ \sin( \frac{\pi}{6}) = \frac{1}{2} $$.

Basic Math

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Question:

John has a business that creates notebooks. It cost 10 cents to create a notebook that he can the sell for $1.10. He can make 100 notebooks a day 5 days a week. How much money can his business make in a week if he sells all of the notebooks?

Zach N.

Answer:

If each notebook cost 10 cents to make, then the total profit for one notebook is $1 because $1.10 - $0.10 = $1. That means his notebooks can produce a profit of $100 a day and if the business is open 5 days a week, $100 x 5 days = $500 a week.

Calculus

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Question:

Show that the derivative of $$x^{3}+x+6 = 3x^{2} + 1 $$ using the equation $$ lim_{h->0} \frac {f(x+h) - f(x)}{h}$$.

Zach N.

Answer:

$$ lim_{h->0} \frac {[(x+h)^{3} + (x+h) + 6] - (x^{3} + x + 6)}{h} $$ Substitution $$ lim_{h->0} \frac {[(x+h)(x+h)(x+h) + (x+h) + 6] - (x^{3} + x + 6)}{h} $$ Seperation of $$ (x+h)^{3}$$ $$ lim_{h->0} \frac {[(x^{2}+2xh+h^{2})(x+h) + (x+h) + 6] - (x^{3} + x + 6)}{h} $$ FOIL $$ lim_{h->0} \frac {x^{3}+3x^{2}h+3h^{2}x+h^{3} + x+h + 6 - x^{3} - x - 6}{h} $$ FOIL $$ lim_{h->0} \frac {3x^{2}h+3h^{2}x+h^{3} +h}{h} $$ Simplify $$ lim_{h->0} \frac {h(3x^{2}+3hx+h^{2} +1)}{h} $$ Factor out $$h$$ $$ lim_{h->0} 3x^{2}+3hx+h^{2} +1 $$ Cancel $$h$$ $$ 3x^{2}+3(0)x+0^{2} +1 $$ Plug in 0 $$ 3x^{2}+1 $$ Simplify

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