David R.

Master's Student in Physics and Astronomy

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GRE

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Question:

The sum of all interior angle of a regular polygon is 1800°. What is the measure of one of the polygon's interior angles?

David R.

Answer:

Most GRE test takers freak out about geometry. I know I did, because it had been so long since I had studied it. However a surprising amount of geometry questions can be reasoned out. An essential fact of geometry is that the interior angles of a triangle (3 sides, 3 angles) add up to 180. That is to say if you cut off all three corners of a triangle and put them together they would make a straight line, 180 degrees. Similarly, we could figure out that a square's 4 right angles add up to 360 degrees ($$4*90=360$$). Now two numbers (180, 360) isn't enough to figure out the sequence for sure, but I bet you could take a guess. Sure enough, each additional side of a regular polygon adds 180 degrees to the total interior angle. To be official: the sum of all interior angles of a regular polygon equals 2 less than the number of sides (n) multiplied by 180. $$(n-2)*180$$. Thus for this problem $$1800=(n-2)*180$$. After some arithmetic we find the number of sides is 12. Thus the number of angles is also 12. To find the measure of one angle take the total interior angle divided by the number of interior angles to get a measure of 150 degrees.

Algebra

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Question:

Factor the quadratic equation: $$12a^{2}-10ab-8b^{2}$$

David R.

Answer:

In every factoring problem, we need to first look for a Greatest Common Factor. By taking one out we can greatly simplify the problem. Here we can see 2 is a factor of 12, 10 and 8. By factoring the 2 out (we'll put it back in at the end) we are left with $$6x^{2}-5x-4$$. This is a quadratic with a leading coefficient in the form $$ax^{2}+bx+c$$. There are many methods to factor a quadratic, but I'd like to share a recent way I learned which I call slide and divide. First slide by multiplying the lead coefficient by the constant $$(a*c)$$ giving us 24. Now we must find two numbers that multiply to 24 and add to -5. List the factors of 24 if needed. We find that -8 and 3 satisfy those requirements. Now here is the simple steps of slide and divide which actually have a lot of difficult math hidden in them. Divide the numbers by the lead coefficient (a) and simplify the fractions. $$\frac{-8}{6}=\frac{-4}{3}$$ and $$\frac{3}{6}=\frac{1}{2}$$. We know all quadratics factor into two binomials. The denominator of each fraction is the coefficient of the x term and the numerator is the constant. I like to say you slide up the bottom number. This gives us $$(3x-4)(2x+1)$$. Slide and divide is a strange method, but it is really fast when you get used to it. Last, don't forget to put the GCF we factored out at the beginning! Final answer: $$2(3x-4)(2x+1)$$.

Physics

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Question:

A field goal is made by kicking a football from the ground with a speed $$v_0$$ with no curve and and at an angle $$\theta$$ above the ground. Find the position as a function of time, neglecting air resistance.

David R.

Answer:

This is a typical introductory classical mechanics question. A simple and elegant result that is the basis of all projectile problems is that Newton's second law in three dimensions is equivalent to three one-dimension equations. This allows us to split up our coordinates into separate problems. Here we should define our x coordinate along the ground in the direction of the kick and the y coordinate as the vertical. In the x-direction no forces are at play when ignoring air resistance. The ball remains at the initial speed given in the x-direction. We can split up the initial velocity into two vectors by defining $$v_x=v_0\cos(\theta)$$ and $$v_y=v_0\sin(\theta)$$. Try drawing the picture to see those vectors. We now have what we need for out basic kinematic equation in the x-direction. Defining the projectile's initial position as the origin the horizontal position of the ball will be $$x=v_0t\cos(\theta)$$. In the y-direction the ball experiences an acceleration from the force of gravity which results in the arc. We usually define gravity to be in the negative direction. Plugging into the kinematic equation we see that the vertical position is given by $$y=v_0t\sin(\theta)-\frac{1}{2}gt$$. From these equations it is possible to expand the problem to find properties such as when the ball returns to the ground (set y to 0).

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