NotThatSmart

It's actually not hard to just list all the numbers out. 

We have

\(\frac{1}{3}, \frac{2}{3}, \frac{4}{3},\frac{5}{3}, \frac{7}{3}, \frac{8}{3}, \frac{10}{3}, \frac{11}{3}, \frac{13}{3}, \frac{14}{3}\)

The ones we excluded from this list, like 3/3 and 6/3 are not in simplest form, therefore eliminated. 

Now, adding all these values up, we get \(75/3=25\)

That means 25 is our answer!

Thanks! :)

First, let's calculate the ones inside the parenthesis. 

We have \(16*5*\frac{1}{10} = 8\)

Now, we just have to find \(\log_{1/2} (8)\)

Using logs and bases, we have \(\log_{1/2} (8) = -3\)

So -3 is our answer!

Thanks! :)

There's actually only one triangle that satisfies all the rules stated.

The only one that works is 

2-2-1

Any other triangle would violate the triangle inequality theorems, therefore being invalid. 

So 1 is our answer!

Thanks! :)

First, let's note that PXQ and RXS are similar triangles. 

The scale factor of the two is \( \sqrt{8/4} = \sqrt{2}\)

This means that the height of RXS is \( \frac{h\sqrt{2}}{ 1 + \sqrt{2}}\) where h is the height of the trapezoid. 

We have \(\frac{ h}{1 + \sqrt{2} }\) for the height of PXQ

The base of PXQ is \(\text{base of RXS}/\sqrt{2}\)

Now, using this information from PXQ, we can write

\(\text{Area} = \frac{(1/2) (\text{base of RXS})}{\sqrt{2} * h ( 1 + \sqrt{2})} \\ 4 = (1/2) (\text{base RXS}) * h / ( 2 +\sqrt{ 2}) \\ 8(2 + \sqrt{2}) / \text{base RXS} = h\)

We know the area of the trapezoid is \((1/2) h (\text{sum of bases})\)

We have 

\( 4 [ 2 +\sqrt {2}] [ 2 + \sqrt {2} ] / / 2 \\ 2 [ 2 + \sqrt {2}] ^2 \\ 2 [ 4 + 4\sqrt {2}+ 2] = \\ 2 [ 6 + 4\sqrt {2}] = \\ 12 + 8\sqrt {2} \)

Thanks! :)

Let's set the center of the circle as O. 

Draw a line from O perpendicular to the chord. Let this line be OM. 

\(OA = 15 \\ AM = AB / 2 = 4/2 = 2\)

Triangle AOM is right, with angle AMO being 90 degrees.

By using the pythagorean theorem, we have

\(OM = \sqrt{ OA^2 -AM^2} = \sqrt{ 15^2 - 2^2 } = \sqrt{221}\)

So sqrt221 is our answer. 

Thanks! :)

Well, let's first note something. 

We have \(\frac{x}{x} = 1\) as long as x is not 0. 

We have \(\frac{y}{y} = 1\) as long as y is not 0. 

We have \( \frac{z}{z} =1\) as long as z is not 0. 

We finally have \(\frac{xyz}{xyz} = 1\) as long as x,y,z are not 0. 

So, we have \(1+1+1+1 = 4\) as long as \(x,y,z \neq 0\)

So there is only one possible value. 

Thanks! :)

We can use stars and bars to solve this problem. 

We have 9 stickers in a row (stars) 

We need to split these 9 stickers into two groups. We can use one bar for this. 

We have something like this

X X X | X X X X X X

In this case, one friend gets 3, the other gets 6. 

Now, the problem becomes, how many positions can the bars be put in. 

We have \(10 \choose 1\) = 10, so there are 10 ways. 

Thanks! :)

For this problem, we must know what EPF is. 

However, there is no clear representation of what it actually is. 

In the problem, it never actually defined what P is, meaning that it is impossible to solve. 

Thanks! :)