ilorty

Instead of solving that (which I don't think anyone has the time for), we could just find the units digits of all those numbers and add and subtract them. 

How to we do that? Well, we know that 19^1 has a unit digit 9, and 19^2 has a unit digit 1, 19^3 has unit digit 9, 19^4 has unit digit 1... you see the patern? Every odd power has unit digit of 9, and even power gives unit digit of 1. Do the same thing with 11: 11^1 : 1, 11^2 : 1, 11^3 : 1... we see that every unit digit of a power of 11 gives 1. Now for 6: 6^1 : 6, 6^2 : 6, 6^3 : 6... same thing here, we see that the unit digit is 6 for every power of 6. 

So, now we do 1+1, which is 2. Borrow a 10 from the tenths place, 12-6 is 6. Therefore, the unit's digit is \(\boxed{6}\)

This involves casework. So, let's first see how many options there are for 5 digits:

This is basically 10 to the power of 5 (ten numbers from 0-9, and you can repeat, so 10(10)(10)(10)(10), which is 10^5), which is 100,000

This involves systems of equations. To do this, let's first set up our equation:

x + y = 6

xy = 3

We now can see that x in terms of y is 6-y

Plug that in for x in the second equation, and you get 6y-y^2=3. We can turn this into a quadratic and solve.

-y^2+6y-3=0.

Divide both sides by -1

y^2-6y+3=0

Now, can you solve the quadratic? 

Once you have done that, find x, and then you are all set!

The greatest common factor is 11:

when you alternatedly add and subtract the digits, you will always end up with 0, which is why 11 is the GCF.

To do this, we see that 2^20 is also known as 20 2's. So, we start from 1, and keep on going until we get a total of more than 20 2's:

Number.         How may 2s

1                    0

2                    1

3                    0

4                   2

5                    0

6                  1

7                   0

8                  3

9                   0

10                1

11                  0

12                 2

13                0

...

keep on going until the sum of the collumn "How many 2s" is greater than 20!

The great thing about this problem is that the angles are 120, 120, 60, 60, which means that you can divide this rhombus into two equilateral triangles. 

Using the area formula for euqilateral triangles ,\(\frac{\sqrt{3}}{4} a^{2}\), we see that  the areas of each triangle is \(9\sqrt 3\). Multiply that by 2, and you get your answer: \(\boxed{18\sqrt 3}\)

For this, we can use half angle formulas. Since 45 degrees is a common angle on the unit circle, we can easily see that it's sine and cosine is \(\frac{\sqrt 2}{2}\). Using the half angle formula for cosines, which is: \(\cos \frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}}\), we can get our answer. 

Here is what you should do:

Plug in \(\frac{\sqrt 2}{2}\) for cos A and solve.

Good luck!