CubeyThePenguin

Triangle MKJ is similar to triangle MKL.

\(\frac{MJ}{MK}= \frac{MK}{KL}\)

x/15 = 16/20

x = 12

Optimize by having x > 0 and y > 0.

The pairs that work are (4, 3) and (3, 4) with a sum of 7.

sum of the first 9 squares: 2^1 + 2^2 + ... + 2^9 = 2^10 - 1

10th square: 2^10

ratio: 1023/1024

This is a repost. I answered this a couple days ago.

The center of the circle is at (2, -4) and the radius of the circle is 4. We can write the equation like this and simplify.

\((x-2)^2 + (y+4)^2 = 16\)

\((x^2 - 4x + 4) + (y^2 + 8y + 16) = 16\)

\(x^2 - 4x + y^2 + 8y + 4 = 0\)

\(\frac{(pq^3)^3(4p^2q)^2}{(2pq^2)^3} = \frac{(p^3q^9)(16p^4q^2)}{8p^3q^6} = \frac{16p^7q^{11}}{8p^3q^6}\)

\(\boxed{2p^4q^5}\)

\(\sqrt{14^2 - 3^2} = \sqrt{187}\)

This is about 13.7 cm.

1. Start with a number: y

Subtract 3: y - 3

Double it: 2(y-3)

2. Start from the inside and work outwards.

1 + (1/(1 + 1/((z+1)/z)))= 5

and so on. 

You can solve this without WolframAlpha. Start by adding all the equations together:

4a + 4b + 4c + 4d + 4e + 4f = 326

a + b + c + d + e + f = 81.5

If each equation is called (1), (2), (3), ..., etc, then you can subtract our "master equation" from each like so:

3a + b + c + d + e + f = 43

a + b + c + d + e + f = 81.5

Subtract, so 2a = -38.5 and a = -19.25.

The numbers may be incorrect, but you get the idea.

Let's think about composite numbers instead. Composite numbers can be written as the product of smaller prime numbers. For example, 6 = 2 * 3.

Composite polynomials are kind of the same. They can be written as the product of smaller, indivisible factors. 

For example, x^4 - 16 = (x^2 + 4)(x + 2)(x - 2). 

You can't divide these factors any further, so x^4 - 16 is composite.