tertre

I might as well post a solution to the first problem...

a = 2 (mod 3)

a = 4 (mod 5)
a = 6 (mod 7)

a = 8 (mod 9)

The above is the system of congruences.

The answer is 314, by using the general Chinese Remainder Theorem. https://www.youtube.com/watch?v=xjtx4CaQa_4

1): We can use the Chinese Remainder Theorem.

I can post a solution if you want.

1.  7^(-1) (mod 13). Multiply both sides by two, such that when the result is divisible by thirteen the remainder is one. Thus, the answer is 2.3

2. We have 8^(-1) (mod 35), or 8*k is congruent to 1 (mod 35), and 8*22 is 176, which s 1 (mod 35), so 22 is the answer,

3. 3^(-1) (mod 100)m or 3*k is congruent to 1 (mod 100). Can you find the answer? Give it a try. 

2. The binomial theorem gives us (2x-1)^5=5C0(2x)^5*(-1)^0+5C1(2x)^4*(-1)^1+5C2(2x)^3(-1)^2+5C3+(2x)^2(-1)^3+5C4(2x)^1(-1)^4+5C5(2x)^0(-1)^5=\(32x^5-80x^4+80x^3-40x^2+10x-1\), so \(\boxed{-40}.\)

3. Try to use the binomial expansion.

4. How many men and woman can you choose?

5. Find the total number of paths to A-C, then from C-B.

4(1+2+3+....19)=4*190=760. Add the 19*19 square, so 760+361=1121 dots in the twentieth stellations. 

Can you post the question again? The title is too long...

A 10-inches pizza has a 5-inches radius.

A 12-inches pizza has a 6-inches radius.

The area of (Pizza 1) is 5^2pi=25pi and the area of (Pizza 2) is 6^2pi=36pi

The percent of the increase is \(\frac{36-25}{25}*100=\boxed{44}\% \)

Hint: Try to figure out a pattern with the tessellations!

These problems are actually fun to work with it once you get used to them :).

If you know this...\(r+\frac{1}{r}\cdot[(r+\frac{1}{r})^2-3]=r^3+\frac{1}{r^3}.\)

Plugging the value of \(({r}+\frac{1}{r})^2\) in, we get \(r+\frac{1}{r}\cdot0=r^3+\frac{1}{r^3}.\) and thus the answer is \(\boxed{0}.\)

We have \(x+x^2+\sqrt{x}=2457.\) 

The best thing to do is to set boundary parenthesis.

Let's try an integer value of \(x\), denote 36, which is 6^2. Plugging it in, we get \(36+36^2+6=1338\), so we have to go a bit higher.

Tring 49 gives us: \(49+49^2+7=2457\)-Yes!

Thus, the answer is \(\boxed{49}.\)

In these type of problems...it's best to set boundaries to minimize the range of the values. You could try moving the \(x\)'s to the other side, but that could go ugly.