BuilderBoi

We have the equation: \(2(n+1)!+6n!=4(n+1)!\)

We can subtract \(2(n+1)!\) from both sides, which yields: \(6n!=2(n+1)!\)

Note that \((n+1)! = n! \times (n+1)\)

Substituting this gives us: \(6n!=2(n+1)n!\)

Now, we just have to divide by \(n!\), then solve for n. 

Can you take it from here?

Note that \(\large{0.125 = {1 \over 8}}\). This can be written as \(\large{1 \over 2^3}\) or \(\large{2^{-3}}\)

We now have \((2^{-3})^3\). We can simplify this by multiplying the exponents. 

Can you solve it now?

Let's count by case:

1 x 1 square: 1 square

2 x 2 squares: 4 squares

3 x 3 squares: 4 squares

4 x 4 squares: 4 squares

5 x 5 squares: 1 square

Now, can you tell me how many squares contain the black square?

Adding the two equations cancels out the t. This gives: \(2.5s=2\)

Solving, we find that \(s = 0.8\). 

Now substitute this into one of the 2 equations and solve for t. 

It is 13. Because a prime number has 0 factors, it can't be any bigger. 

Re-write as \(a+b+8\)

By Vieta's, we know that \(a+b= -{b \over a} \). Plugging the values in, we find that \(a+b=-{1 \over 7}\)

We now have to add 8 to this total, and we find the final answer is \(\color{brown}\boxed{7 {6 \over 7}}\)

There are \(50^2 = 2500\) cases

To solve this problem, we will count the ways of getting at least 1 multiple of 8. 

1 multiple of 8: 6 ways for the multiple of 8, 44 ways for the non-multiple, and 2 ways to order multiple and non-multiple, so \(44 \times 6 \times 2 = 528\)

2 multiples of 8: 6 ways for the first number, 6 ways for the second (we don't order because they are the same case), so \(6 \times 6 = 36\)

Thus, the total probability is \({{528+36} \over 2500 }= \color{brown}\boxed{0.2256}\)

I googled it, and apparently, there's an operator called used in coding that would work. It's called operator \(\text{%}\), and it finds the remainder when division is performed. 

It would work here, because \(3 \times 4 = 12\), and the remainder when 12 is divided by 5 is 2.

Thus, the missing operator is \(\color{brown}\boxed{\text{%}}\)

Here is a link to learn more: https://www.sciencedirect.com/topics/engineering/arithmetic-operator

That's smart actually! I'll do that next time!!

Because \(\triangle ABC\) is isoceles, \(AB = 18 \sqrt 2 \)

Because it is an isceles right triangle, \(\angle B = 45\), meaning that \(BY = {12 \over \sqrt2} = 6 \sqrt 2 \)

We also know that \(AY = 12 \sqrt 2\), because \(BY + YA = AB\)

Thus, the ratio of \(BY: YA = 6 \sqrt 2: 12 \sqrt 2 = \color{brown}\boxed{1 \over 2} \)