BuilderBoi

Let \(\sqrt{x} = m\) and let \( \sqrt{y}=n\)

We now have a new system: 

\(4m^2 + 9n^2 = 793 \ \ \ \ \ \ \ \ (i)\)

\(m + n = 13 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)\)

From \((ii)\), we have \(n = 13 - m \).

Plugging this into \((i)\) gives us \(4m^2 + 9(13 - m)^2 = 793\), which simplifies to \(13m^2 + 1521 - 234m=793\).

Subtracting \(793\) from both sides gives us \(13m^2 - 234m + 728 = 0\).

Now, solve for \(m\), square it, and plug it into both equations to make sure it works.

Good luck!

\(x = \sqrt{6^2 + 8^2}\)

Let the common difference be \(d\).

Because each term is the sum of the previous term and the common difference, we can write a system of equations: 

\(x + y + d = x - y \ \ \ \ \ \ \ \ \ \ \ \ \ (i)\)

\(x - y + d = x + 3y \ \ \ \ \ \ \ \ \ \ \ (ii)\)

\(x + 3y + d = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iii)\)

From \((i)\), we have: \(y + d = -y\), meaning \(d = -2y\)

Plugging this into \((ii)\) and \((iii)\) gives us a system with 2 equations: 

\(x - y - 2y = x + 3y\)

\(x + 3y + 2y = 1 \)

Solving the system, we find that \(x = 1\) and \(y = 0 \), meaning \(d = 0\), so the fifth term is \(1 + 0 = \color{brown}\boxed{1}\) 

There are 2 cases for number 2: the one digit is a 0, and the one's digit is a 5

For the former case: 1 choice for the one's digit, 5 choices for the 10s digit, and 8 choices for the hundreds digit, for \(8 \times 5 \times 1 = 40\)

For the latter case: 1 choice for the one's digit, 4 choices for the 10s digit (keep in mind we can't re-use 5), and 8 choices for the hundreds digit, for \(1 \times 4 \times 8 = 32\) numbers. 

But, we overcount for 015, 035, 075, and 095, making the total \(40 + 32 - 6 = \color{brown}\boxed{68}\)

Note that it is the same as the previous answer divided by 2. Can you figure out why?

I'll do number 1, and let you solve for number 2...

For number 1, there are 2 choices for the units digit (0 or 5), 9 choices for the tens digit (anything but the units digit), and 8 choices for the hundreds digit (anything but the tens or unit digit).

This makes for \(2 \times 9 \times 8 = 144\)choices, but we aren't done yet...

Our answer includes numbers like \(085\), which isn't a 3-digit number. 

There is 1 choice for the hundreds digit, 8 choices for the tens digit, and 1 choice for the one's digit, which makes for \(1 \times 8 \times 1 = 8\) numbers. 

So, there are \(144 - 8 = \color{brown}\boxed{136}\) numbers that work

The equation of the first circle is \((x - 2)^2 + (y + 1)^2 = 16\), and the equation of the second circle is \((x - 2)^2 + (y - 5)^2 = 12\)

Subtracting the second equation from the first gives us \((y + 1)^2 - (y - 5)^2 = 4\)

Solving for y, we find that \(y = -{7 \over 3}\). 

Now, plug this into the first equation (either equation works), and solve for x: \((x-2)^2 + ({-4 \over 3})^2 = 16\)

With your x values, calculate the distance between them, and square it...

Good luck!

I don't see 2 variables...

Find the slope of equation 1 as follows: 

\(3x - 4y = 7 \)

\(-4y = 7 - 3x \)

\(4y = -7 + 3x \)

\(4y = 3x - 7\)
\(y = {3 \over 4} x - {7 \over 4}\)

The slope of the line is \({3 \over 4}\), so the slope of the other line is \(-{4 \over 3}\)(negative reciprocal)

Now, convert the second equation into slope-intercept form:

 \(8x + Ay = B\)

\(Ay = B - 8x\)

\(y = {B \over A} - {8 \over A}x\)

\(y = -{8 \over A} x + {B \over A}\)

The slope of this line is \({-8 \over A}\), and we can solve for A: \(-{8 \over A} = -{4 \over 3} \Rightarrow a = 6\)

Now, plugging this back into the equation gives us: \(8x + 6y = B\)

Plug in the given coordinates: \(8 \times 5 + 6 \times 2 = 52\)

So, \(a = 6\) and \(b = 52\), meaning \(a + b = 6 + 52 = \color{brown}\boxed{58}\)

Let the side of the triangle be \(2s\)

The area of "success" is a \(60 ^ \circ\) sector of a circle centered at Point A, and the radius is half the side length of the triangle. 

The total area is the area of an equilateral triangle. 

Can you take it from here?

It simplifies to \(a + b - 2\), but note that \(a + b = {- b \over a}= {7 \over 2}\). 

Can you take it from here?