BuilderBoi

The roots of the quadratic are \({15 \pm \sqrt{225-4t} \over 2}\)

Note that \(\sqrt {225 - 4t}\) must be an odd, positive integer. 

To solve, set \(\sqrt{225 - 4t}\) equal to 13, 11, 9, 7, 5, 3, and 1.

Now, solve for t in all cases and find the average.

Simplify: \(x^2 + y^2 + 5x - 3y - 24 = 0 \)

Add \(1.5^2 \) and \(2.5^2\) to both sides to complete the square: \(x^2 + y^2 + 5x - 3y - 24 + 1.5^2 + 2.5^2 = 1.5^2 + 2.5^2\)

Now, add 24 to both sides: \(x^2 + y^2 + 5x - 3y + 1.5^2 + 2.5^2 = 24 + 1.5^2 + 2.5^2\)

Complete the square on x: \((x+2.5)^2 + y^2 - 3y - 24 = 1.5^2 + 2.5^2\)

Complete the square on y: \((x+2.5)^2 + (y-1.5)^2 = 1.5^2 +2.5^2 + 24\)

Simplify the right-hand side: \((x+2.5)^2 + (y - 1.5)^2 = 32.5\)

The radius of this circle is \(\sqrt{32.5}\), so the area is \(\color{brown}\boxed{32.5 \pi}\)

\(\sqrt{160} = \sqrt{16} \times \sqrt {10} = 4 \sqrt{10}\)

\(\sqrt{252} = \sqrt{36} \times \sqrt 7 = 6 \sqrt 7\)

\(\sqrt{245} = \sqrt{49} \times \sqrt{5} = 7 \sqrt 5 \)

\(\sqrt{180} = \sqrt{36} \times \sqrt{5} = 6 \sqrt 5\)

\({4 \sqrt {10} \over 6 \sqrt 7} \times {7 \sqrt 5 \over 6 \sqrt5} = {4 \sqrt {10} \over 6 \sqrt 7} \times {7 \over 6} = {28 \sqrt {10} \over 36 \sqrt 7} = {28 \sqrt {10} \over 36 \sqrt 7} \times {\sqrt 7 \over\sqrt 7} = {28 \sqrt {70} \over 252} = \color{brown}\boxed{\sqrt{70} \over 9} \)

Let \(l \) = local stamps and let \(f\) = foreign stamps. 

We have the system: 

\(l = {4 \over 5}(l + f)\)

\(l - 20 = {8 \over 15} (f + l) \)

Now, just solve for l and f...

Oh, wait that's a cube root...

Yea, you're right, I thought it meant \(n\sqrt{3}\) not \(\sqrt[3]{n}\)

"There are three times as many th graders as there are students with blue eyes"

Your question is missing information...

Only 3 does. 

Note that the floor part of x will always be an integer, so x must end in .8.

Now, let \(x = a + 0.8\), where \(a = \lfloor x \rfloor\)

We now have the equation \(2a + 0.8 = 4.8\), meaning \(a = 2\), so \(x = 2 + 0.8 = \color{brown}\boxed{2.8}\)

https://web2.0calc.com/questions/circles_101

https://web2.0calc.com/questions/simplify-1-3-5-199-2-4-6-200-202