neworleans06

Yes, you stop when there are all positive values. Feel free to message me at any time!

Tertre might have typoed, yet I think he forgot the 0x^2. This is really important! Check three just to make sure. Three should work.

You're correct, six does work. 

Drop a perpendicular height from the right of the side length of 5 to the base of 8. This would create a right-triangle with a hypotenuse of 7 units and a measure of one of the legs, 3 units. Using the Pythagorean theorem, we have $7^2-3^2=49-9=\sqrt{40}=2\sqrt{10}.$

Multiply the section of the chord, they should be equal to each other.

That looks great to me! In general, the formula for a hexagon is $\frac{3\sqrt{3}}{2}s^2$ . 

Like what guest had said, drop two perpendicular heights from the left and right to the bottom of the trapezoid. Now, we have two 30-60-90 triangles and using those properties, the angle opposite to 90 degrees is two, so the angle opposite to 30 degrees is 1. Adding up all the side lengths, the perimeter is thus $2+2+2+1+2+1=2(4)+2=\boxed{10}.$

Nice solution tertre, but I think you meant to say $a^2+b^2=49?$

Yeah, that is right. In general, for two points, the slope is:$\frac{y-2-y_1}{x_2-x_1}.$

$\frac{108}{100}*43000=46440.$

Yes, we have to see.

Hmm, this looks very tricky.

Since $k$ is odd, we denote $f(k)=k+3.$ Any odd number plus $3$ is even, so $f(k + 3) = \frac{k + 3}{2}$.

But, if $\frac{k + 3}{2}$ is odd, therefore $f \left( \frac{k + 3}{2} \right) = \frac{k + 3}{2} + 3 = 27.$ Then this leads for $k=45$, and $f(f(f(45))) = f(f(48)) = f(24) = 12$, so it is even. $f \left( \frac{k + 3}{2} \right) = \frac{k + 3}{4} = 27$ leads to $105$, so this is your answer?

Check: $f(f(f(105))) = f(f(108)) = f(54) = 27.$