MathProblemSolver101

$r^2 + 18^2 = (r+6)^2$

$r^2 + 324 = r^2 + 12r + 36$

$12r = 288$

$\boxed{r = 24}$

$\sum_{n=1}^5 n^2$

(1) converges

Updated #2, it was initially wrong I updated

(1) $g(f(4)) = g(-1) = \boxed{3}$

(2) $g(h(-8)) = g(-6) = \boxed{5}$

(3) $f(g(3)) = f(7) = \boxed{-7}$

(4) $f(g(-2)) = f(-8) = \boxed{-127}$

(5) $f(g(8)) = f(4) = \boxed{48}$

(6) $f(g(8)) = f(4) = \boxed{80}$

(7) $f(g(-3)) = f(13) = \boxed{2i \sqrt{3}}$

(8) $g(h(4)) = g(63) = \boxed{3 \sqrt{7}}$

$(y+6) + n = 12$

$12 + n = 3y$

$y + n = 6$

$n = 3y - 12$

$y + (3y - 12) = 6$

$4y - 12 = 6$

$4y = 18$

$\boxed{y = 4.5}$

$\frac{67 + 52 + x}{3} = 70$

$119 + x = 210$

$x = 210 - 119$

$\boxed{x = 91}$

Because $2^{10} = 1024,$ and $1000^2 = 1,000,000,$ we can see that $(2^{10})^2 = 1024^2$ which is obviously greater than $1,000,000.$

If we have $2^{19},$ that is $2^10 \cdot 2^9 = 1024 \cdot 512$ which is obviously less than $1,000,000.$

Thus, $2^{19} < 1,000,000 < 2^{20}$

Note $8^{n} = 2^{3n}.$

Since we know that $2^{19} < 1,000,000$ and $2^{3n} < 1,000,000, n = \left \lfloor{\frac{19}{3}}\right \rfloor = 6.$

Thus, $8^6 < 1,000,000 < 8^7.$

We have the answer to be $19 - 6 = \boxed{13}$

$\frac{1^2 \pi}{(1 - (-1))(1-(-1))} = \frac{\pi}{4}$