MathProblemSolver101

$b_p + b_g = 54$

$b_p = 5 b_g$

$6 b_g = 54$

$b_g = 9$

$b_p = 5 \cdot b_g = 5 \cdot 9 =45$

$45 - 26 = \boxed{19}$

$\frac{40}{25} \cdot 225 = 40 \cdot 9 = \boxed{360}$

$x = \pi k  - \frac{\pi}{2}, k \in \mathbb{Z}$

By Vieta's, we have $r_1 + r_2 = - \frac{8}{1} = -8$ and $r_1 r_2 = \frac{4}{1} = 4.$

We also have $r_1 + r_2 = - \frac{16}{A}$ and $r_1 r_2 = \frac{B}{A}.$

Thus, we have $- \frac{16}{A} = - \frac{8}{1} \Rightarrow \frac{2}{A} = 1 \Rightarrow A = 2.$

We also have $4 = \frac{B}{A} \Rightarrow 4 = \frac{B}{2} \Rightarrow B = 8.$

$2+8 = \boxed{10}.$

I could've noted the fact that $a_{n-1} = \frac{b_{n-1}}{2},$ but visualizations are somewhat unreliable. 

This doesn't make sense. How can an x-intercept have a $y \neq 0$?

In latex, it is \div for $\div$, but in normal text, on a mac, you can do option/alt and hold and press / for ÷.

$3x^2 - 7x = -1$

$3x^2 - 7x + 1 = 0$

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 3} = \frac{7 \pm \sqrt{37}}{6}.$

$7+37+6 = \boxed{50}$

 - Jimmy

$a_0 = 54 \div 3^3 = \frac{2}{3} = 2 \cdot 3^{-1}$

$a_1 = \frac{2}{3} \cdot 3 = 2 = 2 \cdot 3^0$

$a_2 = 2 \cdot 3 = 6 = 2 \cdot 3^1$

$a_3 = 6 \cdot 3 = 18 = 2 \cdot 3^2$

$a_4 = 18 \cdot 3 = 54 = 2 \cdot 3^3$

It should be clear by now that for $a_n$, it will be $\boxed{2 \cdot 3^{n-1}}.$

I'm not looking for the answer. I'm looking for a well approach that can be used in similar problems. That way I learn. If I just scavenge the answer, do I learn?