Logarhythm

a * b = 2a - 3b + ab

3 * 5 = 2(3) - 3(5) + (3)(5)

         = 6 - 15 + 15

         = 6

5 * 3 = 2(5) - 3(3) + (5)(3)

         = 10 - 9 + 15

         = 1 + 15

         = 16

3 * 5 + 5 * 3 = 6 + 16 = 22 

The answer is B.

First find the hypotenuse of the first triangle.
\(\sqrt{8^2 + 6^2} =10\)

Now since we know they're similar triangles, we can see that the relationship between the triangles are that they are doubled. 
10 ÷ 2 = 5, which means x = 5. 

If I'm not mistaken, this question is saying that Hunter gave a quarter of his money and an additional 7 dollars to Shenese. 
Let x represent the amount of money Hunter had originally

\(x - \frac{1}{4}x - 7 = 23\)

\(\frac{3}{4}x = 30\)

\(3x = 120\)

\(x = 40\)

Hopefully it's correct?

x - 90 = 45

x = 135 

 

No problemo :D

Plug in the values:
\(\sqrt[3]{4a^4b^5} + \frac{a-c}{(b+c)^2}\)

\(= \sqrt[3]{4(4)^4(2)^5} + \frac{(4)-(-5)}{((2)+(-5))^2}\)

\(= \sqrt[3]{1024 \cdot 32} + \frac{9}{(-3)^2}\)

\(= \sqrt[3]{32768} + \frac{9}{9}\)

\(= 32 + 1\)

\(= 33\)

\(\frac{14x + 22}{x + 2} = -3x + 1\)

\(14x + 22 = (-3x + 1)(x + 2)\)

\(14x + 22 = -3x^2 -6x + x + 2\)

\(3x^2 + 6x + 14x - x + 22 - 2 = 0\)

\(3x^2 + 19x + 20 = 0\)

\(x = {-(19) \pm \sqrt{(19)^2-4(3)(20)} \over 2(3)}\)

\(x = {-19 \pm \sqrt{361-240} \over 6}\)

\(x = {-19 \pm \sqrt{121} \over 6}\)

\(x = {-19 \pm 11 \over 6}\)

\(x = -5\)

\(x = -\frac{4}{3}\)

\(-5 + -\frac{4}{3} = \frac{-15 - 4}{3} = -\frac{19}{3}\)

3x² + 4x + 12 = x² - 5x + 7

2x² + 9x + 5 = 0

x² + \(\frac{9}{2}\)x + \(\frac{5}{2}\) = 0

x² + \(\frac{9}{2}\)x + \(\frac{81}{16}\) - \(\frac{81}{16}\) + \(\frac{5}{2}\) = 0

(x + \(\frac{9}{4}\) )² - \(\frac{81}{16}\)+ \(\frac{40}{16}\) = 0

(x + \(\frac{9}{4}\) )² = \(\frac{41}{16}\)

x + \(\frac{9}{4}\) = ± \(\frac{\sqrt{41}}{4}\)

x = ± \(\frac{\sqrt{41} - 9}{4}\)

r = \(\frac{\sqrt{41} - 9}{4}\)

s = \(\frac{9 - \sqrt{41}}{4}\)

r² + s² = \((\frac{\sqrt{41} - 9}{4})^2 + (\frac{9 - \sqrt{41}}{4})^2\)

= \(\frac{(\sqrt{41} - 9)^2 + (9 - \sqrt{41})^2}{16}\)

= \(\frac{122 - 18\sqrt{41} + 122 - 18\sqrt{41}}{16}\)

= \(\frac{2(122 - 18\sqrt{41})}{16}\)

= \(\frac{122 - 18\sqrt{41}}{8}\)

= \(\frac{61 - 9\sqrt{41}}{4}\)

tell me if I did anything wrong :)