Alan

"A basketball is shot at 65 degrees above horizontal with an initial velocity of 8.8m/s. ... What is the maximum height the ball reached above its starting point?"

Using the constant acceleration equation:  v² = u² + 2as where: 

v = final vertical velocity  (0 m/s)

u = initial vertical velocity (8.8*sin(65°) m/s)

a = acceleration ( -9.8 m/s²)

s = vertical distance travelled

0 = [8.8*sin(65°)]² - 2*9.8*s

s = [8.8*sin(65°)]² /(2*9.8) or s = 3.245 m

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Top row, 3rd column from the left.

Like so:

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Perhaps the expression is meant to be:  [(x^2 - 2y)/x]^8  in which case:

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Asked and answered here: https://web2.0calc.com/questions/geometry_133

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Use the fact that   log_a b = c means  b = a^c. 

(a)  log₃ (2x-1) = 2   means  2x - 1 = 3²,   or  x = (3² + 1)/2  or  x = 5

(b)  log_x 16 = 2   means   16 = x²,   so x = 4

Delilah noticed that 16 could be expressed as 2*8 and that 10 could be expressed as 2 + 8.

Therefore, the equation can be written as (x + 2)(x + 8) = 0

Hence x = -2 and x = -8

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Hmm!  

In polar notation the direction is given by \(\theta\) 

r gives the magnitude   It's the negative magnitude that I have difficulty visualizing!

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Have a look at:  http://www.embedded.com/design/other/4216719/Performing-efficient-arctangent-approximation