(Will give Brainliest/5 Stars if correct answer)Find the angle between the given vectors to the nearest tenth of a degree. u = <8, 4>, v = <9, -9>A) 81.6°B) 25.8°C) 35.8°D) 71.6°

Answer: Choice D) 
The angle between the two vectors is approximately 71.6 degrees

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Work Shown: 

Each time I write the word "dot" I mean "dot product". 

|u| = length of vector u
|u| = sqrt(u dot u)
|u| = sqrt(<8,4> dot <8,4>)
|u| = sqrt(8*8 + 4*4)
|u| = sqrt(64 + 16)
|u| = sqrt(80)
|u| = sqrt(16*5)
|u| = sqrt(16)*sqrt(5)
|u| = 4*sqrt(5)

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|v| = length of vector v
|v| = sqrt(v dot v)
|v| = sqrt(<9,-9> dot <9,-9>)
|v| = sqrt(9*9 + (-9)*(-9))
|v| = sqrt(81+81)
|v| = sqrt(2*81)
|v| = sqrt(2)*sqrt(81)
|v| = sqrt(2)*9
|v| = 9*sqrt(2)

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u dot v = <8,4> dot <9,-9>
u dot v = 8*9 + 4*(-9)
u dot v = 72-36
u dot v = 36

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cos(theta) = (u dot v)/(|u|*|v|)
cos(theta) = (36)/(4*sqrt(5)*9*sqrt(2))
cos(theta) = (36)/(36*sqrt(10))
cos(theta) = 1/(sqrt(10))
cos(theta) = sqrt(10)/10
theta = arccos(sqrt(10)/10)
theta = 71.56505
which rounds to 71.6 when rounding to one decimal place (nearest tenth)

That's why the approximate answer is roughly 71.6 degrees