If one inch equals 25.4 millimeters, you would have to divide 25.4 by 1/8 which equals to 3.2 (3.175 before rounding). Subtract 3.2 from 25.4 = 22.2 millimeters
Are these two separate questions?
Answer:
Step-by-step explanation:
we know that
---> by addition angle postulate
we have
----> given problem
----> given problem
substitute in the expression above
Divide by 2 both sides
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)
-----------------------
|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)
-----------------------
u dot v = <8,4> dot <9,-9>
u dot v = 8*9 + 4*(-9)
u dot v = 72-36
u dot v = 36
-----------------------
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
Answer:
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Step-by-step explanation:
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