What are the zeros of the function f(x) = x^2 + 5x + 5 written in simplest radical form?
1 answer:
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Answer:
∠J = 60°
Step-by-step explanation:
The Law of Cosines tells you ...
j² = k² +l² -2kl·cos(J)
Solving for J gives ...
J = arccos((k² +l² -j²)/(2kl))
J = arccos((14² +80² -74²)/(2·14·80)) = arccos(1120/2240) = arccos(1/2)
J = 60°
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<em>Additional comment</em>
It is pretty rare to find a set of integer side lengths that result in one of the angles of the triangle being a rational number of degrees.
