Let f(x) = 2x2 and g(x) = x2-1. Then f(g(2)) =
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This problem is better understood with a given figure. Assuming that the flight is in a perfect northwest direction such that the angle is 45°, therefore I believe I have the correct figure to simulate the situation (see attached).
Now we are asked to find for the value of the hypotenuse (flight speed) given the angle and the side opposite to the angle. In this case, we use the sin function:
sin θ = opposite side / hypotenuse
sin 45 = 68 miles per hr / flight
flight = 68 miles per hr / sin 45
<span>flight = 96.17 miles per hr</span>
Step-by-step explanation:
We have been given that AE=BE and .
We can see that angle CEA is vertical angle of angle DEB, therefore, as vertical angles are congruent.
We can see in triangles CEA and DEF that two angles and included sides are congruent.
or
Therefore, by ASA postulate.
Since corresponding parts of congruent triangles are congruent, therefore CE must be congruent to DE.