with that template in mind, let's see
so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.
Jason solved for the quotient of 133.5 ÷ 5.
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with that template in mind, let's see
so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
B = 1/2, thus the graph is stretched by twice as much.
This is adding two vectors, one magnitude 300 direction east, and one magnitude 100 direction 30 degrees northeast.
For the resulting sum we can use the Law of Cosines or just add up the components. Let's try it both ways.
Components first. In the east direction we have
300 + 100 cos 30° = 300 + 100(√3/2) = 386.60 miles per hour
In the north direction we have
0 + 100 sin 30° = 100 (1/2) = 50 miles per hour
So a total speed of
sqrt{386.6^2 + 50^2} = 389.82 miles per hour
Answer: 389.82
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OK let's try the Law of Cosines. We have sides 300 and 100 and included angle 150°.