Angle EBC=3r+10 Angle ABE=2r-20 Find angle EBF
1 answer:
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The function in the graph has the name of square function.
The domain of a function is all values of x the function can have. The domain of this function is all real numbers:
The range of a function is all values of y the function can have. The range of this function is all positive numbers, including zero:
In order to graph f(x) + 2, we just need to translate the graph 2 units up. To find the new points, we need to increase all y-coordinates by 2:
(-2, 6), (-1, 3), (0, 2), (1, 3), (2, 6)
Domain: {x ∈ ℝ}
Range: {y ∈ ℝ | y ≥ 2}
Then, in order to graph f(x) - 2, we just need to translate the graph 2 units down. To find the new points, we need to decrease all y-coordinates by 2:
(-2, 2), (-1, -1), (0, -2), (1, -1), (2, 2)
Domain: {x ∈ ℝ}
Range: {y ∈ ℝ | y ≥ -2}
Answer:
a) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to :
For this case we know that
So then
b) We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to :
For this case we know that
And we know that is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then
c) For this case that any factor of the sin function is equal to 0.
So from definition of tan we have this:
Step-by-step explanation:
a. sin (17pi / 4 )
We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to :
For this case we know that
So then
b. cos (19pi / 6 )
We can remove the complete rotations around the unitary circle like this, because we know that one complete revolution is equivalent to :
For this case we know that
And we know that is on the III quadrant since is equivalent to 210 degrees. And on the III quadrant the cosine is negative. So then
c. tan(450pi)
For this case that any factor of the sin function is equal to 0.
So from definition of tan we have this: