Point Q lies on the circle and has an x-coordinate of 4.
2 answers:
Answer:
2 square root 5
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
The equation of the circle is equal to
where
(h,k) is the center and r is the radius of the circle
In this problem we have
center at (0,0) and radius equal 6 units
so
Remember that
If a point lie on the circle, then the point must satisfy the equation of the circle
Substitute the x-coordinate of point Q in the equation of the circle and solve for the y-coordinate of point Q
For x=4
simplify
The point Q lie on the first Quadrant (see the picture)
The y-coordinate is positive
therefore
The y-coordinate of point Q is
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Answer:
C,
Step-by-step explanation:
To remind you, Euler's formula gives a link between trigonometric and exponential functions in a very profound way:
Given the complex number , we want to try to get it in the same form as the right side of Euler's formula. As things are, though, we're unable to, and the reason for that has to do with the fact that both the sine and cosine functions are bound between the values 1 and -1, and 2√2 and -2√2 both lie outside that range.
One thing we could try would be to factor out a 2 to reduce both of those terms, giving us the expression
Still no good. √2 and -√2 are still greater than 1 and less than -1 respectively, so we'll have to reduce them a little more. With some clever thinking, you could factor out another 2, giving us the expression , and <em>now </em>we have something to work with.
Looking back at Euler's formula , we can map our expression inside the parentheses to the one on the right side of the formula, giving us
and
, or equivalently:
At this point, we can look at the unit circle (attached) to see the angle satisfying these two values for sine and cosine is 7π/4, so , and we can finally replace our expression in parentheses with its exponential equivalent:
Which is c on the multiple choice section.
