Answer:
100 i think
Step-by-step explanation:
Part a)
Answer: 5*sqrt(2pi)/pi
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Work Shown:
r = sqrt(A/pi)
r = sqrt(50/pi)
r = sqrt(50)/sqrt(pi)
r = (sqrt(50)*sqrt(pi))/(sqrt(pi)*sqrt(pi))
r = sqrt(50pi)/pi
r = sqrt(25*2pi)/pi
r = sqrt(25)*sqrt(2pi)/pi
r = 5*sqrt(2pi)/pi
Note: the denominator is technically not able to be rationalized because of the pi there. There is no value we can multiply pi by so that we end up with a rational value. We could try 1/pi, but that will eventually lead back to having pi in the denominator. I think your teacher may have made a typo when s/he wrote "rationalize all denominators"
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Part b)
Answer: 3*sqrt(3pi)/pi
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Work Shown:
r = sqrt(A/pi)
r = sqrt(27/pi)
r = sqrt(27)/sqrt(pi)
r = (sqrt(27)*sqrt(pi))/(sqrt(pi)*sqrt(pi))
r = sqrt(27pi)/pi
r = sqrt(9*3pi)/pi
r = sqrt(9)*sqrt(3pi)/pi
r = 3*sqrt(3pi)/pi
Note: the same issue comes up as before in part a)
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Part c)
Answer: sqrt(19pi)/pi
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Work Shown:
r = sqrt(A/pi)
r = sqrt(19/pi)
r = sqrt(19)/sqrt(pi)
r = (sqrt(19)*sqrt(pi))/(sqrt(pi)*sqrt(pi))
r = sqrt(19pi)/pi
Given:
The point is (-5,7).
To find:
The image of the given point after 90 degrees counter clockwise rotation.
Solution:
If a point rotated 90 degrees counter clockwise, then

Let the given point is P(-5,7). Using the above rule, we get

Therefore, the image of the given point after 90 degrees counter clockwise rotation is (-7,-5).