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so you would get whatever cos^-1(sqrroot:57) would be for the answer to #7... make sure your calc is in radians. Dont forget that the angle is between pi/2 and pi.
For #5: Circumference is 2(pi)(radius) or pi(diameter). Area is pi(radius^2). isolate r in the area equation, so r = sqrroot:(area/pi). Now that you know what r equals, plug that in to the circumference equation, so you get: C = 2(pi)(sqrroot:(area/pi)).
^ I always make the angle positive (not always required). If it isnt, add 2pi to the angle (in radians). make sure the number is in Quad 2 by being sure that the angle is between pi/2 (somewhere around 1.57) and pi (3.14...). if it isnt, subtract the angle from pi (to see the angle's proximity to the x-axis), then add that difference back to pi (to match that proximity, but on the other side of the x-axis).
^oh and make sure you make the answer 0+2(pi)k so the answer is applicable to an infinite domain.
for 6a, since we know that the tangent inverse of a number is equal to the angle where its tangent equals the number in the inverse (tan^-1(y)= x, tan x= y), x is an angle, and y is its tangent value. So the inverse of the tangent value of x will equal x. Because this equation is valid for all real numbers x (except at pi/2 and 3pi/2, because then tanx would be undefined), x belongs to the set of all real numbers. so you could say x does not equal (pi/2) + pi(k), so that the answer is applicable to an infinite domain.Hopefully that made sense.....
Here are a few things that will help you out
Sin = (1/Csc)
Csc = (1/Sin)
Cos = (1/Sec)
Sec = (1/Cos)
Tan = (1/Cot)
Cot = (1/Tan)
Pi = 180 Degrees
(Pi / 2) = 90 Degrees
Sin is positive in the 1st and 2nd quadrents, therefore, Csc is positive in the 1st and 2nd quadrents
Oh and that "O" the line through it is the greek letter theta, which is just a variable, like x or y
That should help
C=2 x pi x r
A=pi x r x r
r= A/ (pi x r)
C= 2 x pi x A/(pi x r)
C=2A/r
r x r=A/pi
r = + or - sqroot(A/pi)
C= 2 x pi x A(pi x sqroot(A/pi))
Did I confuse you?
