Answer:
C
Step-by-step explanation:
the angle measures of triangles must add up to 180 degrees. Since we are only given two angles, we can assume that all of the answers that add up to be below 180 degrees fulfill this requirement with the third angle. However, answer choice C adds up to 190 degrees on the first two angle, which means that no matter how low the third angle is, it will never add up to 180 degrees.
Take the derivative with respect to t

the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero

divide by w

we add sin(wt) to both sides

divide both sides by cos(wt)

OR

(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

since 2npi is just the period of cos

substituting our second soultion we get

since 2npi is the period

so the maximum value =

minimum value =
X is 90 degrees. If AB is the diameter, then it passes through the centre. X is 90 degrees because the angle in a semicircle is always a right angle.
Answer:
43/50
Step-by-step explanation:
Answer: x = -2
Step-by-step explanation: vertical lines are the x axis