I don't know, what did the policeman shout to the math professor as a mob of excited calculus students crowded around these displays on his graphing calculator?
360 - 250 = 110
m<1 = 110/2 = 55
answer
55
Area = perimeter + 132.
Let each side of the city be x miles long, then:-
x^2 = 4x + 132
x^2 - 4x - 132 = 0
x = [-(-4) +/- sqrt((-4)^2 - 4 * 1 *-132)] / 2
x = 13.66, -9.66 We ignore the negative
So the city has dimension of 13.66 * 13.66
13.7 * 13.7 to nearest 10th
Answer:
which one do you need help on
Answer:
Option 3 is right.
Step-by-step explanation:
Reference angle of x is obtained by either 180-x, 180+x. or 360-x depending on the posiiton of terminal whether II quadrant or iv quadrant, or iii quadrant, etc.
In whatever way we find reference angles,
cos will remain cos only and sin will remain sin only there may be only changes in sign.
Of all the ordered pairs given, we find that I, II, and Iv there is a switch over form cos to sine and sin to cos. Hence these options cannot be for reference angles.
III option is 
show that both sign and cos changed sign. This is possible only in III quadrant.
ie reference angle of orignal angle t = 180+t
SO this option is right.