There is a theorem that says that this angle (formed by thre points on the circumference) is half such arc called EF.
We can deduct it as well.
1) Call O the center of the circle. The central angle (EOF) = arc EF = 151°
2) Draw a chord from F to D and other chord from D to E.
You have constructed two triangles, FOD and DOE.
3) Triangle FOD has two sides equals (because both are the radius of the circcle). Then, this is an isosceles triangle and it has two angles of the same measure. Call this measure a.
4) Triangle DOE is also isosceles, for the same reason explained in the poiint 3. Call the measure of its equal angles b.
5) The sum of the angles of the two triangles is 180° + 180 ° = 360°.
This is a + a + b + b + (360 - central angle) = 360°
=> 2a + 2b = central angle
=> 2(a+b) = cantral angle
=> (a+b) = central angle / 2 = 151° / 2 =71.5°
(a+b) is the angle that you are looking for.
Then the answer is 71.5°
There’s no image attached, ask the question again
Find the height of it with pythagoras
Root(100-36)=8
Volume = 1/3 pi r^2 h
1/3(36pi x 8)
=96pi
about 301.59ft^3
For this case we have the following function transformation:
Vertical displacement.
Assume k> 0
If f (x) is the original function, f (x) + k is the original function with a vertical displacement k units up.
Answer:
Part 1:
C. The function is not shifted horizontally from g (x)
Part 2:
A. 3 units up from g (x)
% of the day is equal to
