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 =
B. 1 and 5 are factors of all three numbers :)
Answer:
See below
Step-by-step explanation:
a = 15 sqrt 2 sin 45
= 15 sqrt 2 * sqrt( 2 ) /2 = 15 * 2 /2 = 15
sin 60 = a/c
sqrt(3)/2 = 15/c
c = 15 / ( sqrt(3)/2) = 30 / sqrt 3 = 30 sqrt 3 /3 = 10 sqrt 3
Answer:
335.115
Step-by-step explanation:
First you find the area of the circles to do that you do pi times radius squared
For the top square do 3.14(pi)time 3²=28.26
For the middle square do 3.14 times 4.5²=63.585
For the bottom square do 3.14 times 6²=113.04
Now find the area of the entire thing do 18×12+9+6+3=18×30=540
Finally do 540-the area of the squares 28.26-63.585-113.04= 335.115
So the shaded region is 335.115
Hope this helps and have a great day!
Answer:
1. Add 6 to both sides, cause that is the inverse of subtraction, so you get 5x = 50 as 6 cancels 6 out.
2. Divide 5 from both sides and get x = 10