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 =
Answer:
3
Step-by-step explanation: if he has 2 rows of four on each page and 13 pages 8 * 13 is 104 and 107 - 104 is 3
Answer:
D
Step-by-step explanation: All the numbers have to be less than 7 but greater than -45
Answer:
x=28+n
Step-by-step explanation:
So, the teacher is giving Jonathan papers so we know its addition.
28+n
So lets have x equal the amount of papers he has on his desk after the new papers are added.
x=28+n
You need to divide the amount from the quantity
So 5.92 divide by 166= $0.37 per pound.