Step-by-step explanation:
(secx-tanx)(secx+tanx)
=sec^2-tan^2
=1
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:
write 16 in the soup column and 27 in the peanut butter column
Answer:
Step-by-step explanation:
The common difference (d) can be found using the first and 4th terms:
a1 = 3
a4 = a1 +d(4 -1)
-9 = 3 +3d . . . . . simplify
-3 = 1 + d . . . . . . divide by 3
-4 = d . . . . . . . . . subtract 1
Then ...
x = a1 + d = 3 -4 = -1
y = x + d = -1 -4 = -5
The values of x and y are -1 and -5, respectively.
Answer:
±0.06
Step-by-step explanation:
The range is 0.28 - 0.16 = 0.12. So half of that is 0.06.