Melody

How do I find the equation of a cosine function with a maximum of 3, a minimum of -3, and a period of 4?

lets see

y=3cos b(x+c)

c can be any constant becasue you have not mentioned phase shift.

where period = 2pi/b

\(\frac{2\pi}{b}=4\\ \frac{\pi}{b}=2\\ b=\frac{\pi}{2}\)

so it is

\(y=3cos[\frac{\pi }{2}(x+c)]\)

check (here is your graph )

By the way, simplifying these expressions has nothing to do with calculus.

Do you mean something different?  Maybe you want them differentiated or integrated?

I think I will let you comment before I play with any more of them.

Hi,

Firstly you cannot get a solution becasue there is no equal sign.

All you can do it try to simplify them :)

I am confused about what the actual questions are.  

I assume there are a number of powers here.

23: sec4 (x) - tan4 (x)

I assume

\(sec^4 (x) - tan^4 (x)\\ \qquad \text{Difference of 2 squares}\\ =(sec^2 (x) + tan^2 (x))(sec^2 (x) - tan^2 (x))\\ \qquad sec^2x=\frac{1}{cos^2x}=\frac{sin^2x+cos^2x}{cos^2} =tan^2x+1 \\ =(1+tan^2(x) + tan^2 (x))(1+tan^2x- tan^2 (x))\\ =1+2tan^2(x) \\ \)

I do not understand what you are asking :/

3825=1750e^.23t

\(3825=1750e^{0.23t}\\ \frac{3825}{1750}=e^{0.23t}\\ 2.187514286\approx e^{0.23t}\\ ln{(2.187514286)}\approx ln{e^{0.23t}}\\ ln{(2.187514286)}\approx 0.23t\\ t \approx \frac{ln(2.187514286)}{0.23}\\ \)

t= ln(2.187514286)/0.23 = 3.4033298694398451

Now you already know CPhil, after a couple of milllion years Red got tired of being red, he wanted a new look so together we died him green!  Green goes better with MY red glasses that Red stole from me too!

I bit of trivia here:

In Australia red headed men are often nicknamed 'bluey'. 

May, you cannot multiply two dollar amounts together. It does not make sense. 

what is log10

log 10 base 10 is 1 because   \(10^{1}=10\)

 \(log_{10}10=1\)

only negative numbers are less than negative numbers.

Think about a number line.

It is a letter of the alphabet :)