Math help. +rep

[Zachafer]

No problem. If you need any help in math (or other subjects) around the same level of difficulty you can pm me!

It's better to post on the forums. If you make a mistake then we can all point it out and visa versa

[Ray-Chill]

You go to the doctor and he gives you 15 milligrams of radioactive dye. After 20 minutes, 4.5 milligrams of dye remain in your system. To leave the doctor's office, you must pass through a radiation detector without sounding the alarm. If the detector will sound the alarm if more than 2 milligrams of the dye are in your system, how long will your visit to the doctor take, assuming you were given the dye as soon as you arrived? Give your answer to the nearest minute.

Please =]

[bamag]

A(T) = Ao * e^(kT)
4.5 = 15 * e^(20k)
.3 = e^(20k)
ln .3 = 20k
k = (ln .3)/20 = -.06

2 = 15 * e^(-.06t)
.133 = e^(-.06t)
ln .133 = -.06t
t = (ln .133)/-.06 = 33.62

Rounds to 34 minutes.

Haven't done this in a long time so it'll be cool if someone can verify.

By the way, A(T) refers to the amount after a certain amount of time has passed, Ao is the initial amount, e is euler's constant, ln is the logarithmic log (the power that e is raised to), k is a constant, and T is time.
The general idea is to solve for the constant with the information you have of one situation and then since the constant remains the same, use it for solve for time in the other situation.

[Tachyon]

A(T) = Ao * e^(kT)
4.5 = 15 * e^(20k)
.3 = e^(20k)
ln .3 = 20k
k = (ln .3)/20 = -.06

2 = 15 * e^(-.06t)
.133 = e^(-.06t)
ln .133 = -.06t
t = (ln .133)/-.06 = 33.62

Rounds to 34 minutes.

Haven't done this in a long time so it'll be cool if someone can verify.

I'll second this answer, if you need notes on how to do this for a test or something, I can get you a picture of mine.

[Ray-Chill]

=] Thanks you two.
I was able to figure it out. =] And did awesome (I hope) on my exam!

@(you need an account to see links) - It says I need to spread some rep around first before I can give you some. =] So will do soon.

[Ray-Chill]

Ah!! Help!!

I'm so freaking lost...

I have (1+cos(y))/(1+sec(y))

I need to simplify it in terms of cos/sin..

I'm stuck.. I don't understand how to get the fraction off...

[Tachyon]

(1+cos y)/(1+1/cos y) -> multiply top and bottom by cos y
cos y (cos y + 1)/(cos y + 1) -> Cross of (cos y + 1) Can somebody verify that you can do this please, I forget
Answer = cos y

I am not 100% sure on this so somebody will have to back me up, I always hated doing these.

[Zachafer]

(1 + cos(y)) / (1 + sec(y)) --> original
(1 + cos(y)) / (1 + 1 / cos(y)) --> sec(y)=cos(y)
(1 + cos(y)) / (cos(y) / cos(y) + 1 / cos(y)) --> prepared denominator for adding via LCD
(1 + cos(y)) / ((cos(y) + 1) / cos(y)) --> simplified denominator 
(1 + cos(y)) * (cos(y) / (cos(y) + 1)) --> division is multiplication of inverse
cos(y) --> (1 + cos(y)) and (cos(y) + 1) reduce to 1

Sorry, I don't have access to scanner so I had to type up my work.