A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
can someone help me with my problem?
anonymous
 5 years ago
can someone help me with my problem?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and what is your problem jmfranco?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02) We often use radioisotopes to estimate the age of things (such as the Earth). Assuming the halflife of some isotope is 1000 years, and you found a rock of unknown age near the Mid Atlantic Ridge. You measured the amount of isotope in the rock and determined that 12.5% of the original amount when the rock formed was still present. How old is the rock

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let's have a chat here franco. This is a growth and decay problem. Is that the topic you're studying?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0great. now, sometimes growth and decay have been modeled by exponential functions. This is the model common model and the one you'll be using here. Do you know the form of the exponential function that is used to model growth and decay? is so, type it in here...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I hope you don't mind that I'm not just giving you the answer, but you'll be better able to solve all the other problems you'll have this way. it will only take a few minutes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay i don't really know exponential. i understand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my connection just failed me for a bit, but i'm back

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this website is having a lot of traffic and seems slow, check out this site has a good explanation: http://serc.carleton.edu/quantskills/methods/quantlit/expGandD.html

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0take a look and let me know if you have any questions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please let me know once you have briefly looked at the page...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then we can choose between those two equations and use the information you were given to solve the problem. it's not hard at all and you'll see.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok looked at it its sort of complicated can u go step by step

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the two functions are n(t) = n0*e^(kt)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, the exponent of e is that decides whether we have a growth or decay. If it is negative, then the graph starts high and comes down (decay) and the opposite is true if the exponent is positive.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0by the way, what grade you in? so I know what you sort of should know...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hehe, no worries. I just graduated a year ago myself. any how...i hope you're not rushed cuz I'm definitely not moving very quickly...let me know if you're rushed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so let's go over the equation in more detail...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the site gives good description for each variable involved.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol cool. well i got this question and another, these questions r complicated and due at 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0today sorry don't rush its okay.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm gonna ask you to rephrase them for me here: what does N(t) mean? and the N_0(t) (that is the subscript zero), the k? the t?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0once you have those, we can replace the variables with the information in your problem. I just wanna make sure you're with me...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nt is quatity in time, t is time,n0 is initial quantity, k is a constant and eX is exponential function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, it's N(t) and it is a function of time. It's very important that you understand that. and eX is e^(x) which is the mathematically constant e (approx 2.71..) to power x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what the formulate says is N(t) [the amount as a function of time] = n0*e^(k*t) [the original amount times the exponential function to power k times t]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k here is the rate of change. the larger is it, the faster the amount grows or, in this case, decays.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the question asking of us? what do you think will be the variable that will be the final answer?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0stay with me franco, we're almost there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not sure ive never done a prblem like this one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the question asks you how old is the rock when only 12.5% of it is left..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we're looking for time, which will be contained in our variable t.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a clue that is give to you without knowing it is the term "halflife"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how long is the halflife of this rock?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so that means that in a 1000 years, we'll only have 50% of the original material left. that IS what half life means.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so what do i do??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so let's set up the equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually, I'll let you do it: in 1000 years, we will only have 50% of the original material. can you set that up?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'll prompt with good questions: first: what is t?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nooo :( i don't kno how this is complicateddd!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay: time in this case is 1000 years. second: notice how you can divide both sides by n0 (the original material)  once we do this, what do we have on the left side of the equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is only solving the first component of the problem.any way, when we divide both sides by n0, we get the equation: N(t)/n0 = e^(kt)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in a thousand years, we know we will only have half left, so what will N(t)/n0 equal? think about this one cause it solves 50% of the problem!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Franco? are you still with me?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i didvide both sides by that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0n0 is the original amount. was that given in the probem? 1000 is the number of years...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0before i loose you here, we divide the variable n0 by both sides so that we can remove it from the right side and put it on the left. you understand what this results in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I miss half life problems, what is that calculus 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no what is the asnwer/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now the equation looks like this: \[n(t) \div n _{0} = e ^{k*t}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, there is no calculus involved. this is still precalc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you understand how we get that equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you know so much about math thierryu

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I studied engineering in college, i just graduated last year

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you here to help us Amanda?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Franco, do you see how we got that last equation I put in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, if it has been a 1000 years and we only have 50% of the original materia.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now if t=1000, we're gonna look at n0/N(1000). All we know about this fraction is that N(t) at t=1000 is 50% of n0, the original amount.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0therefore, N(t)/n0 when t=1000 years should give us 1/2. do you agree?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I meant to say N(t)/n0 all along. sorry for the confusion.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bradley, you found me here!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0should just got to wolfram mathmatica, and type in halflife

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Franco, are you with me? do you see why N(t)/n0 at the halftime (t=1000) will give us 1/2 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so wats the answer??? yea i see

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did but i will let you help him first

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright. so now we have 1/2 = e^(k*t)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and t= 1000 so we really have .5=e^(1000k)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0franco plug in your time and your values for n and No

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we solve for k by first takiing the natural log of both sides: ln(.5) = ln(e^(1000k))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we do this because we know that the ln(e^(x)) equals x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so ln(e^(1000k)) reduces to simply 1000k!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you can use your calculator to solve for k from teh equation: \[\ln(.5) = 1000*k\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0once you find k, we go back to the board, but now we have everything we need expect time which is exactly what we're looking for!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why? because we only have 12.5% left. so the ratio of N(t)/n0 = .125

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you're not sure about k?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there you'd just use a calculator; get the natural log of .5 and then divide it by 1000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know this was elongated by if you go over this chat and solve the same or a similar problem, you'll get a hang of it. don't be discouraged.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0great. i didn't check but it looks about right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so on to solving the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i really havve to go can u just solve for me?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we only have 12.5% left; which means the ratio of N(t)/n0 = 12.5% = .125.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so since we now have k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we end up with \[.125 = e^(.0003t)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0remember how we solved for k? we do the same for t now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln(.125) = ln(e^(.0003*t)) = .0003*t

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so \[\ln(.125) = .0003*t\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you solve for t? and that's gonna be the number of years that has passed for us to only have 12.5% of the rock left.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0save this chat and go over it when you get a chance and you should be able to do this types of problems on your own.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I, for one, would love to know what it is...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0divde ln(.125) by .0003

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, by negative .0003

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so how long will have passed franco?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sounds about right. definitely way more than a thousand because it is way past the 50% mark. so that makes sense.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bradley, did you solve yours?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope, i am still lost

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would you care to chat on the other thread

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since you're not rushed and i just got home, give me about half an hour to grab a snack and start dinner and I'll be back. I'm putting on a timer for 30 minutes sharp.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I may also have to review a little calculus ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha okay i will be around

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0could you post the entire problem for me here again please?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0using lagrange multipliers find the maxium and minimum values of the function f(x,y,z)=xy^2z^3 subject to the contraints x^2+y^2+2z^2=25
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.