Recent content by x0hkatielee

it's the negative sign, right? I noticed that.

okay so it looks like what I'm getting for part b is k= (ln 3)/-2 ...? Is this right or am I completely off.

I think I finally got it right. so like A(0)e^-(5k)=(1/3)A(0)e^-(3k)... and the A(0)'s cancel out. So I get e^-(5k)=(1/3)e^-(3k) . and then once I simplify more I can find k

Okay so I think I understand ! If I set A(5)=(1/3)A(3) in the equations then the A(0) values cancel and I can solve for k=(ln 3)/2.. hope that's right!

the only thing that they give me towards figuring that out is that A(5)=(1/3)A(3)... ? Not sure if that's what you meant. I wouldn't be able to solve for them using the equation for part a without knowing K or A(0).

I had dA/dt=-kA and rearranged it and integrated both sides to get to the equation A(t)=A(0)e^(-kt). now I know I'm supposed to plug the values for A(5) and A(3) into part b to solve for k, but how do I get those values if I don't know A(0) or k.

I got the equation the way you said to, but the problem doesn't give me any values for A(0), A(3), or A(5). And this is really confusing me. This problem has been stumping me for a long time because it doesn't give me much to work with. Sorry if I'm bugging you!

Thanks for the reply! What is throwing me off is not having numbers or any values for A(3) or A(5).

Thanks for the reply! I'm a little rusty at calc because I haven't taken it in a while. I'm not sure exactly how to get the formula from the equation given.

I've just been really thrown off by what this problem is asking me. Given: The decay of a radioactive material may be modeled by assuming that the amount A(t) of material present (in grams) at time t (minutes) decays at a rate proportional to the amount present, that is dA/dt= -kA for some...