Question

really could use some help
The half-life of a certain radioactive material is 32 days. An initial amount of the material has a mass of 361 kg. Write an exponential function that models the decay of this material. Find how much radioactive material remains after 5 days. Round your answer to the nearest thousandth.

Answer 1

;0.199KG

Answer 2

;323.945KG

Answer 3

Lets look at what half life means:
A half-life of 32 days means that after 32 days a mass of 1 will reduce to 1/2.
\[e^{k\cdot 32}=\frac{ 1 }{ 2 }\] take the natural log of both sides.\[k \cdot 32=\ln (1/2)=-\ln 2\]
divide by 32 on both sides.\[k=\frac{- \ln 2 }{ 32 }\]

Answer 4

So this gives us our formula for the fraction of material after time "t" as:\[e^{-\frac{ \ln 2 }{ 32 } t}\]

Answer 5

;0.797kg

Answer 6

;0kg

Answer 7

that's abcd I really don't get it at all

Answer 8

abcd?

Answer 9

THATS WHAT I HAVE TO CHOOSE FROM

Answer 10

I don't understand it at all

Answer 11

So for exponential growth or decay the equations are like this.
\[(\text{initial stuff})\cdot e^{k \cdot t}=\text{(stuff after time t)}\]
We just have to figure out what "k" is.

"k" will be negative when it is decaying.
"k" will be positive when it is growing.

Answer 12

ok so what is the e for? an what about the g

Answer 13

e is just the natural exponent it is a constant for all of these equations
e is approximately 2.71828

Answer 14

For half life situations (exponential decay)
K is always as follows:\[k=-\frac{ \ln 2 }{ \text{half life} }\]

Answer 15

half-life of the substance : A = 391*2^(-5/32) using a calc find 2^(-5/32) A = 391 * .89735 A = 350.866 I keep getting that an I don't no why because its not on there

Answer 16

In our situation what is k equal to?

Answer 17

so how do I find the answer

Answer 18

I'm trying to walk you through it.

Answer 19

ok so after the For half life situations (exponential decay)
K is always as follows what is next

Answer 20

So what class is this for?

I guess maybe you aren't using real half lifes but approximate ones.

So your expressions should be like this:\[\text{initial stuff}\cdot (1/2)^{\frac{ 1 }{ \text{halflife} }\cdot t}=\text{stuff after time t}\]

Answer 21

yea I don't get how they wrote it but thanks for helping anyways an after that

Answer 22

Sorry for the confusion.