Question

Help with what I did wrong here

Answer 1

I started doing the corrections but I don't know whats next

Answer 2

sorry but i dont understand your work on the first one

Answer 3

but this is very easy

Answer 4

The first one is wrong. I just put a picture so you could see the problem. the second picture is the beginning of my corrections but I do not know where to go from there

Answer 5

ok. now you talk about what question ? i ve said question 3 /a

Answer 6

no question 4

Answer 7

?

Answer 8

in the 4th line where you elimined the exponent of e ?

Answer 9

@SmokeyBrown

Answer 10

Ok, so you want to find the half-life, right?
So, we'd need to find the value of t when W(t) = (W0)/2, right? In other words, the amount of time it takes for the amount of isotope to be cut in half from the starting point

Answer 11

Okay and the starting point would be 0.002?

Answer 12

To start out, we can write our equation as follows:
\[W _{0}/2 = W _{0} \times e ^{-0.002t}\]
We could divide by W0 on both sides, leaving us with
\[1/2 = e ^{0.002t}\]
Then, you'd just have to solve for t. I think you use logs for this?

Answer 13

=-346.57

Answer 14

would the units be meter

Answer 15

Nah, my dude, half-life is a time, so the units would be years

Answer 16

oh okay but am I right with the positive and btw Imma girl lol

Answer 17

with the negative I meant?

Answer 18

because it is a time frame

Answer 19

Oh, true

Answer 20

like would it still be -346.57 yrs

Answer 21

Haha, no that wouldn't really make sense. Uh, let me check that calculation

Answer 22

okay

Answer 23

So we'd start out by taking the natural log of both sides. That gives us
-0.693 = - 0.002t
If we divide both sides by -0.002, we get 346.5 = t
I see that I forgot a negative sign in my second equation. It should be -0.002t, not 0.002t. That's my bad

Answer 24

okay so it would be 345.5 = t which is also years?

Answer 25

I mean this would be the half life?

Answer 26

You there

Answer 27

Yup, that's right