Question

When radioactive substances​ decay, the amount remaining will form a geometric sequence when measured over constant intervals of time. The table shows the amount of a radioactive isotope initially and after 2 hours. What are the amounts left after 1​ hour, 3​ hours, and 4​ hours?
Hours elapsed 0 1 2 3 4
Grams of the isotope 1926 501
How much of the isotope is left after 1 hour?

Answer 1

did you get anywhere yet?

Answer 2

No, I have 3 chances to get it right and I have gotten it wrong twice already. I set up an equation and everything but I'm getting no where with it.

Answer 3

Recall, for geometric sequence, each next term is a constant multiple 'r' times the last term. So the general form is...

\[\large y=A*r^t\]

where A is the initial value, when t=0

Answer 4

they give you two points, (t,y)
(0 , 1926)
(2 , 501)

Answer 5

you can use those to figure out what the values of A and r will be...

\[y=A*r^t\]
\[\large 1926=A*r^0=A\]

With that, A=1926, now use the second point..

\[y=1926*r^t\]
\[\large 501 = 1926*r^2\]

Answer 6

then you would get 0.2601 right?

Answer 7

the square root of that, so r=0.51, and the function for grams left y in terms of hours passed t, is
for t=0,1,2,3....

\[\huge y(t)=1926*(0.51)^t\]

Answer 8

you get it all?

Answer 9

so then the answer for 1 gram would be 982.26?

Answer 10

yeah, 1 hour elapsed, grams left 982.26

Answer 11

thank you so so much

Answer 12

welcome

Answer 13

I have one more math question I need help with.

You are offered a job that pays ​$39,000 during the first​ year, with an annual increase of 10​% per year beginning in the second year. That​ is, beginning in year​ 2, your salary will be 1.1 times what it was in the previous year. What can you expect to earn in your fourth year on the​ job? Round your answer to the nearest dollar.

In your fourth year on the​ job, you can expect to earn ​$_

Answer 14

similar to the first question...

Answer 15

If you want to make the first year n=1, then you need to alter it a litttle bit, to make n=1 the initial value A

\[\large y=A*r^{t-1}\]

Answer 16

you have A=39000, and r=1.1

\[\huge y(t)=39000*(1.1)^{t-1}\]

Answer 17

you get that?

Answer 18

Yes i do. That makes so much more sense. So then to do year 4 i would put 4 in for t and then solve and get the answer for the earning of the fourth year? you don't do anything with the 10%?

Answer 19

the r=1.1 value adds 10% to each previous year

Answer 20

Ahhhh okay.

Answer 21

So for the 4th year you'd pay $57,098.9? correct?

Answer 22

no, remember we had to change the exponent to make the first term A=39000, the start value after t=1. so the exponent is t-1

Answer 23

so then the equation wouldn't be 39000(1.1)4-1 ?

Answer 24

because wouldn't the t be replaced by 4 since it would be the 4th year?

Answer 25

are you still there?

Answer 26

yeah sorry...

Answer 27

Yes, for the 4th year, you use t=4, which gives

\[\large y(4)=39000*(1.1)^{4-1}\]

the exponent becomes 3, you still used 1.1^4 in your calculation

should be
\[\large y(4)=39000*(1.1)^{4-1}=51909\]