One hundred grams of radium are stored in a container. the amount rr (in grams) of radium present after tt years can be modeled by r=100e−0.00043tr=100e−0.00043t. after how many years will only 5 grams of radium be present? round your answer to the nearest whole year.

Ending Amount = Beginning Amount * e^k*t
5 grams = 100 grams * e^-.00043 * years

We have to solve the equation for "years"
(You can find the formula here: http://www.1728.org/halflif2.htm
or you can just read the next line

time = natural log (bgng amt / endg amt) / k
time = natural log (100 / 5) / .00043
time = natural log (20) / .00043
time = 2.9957322736 / .00043
time =  6,967 years

We should double-check this.
First, we need the half-life
k = ln(.5) / half-life
half-life = -.693147 / -.00043
half-life = 1,612 years
Now let's see how many half-lives that is:
6,967 / 1,612 years = 4.2 half-lives
So basically, after 4 half-lives the mass should go from

100
50  1 half life
25  2 half lives
12.5 3 half lives
 6.25 4 half lives

6.25 is very close to 5 grams so we can assume we have calculated correctly.

The equation is y=17(7)+600 and it equals 717

Xis equal to three