Dichloro- is a peticide that was used in the middle decades of the twenith century to control maleria. After 1945, It was widely used on crops in the US, AND AS MUCH AS one ton might be sprayed on a single field. However, after the toxic effects of Dichloro on the environment began to appear, the chemical was banned in 1972.
A common estimate for the half-life of Dichloro in the soil is 15 years. Write a decay law for Dichloro in the soil?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- toxicsugar22

- schrodinger

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Proportional growth and decay are governed by the differential equation,\[\frac{dN}{dt}=k(N-P)\]where k and P are constants. It leads to an expression for N of the form,\[N(t)=N(0)e^{kt}\]where P is zero here. You'll see that, at time t=0, the amount of substance hanging around is N(0). If you're given initial data, you can substitute it in here.
Also, you're told that 15 years is the half life. This means, if you start with N(0) amount of dichloro at time t=0, then at time t=15, you'll have N(15)=N(0)/2. But that means, looking at the formula,\[\frac{N(0)}{2}=N(0)e^{15k}\rightarrow \frac{1}{2}=e^{15k}\rightarrow \log \frac{1}{2} =15k\]that is,\[k=-\frac{\log 2}{15}\]So your equation is\[N(t)=N(0)e^{-\frac{\log 2}{15}t}\]

- anonymous

I use log for the natural logarithm, ln.

- anonymous

You can also note, using exponent laws and definition of the log, that\[e^{-\frac{\log 2}{15}t}=\left( e^{\log 2} \right)^{-\frac{t}{15}}=2^{\frac{-t}{15}}\]so that\[N(t)=N(0)2^{-t/15}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- toxicsugar22

OK nOW AS A PART OF THIS QUESTION In 1970, many soils samples in the US contained about 0.5 mg of Dichloro per kg of soil. The Mational safe level for Dichloro in the soil is 0.008 mg/kg. when will Dichloro content in the soil be reduced to a safe level?

- toxicsugar22

hurry up

- anonymous

Take a basis of 1kg of soil. Then N(0)=0.5mg. You need t such that N(t)=0.008mg. \[N(t)=N(0)2^{-t/15} \rightarrow \log N(t)=\log [N(0)2^{-t/15}]=\log N(0)+\log2^{-t/15}\]\[=\log N(0)-\frac{t}{15}\log 2\]so\[\log N(t)=\log N(0)-\frac{t}{15}\log 2 \rightarrow t=\frac{15(\log N(0)-\log N(t))}{\log 2}=\frac{15(\log 0.5-\log 0.008)}{\log 2}\]\[\approx89.5 yrs\]

- anonymous

and that's f***ing rude.

- toxicsugar22

srry

- toxicsugar22

do u forgive

- anonymous

ok

- toxicsugar22

the number of websites is growing by 15% each month. In June 2005, there were 4000 websites.
Write a formula for the number of websites as a function of months since june 2005.

- anonymous

\[N(t)=4000(1.15)^t\]

- toxicsugar22

Now with that question how long will it take fr the number of sites to reach 8000? Give the answer in exact value and decimal form

Looking for something else?

Not the answer you are looking for? Search for more explanations.