The amount of substance left of a radioactive element of half life,
![t_{\frac{1}{2}}](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D)
after a time, t, is given by:
![N(t)=N_0\left( \frac{1}{2} \right)^ \frac{t}{t_{ \frac{1}{2} }}](https://tex.z-dn.net/?f=N%28t%29%3DN_0%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E%20%5Cfrac%7Bt%7D%7Bt_%7B%20%20%5Cfrac%7B1%7D%7B2%7D%20%7D%7D)
Given that <span>potassium-40 has a half life of approximately 1.25 billion
years.
The number of years it will take for 0.1% of potassium-40 to remain is obtained as follows:
![0.1=100\left( \frac{1}{2} \right)^ \frac{t}{1.25}} \\ \\ \Rightarrow\left( \frac{1}{2} \right)^ \frac{t}{1.25}}=0.001 \\ \\ \Rightarrow\frac{t}{1.25}\ln\left( \frac{1}{2} \right)=\ln(0.001) \\ \\ \Rightarrow \frac{t}{1.25}= \frac{\ln(0.001)}{\ln\left( \frac{1}{2} \right)} =9.966 \\ \\ t=9.966(1.25)=12.5](https://tex.z-dn.net/?f=0.1%3D100%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E%20%5Cfrac%7Bt%7D%7B1.25%7D%7D%20%5C%5C%20%20%5C%5C%20%5CRightarrow%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E%20%5Cfrac%7Bt%7D%7B1.25%7D%7D%3D0.001%20%5C%5C%20%20%5C%5C%20%5CRightarrow%5Cfrac%7Bt%7D%7B1.25%7D%5Cln%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%3D%5Cln%280.001%29%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20%5Cfrac%7Bt%7D%7B1.25%7D%3D%20%5Cfrac%7B%5Cln%280.001%29%7D%7B%5Cln%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%7D%20%3D9.966%20%5C%5C%20%20%5C%5C%20t%3D9.966%281.25%29%3D12.5)
Therefore, </span><span>the maximum age of a fossil that we could date using 40k is
12.5 billion years.</span>
Its not dos its factorial
Your answer is B. 1 and 3.
This is because adjacent angles are angles that could form one angle if the line inbetween them was removed, which means they have a common side. Angles 1 and 3 are the only angles out of the options that have a common side, so they must be the answer.
I hope this helps!
1) extract the info: You have two joggers. Jogger 1, lets call that J1 is running at 4 miles per hours. Jogger 2, lets call that J2 leaves 10 minutes later running at 7 miles per hour. We want to know how long it will take for J2 to overtake J1 and how far they ran at that point. 2) Now we need to set up the equations: We know that rate x time equals distance. We know that J1 is running at 4 mph, so we have the rate part for J1, we don't have the distance, or the time, so lets call time t. So your equation will look like D(istance)=4mph x t. J2, is running at 7 miles per hour, that jogger leaves 10 minutes later. Since hour rate is in mph, we have to convert 10 minutes to hours. There is 60 minutes in 1 hour so that will need to be 10/60, or 1/6 of an hour. J2's equation will be D(istance)= 7 MPH x (t-1/6hrs). We want to find out when they are at the same distance. So set the equations equal, 7t-7/6=4t; solving for t: 7t-4t=7/6, 3t=7/6, t=7/6 * 1/3, t=7/18, or t=0.3889 hours. Then you can just sub that into the first equation to find the distance. D=4*7/18, D=14/9 or 1.5556 miles.