Answer:
2
Step-by-step explanation:
cuz
<u>Answer:</u>
The coordinates of endpoint V is (7,-27)
<u>Solution:</u>
Given that the midpoint of line segment UV is (5,-11) And U is (3,5).
To find the coordinates of V.
The formula for mid-point of a line segment is as follows,
Midpoint of UV is
, ![\frac{y_{1}+y_{2}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7By_%7B1%7D%2By_%7B2%7D%7D%7B2%7D)
As per the formula,
=5,
=-11
Here ![x_{1}=3; y_{1}=5](https://tex.z-dn.net/?f=x_%7B1%7D%3D3%3B%20y_%7B1%7D%3D5)
Substituting the value of
we get,
=5
![3+x_{2}=5\times2](https://tex.z-dn.net/?f=3%2Bx_%7B2%7D%3D5%5Ctimes2)
![x_{2}=10-3](https://tex.z-dn.net/?f=x_%7B2%7D%3D10-3)
![x_{2}=7](https://tex.z-dn.net/?f=x_%7B2%7D%3D7)
Substituting the value of
we get,
=-11
![5+y_{2}=-11\times2](https://tex.z-dn.net/?f=5%2By_%7B2%7D%3D-11%5Ctimes2)
![y_{2}=-22-5](https://tex.z-dn.net/?f=y_%7B2%7D%3D-22-5)
![y_{2}=-27](https://tex.z-dn.net/?f=y_%7B2%7D%3D-27)
So, the coordinates of V is (7,-27)
Answer:
6 and 11
Step-by-step explanation:
n1 = n + 5
ns^2 = 3 + 3nb
6^2 = 3 + 3(11)
36 = 3 + 33
11 = 6 + 5
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>