-3/4 also know as negative 3/4
Answer:
3.12 kg.
Step-by-step explanation:
Mass after t years:
The mass of the elements after t years is given by the following equation:
![M(t) = M(0)(1-r)^t](https://tex.z-dn.net/?f=M%28t%29%20%3D%20M%280%29%281-r%29%5Et)
In which M(0) is the initial mass and r is the decay rate, as a decimal.
The half life of Cs-137 is 30.2 years.
This means that:
![M(30.2) = 0.5M(0)](https://tex.z-dn.net/?f=M%2830.2%29%20%3D%200.5M%280%29)
We use this to find r.
![M(t) = M(0)(1-r)^t](https://tex.z-dn.net/?f=M%28t%29%20%3D%20M%280%29%281-r%29%5Et)
![0.5 = M(0)(1-r)^{30.2}](https://tex.z-dn.net/?f=0.5%20%3D%20M%280%29%281-r%29%5E%7B30.2%7D)
![(1-r)^{30.2} = 0.5](https://tex.z-dn.net/?f=%281-r%29%5E%7B30.2%7D%20%3D%200.5)
![\sqrt[30.2]{(1-r)^{30.2}} = \sqrt[30.2]{0.5}](https://tex.z-dn.net/?f=%5Csqrt%5B30.2%5D%7B%281-r%29%5E%7B30.2%7D%7D%20%3D%20%5Csqrt%5B30.2%5D%7B0.5%7D)
![1 - r = 0.5^{\frac{1}{30.2}}](https://tex.z-dn.net/?f=1%20-%20r%20%3D%200.5%5E%7B%5Cfrac%7B1%7D%7B30.2%7D%7D)
![1 - r = 0.9773](https://tex.z-dn.net/?f=1%20-%20r%20%3D%200.9773)
So
![M(t) = M(0)(0.9773)^{t}](https://tex.z-dn.net/?f=M%28t%29%20%3D%20M%280%29%280.9773%29%5E%7Bt%7D)
If the initial mass of the sample is 100kg, how much will remain after 151 years?
This is M(151), with M(0) = 100. So
![M(t) = 100(0.9773)^{t}](https://tex.z-dn.net/?f=M%28t%29%20%3D%20100%280.9773%29%5E%7Bt%7D)
![M(151) = 100(0.9773)^{151} = 3.12](https://tex.z-dn.net/?f=M%28151%29%20%3D%20100%280.9773%29%5E%7B151%7D%20%3D%203.12)
The answer is 3.12 kg.
To solve this problem you must apply the proccedure shown below:
1. You have the following quadratic equation, which is given in the problem above:
<span>2y^2+6y-8=0
2. So, to solve it, you can use the quadratic formula, which is:
y=(-b</span><span>±</span>√(b²-4ac))/2a
<span>
a=2
b=6
c=-8
3. When you susbtitute these values into the formula, you obtain:
y1=-4
y2=1</span>