Answer:
-21p - 7c
Step-by-step explanation:
I think x=1
First you have to get rid of an variable on one side so you so 2x-3x than you are left with 4=x+3 than you want to get the variable alone so you subtract 3 from 4 than you are left with 1=x
21 and 2/3 all divided by 1 and 2/3 = answer
answer = 13
Answer:
x=0.5
Step-by-step explanation:
Answer:
After about 9 minutes.
Step-by-step explanation:
We can write an exponential function to model the decay of Herodium-100.
We are given that it decreases by half every minute.
The standard exponential function is given by:
![f(m)=A(r)^m](https://tex.z-dn.net/?f=f%28m%29%3DA%28r%29%5Em)
Where A is the initial amount, r is the rate, and m is the rate (in this case, in minutes).
Mathman initially has 2000 mL of Herodium-100. Therefore, A = 2000.
And since it decreases by half every minute, r = 1/2. Thus:
![\displaystyle f(m)=2000\Big(\frac{1}{2}\Big)^m](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28m%29%3D2000%5CBig%28%5Cfrac%7B1%7D%7B2%7D%5CBig%29%5Em)
Mathman needs to rest when the Herodium-100 levels drop to 4 mL. Therefore, we can substitute 4 for f(m) and solve for m:
![\displaystyle 4=2000\Big(\frac{1}{2}\Big)^m](https://tex.z-dn.net/?f=%5Cdisplaystyle%204%3D2000%5CBig%28%5Cfrac%7B1%7D%7B2%7D%5CBig%29%5Em)
Solve for m. Divide both sides by 2000:
![\displaystyle \frac{4}{2000}=\frac{1}{500}=\Big(\frac{1}{2}\Big)^m](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B4%7D%7B2000%7D%3D%5Cfrac%7B1%7D%7B500%7D%3D%5CBig%28%5Cfrac%7B1%7D%7B2%7D%5CBig%29%5Em)
We can take the natural log of both sides:
![\displaystyle \ln\Big(\frac{1}{500}\Big)=\ln\Big(\frac{1}{2}^m\Big)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cln%5CBig%28%5Cfrac%7B1%7D%7B500%7D%5CBig%29%3D%5Cln%5CBig%28%5Cfrac%7B1%7D%7B2%7D%5Em%5CBig%29)
By logarithm properties:
![\displaystyle \ln\Big(\frac{1}{500}\Big)=m\ln\Big(\frac{1}{2}\Big)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cln%5CBig%28%5Cfrac%7B1%7D%7B500%7D%5CBig%29%3Dm%5Cln%5CBig%28%5Cfrac%7B1%7D%7B2%7D%5CBig%29)
Therefore:
![\displaystyle m=\frac{\ln(1/500)}{\ln(1/2)}\approx 8.9657\approx 9\text{ minutes}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20m%3D%5Cfrac%7B%5Cln%281%2F500%29%7D%7B%5Cln%281%2F2%29%7D%5Capprox%208.9657%5Capprox%209%5Ctext%7B%20minutes%7D)
Mathman will have to rest and replenish after 9 minutes.