Answer:
-1
Step-by-step explanation:
7+(-1)= 6
6/3 = 2
There are no other solutions I can find to it. It says sum but technically you can add a negative number.
Don´t listen to me if you think that I am wrong. But I hope this helps!!
Answer:
Time(t) = 11.61 hours (Rounded to two decimal place)
Step-by-step explanation:
Given: The antibiotic clarithromycin is eliminated from the body according to the formula:
......[1]
where;
A - Amount remaining in the body(in milligram)
t - time in hours after the drug reaches peak concentration.
Given: Amount of drug in the body is reduced to 100 milligrams.
then,
Substitute the value of A = 100 milligrams in [1] we get;
![100= 500e^{-0.1386t}](https://tex.z-dn.net/?f=100%3D%20500e%5E%7B-0.1386t%7D)
Divide both sides by 500 we get;
![\frac{100}{500}=\frac{ 500e^{-0.1386t}}{500}](https://tex.z-dn.net/?f=%5Cfrac%7B100%7D%7B500%7D%3D%5Cfrac%7B%20500e%5E%7B-0.1386t%7D%7D%7B500%7D)
Simplify:
![\frac{1}{5} = e^{-0.1386t}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D%20%3D%20e%5E%7B-0.1386t%7D)
Taking logarithm both sides with base e, then we have;
![\log_e (\frac{1}{5})= \log_e (e^{-0.1386t})](https://tex.z-dn.net/?f=%5Clog_e%20%28%5Cfrac%7B1%7D%7B5%7D%29%3D%20%5Clog_e%20%28e%5E%7B-0.1386t%7D%29)
[ Using
]
or
![\log_e (0.2)=-0.1386t](https://tex.z-dn.net/?f=%5Clog_e%20%280.2%29%3D-0.1386t)
[using value of
]
then;
![t = \frac{-1.6094379124341}{-0.1386}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-1.6094379124341%7D%7B-0.1386%7D)
Simplify:
t ≈11.61 hours.
Therefore, the time 11.61 hours(Rounded two decimal place) will pass before the amount of drug in the body is reduced to 100 milligrams
<span>4(-2)^2+8(-2)+3(-2)+6=0</span>
Hi there!
The answer is 104/25.
Could you Brainliest me please?
To know this answer, we have to divide the amount each one earns between the number of hours they work. So: For Nina: 117.60 / 1414 = 0.083168317 For Ty: 148.50 / 1818 = 0.081683168 For Kim: 137.60 / 1616 = 0.085148515 Then Kim has the highest pay rate in dollars per hour with a total of 0.085148515 $ / h.