The answer is 4 hope this helps kindergartner
Answer:
A.
Step-by-step explanation:
A slope is 2
B slope is -1/2
hope this helps :3
if it did pls mark brainliest
Rachel would take 30 minutes to run in 4miles
Answer:
Step-by-step explanation:
carnival A : $ 6 dollar admission and $ 1.50 per ride
A(x) = 1.5x + 6
carnival B : $ 2.50 admission and $ 2 per ride
B(x) = 2x + 2.5
how many rides can Marnie go on so that the total cost of attending both is the same...
1.5x + 6 = 2x + 2.5
6 - 2.5 = 2x - 1.5x
3.5 = 0.5x
3.5 / 0.5 = x
7 = x <=== she would have to go on 7 rides for them to both cost the same
Answer:
the maximum concentration of the antibiotic during the first 12 hours is 1.185
at t= 2 hours.
Step-by-step explanation:
We are given the following information:
After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in ![\mu g/mL](https://tex.z-dn.net/?f=%5Cmu%20g%2FmL)
![C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})](https://tex.z-dn.net/?f=C%28t%29%20%3D%208%28e%5E%7B%28-0.4t%29%7D-e%5E%7B%28-0.6t%29%7D%29)
Thus, we are given the time interval [0,12] for t.
- We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
- The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.
First, we differentiate C(t) with respect to t, to get,
![\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28C%28t%29%29%7D%7Bdt%7D%20%3D%208%28-0.4e%5E%7B%28-0.4t%29%7D%2B%200.6e%5E%7B%28-0.6t%29%7D%29)
Equating the first derivative to zero, we get,
![\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28C%28t%29%29%7D%7Bdt%7D%20%3D%200%5C%5C%5C%5C8%28-0.4e%5E%7B%28-0.4t%29%7D%2B%200.6e%5E%7B%28-0.6t%29%7D%29%20%3D%200)
Solving, we get,
![8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2](https://tex.z-dn.net/?f=8%28-0.4e%5E%7B%28-0.4t%29%7D%2B%200.6e%5E%7B%28-0.6t%29%7D%29%20%3D%200%5C%5C%5Cdisplaystyle%5Cfrac%7Be%5E%7B-0.4%7D%7D%7Be%5E%7B-0.6%7D%7D%20%3D%20%5Cfrac%7B0.6%7D%7B0.4%7D%5C%5C%5C%5Ce%5E%7B0.2t%7D%20%3D%201.5%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7Bln%281.5%29%7D%7B0.2%7D%5C%5C%5C%5Ct%20%5Capprox%202)
At t = 0
![C(0) = 8(e^{(0)}-e^{(0)}) = 0](https://tex.z-dn.net/?f=C%280%29%20%3D%208%28e%5E%7B%280%29%7D-e%5E%7B%280%29%7D%29%20%3D%200)
At t = 2
![C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185](https://tex.z-dn.net/?f=C%282%29%20%3D%208%28e%5E%7B%28-0.8%29%7D-e%5E%7B%28-1.2%29%7D%29%20%3D%201.185)
At t = 12
![C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059](https://tex.z-dn.net/?f=C%2812%29%20%3D%208%28e%5E%7B%28-4.8%29%7D-e%5E%7B%28-7.2%29%7D%29%20%3D%200.059)
Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185
at t= 2 hours.