Length (L): w + 3 ⇒2(w + 3)
width (w): w ⇒ w - 1
Area (A) = L x w
A = (w + 3)(w)
A = w² + 3w
*******************************************
A + 176 = 2(w + 3)(w - 1)
(w² + 3w) + 176 = 2(w + 3)(w - 1)
w² + 3w + 176 = 2w² + 4w - 6
3w + 176 = w² + 4w - 6
176 = w² + w - 6
0 = w² + w - 182
0 = (w - 13) (w + 14)
0 = w - 13 0 = w + 14
w = 13 w = -14
Since width cannot be negative, disregard -14
w = 13
Length (L): w + 3 = (13) + 3 = 16
Answer: width = 13 in, length = 16 in
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.
Answer:
![\left \{ {{y=2} \atop {x=2}} \right. > \lim_{n \to \infty} a_n](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20%3E%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n)
Step-by-step explanation:
y = 2
x = 2
lim = an
infinity = x-0-0
Answer:
The answer is 7000
Step-by-step explanation:
trapezoid area is 9600 - 2600 (square area) = 7000
Use the substitution method
-x+4x when x=-2
-(-2)+4(-2) Positive number * ( multiplying)Negative number=Negative number
2-8
=-6
Answer is -6