Answer:the number of gallons of water left is 3.5
Step-by-step explanation:
The initial amount of water in the container is 4 3/8 gallons. Converting 4 3/8 gallons to improper fraction, it becomes 35/8 gallons of water.
A hole in the container causes 7/8 gallon to leak out. The number of gallons of water left in the container would be
Initial amount - amount that leaked. It becomes
35/8 - 7/8 = 28/8 = 3 4/8 = 3.5 gallons of water
<h2>
Explanation:</h2>
The yard, whose symbol is yd, is standardized as 0.9144 meters since 1959. The unit system that use it is both the imperial and the US units. On the other hand, the foot, whose symbol is ft, is standardized as 0.3048 meter. It is also used by the imperial and US units. The relationship between these two units is:
![1yd=3ft \\ \\ 1ft=1/3yd](https://tex.z-dn.net/?f=1yd%3D3ft%20%5C%5C%20%5C%5C%201ft%3D1%2F3yd)
Well u have to have 2 cases a day witch is 300 150+150 then times it by 7 days its 2100 bars so 2 cases would be 1050 ? I think I'm sure
Answer:
A) f(x) = |x|
B) f(x) = -|x+3| + 3
Step-by-step explanation:
Absolute value functions
The general form of an absolute value function is f(x) = a|x-b| +c
a indicates the 'amplitude' or multiplier
b indicates horizontal shift
c indicates vertical shift
The turning point of f(x) = |x| is (0, 0)
a = y/x when b = 0 and c = 0
A) f(x) = |x|
Since the turning point is still (0, 0) there is no horizontal nor vertical shift, meaning b = 0 and c = 0
We have the point (2, 2), thus a = 2/2 = 1
f(x) = a|x-b| + c
f(x) = |x|
B) f(x) = -|x+3| + 3
The turning point is at (-3, 3) b = -3 and c = +3
We have the point (3, -3), thus a = y/x = -3/3 = -1
f(x) = a|x-b| + c
f(x) = (-1) (|x-(-3)|) + (+3)
f(x) = -|x+3| + 3
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.