Answer:
1. 24 2. 20 3. 1.75(I think) 4. 11
Step-by-step explanation:
3 might be incorrect but The others seem more correct than ever :)
Answer:
about 0.177 mg/mL
Step-by-step explanation:
The maximum is found where the derivative of C(t) is zero.
dC/dt = 1.35e^(-2.802t) -(1.35t)2.802e^(-2.802t) = 0
Solving for t gives ...
t = 1/2.802
So, the maximum C(t) is ...
C(1/2.802) = 1.35/2.802e^(-1) ≈ 0.177 . . . . . mg/mL
The maximum average BAC during the first 6 hours is about 0.177 mg/mL.
_____
The maximum occurs about 21 minutes 25 seconds after consumption.
Answer:
3
Step-by-step explanation:
note that zeros and x-intercepts are the same with different names
There are 2 zeros from the quadratic factor and 1 from the linear factor.
To find them equate the function to zero, that is
(x² - x - 2)(3x - 2) = 0
(x - 2)(x + 1)(3x-2) = 0 ← factoring the quadratic
equate each factor to zero and solve for x
x - 2 = 0 ⇒ x = 2
x + 1 = 0 ⇒ x = - 1
3x - 2 = 0 ⇒ x = 
Answer:
198
Step-by-step explanation:
2(9)(11)
Since you know that r = 9 and t = 11, you can substitute the variables to numbers. Then you just multiply to get 198. Hope this helps :)
Answer:
i think ur answer is a im not 100 percent sure let me know if im wrong
Step-by-step explanation: