The answer
<span>the third rope to counterbalance Sam and Charlie is F
and vectF +vectF1 +vectF2 =vect0
let's consider axis
y'y </span>vectF = -F
vectF 1= F1cos60
vectF 2= F2cos45
-F = -F1cos60-F2co45
so F= F1cos60+F2co45= 350x0.5+400x0.7=457.84 pounds
Answer:
sorry i was late but the answer is 100% A
Step-by-step explanation:
i just took the test made 100
Answer:
4 Units
Step-by-step explanation:
Just count the squares from X to Y or You can do just mental math
Answer:
42
Step-by-step explanation:
90 (the angle that was got split)
if the right side got 48 degree, simply find jow much is left on tge left side.
90 - 48 = 42°
Answer:
a. dQ/dt = -kQ
b. 
c. k = 0.178
d. Q = 1.063 mg
Step-by-step explanation:
a) Write a differential equation for the quantity Q of hydrocodone bitartrate in the body at time t, in hours, since the drug was fully absorbed.
Let Q be the quantity of drug left in the body.
Since the rate of decrease of the quantity of drug -dQ/dt is directly proportional to the quantity of drug left, Q then
-dQ/dt ∝ Q
-dQ/dt = kQ
dQ/dt = -kQ
This is the required differential equation.
b) Solve your differential equation, assuming that at the patient has just absorbed the full 9 mg dose of the drug.
with t = 0, Q(0) = 9 mg
dQ/dt = -kQ
separating the variables, we have
dQ/Q = -kdt
Integrating we have
∫dQ/Q = ∫-kdt
㏑Q = -kt + c

when t = 0, Q = 9

So, 
c) Use the half-life to find the constant of proportionality k.
At half-life, Q = 9/2 = 4.5 mg and t = 3.9 hours
So,

taking natural logarithm of both sides, we have
d) How much of the 9 mg dose is still in the body after 12 hours?
Since k = 0.178,

when t = 12 hours,
