Don’t touch my truck!!!…….
<span>So we need to find a number that is between 8/11 and 0.85. This will be easy if we write all numbers in decimal form. 8/11=0.7272727273 and 0.85 is the interval in which we are looking for. The numbers are: A 23/33=0.69696969, B 39/66=0.590909090 C 157/198= 0.792929292 D 88/99=0.888888888. The only number in the interval is C 157/198=0.792929292.</span>
Answer:
so you you multiply the z value to the x value witch you should already know sothen you have a fraction so my answer would be the second.
Step-by-step explanation:
up there
Answer:
x(15) = 21 lb
Step-by-step explanation:
Rate of change in volume of salt water solution = rate of volume incoming - rate of volume outgoing
dV/dt = 4 - 2 =2gal/min
So, the equation for volume at cetain time t at given conditions and values becomes,
V(t) = 2t + V
V(t) = 2t + 20 gal-------------------euqation (1)
Rate of change in amount of salt = rate of salt in - rate of salt out
dx/dt = {0.5*4} - {[x(t)/V(t)]*2}
dx/dt = 2-2[(x(t))/(2t+20)]
dx/dt = 2-[(x(t))/(v(t))] lb/min
Now, with integrating factor, we get
exp[∫(1/(1+10))dt)] = t+10
the equation becomes
(t + 10)*x' + x = 2*(t+10)
((t+10)*x') = 2*(t+10)
(t+10)*x = t² + 20t + C
As x(0) = 0,
x(t) = (t²+20t)/(t+10)
x(15) = (15²+20*15)/(15+10)
x(15) = 21 lb
