Answer:
Second choice:


Fifth choice:


Step-by-step explanation:
Let's look at choice 1.


I'm going to subtract 1 on both sides for the first equation giving me
. I will replace the
in the second equation with this substitution from equation 1.

Expand using the distributive property and the identity
:




So this not the desired result.
Let's look at choice 2.


Solve the first equation for
by dividing both sides by 2:
.
Let's plug this into equation 2:



This is the desired result.
Choice 3:


Solve the first equation for
by adding 3 on both sides:
.
Plug into second equation:

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:



Not the desired result.
Choice 4:


I'm going to solve the bottom equation for
since I don't want to deal with square roots.
Add 3 on both sides:

Divide both sides by 2:

Plug into equation 1:

This is not the desired result because the
variable will be squared now instead of the
variable.
Choice 5:


Solve the first equation for
by subtracting 1 on both sides:
.
Plug into equation 2:

Distribute and use the binomial square identity used earlier:



.
This is the desired result.
Part A= $27.32
Part B= $32.24
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
Answer:
615.4³
Step-by-step explanation: