Equations with absolute value:

Where k is a positive number; if k is a negative number, the equation is impossible (absolute value is always positive).
How to solve:


Then:
1. |x+7|=12
x+7=12 V -x-7=12
x=5 V -x=19
x=5 V x=-19
{-19, 5}
2. |2x+4|=8
2x+4=8 V -2x-4=8
2x=4 V -2x=12
x=2 V x=-6
3. 3|3k|=27
3×3k=27 V 3×(-3k)=27
9k=27 V -9k=27
k=3 V k=-3
{-3, 3}
4. 5|b+8|=30
5×(b+8)=30 V 5×(-b-8)=30
5b+40=30 V -5b-40=30
5b=-10 V -5b=70
b=-2 V b=-14
{-14, -2}
5. |m+9|=5
m+9=5 V -m-9=5
m+9=5 V m+9=-5
Answer:
Step-by-step explanation:
This is a differential equation problem most easily solved with an exponential decay equation of the form
. We know that the initial amount of salt in the tank is 28 pounds, so
C = 28. Now we just need to find k.
The concentration of salt changes as the pure water flows in and the salt water flows out. So the change in concentration, where y is the concentration of salt in the tank, is
. Thus, the change in the concentration of salt is found in
inflow of salt - outflow of salt
Pure water, what is flowing into the tank, has no salt in it at all; and since we don't know how much salt is leaving (our unknown, basically), the outflow at 3 gal/min is 3 times the amount of salt leaving out of the 400 gallons of salt water at time t:

Therefore,
or just
and in terms of time,

Thus, our equation is
and filling in 16 for the number of minutes in t:
y = 24.834 pounds of salt
Answer:
BCF is the correct answer
Area shaded = Area big circle- Area of small circle;
200 pi= pi•(2x)^2 -pi•6^2;
200pi= pi•4x^2 -pi•36;
200pi=pi•4(x^2 -9) divide both sides by 4pi;
50=x^2 -9; So x=sqrt(59)~7.68cm