Answer:

Step-by-step explanation:
Volume of water in the tank = 1000 L
Let y(t) denote the amount of salt in the tank at any time t.
Initially, the tank contains 60 kg of salt, therefore:
y(0)=60 kg
<u />
<u>Rate In</u>
A solution of concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min.
=(concentration of salt in inflow)(input rate of solution)

<u>Rate Out</u>
The solution is mixed and drains from the tank at the same rate.
Concentration, 
=(concentration of salt in outflow)(output rate of solution)

Therefore, the differential equation for the amount of Salt in the Tank at any time t:

Answer:
3.875
Step-by-step explanation:
8(m+5)=16
8m+40=16
8m=24
m=3
Answer:
the first one is false
second one is True
Third one is false
fourth one is false
Step-by-step explanation:
I just looked at the chart please mark me brainlest answer
Answer:
Step-by-step explanation:
t5 = a + (5-1)d
t5 = a + 4d
40 = a + 4d
t3 = a + 2*d
t7 = a + 2d + 28 = a + 6d
Start with the seventh term which has 2 equal answers
a + 2d + 28 = a + 6d Subtract a from both sides
2d + 28 = 6d Subtract 2d from both sides
28 = 4d Divide by 4
28/4 = d
d = 7
==================
t5 = a + 4d = 40
a + 4*7 = 40
a = 40 - 28
a = 12
===================
First term is 12
10th term is
t10 = 12 + (10 - 1)*7
t10 = 12 + 9*7
t10 = 75
Interesting question. Thanks for posting