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Answer:
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Step-by-step explanation:
Given the data in the question;
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?
dA/dt = rate in - rate out
first we determine the rate in and rate out;
rate in = 3pound/gallon × 5gallons/min = 15 pound/min
rate out = A pounds/1000gallons × 5gallons/min = 5Ag/1000pounds/min
= 0.005A pounds/min
so we substitute
dA/dt = rate in - rate out
dA/dt = 15 - 0.005A
Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
The equation, when rewritten would be written as:
q = ( c - 54293 ) / 3.59
<h3>How to rewrite the equation in order to make q the subject</h3>
C = 3.59q + 54,293
We have the equation above, what we would have to do now would be to write the equation in such a way that q would be the subject of the formula.
This would be
3.59q = C - 54,293
Next we have to divide through by the value of q
q = 
Read more on subject of formula here: brainly.com/question/21140562
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Answer:
Every 18 miles he uses 3 gallons
divide this:
18 ÷ 3 = 6 miles per gallon
30 miles = 30 ÷ 6 = 5 gallons
7 gallons = 7 x 6 = 42 miles
48 miles = 48 ÷ 6 = 8 gallons
10 gallons = 10 x 6 = 60 miles
Hope this helps :D
Answer:
3/5
Step-by-step explanation:
1/2 ÷ 5/6 = 1/2 × 6/5 = 6/10 = 3/5
