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    two liquids x and y are flowing into a trough at the constant rate of 10 and 20 litres per minute respectively. the liquid in the trough is stirred continuously and pumped out at the rate of 30 litres per minute. initially the trough contains 200 litres of x and 100 litres of y. After t minutes the tank contains x litres of x. by considering the change in x in a small interval of 'lamda' time, show that

    dx/dt = 10 - (x/10)
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    (Original post by samnb175)
    two liquids x and y are flowing into a trough at the constant rate of 10 and 20 litres per minute respectively. the liquid in the trough is stirred continuously and pumped out at the rate of 30 litres per minute. initially the trough contains 200 litres of x and 100 litres of y. After t minutes the tank contains x litres of x. by considering the change in x in a small interval of 'lamda' time, show that

    dx/dt = 10 - (x/10)
    What have you tried?
 
 
 
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