Help please !!!!!!!!!!!
2 answers:
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So... the radiator has 15 liters of 70% antifreeze.. but needs an 80% antifreeze
well, so, you need to drain some and put some with higher percentage, seems to be, you will end up at the same 15 liters, possible the radiator's capacity, of 80% antifreeze
so, the same amount going out, of 70% is the same amount going in, of 100% antifreeze
now.. let's use the decimal format for the percents, or 70% is 70/100 or 0.7 and so on
so.. let's subtract, from the current solution, 0.7x and add 1x or x, our antifreeze concentration amount, should be 12 though
10.5 - 0.7x + x = 12
solve for "x"
well, so, you need to drain some and put some with higher percentage, seems to be, you will end up at the same 15 liters, possible the radiator's capacity, of 80% antifreeze
so, the same amount going out, of 70% is the same amount going in, of 100% antifreeze
now.. let's use the decimal format for the percents, or 70% is 70/100 or 0.7 and so on
so.. let's subtract, from the current solution, 0.7x and add 1x or x, our antifreeze concentration amount, should be 12 though
10.5 - 0.7x + x = 12
solve for "x"
Answer:
A=152
K= -Ln(0.5)/14
Step-by-step explanation:
You can obtain two equations with the given information:
T(14 minutes) = 114◦C
T(28 minutes)=152◦C
Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:
Applying the following property of exponentials numbers in (II):
Therefore can be written as
Replacing (I) in the previous equation:
Solving for k:
Subtracting 190 both sides, dividing by -76:
Applying the base e logarithm both sides:
Ln(0.5)= -14k
Dividing by -14:
k= -Ln(0.5)/14
Replacing k in (I) and solving for A:
Dividing by 0.5
A=152