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Tasya [4]
3 years ago
13

Write 0.05 as a fraction. A) 1 12 B) 1 15 Eliminate C) 1 18 D) 1 9

Mathematics
1 answer:
zheka24 [161]3 years ago
6 0

x=0.0\overline{5}\\\\x=0.0555...\qquad\text{multiply both sides by 10}\\\\10x=0.555...\qquad\text{multiply both sides by 10}\\\\100x=5.555...\\\\\text{Make the difference:}\\\\100x-10x=5.555...-0.555...\\\\90x=5\qquad\text{divide both sides by 90}\\\\x=\dfrac{5}{90}\\\\x=\dfrac{5:5}{90:5}\\\\x=\dfrac{1}{18}\\\\Answer:\ \boxed{0.0\overline{5}=\dfrac{1}{18}}\to\boxed{C)}

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Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
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Answer:

The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t

where, \text{R}_i_n = concentration of salt in the inflow \times input rate of brine solution

and \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

So, \text{R}_i_n = 4 lb/gal \times 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

                                                = 3 gal/min - 2 gal/min

                                                = 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

             = \frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min} = \frac{2A(t)}{500+t} \text{ lb/min }

Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;

= \frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }

or \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

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