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Softa [21]
2 years ago
15

Sue Mitchell weighed 8 lb. 13 oz. when she was born. She now weighs 93 lb. 5

Mathematics
1 answer:
ra1l [238]2 years ago
8 0

Answer:

Weight gain = 84 lb 5 oz

Average weight gain = 6 lb 5 oz

Step-by-step explanation:

lb means pounds

16 oz = 1 pound

Convert the weight to pounds

Weight when she was born = 8 lb. 13 oz

= 8 pounds + 0.8125 pounds

= 8.8125 pounds

Present weight = 93 lb. 5

oz

= 93 pounds + 0.3125 pounds

= 93.3125 pounds

How much has she gained since birth

Weight gain = Present weight - Weight when she was born

= 93.3125 pounds - 8.8125 pounds

= 84.5 pounds

= 84 lb 5 oz

Weight gain = 84 lb 5 oz

what has been her average weight for each year if she is 13 year old

Average weight gain = Weight gain / 13

= 84.5 pounds / 13

= 6.5 pounds

= 6 lb 5 oz

Average weight gain = 6 lb 5 oz

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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}.

4 0
3 years ago
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