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Yuki888 [10]
2 years ago
11

At the beginning of the day the stock market goes down 30 3/4 points and stays at this level for most of the day. at the end the

stock market goes up 120 1/2 points from the low at the beginning of the day. what is the total change i the stock market from the beginning of the day to the end of the day
Mathematics
1 answer:
tia_tia [17]2 years ago
8 0

The computation shows that the change in points is 89 3/4 points.

<h3>How to compute the value?</h3>

From the information, at the beginning of the day the stock market goes down 30 3/4 points and stays at this level for most of the day and at the end the stock market goes up 120 1/2 points.

Therefore, the change will be:

= 120 1/2 - 30 3/4

= 89 3/4

Learn more about computations on:

brainly.com/question/4658834

#SPJ4

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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
Lina20 [59]

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

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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

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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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3 years ago
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