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Serggg [28]
2 years ago
14

See your levels

Mathematics
1 answer:
Crazy boy [7]2 years ago
5 0

See attachment of the graph of the points on the coordinate plane.

<h3>What are linear equations?</h3>

Linear equations are equations that have constant average rates of change. Note that the constant average rates of change can also be regarded as the slope or the gradient

<h3>How to determine the graph of the points on the coordinate plane.?</h3>

A system of linear equations is a collection of at least two linear equations.

In this case, the points are given as

(7, 3) and (1, 5)

The point (7, 3) means that:

x = 7 and y = 3

i.e. the point is located 7 units to the right of the origin and 3 units up the origin

Similarly,

The point (1, 5) means that:

x = 1 and y = 5

i.e. the point is located 1 unit to the right of the origin and % units up the origin

Next, we plot the points on the coordinate plane.

See attachment of the graph of the points on the coordinate plane.

Read more about coordinate plane at:

brainly.com/question/17206319

#SPJ1

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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

                                                = 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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Answer:

$25.44

Step-by-step explanation:

35% of $39 is $13.65

$39-$13.65=$25.44

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