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Travka [436]
4 years ago
13

X +y = 0 4x + 3y = 10 solve the systems by the method of substitution

Mathematics
1 answer:
Anna [14]4 years ago
8 0

Answer:Solve for the first variable in one of the equations, then substitute the result into the other equation.

X=4x/3−10/3

y=−4x/3+10/3

Step-by-step explanation:

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3 years ago
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

                                                = 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
Kristin is 20 km behind a truck that is driving 90 km/h how long will it take her to catch up to the truck? how far will she go
malfutka [58]
Time taken for Kristin to catch up with the truck will be found using the formula
time=(distance)/(speed)
distance=20 km
speed=90 km/h
thus
time=20/90=2/9 hours

Answer: 2/9 hours
5 0
3 years ago
Will give brainliest ! 15 points please help <3
andrey2020 [161]

Answer:

1. C

2. G, H

3. A, H

4. x=-4.5

5. B, D, F, G

Step-by-step explanation:

pls Mark Brainliest

8 0
3 years ago
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