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Sunny_sXe [5.5K]

2 years ago

6

At the animal shelter, there are 4 dogs for every 3 cats. Only dogs and cats are accepted (let in) at the shelter. If there are

24 dogs currently at the shelter, how many animals are at the shelter?

1 answer:

artcher [175]2 years ago

7 0

Answer:

There are 42 animals at the shelter

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How to solve x+3y=3 and 2x-3y=-12

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Get x on one side then you can use substitution
So, change x+3y=3 to x= -3y+3
Then plug in x to other equation to find y
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ank A contains 80 gallons of water in which 20 pounds of salt has been dissolved. Tank B contains30 gallons of water in which 5

adell [148]

Answer:

$\frac{dx}{dt}=2-\frac{3}{40}x+\frac{y}{15}$

$\frac{dy}{dt}=\frac{3}{40}x-\frac{y}{5}$

x(0)=20,y(0)=5

Step-by-step explanation:

We are given that

Tank A contains water=80 gallons

x(0)=20,y(0)=5

Tank B contains water=30 gallons

Rate=4 gallon/min

Concentration of salt is pumped into tank=0.5 pound /gallon of water

Solution pumped from tank A to tank B at the rate=6 gallons/min

Solution pumped from tank B to tank A at the rate=2gallon/min

Solution from tank B is pumped out of the system at the rate=4 gallon/min

We have to find the DE at time t

For x

Rate in= $0.5\times 4+y(t)/30\times 2$

Rate in= $2+y/15$

Rate out= $x/80\times 6$

Rate out=3x/40[/tex]

$\frac{dx}{dt}=$ Rate in-Rate out

$\frac{dx}{dt}=2+y/15-3x/40$

$\frac{dx}{dt}=2-\frac{3}{40}x+\frac{y}{15}$

For y

Rate in= $x/80\times 6$

Rate in=3/40x

Rate out= $y/30\times (4+2)$

Rate out=y/5

$\frac{dy}{dt}=$ Rate in-Rate out

$\frac{dy}{dt}=\frac{3}{40}x-\frac{y}{5}$

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