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2 years ago

CAT FOODA grocery store is selling 6 cans of cat food for $3. Write and solve an equation to find the cost of a can of cat food.

Mathematics

2 answers:

Digiron [165]2 years ago

4 0

Answer:

6 * 3 = 18

Step-by-step explanation:

NemiM [27]2 years ago

3 0

Answer:

6 divided by 3= 2

t Step-by-step explanation: I'm not sure so probaly don't listen to me.

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What is the slope for this table?

sdas [7]

<h3>Answer: Slope = -3</h3>

=======================================================

Explanation:

Pick any two rows you want. I'll pick the middle two rows where every value is positive. So I'll pick these two rows:

x = 0 and y = 4
x = 1 and y = 1

They lead to the points (0,4) and (1,1). Apply the slope formula for these rows.

m = (y2-y1)/(x2-x1)

m = (1-4)/(1-0)

m = -3/1

m = -3

The slope is -3

6 0

2 years ago

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Find population density of California at 36,500,000 and 163,707 square miles

vlabodo [156]

There would be 223 people every square mile

7 0

3 years ago

– 7р — 4 >26р – 94<br> I don’t understand how to solve, what is the answer?

pishuonlain [190]

Answer:

I don't know either but thanks for asking me

3 0

3 years ago

Grandma has 1/2 of her birthday cake left . she wants her 12 grandchildren to each get the same size piece of cake to take home

eimsori [14]

Answer: 1/24

Step-by-step explanation:

you need to use kcf ( keep the first fraction change division into multiplication and flip the other fraction)

turn 12 into 12/1 and then just divide.

1/2 x 1/12

= 1/24

6 0

3 years ago

Read 2 more answers

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

3 years ago