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4 years ago

Six donuts cost a total of $3.54 what is the cost for one donut? Six donuts cost a total of $3.54 what is the cost of one donut?

Mathematics

2 answers:

UkoKoshka [18]4 years ago

6 0

Answer:

one doughnut costs 59 cents

Step-by-step explanation:

Sindrei [870]4 years ago

4 0

Answer:

Step-by-step explanation: the cost of one donut is $0.59 because when you divide 3.54 by 6, you get 0.59.

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Someone explain it please

Alina [70]

9514 1404 393

Answer:

∠A = 44°

Step-by-step explanation:

In order to find the measure of angle A, you need to know the value of the variable x. This means you need some relation that you can solve to find x.

Happily, that relation is "the sum of angles in a triangle is 180°." This means ...

84° +(x +59)° +(x +51)° = 180°

(2x + 194)° = 180° . . . collect terms

2x = -14 . . . . . . . . . . divide by °, and subtract 194

x = -7 . . . . . . . . . . . .divide by 2

Now, the measure of angle A is ...

∠A = (x +51)° = (-7 +51)°

∠A = 44°

4 0

3 years ago

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