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vovangra [49]
3 years ago
10

Calculate the value of the expression 18/-3

Mathematics
2 answers:
alexgriva [62]3 years ago
7 0

Answer:

-6

Step-by-step explanation:

Any number that is divided by a negative number is a negative number as an answer which is -6.

vampirchik [111]3 years ago
3 0
The answer is B. -6

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A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
Novosadov [1.4K]

Answer:

A(t)=240-220e^{-\frac{t}{40}}

Step-by-step explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

  • Volume of the tank = 240 liters
  • Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

R_{in}=(concentration of salt in inflow)(input rate of fluid)

R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}

R_{out}=(concentration of salt in outflow)(output rate of fluid)

R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}

Rate of change of the amount of salt in the tank:

\dfrac{dA}{dt}=R_{in}-R_{out}

\dfrac{dA}{dt}=6-\dfrac{A}{40}

We then solve the resulting differential equation by separation of variables.

\dfrac{dA}{dt}+\dfrac{A}{40}=6\\$The integrating factor: e^{\int \frac{1}{40}dt} =e^{\frac{t}{40}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{40}}+\dfrac{A}{40}e^{\frac{t}{40}}=6e^{\frac{t}{40}}\\(Ae^{\frac{t}{40}})'=6e^{\frac{t}{40}}

Taking the integral of both sides

\int(Ae^{\frac{t}{40}})'=\int 6e^{\frac{t}{40}} dt\\Ae^{\frac{t}{40}}=6*40e^{\frac{t}{40}}+C, $(C a constant of integration)\\Ae^{\frac{t}{40}}=240e^{\frac{t}{40}}+C\\$Divide all through by e^{\frac{t}{40}}\\A(t)=240+Ce^{-\frac{t}{40}}

Recall that when t=0, A(t)=20 (our initial condition)

20=240+Ce^{-\frac{0}{40}}\\20-240=C\\C=-220\\$Therefore, the number A(t) of grams of salt in the tank at time t\\A(t)=240-220e^{-\frac{t}{40}}

3 0
3 years ago
Will give brainliest
fiasKO [112]

Answer:

either 40 or 140

Step-by-step explanation:

5 0
3 years ago
Two trains travel a 160 km track each day. The express travels 10 km h^-1 faster and take 30 minutes less time than the normal t
aniked [119]
Let s = the speed of the slower train in km/hrs+10 = the speed of the faster train in km/hrLet t = time in hrs for the slower train to go 160 kmt-5 = time in hrs for faster train to go 160 km---------------Slower train:(1)
160=s.t

Faster train:(2) 160=(s+10) .(t-.5)From (1)(1) t=160/s
use the relations given in following screenshot and then use quadratic equation.

4 0
3 years ago
Helooo can some one help me please
oee [108]
The answer is 184. 2711
Round it off is 184
Hope that helped :D
5 0
3 years ago
If 1900 square centimeters of material are available to make a box with a square base and an open top, find the largest possible
Evgen [1.6K]

Answer:

Volume = 7969 cubic centimeter

Step-by-step explanation:

Let the length of each side of the base of the box  be A and the height of the box be H.

Area of material required to make the box  is equal to  is A^2 + 4*A*H.

A^2 + 4*A*H = 1900

 Rearranging the above equation, we get -  

`H = \frac{(1900 - A^2)}{(4*A)}

Volume of box is equal to product of base area of box and the height of the box -  

V = A*A* H

Substituting the given area we get -

\frac{A^2*(1900 - A^2)}{4A} = \frac{(1900*A - A^3)}{4}

For maximum volume

\frac{dV}{dA} =0

\frac{ 1900}{4} - \frac{3*A^2}{4} = 0

A^2 = \frac{1900}{3}

Volume of the box

= \frac{\frac{1900}{3}*(1900 - \frac{1900}{3}) }{4 * \sqrt{\frac{1900}{3} } }

= 7969 cubic centimeter

3 0
2 years ago
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