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2 years ago

Y = 2x – 6 y = –4x + 3 will give brainliest

Mathematics

1 answer:

sammy [17]2 years ago

3 0

Answer:

3/2

Step-by-step explanation:

2x-6 = -4x+3

6x = 9

x = 3/2

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2 years ago

3. Solve each equality or inequality:<br> a. |8x-6|= 0

larisa [96]

Answer:

x = 3/4

Step-by-step explanation:

Step 1: Find possibilities.

We know one possibility: 8x - 6 = 0. This is because 0 cannot be positive or negative.

Step 2: Solve possibility 1.

$8x - 6 = 0$
$8x - 6 + 6 = 0 + 6$ (Add 6 to both sides)
$8x = 6$
$\frac{8x}{8} =\frac{6}{8}$ (Divide both sides by 8)
$x = \frac{3}{4}$

Therefore, x = 3/4.

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2 years ago

How many deciliters are equivalent to 9 cups? Use the conversion table to find the answer. Round to the nearest tenth.

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You would have 21.29 dL or rounded up 21.3 dL.

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2 years ago

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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}}$

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2 years ago