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3 years ago

Find the solution to the equation 4r + 2(r - 8) = 4(3 + r).

Mathematics

2 answers:

BlackZzzverrR [31]3 years ago

7 0

Answer:

r = 14

Step-by-step explanation:

4r + 2(r - 8) = 4(3 + r)

4r + 2r - 16 = 12 + 4r

6r - 16 = 12 + 4r

6r - 4r - 16 = 12 + 4r - 4r

2r - 16 = 12

2r - 16 + 16 = 12 + 16

2r = 28

2r ÷ 2 = 28 ÷ 2

r = 14

o-na [289]3 years ago

3 0

Answer:

r = 14

Step-by-step explanation:

Use distributive property:

4r + 2r - 16 = 12 + 4r

Combine like terms:

6r - 16 = 12 + 4r

Isolate the variable:

6r = 28 + 4r

2r = 28

r = 14

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4 years ago

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Read 2 more answers

A tank is filled with 1000 liters of pure water. Brine containing 0.04 kg of salt per liter enters the tank at 9 liters per minu

klemol [59]

Answer:

The differential equation which describes the mixing process is $\frac{dc_{salt,out}}{dt} + \frac{2}{125}\cdot c_{salt,out} = \frac{71}{100000}$ .

Step-by-step explanation:

The mixing process within the tank is modelled after the Principle of Mass Conservation, which states that:

$\dot m_{salt,in} - \dot m_{salt,out} = \frac{dm_{tank}}{dt}$

Physically speaking, mass flow of salt is equal to the product of volume flow of water and salt concentration. Then:

$\dot V_{water, in, 1}\cdot c_{salt, in,1} + \dot V_{water, in, 2} \cdot c_{salt,in, 2} - \dot V_{water, out}\cdot c_{salt, out} = V_{tank}\cdot \frac{dc_{salt,out}}{dt}$

Given that $\dot V_{water, in, 1} = 9\,\frac{L}{min}$ , $\dot V_{water, in, 2} = 7\,\frac{L}{min}$ , $c_{salt,in,1} = 0.04\,\frac{kg}{L}$ , $c_{salt, in, 2} = 0.05\,\frac{kg}{L}$ , $\dot V_{water, out} = 16\,\frac{L}{min}$ and $V_{tank} = 1000\,L$ , the differential equation that describes the system is:

$0.71 - 16\cdot c_{salt,out} = 1000\cdot \frac{dc_{salt,out}}{dt}$

$1000\cdot \frac{dc_{salt, out}}{dt} + 16\cdot c_{salt, out} = 0.71$

$\frac{dc_{salt,out}}{dt} + \frac{2}{125}\cdot c_{salt,out} = \frac{71}{100000}$

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3 years ago