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Law Incorporation [45]

3 years ago

6

The sum of a number times 6 and 19 is at least -29

1 answer:

Aleksandr [31]3 years ago

5 0

The missing number is -16

6x + 19 = -29
Subtract 19 on both sides
6x = -29 - 19

6x = -48
Divide by 6 on both sides

x= -16

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-1(-2x + 4) = 6<br><br> Hellooooooooooo

butalik [34]

Answer:

x=5

Step-by-step explanation:

First we distribute

2x-4=6

isolate x

2x=10

divide both sides by 2

x=5

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3 years ago

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ollegr [7]

The solution is: 252. 14x18

7 0

3 years ago

F (x) = –2x+1<br> If<br> f (x) = 11<br> Find the value of x:

a_sh-v [17]

Answer:

Now x is equals to 11

-2(11) + 1 = - 22 + 1 = - 21

The answer is - 21

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8 0

3 years ago

Q # 20 please help to solve

spin [16.1K]

If submarine is at the surface (the initial position) and then dives 375 ft, the coordinate of submarine becomes -375 ft <span>relatively to the initial position. When the submarine dives another 175 ft down, the coordinate of submarine becomes -375-175=-550 ft relatively to the initial position.
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Answer: Correct choice is A.
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8 0

3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago