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3 years ago

Evaluate 32 + (4c − 6c) + (2 + 6c) for c = 5. A.22 B.64 C.−10 D.54

Mathematics

2 answers:

Butoxors [25]3 years ago

7 0

Answer:

D. 54

Step-by-step explanation:

32 + (4c - 6c) + (2 + 6c)

Put c as 5 and evaluate.

32 + 4(5) - 6(5) + 2 + 6(5)

Multiply the terms.

32 + 20 - 30 + 2 + 30

Add or subtract the terms.

32 - 10 + 2 + 30

22 + 32

= 54

sweet-ann [11.9K]3 years ago

5 0

Answer:

The answer is option D

32 + (4c − 6c) + (2 + 6c)

c = 5

32 + ( 4(5) - 6(5)) + (2 + 6 (5)

= 32 + (20 - 30) + ( 2 + 30)

= 32 - 10 + 32

= 54

Hope this helps

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Jason’s savings account has a balance of $2179. After 5 years , what will the amount of interest be at 6% compounded quarterly?

satela [25.4K]

Answer:

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Step-by-step explanation:

Determine the compound amount first and then subtract the principal from it, to find the amount of interest.

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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.

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