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Tpy6a [65]
2 years ago
14

Carla has a scooter worth $3,890. Her computer is worth $600. She owes $700 on her scooter and $497 on her credit card. Her pers

onal property has a value of $2,100. She has $3,000 in her bank accounts and a mutual fund valued at $4,800. She wants to take out a $2,600 loan to buy some new appliances. The value of the appliances after purchase will go down 20%. Complete the explanation of how the purchase of the appliances will change Carla's net worth.
Mathematics
2 answers:
Zina [86]2 years ago
7 0

Answer:

will change by 12% because of the loan.

Step-by-step explanation:

zhuklara [117]2 years ago
4 0

Answer:

a problem with saving money

Step-by-step explanation:

i just foudn out you lips is the same skin as your eyes

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Which is the graph of y – 3 = (x + 6)? A coordinate grid with a line passing through the points at (negative 3, 1), (0, negative
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A and D

If you take all the numbers and add them together they all equal and equivalence to 1.6a and 1.7d. So, if you take the numbers then add them it is A and D

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Adam has 16.45 kg of flour, he uses 6.4 kg to make hot cross buns. The remaining flour is exactly enough to make 15 batches of s
Doss [256]

Answer:

Each batch of scones needed 0.67 kg of flour

Step-by-step explanation:

16.45 - 6.4 = 10.05kg (remaining)

15 batches = 10.05kg

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How do you find the area of a square
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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
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