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Nitella [24]
2 years ago
12

The present value of a house is $242,000. The value of the house increases at the rate of 4% every year. A function is shown bel

ow: f(x) = 242,000(1.04)x What does f(x) represent?
The original value of the house
The future value of the house after x years
The increase in the value of the house each year
The increase in the value of the house after x years
Mathematics
2 answers:
spayn [35]2 years ago
6 0
It will be the value of the house in x years.

DiKsa [7]2 years ago
5 0
The future value of the house after x years
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write an equation, in slope-intercept form, to model each sitution. : the population of pine bluff is 6791 and is decreasing at
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6791 of per year answer
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2 years ago
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ILL GIVE BRAINLIEST IF CORRECT!! The average of x+6, 6x+2, and 2x+7 is 4x-7. What is x?
mina [271]

Answer:

35/3

Step-by-step explanation:

\frac{x+6 + 6x+2+x+6}{3} =4x-7

Simplify to get

9x+14=3(4x-7)

35=3x

x=35/3 or 11.66

6 0
3 years ago
She earns $3.10 for each hour she works and $2.30 for each bushel of apples she picks. Her goal is to earn at least $100 this we
nika2105 [10]

The inequality would be 3.1h + 2.3b ≥100

Step-by-step explanation:

Earning of Tammy for each hour of work= $3.1

Total hours she worked= h

Earning for picking each bushel of apple= $2.3

Total bushel she picks= b

Her goal = minimum of $100

Hence total earning from working= earning from each hour* no of hours worked

Total earning= 3.1h

Similarly, total earning from picking up apple bushels= earning from picking one bushel*no of bushels picked

Total earning= 2.3b

According to question total earning must be greater than $100

Hence, 3.1h + 2.3b ≥100

3 0
3 years ago
A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe
adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

\frac{dC}{dt}=Ci*Qi-Co*Qo

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

\frac{dC}{dt}=-Co*Qo

Rearranging the equation, it becomes

\frac{dC}{C}=-Qo*dt

Integrating both sides

\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\

The final equation for the concentration of salt at any given time is

C=exp^{-3*t-0.693}

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}=  -\frac{-0.266}{3}=0.088

5 0
3 years ago
I need help on this please.
Pachacha [2.7K]

Answer:

Write in the box :....

30

8 0
2 years ago
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