1. Traci purchases a car for $16,000. If its depreciates by 12% per year, what will be the car's value in 5 years?

2. The population of flies in Mr. Bunn's classroom doubles every 5 days. If there where 10 flies on the first day of school, when will the population reach 2,000,000?

I don't understand how to do it .Is there a formula that I just, && what is this called? decay?

Yes, the formula is:

A(t) = A(0)*R^t
where
A(t) is the quantity at time t,
A(0) is the initial quantity
R=compounding rate (R>1), or decay rate(R<1)
t = time in appropriate units.

Take
1a.
Traci purchases a car for $10,000. If its depreciates by 11% per year, what will be the car's value in 5 years?
A(0)=10000,
t=5 years
R=1.0 - 0.11= 0.89
A(t)=A(5)=10000*0.89^5=10000*0.5584=$5584

2a.
The population of flies in Mr. Bunn's classroom doubles every 6 days. If there were 100 flies on the first day of school, when will the population reach 2,000,000?
A(0)=100
A(t)=2,000,000
R=2^(1/6)=1.12246 (because it doubles every 6 days).
Then
2000000=100*1.12246^t
1.12246^t=2000000/100=20000
t*log(1.12246)=log(20000)
t=log(20000)/log(1.12246)
=85.7 days

Post if you need further help.

1. To find the car's value in 5 years after depreciating by 12% per year, we can use the formula for exponential decay:

Value after depreciation = Initial value * (1 - rate of depreciation/100)^number of years

Given that the car was purchased for $16,000 and depreciates by 12% per year, we can calculate its value after 5 years:

Value after depreciation = $16,000 * (1 - 12/100)^5
Value after depreciation = $16,000 * (0.88)^5
Value after depreciation ≈ $10,736.89

Therefore, the car's value in 5 years will be approximately $10,736.89.

2. Since the population of flies doubles every 5 days, we can use the formula for exponential growth:

Population = Initial population * (growth rate)^number of intervals

Given that there were 10 flies on the first day of school and the population doubles every 5 days, we want to find when the population reaches 2,000,000. Let's count the number of 5-day intervals it takes for the population to reach that level.

Starting with an initial population of 10 flies, the population after n 5-day intervals can be calculated as:

Population = 10 * 2^n

We want to find the value of n when the population reaches 2,000,000:

2,000,000 = 10 * 2^n

Dividing both sides by 10, we get:

200,000 = 2^n

To solve for n, we can take the logarithm base 2 of both sides:

log2(200,000) = log2(2^n)
log2(200,000) = n * log2(2)
log2(200,000) ≈ 17.6096404744

Therefore, the population of flies will reach 2,000,000 after approximately 17.61 (rounded) 5-day intervals or days.

1. To find the car's value in 5 years, we need to calculate the annual depreciation and then multiply it by the number of years.

The car's value depreciates by 12% each year. We can calculate the value after one year by subtracting 12% of $16,000 from $16,000:
Value after 1 year = $16,000 - (12/100) * $16,000

To find the value after 5 years, we need to apply this formula five times:
Value after 5 years = $16,000 - (12/100) * $16,000 - (12/100) * ($16,000 - (12/100) * $16,000) - ... (five times)

Solving this equation step by step, we can use the following pattern:
Value after n years = $16,000 - (12/100) * (previous value)

So, plugging in the values for our case:
Value after 1 year = $16,000 - (12/100) * $16,000 = $16,000 - $1,920 = $14,080
Value after 2 years = $14,080 - (12/100) * $14,080
Value after 3 years = $14,080 - (12/100) * ($14,080 - (12/100) * $14,080)
Value after 4 years = $14,080 - (12/100) * ($14,080 - (12/100) * ($14,080 - (12/100) * $14,080))
Value after 5 years = $14,080 - (12/100) * ($14,080 - (12/100) * ($14,080 - (12/100) * ($14,080 - (12/100) * $14,080)))

Calculating this value step by step will give us the car's value in 5 years.

2. The population of flies doubles every 5 days, starting with 10 flies on the first day. To find the time it takes for the population to reach 2,000,000 flies, we can set up a exponential growth equation.

Let's express the population at any given day, 'n', as P(n). We know that the population doubles every 5 days, so we have the following equation:

P(n) = 10 * 2^(n/5)

We want to find when P(n) equals 2,000,000. We can set up the following equation and solve for 'n':

2,000,000 = 10 * 2^(n/5)

To solve this equation, we need to isolate the exponent. Divide both sides by 10:

200,000 = 2^(n/5)

Now we can rewrite 2 as 2^1 and use logarithms to solve for 'n'. Taking the logarithm (base 2) of both sides gives us:

log2(200,000) = n/5

Simplify and solve for 'n':

n = 5 * log2(200,000)