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Exponential function are functions that use exponents or power of an argument
The remaining amount after one year
The given parameters are:
Initial, a = 200
Rate, r = 2% loss
The remaining after a year is calculated as:
Remaining = a * (1 - r)
This gives
Remaining = 200 * (1 - 2%)
Evaluate the difference
Remaining = 200 * 0.98
Evaluate the product
Remaining = 196
Hence, the remaining amount after one year is 196 kg
The single number of the product in (a)
In (a), we have:
Remaining = 200 * 0.98
Remove the initial value
Single = 0.98
This represents the multiplying factor.
Hence, the single number of the product in (a) is 0.98
The exponential function
We have:
Initial, a = 200
Multiplying factor, b = 0.98
An exponential function is represented as:
A(t) = ab^t
So, we have:
A(t) = 200 * 0.98^t
Hence, the exponential function is A(t) = 200 * 0.98^t
Time to reach half life
In (c), we have:
A(t) = 200 * 0.98^t
The half life is when
A(t) = 0.5 * A = 0.5 * 200 = 100
So, we have:
100 = 200 * 0.98^t
Divide both sides by 200
0.5 = 0.98^t
Take the logarithm of both sides
log(0.5) = t * log(0.98)
Divide both sides by log(0.98)
t = 34 (approximated)
Hence, the time to reach the half life is 34 years
The given parameters are:
An exponential function is represented as:
A(t) = a(1 - t)^t
So, we have:
A(t) = 12500(1 - 4%)^t
In 10 years, t = 10.
So, we have:
A(10) = 12500(1 - 4%)^10
Evaluate
A(10) = 8310
Hence, the population of the town in 10 years is (d) 8310
The equation of the amount of water
The given parameters are:
An exponential function is represented as:
A(t) = a(1 - t)^t
So, we have:
A(t) = 4(1 - 30%)^t
Evaluate the difference
A(t) = 4 * 0.7^t
Hence, the equation of the water is A(t) = 4 * 0.7^t
The graph of the function
The domain is given as:
[0,10]
See attachment for the graph of A(t) = 4 * 0.7^t in the domain [0,10]
Read more about exponential functions at:
https://brainly.com/question/2456547
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