Dr. Horton is studying an endangered fox whose population is declining. The population decrease is represented by the function f(x) = 6500(1 - 0.10)^x - 40. She is using this equation to study the population decline from year to year, with x representing each year.
a) Calculate how many foxes there were in year 3 and in year 8. You must show all your work and explain each step.
(please someone help me with this!)
Dr. Suzuki is studying an endangered fox whose population is declining by 20% every year. If there are currently 3,300 foxes, how many will remain in 4 years?
If necessary, round your answer to the nearest whole number.
Alright, let me see if I can help you with this! First, let's tackle year 3:
To find the number of foxes in year 3, we need to substitute x = 3 into the equation f(x) = 6500(1 - 0.10)^x - 40.
f(3) = 6500(1 - 0.10)^3 - 40
Now, let's simplify the equation:
f(3) = 6500(0.90)^3 - 40
f(3) = 6500(0.729) - 40
f(3) = 4738.5 - 40
f(3) = 4698.5
So, in year 3, there were approximately 4698.5 foxes.
Next, let's move on to year 8:
Following the same process, we substitute x = 8 into the equation f(x) = 6500(1 - 0.10)^x - 40.
f(8) = 6500(1 - 0.10)^8 - 40
Now, let's simplify the equation:
f(8) = 6500(0.90)^8 - 40
Now, it's time for some math!
f(8) = 6500(0.43046721) - 40
f(8) = 2797.955365 - 40
f(8) = 2757.955365
So, in year 8, there were approximately 2757.955365 foxes.
To calculate the number of foxes in year 3 and year 8, we need to substitute the respective values of x into the given function f(x) = 6500(1 - 0.10)^x - 40.
a) Year 3:
Substitute x = 3 into the function f(x) = 6500(1 - 0.10)^x - 40.
f(3) = 6500(1 - 0.10)^3 - 40
Now, we need to calculate the exponent (1 - 0.10)^3 first:
(1 - 0.10)^3 = (0.90)^3 = 0.729
Now substitute this value back into the original equation:
f(3) = 6500(0.729) - 40
Multiply 6500 by 0.729:
f(3) = 4738.5 - 40
Subtract 40 from 4738.5:
f(3) = 4698.5
Therefore, in year 3, there were approximately 4,698.5 foxes.
b) Year 8:
Substitute x = 8 into the function f(x) = 6500(1 - 0.10)^x - 40.
f(8) = 6500(1 - 0.10)^8 - 40
Now, calculate the exponent (1 - 0.10)^8 first:
(1 - 0.10)^8 = (0.90)^8 = 0.43046721
Now substitute this value back into the original equation:
f(8) = 6500(0.43046721) - 40
Multiply 6500 by 0.43046721:
f(8) = 2802.955665 - 40
Subtract 40 from 2802.955665:
f(8) = 2762.955665
Therefore, in year 8, there were approximately 2,762.955665 foxes.
Hence, the population of the endangered foxes was approximately 4,698.5 in year 3 and 2,762.955665 in year 8.
huh? You have the formula. Just plug in the values for x.
For example,
f(5) = 6500 * 0.9^5 - 40 = 3798
I have no idea which "steps" you need to show. Just evaluate the expression.