Answer:
2598 square kilometers
Step-by-step explanation:
Hello
Step 1
year one
using a rule of three is possible to find how much is 8.75 od 4500 km2
Let
if
4500 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
![4500:100\\x:8.75\\\frac{4500}{100}=\frac{x}{8.75}\\x=\frac{4500*8.75}{100} \\x=393.75\\](https://tex.z-dn.net/?f=4500%3A100%5C%5Cx%3A8.75%5C%5C%5Cfrac%7B4500%7D%7B100%7D%3D%5Cfrac%7Bx%7D%7B8.75%7D%5C%5Cx%3D%5Cfrac%7B4500%2A8.75%7D%7B100%7D%20%5C%5Cx%3D393.75%5C%5C)
at the end of the year one, the area will be
4500-393.75=4106.25
this will be the initial area for the year 2.
Step 2
repite the step 1 with area initial =4106.25 km2
4106.25 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
![4106.25:100\\x:8.75\\\frac{4106.25}{100}=\frac{x}{8.75}\\x=\frac{4106.25*8.75}{100} \\x=359.29\\](https://tex.z-dn.net/?f=4106.25%3A100%5C%5Cx%3A8.75%5C%5C%5Cfrac%7B4106.25%7D%7B100%7D%3D%5Cfrac%7Bx%7D%7B8.75%7D%5C%5Cx%3D%5Cfrac%7B4106.25%2A8.75%7D%7B100%7D%20%5C%5Cx%3D359.29%5C%5C)
at the end of the year 2, the area will be
4106-359.29=3746.70
this will be the initial area for the year 3.
Step 3
repite the step 1 with area initial =4106.25 km2
3746.70 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
![3746.70:100\\x:8.75\\\frac{3746.70}{100}=\frac{x}{8.75}\\x=\frac{3746.7*8.75}{100} \\x=327.83\\](https://tex.z-dn.net/?f=3746.70%3A100%5C%5Cx%3A8.75%5C%5C%5Cfrac%7B3746.70%7D%7B100%7D%3D%5Cfrac%7Bx%7D%7B8.75%7D%5C%5Cx%3D%5Cfrac%7B3746.7%2A8.75%7D%7B100%7D%20%5C%5Cx%3D327.83%5C%5C)
at the end of the year 3, the area will be
3746.70-327.83=3419.09
this will be the initial area for the year 4.
Step 4
year four
repite the step 1 with area initial =3419.09 km2
3419.09 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
![3419.09:100\\x:8.75\\\frac{3419.09}{100}=\frac{x}{8.75}\\x=\frac{3419.09*8.75}{100} \\x=299.17\\](https://tex.z-dn.net/?f=3419.09%3A100%5C%5Cx%3A8.75%5C%5C%5Cfrac%7B3419.09%7D%7B100%7D%3D%5Cfrac%7Bx%7D%7B8.75%7D%5C%5Cx%3D%5Cfrac%7B3419.09%2A8.75%7D%7B100%7D%20%5C%5Cx%3D299.17%5C%5C)
at the end of the year 4, the area will be
3419.09-299.173=3119.82
this will be the initial area for the year 5.
Step 5
year five
repite the step 1 with area initial =3119.82 km2
3119.82 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
![3119.82:100\\x:8.75\\\frac{3119.82}{100}=\frac{x}{8.75}\\x=\frac{3119.82*8.75}{100} \\x=272.99\\](https://tex.z-dn.net/?f=3119.82%3A100%5C%5Cx%3A8.75%5C%5C%5Cfrac%7B3119.82%7D%7B100%7D%3D%5Cfrac%7Bx%7D%7B8.75%7D%5C%5Cx%3D%5Cfrac%7B3119.82%2A8.75%7D%7B100%7D%20%5C%5Cx%3D272.99%5C%5C)
at the end of the year 5, the area will be
3119.82-272.99=2846.92
this will be the initial area for the year 6.
Step 6
year six
repite the step 1 with area initial =2846.92km2
2846.92 km2 ⇒ 100$
x?km2 ⇒8.75
do the relation and isolate x
![2846.92:100\\x:8.75\\\frac{2846.92}{100}=\frac{x}{8.75}\\x=\frac{2846.92*8.75}{100} \\x=249.10\\](https://tex.z-dn.net/?f=2846.92%3A100%5C%5Cx%3A8.75%5C%5C%5Cfrac%7B2846.92%7D%7B100%7D%3D%5Cfrac%7Bx%7D%7B8.75%7D%5C%5Cx%3D%5Cfrac%7B2846.92%2A8.75%7D%7B100%7D%20%5C%5Cx%3D249.10%5C%5C)
at the end of the year six, the area will be
2846.92-249.10=2597.82 square kilometers
Have a great day.