Colin buys a car for £15500. It depreciates at a rate of 1% per year.
How much will it be worth in 4 years?
Give your answer to the nearest penny where appropriate.

£14889.30

Step-by-step explanation:

The cost of the car = £15500.

For the first year,

depreciation = £15500 x 1%

                     = £155

Its worth after the first year = £15500 - £155

                             = £15345

For the 2nd year,

depreciation = £15345 x 1%

                     = £153.45

Its worth after the second year = £15345 - £153.45

                                 = £15191.55

For the 3rd year,

depreciation = £15191.55 x 1%

                      = £151.9155

Its worth after the third year = £15191.55 - £151.9155

                                            = £15039.6345

For the 4th year,

depreciation = £15039.6345 x 1%

                     = £150.3963

Its worth after the fourth year = £15039.6345 - £150.3963

                                             = £14889.2382

Thus, the worth of the car in 4 years would be £14889.30

d

gimmee one of those gold crowns!

~coco

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.