Car A has an initial value of 30,000 and depreciates at a rate of 20% per year. Car B has an initial value of 20,000 and depreciates at a rate of 15% per year. Car B will become greater than the value of car A during the ninth, seventh, third, or fifth year

The value of Car B will become greater than the value of car A during the fifth year.

Step-by-step explanation:

Note: See the attached excel file for calculation of beginning and ending values of Cars A and B.

In the attached excel file, the following are used:

Annual Depreciation expense of Car A = Initial value of Car A * Depreciates rate of Car A = 30,000 * 20% = 6,000

Annual Depreciation expense of Car B from Year 1 to Year 6 = Initial value of Car B * Depreciates rate of Car B = 20,000 * 15% = 3,000

Annual Depreciation expense of Car B in Year 7 =  Beginning value of Car B in Year 7 = 2,000

Conclusion

Since the 8,000 Beginning value of Car B in Year 5 is greater than the 6,000 Beginning value of Car A in Year 5, it therefore implies that the value Car B becomes greater than the value of car A during the fifth year.

c

step-by-step explanation:

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.