Answer:
=k= -2.2/6.7
= k=(.328)
Step-by-step explanation:
2.2=k+6.7/-1
=2.2×-1=6.7+k
=k= -2.2/6.7
= k=(.328)
Hey mate hope it's help you.....
Answer:
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.
Answer:
Step-by-step explanation:
there are 4 sixes soo
Assuming you can’t use the same topping twice,
10 x 9 x 8 = 720
X^2 + 8x - 5 = 0
a = 1, b = 8, c = -5
x = ( - 8 + - sqrt(64 - 4*1*-5))/2
x = ( -8 + - sqrt(64+20))/2
x = ( -8 + - sqrt (84))/2
x = ( - 8 + - 2sqrt21))/2
x = -4 + - sqrt21
