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:
C
Step-by-step explanation
You have to do the rainbow move which is basically multiplying
4x*2x, 4x*3, -2*2x, -2*3
which then simplified would equal to 8x^2+8x-6
Answer:
if the height is h and the base is 2b
then we have:
½(2bh)=60→2bh=120
b²+h²=13²=169
so
(b+h)²=b²+h²+2bh=120+169=289
(b-h)²=b²+h²-2bh=169-120=49
so
b+h=17
|b-h|=7
so b can be 12 and h can be 5
or b can be 5 and h can be 12
which means the base is either 24cm or 10cm
