JT Engineering is deciding between two machines. Machine A costs $352,000, with inflows of $209,000 and outflows of $154,000. Machine B costs $380,000, with inflows of $231,000 and outflows of $166,000. Both have a 10-year life and no salvage value. JT uses the straight-line method for depreciation and requires a return of 12%. How desirable are the machines? Use annual rate of return to determine the answer.

Machine B is preferrable with annual rate of return of 14.21%,higher than the required annual rate of return of 12%

Explanation:

Annual rate of return of both machine needs to ascertained ,then compared with the required annual rate of rate of 12% in order to determine  which machine gives at least 12% annual rate of return and worth investing in.

Annual rate of return=net income/average investment

net income=inflows-outflows-depreciation

Machine A average investment=$352,000/2=$176,000

Machine B average investment=$380,000/2=$190,000

Machine A net income=$209,000-$154,000-($355,000/10)

                                      =$209,000-$154,000-$35,500

                                       =$19,500

Machine B net income=$231,000-$166,000-($380,000/10)

                                      =$231,000-$166,000-$38,000

                                       =$27,000

Machine A annual rate of return=$19,500/$176,000

                                                     =11.08%

Machine B annual rate of return=$27,000/$190,000

                                                     =14.21%

explanation:

kool beans

start by isolating the variable on one side by adding 43w^2 to both sides.   the equation will read:

25w^2+5=0

*subtract 5 from both sides*

25w^2=-5

*divide both sides by 25*

w^2=-1/5

*take the square root of both sides*

w=√(-1/5)

the square root of -1 is i.   the square root of 5 remains √5

the answer is i/√5.

hope this