A new building that costs $1,000,000 has a useful life of 10 years and a scrap value of $200,000. Using straight-line depreciation, find the equation for the value V in terms of t, where t is in years. (Make sure you use t and not x in your answer.)

V = -80,000*t + 1,000,000

Step-by-step explanation:

Given:

- The original cost V_i = $1,000,000

- Scrap Value V_f = $200,000

- Total number of years n = 10

Find:

Using straight-line depreciation, find the equation for the value V in terms of t, where t is in years.

Solution:

- The straight line depreciation method means their is a linear relationship between the value of building and time elapsed till useful life. The linear relationship takes a form:

                                 V = m*t + C

Where,

V: It is the current value of the building

m: The rate at which value increase or decrease with time t

C: The initial value of the building

- We will compute constants m and C as follows:

                               m = (V_f - V_i) / ( 10 - 0 )

                               m = (200,000 - 1,000,000) / 10

                               m = - $80,000/year

                               C = V_i = $1,000,000

- Plug the values of the constants evaluated above in the linear expression:

                                V = -80,000*t + 1,000,000

60°

step-by-step explanation:

the measure angle of c in an isosceles triangle abc is 60° because in isosceles triangle,the measures of all the angles is 60°.

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