Question

a business uses a straight - line depreciation model to determine the value ( y ) of a piece of machinery over a twenty year period. suppose the original value (when ( t = 0 )) is equal to $50,000 and its value is reduced by $2500 each year. write the linear equation that models the value ( y ) of this machinery at the end of the year ( t ).
( \bigcirc y=-20x )
( \bigcirc y = 20x )
( \bigcirc y=50000t - 2500 )
( \bigcirc y=-2500t + 50000 )
( \bigcirc y=50000t + 2500 )
( \bigcirc y=2500t + 50000 )
question 21 (4 points)
a company finds it can produce 40 heaters for $7900, while producing 45 heaters costs $8500. express the cost, ( y ), as a linear function of the number of heaters, ( x ).
( \bigcirc y=-120x + 2300 )
( \bigcirc y = 120x+3100 )
( \bigcirc y=3100x + 120 )
( \bigcirc y = 0.083x-25.83 )
question 22 (4 points)
the average monthly bill for wireless telephone subscribers from 1985 to 2010 can be modeled by ( b(x)=-1.374x + 97.58 ), where ( x ) is the number of years after 1980. if this model remains valid, in what year will the average monthly bill be $63.23?
( \bigcirc 30 )
( \bigcirc 2005 )
( \bigcirc 20 )
( \bigcirc 2010 )

Explanation:

Step1: Identify linear form for Q1

Linear model: $y=mt+b$

Step2: Define values for Q1

$b=50000$, $m=-2500$

Step3: Build equation for Q1

$y=-2500t+50000$

Step4: Calculate slope for Q21

Slope $m=\frac{8500-7900}{45-40}=120$

Step5: Find intercept for Q21

$7900=120(40)+b \implies b=3100$

Step6: Build cost function for Q21

$y=120x+3100$

Step7: Set up equation for Q22

$63.23=-1.374x+97.58$

Step8: Solve for x in Q22

$1.374x=97.58-63.23 \implies x=\frac{34.35}{1.374}=25$

Step9: Find year for Q22

$1980+25=2005$

Answer:

1. $y = -2500t + 50000$
2. $y = 120x + 3100$
3. 2005