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katrin [286]
4 years ago
14

In 2003, the price of a certain automobile was approximately $32,000 with a depreciation of $1,740 per year. After how many year

s will the car's value be $23,300? a) Write an equation to model the problem. Let t represent the number of years after 2003. For example, the year 2005 would be represented by t = 2.
Mathematics
1 answer:
IRINA_888 [86]4 years ago
7 0

Answer:

The correct answer is 5 years i.e. 2008.

Step-by-step explanation:

Price of the automobile in 2003 is $32000.

Depreciation per year is given by $1740.

Therefore let the car value is depreciated for t number of years.

Value depreciated for t years is given by $ (1740t).

The final value of the car after t years is given to be $23300.

Thus the equation is given by 32000 - 1740t = 23300.

⇒ 1740t = 32000 - 23300

⇒ 1740t = 8700

⇒ t = 5

Thus after 5 years the value of the car is $23300.

Thus in 2008 the depreciated price of the car would be $23300.

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Bullco blends silicon and nitrogen to produce two types of fertilizers. Fertilizer 1 must be at least 40% nitrogen and sells for
leva [86]

Answer:

Maximize

z = 70(Xs1 + Xn1 ) + 40(Xs2 + Xn2 ) - 10 (Xs1 + Xs2 ) - 15(Xn1 + Xn2 )  

Subject to the constraints  

Xs1 + Xs2 ≤ 100

Xn1 + Xn2 ≤ 80

Xn1 ≥ 0.4 ( Xs1 + Xn1 )

Xs2 ≥ 0.7 ( Xs2 + Xn2 )

All Variables ≥ 0

Step-by-step explanation:

Firstly lets consider Xs1 and Xs2 to be the number of pounds of silicon used in fertilizer1 and fertilizer2 respectively

Also let Xn1 and Xn2 be the number of pounds of nitrogen used in fertilizer1 and fertilizer2 respectively

We know that the objective is to maximize the profits of Bullco.

z = [(Selling price of fertilizer1) (Amount of silicon and nitrogen used to produce fertilizer1) + (Selling price of fertilizer2) (Amount of silicon and nitrogen used to produce fertilizer2) - (Cost of silicon) (Amount of silicon used to produce fertilizer I and 2) - (Cost of nitrogen) (Amount of nitrogen used to produce fertilizer I and 2)]

so

z = 70(Xs1 + Xn1 ) + 40(Xs2 + Xn2 ) - 10 (Xs1 + Xs2 ) - 15(Xn1 + Xn2 )  

Now

Constraint 1;  At most, 100 lb of silicon can be purchased

Amount of silicon used to produce fertilizer 1 and 2 ≤ 100

Xs1 + Xs2 ≤ 100

Constraint 2; At most, 80 lb of nitrogen can be purchased

Amount of nitrogen used to produce fertilizer 1 and 2 ≤ 80

Xn1 + Xn2 ≤ 80

Constraint 3; Fertilizer 1 must be at least 40% of nitrogen

Amount of nitrogen used to produce fertilizer 1 ≥ 40% (fertilizer 1)

Xn1 ≥ 0.4 ( Xs1 + Xn1 )

Constraint 4; Fertilizer 2 must be at least 70% of silicon

Amount of silicon used to produce fertilizer 2  ≥ 70% (fertilizer 2)

Xs2 ≥ 0.7 ( Xs2 + Xn2 )  

so the formulization of the given linear program is,  

Maximize

z = 70(Xs1 + Xn1 ) + 40(Xs2 + Xn2 ) - 10 (Xs1 + Xs2 ) - 15(Xn1 + Xn2 )  

Subject to the constraints  

Xs1 + Xs2 ≤ 100

Xn1 + Xn2 ≤ 80

Xn1 ≥ 0.4 ( Xs1 + Xn1 )

Xs2 ≥ 0.7 ( Xs2 + Xn2 )

All Variables ≥ 0

8 0
3 years ago
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