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A local bakery, theprice for abughouts for his employeespurchasedAFX) -0.65- 3.5prie for $3.50 and some doudoughnuts 30.05. Each day the manager at the store buyswhich equation represents the total cosa function of the number of doughtswhich equation on where represents the number of tires produced over resismodels the function ?BFX) - 0.65x + 3.5CX-3.5x + 0.85DX) - 3.5x -0.85Aebire manufacturing plant produces soo tires a day on average. If the production ofAFX) - 500 + xB (x) - 500 -Cx) - 500xDX) - 5006.Bushra purchases a car for $12,900. The car will depreciate at a rate of 15% each year,After how many years will the value of the car bethan $3,000?A 6 yearsB 7 yearsC8 yearsD 9 years

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In order to create a function that represents the cost of the manager as a function of the number of doughnuts he buys, we need to multiply the cost of each doughnut ($ 0.85) by the number of employees the manager has and add the value of the pie ($ 3.5). This is done below:

[tex]f(x)\text{ = }0.85\cdot x\text{ + 3.5}[/tex]

The correct option is the letter B.

The car starts at $ 12,900 and depreciate at a rate of 15% each year. This means that the value of the car on any given year is ruled by the tollowing expression:

[tex]M\text{ = C}\cdot(1-r)^t[/tex]

Where "M" is the value of the car after "t" years, C is the initial value of the car and "r" is the rate at which the car depreciates every year divided by 100. Aplying the data from the problem on the expression gives us:

[tex]3000\text{ = 12900}\cdot(1\text{ - }\frac{15}{100})^t[/tex]

We want to solve for the variable "t", because we want to know how many years it'll take until the car reaches the final value of 3000.

[tex]\begin{gathered} 12900\cdot(1\text{ - }\frac{15}{100})^t\text{ = 3000} \\ (1\text{ - }\frac{15}{100})^t\text{ = }\frac{3000}{12900} \\ (1-0.15)^t\text{ = }\frac{30}{129} \\ (0.85)^t\text{ = }\frac{30}{129} \end{gathered}[/tex]

We have reached an exponential equation. To solve it we need to aply a logarithm on both sides of the equation.

[tex]\begin{gathered} \ln (0.85^t)\text{ = }\ln (\frac{30}{129}) \\ t\cdot\ln (0.85)\text{ = }ln(30)\text{ - ln(129)} \\ t\cdot(-0.1625)\text{ = }3.4\text{ - 4.86} \\ t\text{ = }\frac{-1.46}{-0.1625}\text{ = 8.98} \end{gathered}[/tex]

It'll take approximately 9 years to reach that value. The correct option is the letter "D".

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