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2 years ago

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You bought a new car in 2015 for $32,000 that is depreciating in value by 15% each year. Write a function to represent this situ

ation. What will the car be worth in 2020? Function: ___________________________________
Value of car in 2020: ______________________

1 answer:

Hitman42 [59]2 years ago

4 0

Function: x = years

x = $32,000-15%x

Value of car in 2020:

$8000

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A graph shows the vertices of the feasible region to be (0, 6), (3, 0) and (0, -3). Of these, the one that minimizes f(x, y) is (0, -3). The minimum value is ...

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Here, three regions overlap to form the region where solutions are feasible. By reversing the inequality in each of the constraints, <em>the feasible region shows up on the graph as a white space</em>, making it easier to identify. The corner of the feasible region that minimizes the objective function is the one at the bottom, at (0, -3).

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Andreyy89

Answer:

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx}= x - \ln(1 + e^{x}) + C\end{aligned}$ .

Step-by-step explanation:

The first derivative of the denominator $1 + e^{x}$ is $e^{x}$ . Rewrite the fraction to obtain that expression on the numerator.

$\begin{aligned}\frac{1}{1 + e^{x}} &= \frac{1 + e^{x}}{1 + e^{x}} - \frac{e^{x}}{1+e^{x}}\\&=1-\frac{e^{x}}{1+e^{x}}\end{aligned}$ .

In other words,

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\end{aligned}$ .

Apply $u$ -substitution on the integral $\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx}$ :

Let $u = 1 + e^{x}$ . $u > 1$ .

$du = e^{x}\cdot dx$ .

$\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx} = \int{\frac{du}{u}} = \ln{|u|} = \ln{u} +C = \ln{(1+e^{x})}+C$ .

Therefore

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\\ & = x - \ln{(1 + e^{x})}+C\end{aligned}$ .

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