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Answer:

A. y = 80x

B. g(x) = 80x

C. To graph the equation, plot a point at (0,0) and a point at (1,80). Connect the points. Continue adding points by moving up $80 and over 1 day.

Step-by-step explanation:

Part A:

To write an equation, use y= mx where m is the slope, x is the number of days, and y is the rent cost.

x and y remain the same in the equation.

To find m, use the slope formula with (5,465) and (7, 625).

$m=\frac{y_2-y_1}{x_2-x_1}= \frac{625-465}{7-5} =\frac{160}{2} =80$

It costs $80 a day.

The equation is y = 80x.

Part B:

Function notation replaces Y as g(x). So the equation is g(x) = 80x.

Part C:

To graph the equation, plot a point at (0,0) and a point at (1,80). Connect the points. Continue adding points by moving up $80 and over 1 day.

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Alex Ar [27]

Answer:

$\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C$

General Formulas and Concepts:

<u>Algebra I</u>

<u>Calculus</u>

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration: $\displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C$

Step-by-step explanation:

<u>Step 1: Define</u>

<u /> $\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx$ <u />

<u />

<u>Step 2: Integrate Pt. 1</u>

[Integral] Factor fraction denominator: $\displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx$
[Integral] Integration Property - Multiplied Constant: $\displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx$

<u>Step 3: Identify Variables</u>

<em>Set up u-substitution for the arctan trig integration.</em>

$\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx$

<u>Step 4: Integrate Pt. 2</u>

[Integral] Substitute u-du: $\displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du$
[Integral] Trig Integration: $\displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C$
[Integral] Simplify: $\displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C$
[integral] Multiply: $\displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C$
[Integral] Back-Substitute: $\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C$