Graph the line y=-2x+5

Answer:

Step-by-step explanation:

[tex]\int\limits^{\infty}_0 {A^2_1} (e^{-r/a})r^2dr= {A^2_1}\int\limits^{\infty}_0r^2(e^{-r/a})^2\, dr)[/tex]

[tex]=A_1^2\int\limits^{\infty}_0 r^2e^{-2r/a}\ dr[/tex]

[tex]=A_1^2[\frac{r^2e^{2r/a}}{-2/a} |_0^{\infty}-\int\limits^{\infty}_0 2r\frac{e^{-2r/a}}{-2/a} \ dr][/tex]

[tex]=A^2_1[0+\int\limits^{\infty}_0 a\ r\ e^{-2r/a}\ dr][/tex]

[tex]=A^2_1[\frac{a \ r \ e^{-2r/a}}{-2/a} |^{\infty}_0-\int\limits^{\infty}_0 \frac{a \ e^{-2r/a}}{-2/a} \ dr][/tex]

[tex]=A_0^2[0-0+\int\limits^{\infty}_0 \frac{a^2}{2} e^{-2r/a}\ dr\\\\=A_1^2\frac{a^2}{2} \int\limits^{\infty}_0 e^{-2r/a}\ dr\\\\=A_1^2\frac{a^2}{2} [\frac{e^{-2r/a}}{-2/a} ]^{\infty}_0[/tex]

[tex]=\frac{A_1^2a^2}{2} -\frac{a}{2} [ \lim_{r \to \infty} [e^{-2r/a} -e^0]\\\\=\frac{A_1^2a^2}{2} -(\frac{a}{2}) (0-1)[/tex]

[tex]=\frac{A_1^2a^3}{4}[/tex]

[tex]\therefore A_1^2\int\limits^{\infty}_0 r^2(e^{-r/a}) \ dr =\frac{A_1^2a^3}{4}[/tex]

Find the unique positive value of A1

[tex]=4\pi (\frac{A_1^2a^3}{4} )\\\\=A_1^2a^3\pi\\\\A_1^2=\frac{1}{a^3\pi} \\\\A_1=\sqrt{\frac{1}{a^3\pi} }[/tex]

Answer:

Step-by-step explanation:

For this kind of question you'd be better off if you'd write down and share your own answers to these conceptual questions and then ask for Brainly feedback on what you have written.  You'll need to understand the concept of "function" often in algebra and beyond.

What concepts (only the names) did you need to accommodate the concept of function in your mind?  input, output, rule, domain, range, mapping, variation (direct and inverse)

Simplest function:  y = c  (there's only one x-value and y equals that value)

In your day to day, is there any occurring fact that can be interpreted as a function?  An electronic parking meter:  the amount of time you can park at the meter without risking getting a ticket is dependent upon the number of quarters you insert into the meter, e. g, 15 minutes for 25 centers, 30 minutes for 50 cents, and so on.

Is it possible to view a function?  Sure.  Graph the function.

What strategy are you using to get the graph of a function?  Set up a coordinate plane.  Label the horizontal axis "x" and the vertical axis "y".  Choose x (input) values that are included in the domain of the function.  If the domain includes '0' you will be finding the 'y-intercept' of the function.  Write the input and output as a point:  (x, y).  Plot that point.  Choose other x values within the domain and calculate the corresponding y value for each.  Plot several more points and draw a line or a curve through them.  Of course there are more sophisticated strategies for graphing functions.  Remember:  If you're working with a function, there is never more than one output or y value for any particular input value.