Help me!!
1 answer:
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Step-by-step explanation:
Hello!
I'd love to help you learn.
I see the formula attached is
This problem is explaining you to graph, since algebraically it's a little more harder to try and find the solution.
Since they're both equal to each other, we can assign a variable like y, so we can make them into two individual lines.
You can use your scientific calculator to graph lines like this, or otherwise there are online sources (Desmos helps alot!) which can graph two equations as so.
Now that we have the corresponding system of equations, we can graph both.
Let's graph them on Desmos on the same plane. The answers are attached. Red is the first equation, blue is the second.
What determines the solutions of a system of equations?
-No solution: the two lines will never intersect/does not intersect ever, which means that there are no set point where it satisfies both equations.
-One solution: the two lines intersect once and only once, meaning that there is that one set point where the x and y values both satisfy both equations.
-Multiple solutions: the two lines will intersect each other multiple times, meaning there are multiple set points where the x and y values satisfy both equations. You usually will not have to worry about these problem sets.
-Infinite solutions: the two lines are both the same line, which means every x and y value will satisfy both equations.
Looking at the solution attached, we can see that there are no places where a system of equations intersect, therefore ruling it that they have no solution.
And to answer your last question, an asymptote is a imaginary line in which a equation can approach closely and closely, but will never touch that imaginary line. Think of a line at y=. When you graph it, you can see that the two lines never ever EVER intersect the y-axis, or x=0; giving that x=0 as a vertical asymptote.
The Jacobian for this transformation is
with determinant , hence the area element becomes
Then the integral becomes
where is the unit circle,
so that
Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.
Then