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

Based on the characteristics of <em>linear</em> and <em>piecewise</em> functions, the <em>piecewise</em> function
is shown in the graph attached herein. (Correct choice: A)
<h3>How to determine a piecewise function</h3>
In this question we have a graph formed by two different <em>linear</em> functions. <em>Linear</em> functions are polynomials with grade 1 and which are described by the following formula:
y = m · x + b (1)
Where:
- x - Independent variable.
- y - Dependent variable.
- m - Slope
- b - Intercept
By direct observation and by applying (1) we have the following <em>piecewise</em> function:

Based on the characteristics of <em>linear</em> and <em>piecewise</em> functions, the <em>piecewise</em> function
is shown in the graph attached herein. (Correct choice: A)
To learn more on piecewise functions: brainly.com/question/12561612
#SPJ1
4 in (4,-5) implies positive x-axis.
-5 in (4,-5) implies negative y-axis.
Since x is positive and y is negative in fourth quadrant, so the answer is
QUADRANT IV (or 4th quadrant.
Answer:
56
Step-by-step explanation:
degree of freedom is n - 1, n being the sample size
One sample of 57 participants in a study (given)
For this study, n = 57
Using the degree of freedom formula, 57 - 1 = 56
Hope it helps, Let me know if yoy have any more questions/concerns !
Have a nice rest of your day :)