The second partial derivatives of 
exist everywhere in its domain. By Schwarz's theorem, the mixed second-order partials derivatives are equal:
Then
We don't actually need to compute 
and 
to know that they are both free of 
. So when we differentiate for a second time with respect to , the whole thing cancels out.
First get all the like terms on the same side and combine like terms
4x+2x+2=3x-7
4x+2x-3x=-2-7
3x=-9
Simplify
X=-3
Well, to find a circles circumference, C=D* pi. if d =10, 10 * pi (or 3.14 for now), C= 31.4. so the answer is false. It is measuring distance as a radius instead of diameter.
Answer:
D-Quadrilateral A'B'C'D' will be congruent to quadrilateral ABCD.
Step-by-step explanation:
The scale factor is equal to 1. Hence, the quadrilateral A'B'C'D' will be identical to quadrilateral ABCD, whose vertices are the same.
The algebraic expression of the dilation is:
Besides, the coordinates of the new quadrilateral are A'(3, 3), B'(5, 3), C'(7, 1), and D'(1, 1)
And quadrilateral A'B'C'D' will be congruent to quadrilateral ABCD.
Therefore, the right answer is D.
9514 1404 393
Answer:
x = 2
Step-by-step explanation:
Using the scale factor, we have ...
x/8 = 0.25/1
x = 8(0.25) . . . . multiply by 8
x = 2
_____
<em>Additional comment</em>
I like to write proportions with the variable in the numerator. That way they are easily solved by a single multiplication. The trick is to get corresponding parts in the same positions. Here, the fractions are written ...
B/A = smaller part / larger part = x/8 = 0.25/1
