For this case we have the following definitions:
A function is even if, for each x in the domain of f, f (- x) = f (x). The even functions have reflective symmetry through the y-axis.
A function is odd if, for each x in the domain of f, f (- x) = - f (x). The odd functions have rotational symmetry of 180º with respect to the origin.
We then have the following function:

Applying the definitions we have:

Answer:
The function is not odd because it is fulfilled:

Therefore, the function is even.
A system is inconsistent when there are no solutions between the two equations. Graphically, the lines will be parallel (they never meet!) and the slopes will be the same. But the y-intercepts will be different.
Let's look at the four equations, with each solved as needed, into y = mx + b form.
A: 2x + y = 5
y = 5 - 2x
y = -2x + 5
Compared to y = 2x + 5, the slopes are different, so this system won't be inconsistent. Not a good choice.
B: y = 2x + 5
Compared to y = 2x + 5, the slopes are the same and the y intercepts are the same. This system has infinitely many solutions. Not a good choice.
C: 2x - 4y = 10
-4y = 10 - 2x
-4y = -2x + 10
y = 2/4x -10/4
Here the slopes are different, so, like A this is not a good choice.
D: 2y - 4x = -10
2y = =10 + 4x
2y = 4x - 10
y = 2x - 5
Compared to y = 2x + 5 we have the same slopes and different y intercepts. The lines will be parallel and the system is inconsistent.
Thus, D is the best choice.
2/3 is the correct answer because it’s bigger then a half
<span>Together with triangles, circles comprise most of the GMAT Geometry problems.
A circle is the set of all points on a plane at the same distance from a single point ("the center").
The boundary line of a circle is called the circumference.</span>
Answer:
okay so the first one ull need to kno pemdas so start with the perenthesis (5^2-10) and then the exponent 2^3 and multyply thats by 5 and what ever answer u got in the perenthisis u divide by what u got with the second side of the problem
Step-by-step explanation:
if u need more help ill tutor u on it :)