A set of values A can have a relation with another set B as long as at least one element of A has at least one image in B. Functions are special relations where each element of A (the domain of the function) has one and only one image on B (the range of the function).

By looking at the options, we can see that x=9, x=-8, and x=-1 already have defined values in Y, so if we define another value for any of them the relation will stop being a function. The only possible choice to preserve the function is the option

$\boxed{A)\ (-5,7)}$

8 0

3 years ago

Simplify: −z^3+5k^6−(−z^3+10k^6)

Georgia [21]

-z³ + 5k^6 + z³ -10k^6

(-z³ cancels out with z³)

5k^6 -10k^6

(then subtract)

Answer is -5k^6

6 0

3 years ago

3 to the power of 2+12÷(6-3)×8.

larisa86 [58]

3^2+12/(6-3)*8
3^2+12/3*8
9+12/3*8
9+4*8
9+32
41

Final answer: 41
Remember to follow pemdas, parentheses, exponents, multiplication and division, followed by addition and subtraction going from left to right.

7 0

2 years ago