Group points {(0,1),(0,5)(2,6),(3,3)}is not a function, but group points {(1,4)(2,7)(3,1)(5,7)} is a function. What do you notic
Tasya [4]
<h2>Any value in the domain of the function should have a unique value in codomain.</h2>
Step-by-step explanation:
In the first set of points 
,
value 
maps to two distinct values 
in the codomain.
This violates the property of functions.
The first set of points does not form a function.
In the second set of points 
,
Every value in domain corresponds to unique value in domain.
There is no violation in the property of functions.
The second set of points does form a function.
Answer:
Men it's a really long process and I've just woken up men
Answer:
2 imaginary roots
Step-by-step explanation:
The discriminate is the portion of the quadratic equation under the square root
D = b^2 - 4ac
D = 8^2 - 4(2)(16)
D = 64 - 128
D = -64
Since the discriminate is negative the square root will be imaginary
and there will be 2 of them
2 imaginary roots
Answer:
Step-by-step explanation:
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined. y=6 is a straight line perpendicular to the y-axis at point (0,6) , which means that the range is a set of one value {6} .
Z= -7
So your answer is number 2, -7
