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salantis [7]
3 years ago
11

On a coordinate plane, solid circles appear at the following points: (negative 3, 2), (negative 3, 3), (1, 4), (1, negative 4),

(2, negative 4), (2, negative 2).
How many points need to be removed from this graph so that it will be a function?

1 point
2 points
3 points
4 points
Mathematics
2 answers:
hammer [34]3 years ago
7 0

Answer:

3 points

Step-by-step explanation:

A function is a relation that maps elements of a set for the inputs (the domain, usually denoted with the variable x) into elements of another set, the set of the outputs (the range, usually denoted with the variable y).

So the relation can be represented with points (x, y), which means that point x is being mapped into point y.

Such that the needed condition is that each element of the domain is mapped into only one element in the domain.

So for example, if for a relationship we have the points (a, b) and (a, c)

The point a is being mapped into two different outputs.

Now let's go to our problem, we have the points:

(-3, 2), (-3, 3), (1, 4), (1, -4), (2, -4), (2, -2)

If we have repeated inputs, we need to remove points until we have one of each.

We can see that the inputs -3, 1, and 2 are repeated one time.

Then we need to remove one of each, for example if we remove the second, fourth and sixth points, the set becomes:

(-3, 2), (1, 4), and (2, -4)

This is a function.

Then we need to remove 3 points.

vaieri [72.5K]3 years ago
3 0

Answer:

C) 3 POINTS

Step-by-step explanation:

ON EDGE 2021

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8 0
2 years ago
A) how many ways are there to choose 10 players to take the field?
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A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.

B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4. (13−10)! 3!

C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =

10! 3!

9!1! 1!2! 10! 3!

8!2! 2!1! 10! 3!

7!3! 3!0!

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<span>possible choices.</span><span> 
</span>

  


5 0
4 years ago
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