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Ganezh [65]
3 years ago
12

Which table represents a relation that's a non-function?

Mathematics
1 answer:
bazaltina [42]3 years ago
4 0

Answer:

The table in the attachment is the right option

Step-by-step explanation:

A table that represents a function must have exactly one y-value assigned to every x-value. In other words, a table that is a function cannot have any x-value (input) with two corresponding different y-values (outputs).

The table in the attachment below represents a relation that is non-function because it has two outputs, 5 and 7, that are assigned or corresponding to one input, 7.

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Step-by-step explanation:

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Start at (12, 3). Move 2 units right and 7 units up. Where do you end up?
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Three years ago, Jolene bought $625 worth of stock in a software company. Since then the value of her
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3 years ago
43. Can you place ten lumps of sugar in three empty cups so that
erma4kov [3.2K]

Answer: No.

Step-by-step explanation:

We have 10 lumps of sugar, and we want to divide them into 3 cups, in such a way that there is an odd number of lumps in each cup.

This only can happen if we have 3 odd numbers such that the addition is equal to 10.

Now 10 is an even number, remember that even numbers can be written as:

2*k

where k is an integer number.

And odd numbers can be written as:

2*n + 1

where n is an integer.

Then we have 3 odd numbers, let's call them:

(2*n + 1), (2*k + 1) and (2*p + 1).

Now let's add them:

(2*n + 1) + (2*k + 1) + (2*p + 1).

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2*(n + k + p) + 2 + 1 =

2*(n + k + p + 1) + 1.

Now, the number n + k + p + 1 is an integer number, let's call it X, then we have that the addition of the 3 odd numbers is:

2*X + 1

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So for any 3 odd numbers that we add together, the result will always be an odd number.

Then is impossible to add 3 odd numbers and get 10 as the result (Again, 10 is an even number).

Then is not possible to have an odd number of lumps in each cup.

5 0
3 years ago
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