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kramer
3 years ago
7

Which set of ordered pairs represents a function?

Mathematics
2 answers:
mel-nik [20]3 years ago
6 0
The one that doesn't have an "x" repeated, that is an X-REPEAT
notice, the pair is just an x,y  pair.. .so if the "x" is repeated, is NOT a function

let's see

\bf \{(2, -2), (\boxed{1}, 5), (-2, 2), (\boxed{1}, -3), (8, -1)\}\impliedby \textit{has an "x" rep}eated\\\\
\{(3, -1), (7, 1), (-6, -1), (9, 1), (2, -1)\}\\\\
\{(6, 8), (5, 2), (\boxed{-2}, -5), (1, -3), (\boxed{-2}, 9)\}\impliedby \textit{has an "x" rep}eated\\\\
\{(\boxed{-3}, 1), (6, 3), (\boxed{-3}, 2), (\boxed{-3}, -3), (1, -1)\}\impliedby \textit{an "x" rep}eated
Agata [3.3K]3 years ago
5 0

Answer:

The second set of ordered pairs :

{(3,-1),(7,1),(-6,-1),(9,1),(2,-1)}

Step-by-step explanation:

A function in IR2 (coordinate plane) assigns to a value of ''x'' an unique value of ''y''.

For example, the function f(x)=x+3.

When x=2 ⇒ f(2)=2+3=5

Therefore, the ordered pair (2,5) belongs to the function.

For each value of ''x'' we will have an unique value of ''y''.

For example, the ordered pairs (4,5) and (4,8) don't belong to any function because the value ''4'' can't be assign to ''5'' and ''8'' at the same time.

On the other hand, we can have the same value of ''y'' assigns to different values of ''x''. For example , (3,0) and (7,0) are correct ordered pairs for the function y=0.

In order to answer the question, we will discard the sets in which two or more ordered pairs have the same value of ''x'' assign to a different value of ''y''.

Examining the sets we find that the first set has the ordered pairs (1,5) and (1,-3) so it is not a valid set of ordered pairs which represents a function.

The second set is correct.

The third set has (-2,-5) and (-2,9) so it doesn't represent a function.

We will have the same problem in the fourth set with the ordered pairs (-3,1) , (-3,2) and (-3,-3).

The only correct set is the second one.

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We calculate the average per week, this way:

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In consequence the multiplication equation that represents this situation is:

15w + 30.75w = 183, where w is the number of weeks Mrs. Hawkins' classes sold calendars.

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We already know the average number of calendars that 1st and 2nd period classes sold per week. Now, let's calculate the average for 3nd 4th peiod classes, this way:

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7 0
1 year ago
Show 2 different solutions to the task.
laila [671]

Answer with Step-by-step explanation:

1. We are given that an expression n^2+n

We have to prove that this expression is always is even for every integer.

There are two cases

1.n is odd integer

2.n is even integer

1.n is an odd positive integer

n square is also odd integer and n is odd .The sum of two odd integers is always even.

When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .

2.When n is even positive integer

Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.

When n is negative even integer

n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.

Hence, the given expression is always even for every integer.

2.By mathematical induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then k^2+k is even integer

We shall prove that it is true for n=k+1

(k+1)^1+k+1

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=k^2+k+2k+2

=Even +2(k+1)[/tex] because k^2+k is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.

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