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Irina18 [472]
3 years ago
5

If x>7, then |x|>7. |y|>7, so y=7 Valid or invalid?

Mathematics
1 answer:
Alex3 years ago
7 0

\text{if }x>7\text{, then }|x|>7 is a valid argument

|y|>7\text{, so }y=7 is not a valid argument

For the first argument: \text{if }x>7\text{, then }|x|>7

From the definition of absolute value function

|x|=x   if x\ge0

That is every positive number is its own absolute value. Since

x>7\implies x\ge0,

we can argue that

x>7\implies |x|>7

so the first argument is valid

For the second argument: |y|>7\text{, so }y=7

From the definition of absolute value function

|y|:=\left \{ {y\text{  if }y\ge0}\atop{-y\text{  if }y

This means that

|y|>7:=\left \{ {y>7\text{  if }y\ge0}\atop{-y>7\text{  if }y

or

|y|>7:=\left \{ {y>7\text{  if }y\ge0}\atop{y

no part of the definition allow for the option y=7. So the second argument is not valid.

Learn more here: brainly.com/question/11897796

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Answer:

4032 different tickets are possible.

Step-by-step explanation:

Given : At a race track you have the opportunity to buy a ticket that requires you to pick the first and second place horses in the first two races. If the first race runs 9 horses and the second runs 8.

To find : How many different tickets are possible ?

Solution :

In the first race there are 9 ways to pick the winner for first and second place.

Number of ways for first place - ^9C_1=9

Number of ways for second place - ^8C_1=8

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Therefore, 4032 different tickets are possible.

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3 years ago
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Answer:

The answer is explained below

Step-by-step explanation:

A. How many ways can you distribute 4 different balls among 4 different boxes

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