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Ne4ueva [31]
2 years ago
15

Simplify without the absolute value |x+3| ; if x>2

Mathematics
1 answer:
inna [77]2 years ago
8 0

Answer:

x+3 if x>2

Step-by-step explanation:

Recall some following basic ideas about absolute value:

|0|=0 or I guess -(0)=0

|3|=3

|-3|=-(-3) or just 3 but there is a reason why I wrote it the way I did.

You see that the absolute value returned the same value if the value was positive.

You see that the absolute value returned the opposite value if the value was negative.

You can include 0 in either.

Let's look at |x+3| alone for a second.

*|x+3|=x+3 if x+3 is positive or zero.

When is x+3 positive or zero? [greater than or equal to 0]

x+3\ge 0

Subtract 3 on both sides

x\ge -3

*|x+3|=-(x+3) if x+3 is negative or zero.

When is x+3 negative or zero? [less or equal to 0]

x+3\le 0

Subtract 3 on both sides

x\le -3

So looking at the bullet points (the *), x>2 fits into the first inequality.

Therefore, |x+3|=x+3.

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Answer:

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This is a Combination problem.  

Combination is a branch of mathematics that deals with the problem relating to the number of iterations which allows one to select a sample of elements which we can term "<em>r</em>" from a collection or a group of distinct objects which we can name "<em>n</em>". The rules here are that replacements are not allowed and sample elements may be chosen in any order.

Step-by-step explanation:

Step I

The formula is given as

C (n,r) = \frac{n}{r} = \frac{n!}{(r!(n-r)!)}

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Step 2 - Insert Figures

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Step 3

The total number of ways a recorder, a facilitator and a questioner can be chosen in a club containing 14 members therefore is 364.

Cheers!

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