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Alecsey [184]
3 years ago
7

Brayden is simplifying the expression 4n – (n – 3)2. He begins by distributing the negative to the terms inside the parentheses.

What should he have done instead?
Mathematics
2 answers:
seropon [69]3 years ago
4 0

ANSWER

Brayden should have expanded the parenthesis first.

EXPLANATION

The expression Brayden is trying to simplify is

4n -  {( n - 2)}^{2}

To simplify this expression, the first thing to do is to expand the perfect square to obtain:

4n -  {( n}^{2}  - 4n + 4)

We can now distribute the negative to the terms in the parenthesis.

4n -  { n}^{2}   + 4n  - 4

vladimir2022 [97]3 years ago
3 0

Answer:

He should have simplified (n-3)^2 first then distribute the negative sign.

Step-by-step explanation:

Given that Brayden is simplifying the expression 4n-(n-3)^2. He begins by distributing the negative to the terms inside the parentheses. Which is basically the wrong step for the given problem.

Now we need to find about what should he have done instead.

According to order of operations, we should begin with exponent first.

So that means he should have simplified (n-3)^2 first then distribute the negative sign.

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A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

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In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

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Now we add together the combinations

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Note: There is also an equation for permutations which is:

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Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








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