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3 years ago

How many ways are there to arrange the first five letters of the alphabet?

Mathematics

2 answers:

stealth61 [152]3 years ago

7 0

Answer:

120 ways

Explanation:

Since there are no repeating letters, and there are 5 total letters, there are 5!= 120 ways to arrange them. In other words, there are 5 slots to place the first letter in, then 4 slots for the second letter, 3 for the third, 2 for the fourth, and then the last letter goes in the 1 slot that is left.

I hope this is correct and if it is I hope it helps. I hope you have a great day and remember it doesn't matter what other people say or think about you, you are still and always will be strong, smart, and an amazing person. Have a wonderful day.

I am happy to help with anything.

Nonamiya [84]3 years ago

6 0

In probability, problems involving arrangements are called combinations or permutations. The difference between both is the order or repetition. If you want to arrange the letters regardless of the order and that there must be no repetition, that is combination. Otherwise, it is permutation. Therefore, the problem of arrange A, B, C, D, and E is a combination problem.

In combination, the number of ways of arranging 'r' items out of 'n' items is determined using n!/r!(n-r)!. In this case, you want to arrange all 5 letters. So, r=n=5. Therefore, 5!/5!(505)! = 5!/0!=5!/1. It is simply equal to 5! or 120 ways.

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A classmate simplified a rational expression below

lara31 [8.8K]

<h2>Part a) </h2><h2>Explain the error in this simplification.</h2>

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<u><em>Identifying the Main Error</em></u>

$1-\frac{2}{x-2}=\frac{x+1}{x+2}$

$1-2\left(x+2\right)=\left(x+1\right)\left(x-2\right)$ ← ERROR Starts here

<u><em>Here is the Explanation of the Error </em></u>

<u><em /></u> $\mathrm{The\:equation\:should\:have\:been\:Multiplied\:by\:LCM=}\left(x-2\right)\left(x+2\right)$ <em>. </em>In your case you wrongly multiply the equation.

<em><u>CORRECTION</u></em>

<em>HERE IS HOW YOU SHOULD HAVE MULTIPLIED BY </em><em>LCM = (x-2)(x+2):</em>

<em />

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Here is the complete correction of the error.

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3 years ago

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PilotLPTM [1.2K]

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Read 2 more answers

Use the definition of continuity to determine whether f is continuous at a.

dmitriy555 [2]

$f(x)$ will be continuous at $x=a=7$ if
(i) $\displaystyle\lim_{x\to7}f(x)$ exists,
(ii) $f(7)$ exists, and
(iii) $\displaystyle\lim_{x\to7}f(x)=f(7)$ .

The second condition is immediate, since $f(7)=8918$ has a finite value. The other two conditions can be established by proving that the limit of the function as $x\to7$ is indeed the value of $f(7)$ . That is, we must prove that for any $\varepsilon>0$ , we can find $\delta>0$ such that

$|x-7|$

Now,

$|f(x)-f(7)|=|5x^4-9x^3+x-8925|$

Notice that when $x=7$ , we have $5x^4-9x^3+x-8925=0$ . By the polynomial remainder theorem, we know that $x-7$ is then a factor of this polynomial. Indeed, we can write

$|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|$

This is the quantity that we do not want exceeding $\varepsilon$ . Suppose we focus our attention on small values $\delta$ . For instance, say we restrict $\delta$ to be no larger than 1, i.e. $\delta\le1$ . Under this condition, we have

$|x-7|$

Now, by the triangle inequality,

$|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275$

If $|x|$ , then this quantity is moreover bounded such that

$|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955$

To recap, fixing $\delta\le1$ would force $|x|$ , which makes

$|x-7||5x^3+26x^2+182x+1275|$

and we want this quantity to be smaller than $\varepsilon$ , so

$6955|x-7|$

which suggests that we could set $\delta=\dfrac{\varepsilon}{6955}$ . But if $\varepsilon$ is given such that the above inequality fails for $\delta=\dfrac{\varepsilon}{6955}$ , then we can always fall back on $\delta=1$ , for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set $\delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}$ .

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that $f(x)$ is indeed continuous at $x=7$ .

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3 years ago