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3 years ago

Find the number of permutations in the word “freezer".

2 answers:

7 0

There are seven letters in the word "FREEZER"
so, it would be: ⁷P⁶ = 7! / (7-6)! = 7! / 1!
= 7 * 6 * 5 * 4 * 3 * 2
= 5040

In short, Your Answer would be 5040

Hope this helps!

Paladinen [302]3 years ago

4 0

Answer:

The answer is actually 420. the last person was wrong!

Step-by-step explanation:

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Morgarella [4.7K]

Answer:

$f'(x)=-\frac{2}{x^\frac{3}{2}}$

Step-by-step explanation:

So we have the function:

$f(x)=\frac{4}{\sqrt x}$

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Therefore, our derivative would be:

$\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}$

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

$=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}$

Place the 4 in front:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}$

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})$

Distribute:

$=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}$

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

$=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }$

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

$= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}$

The numerator will use the difference of two squares. Thus:

$=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Simplify the numerator:

$=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Both the numerator and denominator have a h. Cancel them:

$=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Now, substitute 0 for h. So:

$=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})$

Simplify:

$=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})$

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

$=4( \frac{-1}{(x)(2\sqrt{x})})$

Multiply across:

$= \frac{-4}{(2x\sqrt{x})}$

Reduce. Change √x to x^(1/2). So:

$=-\frac{2}{x(x^{\frac{1}{2}})}$

Add the exponents:

$=-\frac{2}{x^\frac{3}{2}}$

And we're done!

$f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}$

5 0

3 years ago

Can someone help me please

docker41 [41]

0 there are no odd number 1-15

5 0

3 years ago

Read 2 more answers

The first three terms of a sequence are given. Round to the nearest thousandth (if necessary).

lions [1.4K]

10, 10, 20 or either of u talking abt about tah the whole either 0-100

7 0

2 years ago

I’m confused on questions c-f

Nadya [2.5K]

I gotchu !! just give me a moment to write it out

4 0

3 years ago

Read 2 more answers

Adrian's CD player can hold six disks at a time and shuffles all of the albums and their songs. If he has thirteen CD's, how man

EastWind [94]

To answer this problem, the utilization of the provided formula is of utmost importance.

The formula to be used would be for the combination with no repetition since it is what is asked and possibly happens when involving the products mentioned in the problem. The formula would be

n! where in
------------- n = number of things to choose from
(n - r)! r! r = number of things needed to form

the symbol ! is called a factorial function which replaces the sequence of multiplying a number in a descending order.

lets substitute and solve

13!
--------------
(13 - 6)! 6!

13!
-------------
(7!) 6!

6,227,020,800
-------------------
5,040 x 720

6,227,020,800
-------------------
3,628,800

1,716

The answer would be 1,716.

6 0

3 years ago