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Helen [10]
3 years ago
5

How many distinguishable permutations of the letters in DECEMBER are possible?

Mathematics
2 answers:
Lyrx [107]3 years ago
7 0
5 percent off the floor in the morning and
Genrish500 [490]3 years ago
7 0

#1

avatar+17746

+10

This can be calculated by using a fraction:

Count the number of letters in the word (in this case, 9) and create the numerator by the factorial of that number.

Since there are 9 letters in the word COMMITTEE, the numerator is  9!

Count the number of times each different letter is used and create the numerator by multiplying together the factorials of those numbers.  OK, that description may be clear as mud, so here's an example:

For COMMITTEE, the C is used once:  1!, the O is used once: 1!, the M is used twice: 2!, the I is used once: 1!, the T is used twice: 2!, and the E is used twice: 2!.

Putting these together, the denominator is  1!·1!·2!·1!·2!·2!  

So the answer is:  9! / [ 1!·1!·2!·1!·2!·2!  ]  =  45360

(A couple things about the denominator:  you really don't need to write down all the 1!, because their values are 1, but if you do, the sum of the numbers used in the denominator is equal to the number of letters in the word.)

Another example:  MISSISSIPPI  =  11! / [ 1!·4!·4!·2! ]


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Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)
vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

x^{\frac{2}{3}}+y^{\frac{2}{3}}=4

We have to find the equation of tangent line to the given curve at point (-3\sqrt3,1)

By using implicit differentiation, differentiate w.r.t x

\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0

Using formula :\frac{dx^n}{dx}=nx^{n-1}

\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}

\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}

\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}

Substitute the value x=-3\sqrt3,y=1

Then, we get

\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}

\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}

Slope of tangent=m=\frac{1}{\sqrt3}

Equation of tangent line with slope m and passing through the point (x_1,y_1) is given by

y-y_1=m(x-x_1)

Substitute the values then we get

The equation of tangent line is given by

y-1=\frac{1}{\sqrt3}(x+3\sqrt3)

y-1=\frac{x}{\sqrt3}+3

y=\frac{x}{\sqrt3}+3+1

y=\frac{x}{\sqrt3}+4

This is required equation of tangent line to the given curve at given point.

8 0
3 years ago
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