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RideAnS [48]
1 year ago
7

A troupe of 12 dancers is going to line up on stage. Of these dancers, 6 are wearing green and 6 are wearing blue. A dancer wear

ing green must be in the first position and they must alternate between colors. In how many ways can the dancers line up?
Mathematics
1 answer:
Wittaler [7]1 year ago
4 0

The number of ways that the dancers can line up is; 518400 ways

<h3>What is the Fundamental Counting Theorem?</h3>

The fundamental counting theorem is a theorem that states that if there are n things, each with  ways to be done, each thing independent of the other, the number of ways they can be done is:

N = n1 * n2 * n3.....*nₙ

Total = 12

Dancers on Green = 6

Dancers on blue = 6

Number of ways for those wearing green = 6!

Dancers wearing green must be in the first position and so 6 places are left for the blue dancers.

Number of ways to arrange those wearing blue = 6!

Thus;

Number of ways that the dancers can line up = 6! * 6! = 518400 ways

Read more about Fundamental Counting Theorem at; brainly.com/question/26814852

#SPJ1

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Blizzard [7]

Answer:

the correct answer is , 4(20+3)

Step-by-step explanation:

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40% =32/ what is the full fraction
Levart [38]
40% = 32/x

Rewrite using a decimal

.40 = 32 / x

Multiply both sides by x
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Divide both sides by .40
x = 32 / .40
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Write the full fraction
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Mikaela purchased a paddleboard originally priced at $699. If all merchandise was on sale for 40% off and the state tax rate is
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299.87

Step-by-step explanation:

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Liula [17]
Your proportion should look something like this

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6 0
3 years ago
Read 2 more answers
Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^
Gemiola [76]

Answer:

\rm \displaystyle y' =   2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x}

Step-by-step explanation:

we would like to figure out the differential coefficient of e^{2x}(1+\ln(x))

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

\displaystyle y =  {e}^{2x}  \cdot (1 +   \ln(x) )

to do so distribute:

\displaystyle y =  {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x}

take derivative in both sides which yields:

\displaystyle y' =  \frac{d}{dx} ( {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x} )

by sum derivation rule we acquire:

\rm \displaystyle y' =  \frac{d}{dx}  {e}^{2x}  +  \frac{d}{dx}   \ln(x)  \cdot  {e}^{2x}

Part-A: differentiating $e^{2x}$

\displaystyle \frac{d}{dx}  {e}^{2x}

the rule of composite function derivation is given by:

\rm\displaystyle  \frac{d}{dx} f(g(x)) =  \frac{d}{dg} f(g(x)) \times  \frac{d}{dx} g(x)

so let g(x) [2x] be u and transform it:

\displaystyle \frac{d}{du}  {e}^{u}  \cdot \frac{d}{dx} 2x

differentiate:

\displaystyle   {e}^{u}  \cdot 2

substitute back:

\displaystyle    \boxed{2{e}^{2x}  }

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

\displaystyle  \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)

let

  • f(x) \implies   \ln(x)
  • g(x) \implies    {e}^{2x}

substitute

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =  \frac{d}{dx}( \ln(x) ) {e}^{2x}  +  \ln(x) \frac{d}{dx}  {e}^{2x}

differentiate:

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =   \boxed{\frac{1}{x} {e}^{2x}  +  2\ln(x)  {e}^{2x} }

Final part:

substitute what we got:

\rm \displaystyle y' =   \boxed{2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x} }

and we're done!

6 0
3 years ago
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