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SpyIntel [72]
1 year ago
11

A restaurant offers a special pizza with any 5 toppings. if the restaurant has 16 topping from which to choose, how many differe

nt special pizzas are possible?
Mathematics
1 answer:
Mariana [72]1 year ago
5 0

By permutation without repetition, there are 524160 possible different special pizzas.

The possible special pizzas can be calculated with permutation without repetition. The formula of permutation without repetition can be written as

P = n! / (n - k)!

where P is all of the possible combinations, n is the number of objects or elements, and k is how many numbers should be chosen.

From the question above, we know that :

n = 16

k = 5

By substituting the parameters, we can determine all of the possible different pizzas

P = n! / (n - k)!

P = 16! / (16 - 5)!

P = 16! / 11!

P = 16 x 15 x 14 x 13 x 12 x 11! / 11!

P = 524160

Hence, there are 524160 possible different special pizzas

Find out more on permutation without repetition at: brainly.com/question/1216161

#SPJ4

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Answer:

The answer is

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<h3>x - 1 =  \frac{2y}{3}</h3>

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Divide both sides by 2 to make y stand alone

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<h3>\frac{2y}{2}  =  \frac{3x - 3}{2}</h3>

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<h3>y =  \frac{3x - 3}{2}</h3>

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andrew-mc [135]

The remainder from the division of the algebraic equation is -53/8.

<h3>What is the remainder of the algebraic expression?</h3>

The remainder of the algebraic expression can be determined by using the long division method.

Given that:
\mathbf{f(x) = \dfrac{x^3 - 6x^2 + 3x - 1}{2x-3}}

where:

  • The divisor = 2x -3

Using the long division method, we have:

\mathbf{= \dfrac{x^2}{2} +\dfrac{-\dfrac{9x^2}{2}+3x -1 }{2x-3}}

\mathbf{= \dfrac{x^2}{2}-\dfrac{9x}{4} +\dfrac{-\dfrac{-15x}{4}-1 }{2x-3}}

\mathbf{= \dfrac{x^2}{2}-\dfrac{9x}{4} -\dfrac{15}{8}+\dfrac{-\dfrac{53}{8} }{2x-3}}

\mathbf{= \dfrac{x^2}{2}-\dfrac{9x}{4} -\dfrac{15}{8}-\dfrac{53 }{8(2x-3)}}

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