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vovangra [49]
3 years ago
5

Name four fractions between 5/11 and 5/6

Mathematics
2 answers:
liq [111]3 years ago
8 0
5/10, 5/9, 5/8, 5/7.

These are all between 5/11 and 5/6.
Allisa [31]3 years ago
7 0

Answer:

6/11,  20/33,  2/3,  25/33  (answers vary)

Step-by-step explanation:

First, we need to convert both of these fractions to the same denominator. To do that, we need to find the least common denominator for both fractions.

6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72

11: 11, 22, 33, 44, 55, 66, 77

The LCM of 6 and 11 is 66. (6 * 11 = 66)

Now, we multiply 5/6 by 11/11 and 5/11 by 6/6.

\frac{5}{6} *\frac{11}{11}=\frac{55}{66}\\\\\\\frac{5}{11} *\frac{6}{6} =\frac{30}{66}

Now, we can just choose four numbers between 30 and 55 and get our answer (remember to simplify!).

\frac{36}{66} =\frac{6}{11}

\frac{40}{66}=\frac{20}{33}

\frac{44}{66} =\frac{2}{3}

\frac{50}{66} =\frac{25}{33}

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A system contains n atoms, each of which can only have zero or one quanta of energy. How many ways can you arrange r quanta of e
My name is Ann [436]

Answer:

\mathbf{a)} 2\\ \\ \mathbf{b)} 184 \; 756 \\ \\\mathbf{c)}  \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}

Step-by-step explanation:

If the system contains n atoms, we can arrange r quanta of energy in

                         \binom{n}{r} = \dfrac{n!}{r!(n-r)!}

ways.

\mathbf{a)}

In this case,

                                n  = 2, r=1.

Therefore,

                    \binom{n}{r} = \binom{2}{1} = \dfrac{2!}{1!(2-1)!} = \frac{2 \cdot 1}{1 \cdot 1} = 2

which means that we can arrange 1 quanta of energy in 2 ways.

\mathbf{b)}

In this case,

                                n  = 20, r=10.

Therefore,

                    \binom{n}{r} = \binom{20}{10} = \dfrac{20!}{10!(20-10)!} = \frac{10! \cdot 11 \cdot 12 \cdot \ldots \cdot 20}{10!10!} = \frac{11 \cdot 12 \cdot \ldots \cdot 20}{10 \cdot 9 \cdot \ldots \cdot 1} = 184 \; 756

which means that we can arrange 10 quanta of energy in 184 756 ways.

\mathbf{c)}

In this case,

                                n = 2 \times 10^{23}, r = 10^{23}.

Therefore, we obtain that the number of ways is

                    \binom{n}{r} = \binom{2\times 10^{23}}{10^{23}} = \dfrac{(2\times 10^{23})!}{(10^{23})!(2\times 10^{23} - 10^{23})!} = \dfrac{(2\times 10^{23})!}{(10^{23}!)(10^{23})!}

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Triangle 4 is <span> is congruent to ΔABC by the ASA

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What binomial must he added to (3r+14) to make the sum of the two polynomials equal (8r-6)
rodikova [14]

Answer:

(5r - 20)

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Answer:

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u have to substitute -3 in the question

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