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notka56 [123]
3 years ago
7

How many five digit numbers can be formed from the digits 1,,3,4,5,6,7 if no digit is repeated

Mathematics
2 answers:
zhuklara [117]3 years ago
6 0

There can be up to 36 i believe

Hope this helps


Viktor [21]3 years ago
3 0

<em>To get all possible permutations of these five spaces, we just multiply these numbers (that is, 8∗7∗6∗5∗4=6,720)</em>

<em>There are 6,720 five digit numbers that can be formed from the digits provided given that no digit may be repeated.</em>

<em>Hope that helped!</em>

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tino4ka555 [31]
We will use demonstration of recurrences<span>1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation </span>10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, <span>for all positive integers n>=1
</span>3) let's show that the equation<span> is also true for n+1, n>=1
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3 years ago
If the simple interest on ​$4000 for 3 years is ​$720​, then what is the interest​ rate?
solong [7]

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0.06% annually

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2 years ago
Solve the systems of equations by the elimination method:<br> -2x + 2y = 2<br> -4y - 2x = -22
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Step by step explanation:
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3 years ago
Help? I can't seem to understand arithmegic sequence.
katrin2010 [14]

A sequence \{a_n\} is arithmetic if the difference between consecutive terms is some fixed number, regardless of which pair of consecutive terms you pick out of the sequence.

For example, the following sequences are arithmetic:

1, 2, 3, 4, 5, 6, ... (difference = 1)

-25, -20, -15, -10, -5, ... (difference = 5)

2. Carla's sequence is not arithmetic, because the differences between consecutive terms are all different:

13 - 11 = 2

17 - 13 = 4

25 - 17 = 8

She can adjust the sequence by changing the last two numbers to 15 and 17, since this makes the difference fixed:

13 - 11 = 2

15 - 13 = 2

17 - 15 = 2

and so on.

3. The sequence

45, 48, 51, 54, ...

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a_n=a_{n-1}+3

By this definition, we can just as easily write the (n-1)th term in terms of the (n-2)th term:

a_{n-1}=a_{n-2}+3

Then, substituting this into the previous equation, we have

a_n=(a_{n-2}+3)+3=a_{n-2}+2\cdot3

We can continue this process to write a_n in terms of a_1:

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a_n=a_1+(n-1)\cdot3

The first term in this sequence is a_1=45, so we have

a. a_n=45+3(n-1)=42+3n

where n=1,2,3,\ldots.

b. You can fill in the blanks by just adding 3 to the previous term:

45, 48, 51, 54, <u>57</u>, 60, <u>63</u>, 66, 69, ...

Then, using the formula found in (a), the 15th term of the sequence is

a_{15}=42+3\cdot15=87

4 0
2 years ago
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