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nevsk [136]
2 years ago
12

Using the digits 2 through 8, find the number of different 5-digit numbers such that, digits can be used more than once.

Mathematics
1 answer:
Solnce55 [7]2 years ago
7 0

Answer:

7 digits can be used for each position

There are a total of 5 positions

N = 7^5 = 16,807 numbers

You have 7 choices for the first position, second position, etc.

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Gala2k [10]

Answer:

(- 2, 4 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) , then the midpoint is

( \frac{x_{1}+x_{2}  }{2} , \frac{y_{1}+y_{2}  }{2} )

Here (x₁, y₁ ) = A (4, 6 ) and (x₂, y₂ ) = B (- 8, 2 )

midpoint = ( \frac{4-8}{2} , \frac{6+2}{2} ) = ( \frac{-4}{2} , \frac{8}{2} ) = (- 2, 4 )

6 0
2 years ago
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Kobotan [32]

Answer:

30,40,50

Step-by-step explanation:

If it is a right triangle then the pythagorean theorum would work

a^2+b^2=c^2

30^2+40^2=50^2

900+1600=2500

2500=2500

this means that 30,40,50 is the right answer

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also 30,40,50 is a pythagorean triplet

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this one came from the original triplet of 3,4,5---just multiply by 10

6 0
2 years ago
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irinina [24]

Answer:

12.35

Step-by-step explanation:

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then you need to move the decimal number 2 spaces, because you removed a total of 2 space from the original numbers.

12.35

4 0
2 years ago
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eduard

We have sequence equation a_n=a(n-1)+4.

In this case n is a natural number (1, 2, 3, ...).

So start inserting and computing value of a_n given that you know the value of n and a.

a_1=2(1-1)+4=4 (first term)

a_2=2(2-1)+4=6 \\a_3=2(3-1)+4=8 \\a_4=2(4-1)+4=10

Hope this helps.

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2 years ago
Select the choice that translates the following verbal phrase correctly to algebra: the product of k and m
romanna [79]
<em>km</em>=<em>x, </em>the product being represented by <em>x.
</em><em>
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5 0
3 years ago
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