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ArbitrLikvidat [17]
2 years ago
10

For which k is it true that the sum of k consecutive integers is always divisible by k?

Mathematics
1 answer:
nevsk [136]2 years ago
5 0

Answer:

Step-by-step explanation:

We have two consecutive even integers:

a and b.

We know that we can write a even integer as:

a = 2*n

where n is an integer.

Then the consecutive integer to a, b, will be:

b = 2*n + 2.

Then the sum of a and b is:

a + b = 2n + 2n + 2 = 4*n + 2 = 2*(2n + 1) = k.

The next two consecutive integers greater than b will be:

c = 2n + 4

d = 2n + 6

then:

c + d = (2n + 4) + (2n + 6) = 4n + 10 = k + 4.

I hope i helped:D

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Let the initial point of the vector be (x,y). Then the magnitude of the vector v can be written as:

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The magnitude of vecor v is given to be 10. So we can write:

10= \sqrt{ (-1-x)^{2} + (5-y)^{2} } \\ \\ 100=(-1-x)^{2} + (5-y)^{2}

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Answer:

10% probability that a given class period runs between 51.25 and 51.75 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

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