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3 years ago

If k stands for an integer, then is it possible for k2 + k to stand for an odd integer? Be prepared to justify your answer.

Mathematics

1 answer:

BartSMP [9]3 years ago

5 0

Answer:

k^2 + k never stands for an odd integer

Step-by-step explanation:

Let us consider either case with which k stands for an odd or even integer;

Case 1: k is an odd integer

For integer a, k = 2a + 1

So, k + 1 = 2a + 2 = 2( a + 1 ) = 2b for integer b

k^2 + k = k ( k + 1 ) = k ( 2b ) = 2kb = 2c for integer c,

Therefore, if k is an odd integer, then k^2 + k is an even integer ;

Case 2: k is an even integer

For an integer a, k = 2a

So, k + 1 = 2a + 1

k^2 + k = k( k+1 ) = 2a( 2a + 1 ) , multiple of 2

Therefore, if k is an even integer, then k^2 + k is an even integer;

This would make k^2 + k never stand for an odd integer

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3 years ago

One mixture contains 6 fluid Ounces of water and 10 fluid Ounces of vinegar. A second mixture

Nadya [2.5K]

Answer:

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its =3

Step-by-step explanation:

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Therefore, the ratio of water to vinegar in the first mixture is 6 : 10 = 3 : 5

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