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.

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

$450 is the answer.

step-by-step explanation:

30% of 450 = 135

450-135=315