Respuesta :
bearing in mind that, if you multiply any integer whatsoever, by 2, you get an even number, 3*2 or 17*2 or 4*2 and so on.
to get an ODD integer, simply hop once forward of backwards from any even integer, 2-1, or 2+1 or 8+1 or 8-1.
to get a consecutive ODD integer, simply hop twice from any odd integer, say 3+2, or 7+2 and so on.
now, let's pick a reference integer, say hmmm "a".
we know that a * 2, or 2a is an even integer, therefore (2a + 1) is an ODD integer, and (2a +1 ) + 2 is a consecutive odd integer. So, there you have it, our first two odd integers, and the next one, is, you guessed it, (2a + 1) + 2 + 2
so, our three odd consecutive integers are (2a+1), (2a+3) and (2a+5)
so, the triangle has a perimeter of 27, so, let's add them up then
[tex]\bf (2a+1)+(2a+3)+(2a+5)=27\implies 6a+9=27 \\\\\\ 6a=18\implies a=\cfrac{18}{6}\implies a=3[/tex]
so, our first odd integer is
2a + 1
2(3) + 1 --> 7
and surely you know what the other two are.
to get an ODD integer, simply hop once forward of backwards from any even integer, 2-1, or 2+1 or 8+1 or 8-1.
to get a consecutive ODD integer, simply hop twice from any odd integer, say 3+2, or 7+2 and so on.
now, let's pick a reference integer, say hmmm "a".
we know that a * 2, or 2a is an even integer, therefore (2a + 1) is an ODD integer, and (2a +1 ) + 2 is a consecutive odd integer. So, there you have it, our first two odd integers, and the next one, is, you guessed it, (2a + 1) + 2 + 2
so, our three odd consecutive integers are (2a+1), (2a+3) and (2a+5)
so, the triangle has a perimeter of 27, so, let's add them up then
[tex]\bf (2a+1)+(2a+3)+(2a+5)=27\implies 6a+9=27 \\\\\\ 6a=18\implies a=\cfrac{18}{6}\implies a=3[/tex]
so, our first odd integer is
2a + 1
2(3) + 1 --> 7
and surely you know what the other two are.
