Consider reflections of AJKL.
1 answer:
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Let
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer
we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0
using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3
the answer is
the first odd integer x is 1
the second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer
we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0
using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3
the answer is
the first odd integer x is 1
the second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5