Answer:
Let x an odd positive integer
Then, according to question
x^2 +(x+2)^2=290
2x^2 +4x−286=0
x^2 +2x−143=0
x ^2+13x−11x−143=0
(x+13)(x−11)=0
x=11 as x is positive
Hence required integers are 11 and 13
Step-by-step explanation:
Hope it is helpful....
p(even) = 3/6
p number 3 = 1/6
p (even then number 3) = 3/6 * 1/6 = 3/36
C
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