If x=5 and y=2 then find (x^y+y^x)^-1 and (x^x+y^y)^-1
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Answered by
3
(5^2+2^5)-1
=25+32-1
=56
5^5+2^2-1
=3125+4-1
=3128
=25+32-1
=56
5^5+2^2-1
=3125+4-1
=3128
Answered by
7
Given,
x = 5 and y = 2
•°• (x^y+y^x)^-1
= 1/(x^y+y^x) [°•° a^-1=1/a]
= 1/(5^2+2^5)
=1/(25+32)
= 1/57
And,
(x^x+y^y)^-1
= 1/(x^x+y^y) [°•°a^-1=1/a]
= 1/(5^5+2^2)
= 1/ 3125+4
=1/3129
x = 5 and y = 2
•°• (x^y+y^x)^-1
= 1/(x^y+y^x) [°•° a^-1=1/a]
= 1/(5^2+2^5)
=1/(25+32)
= 1/57
And,
(x^x+y^y)^-1
= 1/(x^x+y^y) [°•°a^-1=1/a]
= 1/(5^5+2^2)
= 1/ 3125+4
=1/3129
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