if x = 5 and y =2 find the value of : (x^7 + y^x)^-1
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Answered by
1
Hi,
x^2 + y = 7 … 1
x + y^2 = 11 … 2
From equation 1 it is
y = 7 - x^2
Substitute the above in eqn 2
=> x + (7 - x^2)^2 = 11
On solving the above equation we get
=> x^4 - 14x^2 + x + 38 = 0
On applying a technique known as synthetic (polynomial) division we get x =2.
substituting in eqn 1 gives y =3.
x^2 + y = 7 … 1
x + y^2 = 11 … 2
From equation 1 it is
y = 7 - x^2
Substitute the above in eqn 2
=> x + (7 - x^2)^2 = 11
On solving the above equation we get
=> x^4 - 14x^2 + x + 38 = 0
On applying a technique known as synthetic (polynomial) division we get x =2.
substituting in eqn 1 gives y =3.
Answered by
1
(5^7 + 2^5)^-1
= (78125 + 32)^-1
= (78157)^-1
= 1/78157
ps - u could have used a calculator XD
= (78125 + 32)^-1
= (78157)^-1
= 1/78157
ps - u could have used a calculator XD
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