X can be any real number
As long as it is to the power of 0, it will equal one. And because 6^0 = 1, 1^x can be any positive number, and it will still give you one.
As stated by WojtekR, even negative numbers can get 1
x = - 5
(6^0)^-5 = 1^-5 = 1/1 = 1
~ Thanks @WojtekR
Wouldn’t it be D. since the domain is x. (x,y)
6r - 3t = 6 /*8
6r + 8t = - 16 /*3
48r - 24t = 48
48r + 24t = - 48
---------------------- +
96r = 0
r = 0
t = ( -16 + 6r)/8 = - 16/8 = - 2
Answer:
i feel bad for you
Step-by-step explanation:
Answer:
There is no solution to this equation that is an integer for x.
Step-by-step explanation:
x is the first integer, and x+1 the second

