A good place to start is to set 
to y. That would mean we are looking for 
to be an integer. Clearly, 
, because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since 
is a radical, it only outputs values from ![[0,\infty]](https://tex.z-dn.net/?f=%20%5B0%2C%5Cinfty%5D%20)
, which means y is on the closed interval: ![[0,120]](https://tex.z-dn.net/?f=%20%5B0%2C120%5D%20)
.
With that, we don't really have to consider y anymore, since we know the interval that is on.
Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval , which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:
Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.
Answer:
106 and 93
Step-by-step explanation:
The whole thing has to add up to 360
The answer is D- negative correlation
Answer with Step-by-step explanation:
We are given S be any set which is countable and nonempty.
We have to prove that their exist a surjection g:N
Surjection: It is also called onto function .When cardinality of domain set is greater than or equal to cardinality of range set then the function is onto
Cardinality of natural numbers set =
( Aleph naught)
There are two cases
1.S is finite nonempty set
2.S is countably infinite set
1.When S is finite set and nonempty set
Then cardinality of set S is any constant number which is less than the cardinality of set of natura number
Therefore, their exist a surjection from N to S.
2.When S is countably infinite set and cardinality with aleph naught
Then cardinality of set S is equal to cardinality of set of natural .Therefore, their exist a surjection from N to S.
Hence, proved
