Compute the cardinality of the set $\{z \in \Bbb{Z} \mid z>-10, z^3<0\}$
We will also write this with the notation: compute $\bigl|\{z \in \mathbb{Z} \mid z>-10, z^3<0\}\bigr|$. (To show your work, makesure you show the set explicitly before telling us how many elements it has.)
If we're talking about all integers ($\Bbb{Z}$) including negatives. For the first part of $z > -10$. $z$ can only be up to $-9$ otherwise it would be greater then or equal to $-10$. But what would the cardinality be? Since the set would have $\{-9,-8,-7,-6...$ and so on till infinity?$\}$
And for the second part $z^3<0$. $z$ can only be $0$ otherwise anything else for $z^3$ would be grater then $0$. So the set would be $\{0\}$ with cardinality $1$?
The integers satisfying $z>-10$ form the set $\{z \in \mathbb{Z} | z>-10\}=\{-9,-8,-7,...\}$.
The cube of a negative integer is negative, so the numbers satisfying $z^3<0$ are all the negative integers.
In the set $\{z \in \mathbb{Z}|z>-10, z^3<0\}$, you want both of these conditions to be true, so $$\{z \in \mathbb{Z}|z>-10, z^3<0\}=\{-9,-8,...,-2,-1\}$$ which has $9$ elements. That is, the cardinality is $9$.