Given the expression:
[tex]\frac{j^3k^{}}{h^0}[/tex]
Where,
h = 8
j = -1
k = -12
Let's plug in the values to the expression:
[tex]\frac{j^3k^{}}{h^0}[/tex][tex]\frac{(-1)^3(-12)^{}}{8^0}[/tex]
Recall: Any number that is raised to the power of zero is always equal to 1.
We get,
[tex]\frac{(-1)^3(-12)^{}}{8^0}\text{ = }\frac{(-1)(-12)}{1}\text{ = (-1)(-12)}[/tex]
Recall: When multiplying integers of the same sign, the product is always a positive integer.
We get,
[tex](-1)(-12)\text{ = 12}[/tex]
Therefore, the answer is 12, letter A.
