Answer:
$9$
Step-by-step explanation:
Given: Thea enters a positive integer into her calculator, then squares it, then presses the $\textcolor{blue}{\bf\circledast}$ key, then squares the result, then presses the $\textcolor{blue}{\bf\circledast}$ key again such that the calculator displays final number as $243$.
To find: number that Thea originally entered
Solution:
The final number is $243$.
As previously the $\textcolor{blue}{\bf\circledast}$ key was pressed,
the number before $243$ must be $324$.
As previously the number was squared, so the number before $324$ must be $18$.
As previously the $\textcolor{blue}{\bf\circledast}$ key was pressed,
the number before $18$ must be $81$
As previously the number was squared, so the number before $81$ must be $9$.
Answer:
4L-12
Step-by-step explanation:
let's assume the length = L
then the width of the rectangle is L - 6
perimeter = 2× (a+b)
where "a" and "b" are the length and the width of the rectangle.
so perimeter = 2× (L + L - 6)
2 × ( 2L - 6)
4L - 12
