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:
75% left is not taken up so do the Ab to the other
Step-by-step explanation:
0.75 sorry if Im wrong
Answer:
2x^2 + 2y^2 + 2z^2
Step-by-step explanation:
x^2+xy-yx+y^2 + y^2+yz-zy+z^2 + z^2+zx-xz+x^2
First step is to put all the same ones together
( FYI, xy and yx is the same thing, just like how
2 x 3 and 3 x 2 is the same thing) this time I'll bracket the groups to make it easier on the eyes
(x^2 +x^2) + (xy - xy) +( y^2 + y^2) + (yz - yz) +
(z^2 +z^2) + (zx - zx)
2x^2 + 2y^2 + 2z^2
All the others just cancel themselves out, for example, xy-xy
When anything minus themselves it will become 0
Answer:
2x + 3y = 6
Step-by-step explanation: