You didn't give us the choices. It doesn't matter. I think you're trying to write




Answer: 60 x² y⁸
Answer:
3(n + 4) ≤ 10 is the equation and the solution is n ≤ -2/3 Decimal Form: -0.666667
hope this helps you
Step-by-step explanation:
-4/2=-2
(-2)^2=4
the blank is +4
x^2-4x+4
factored
(x-2)^2
Answer: OPTION B.
Step-by-step explanation:
Given the following System of equations:

You can use the Elimination Method to solve it. The steps are:
1. You can mutliply the second equation by -3.
2. Then you must add the equations.
3. Solve for the variable "y".
Then:

4. Now that you know the value of the variable "y", you must substitute it into any original equation.
5. The final step is to solve for "x" in order to find its value.
Then:

Therefore, the solution is:

Answer:
3 one ofx+2 And 4th one x4