Answer:
section covers somplifying algebraic expressions
Step-by-step explanation:
<h3>
(4
)(2
) =
= 2/x⇒</h3>

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

add both amounts, and that's her investment for the 11 years.
Answer: 13x
Step-by-step explanation: Since 8x and 5x are like terms, we can add them together.
8x+5x=13x
Have a nice day! :)
Answer:
3x(4x+5y)
Step-by-step explanation:
and yh.........