Answer:
I think the correct answer to this, (if I'm not wrong), would be D.
Step-by-step explanation:
I hope this helped!! Have a wonderful day and never forget to smile!!
Answer:
6
Step-by-step explanation:
you kna mean?
Answer:
Step-by-step explanation: it doesnt make that much sense tho can you ecplain it again pls
Let's call the width
and the length
based on this, we can make two equations based on the first and second sentences in the problem.


If we substitute
for
into the second equation, we can make the following:


- Use the Distributive Property


- Subtract 4 from both sides

- Divide both sides by 4 and simplify
Now, let's find
:

The length of the rectangle is
and the width of the rectangle is
.
Find common variables. I wrote the ones I found above the polynomial. write them in parentheses the first number will have the variable. now it's just a matter of figuring out how to order it so that the first terms added to the second terms equal to the middle term of the polynomial. ... I think this is right. you could always look on wikiHow if I'm wrong.