Help 40

So just do 3.4*4 and I will get my answer?

No, you must do problems on both sides of the addition symbol add the products and you will get your sum  like this 

\(factor\times factor +factor\times factor\)  

\(Product + Product = sum\) 

\(sum = answer.\)

Yes he would but in math for sixth graders it teaches the basics of pre-algebra equations so this input is along correct but to not confuse him he must do both sides then add the products to derive his answer.

 I figured out 1/2 * 27 do I do that I am confused because that is what you get 

I already do stuff like Emerald Wonder said I use that order.... Hiylink 

If 27 is what you got then you can multiply it by 1/2 like this 

\(\frac{1}{2} * \frac{27}{1}\) 

Multiply the numerator by the numerator and the like wise with the denominator.

so

27

-

2

Yes can   \(\frac{27}{2}\) 

Be simplified further?

But that wouldnt be the correct answer since 27/2 is 13.5 and that isnt an answer. He should use PEMDAS.

183.6

Is that the answer 

This problem works the same way you can apply pemdas on either side of the problem? in the midst of this we have already solved the major parts of pemdas......... after completes (Using Pemdas) the problems on either side he can add the products and this is what the problem is looking for.

Nickolas line those numbers up again 183.6 is not the answer between \(\frac{27}{2} * \frac{1}{2}\)

YOU alone can solve this problem comppletly I am just assiting you its are methods that were looking at.

It is 

27

-

4

and you changed it 

Sorry that was my error simlify \(\frac{27}{1} * \frac{1}{2}\) which is \(\frac{27}{2}\) so simplified that is (Hint) you can divide 27 by two and the remainder has to bein fraction form.

Acurate answer I dont even now what that means I just want a answer with work so I can look at it...............Please 

No, that would be lazy take the assitance your offered you should be able to do this by your self you have had countless people help you.

You were on the right track by solving one side. You were still using PEMDAS because you worked on the side of exponenets you just needed to do the other side.