2x^3y^4z^3a^2
over
4x^2y^5za

Answer choices:

A) axz
over
2y

B) xz^2a
over
2y

C) ax^2z
over
2y

D) xy
over
2

Question

2x^3y^4z^3a^2
over
4x^2y^5za

Answer choices:

A) axz
    over 
    2y

B) xz^2a
     over
      2y

C) ax^2z
     over
      2y

D) xy
  over
    2

kinggeorge · Answer

Just give it a shot. What do you think $$\large \frac{y^4}{y^5}$$simplifies as?

anonymous · Answer

y1

kinggeorge · Answer

Close, you have the right idea though. It should be $$y^{-1}=\frac{1}{y}$$

anonymous · Answer

o okay i was close

kinggeorge · Answer

Let's do another one. Can you simplify $$\large \frac{z^3}{z}?$$Just keep using the same idea.

anonymous · Answer

1^3?

anonymous · Answer

could we try my question?

anonymous · Answer

2x^3y^4z^3a^2
over
4x^2y^5za

Answer choices:

A) axz
    over 
    2y

B) xz^2a
     over
      2y

C) ax^2z
     over
      2y

D) xy
  over
    2

anonymous · Answer

which one would it be?

saranya · Answer

Ans is 

B) xz^2a
     over
      2y

kinggeorge · Answer

It should just be $$z^{3-1}=z^2.$$We're doing your question one part at a time. There's a reason I'm choosing these particular variables and exponents. We just have one left to do before we finish. Give a shot at simplifying $$\frac{a^2}{a}$$

anonymous · Answer

@saranya  thanks so much

anonymous · Answer

@KingGeorge okay thanks

saranya · Answer

no probs dear :)

anonymous · Answer

@saranya  how did you get this answer it was kinda difficult could you explain please?

kinggeorge · Answer

She got it using the method I was showing you. We started with a $$\frac{x^3}{x^2}=x$$the last time you asked this, and here, we moved on to get $$\frac{y^4}{y^5}=\frac{1}{y}\qquad \text{and}\qquad \frac{z^3}{z}=z^2$$That means we had just had to simplify $$\frac{a^2}{a}=a\qquad \text{and}\qquad \frac{2}{4}=\frac{1}{2}$$to get the final solution.

kinggeorge · Answer

You just combine all of those terms to get $$x\cdot\frac{1}{y}\cdot{z^2}\cdot a\cdot\frac{1}{2}=\frac{xz^2a}{2y}$$

saranya · Answer

yea sure :)
It's simple !
Now take the x part
2x^3/4x^2= x/2
Now take the y part
y^4/y^5 = 1/y
Now take z part
z^3/z = z^2
now take a part
a^2/a = a
Now combine all that is simply multiply all the ans : yu will get 
 xz^2a
     over
      2y

anonymous · Answer

@KingGeorge thanks so much you helped alot

anonymous · Answer

thanks to you guys i got an 80 on my test thanks so much

saranya · Answer

ur welcome dear as always :)

kinggeorge · Answer

For the future, posting these questions that were on your test is against the code of conduct. Especially if you were only looking for the answer. http://openstudy.com/code-of-conduct