Jravenv
  • Jravenv
Would someone kindly give their help? screencap in comments.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Jravenv
  • Jravenv
kropot72
  • kropot72
To solve this, you need to remember that when dividing powers of terms with the same base you need to subtract the indices. For example \[\large \frac{w^{6} x^{4}}{w^{2} x^{3}}=w^{(6-2)}x^{(4-3)}=w^{4}x\]
Jravenv
  • Jravenv
I don't know how to do that :/

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kropot72
  • kropot72
\[\large \frac{4x^{3} y^{4}}{-8xy^{2}}=\frac{4}{-8} \times x^{(3-1)}y^{(4-2)}=?\]
Jravenv
  • Jravenv
This is too complicated for me, can you help me break it down?
kropot72
  • kropot72
Can you subtract 1 from 3, and subtract 2 from 4?
Jravenv
  • Jravenv
2 and 2 so yes
kropot72
  • kropot72
Correct. And what is 4/-8?
Jravenv
  • Jravenv
-1/2
UnkleRhaukus
  • UnkleRhaukus
yeah, and what isare (3-1) ? , (4-2) ?
UnkleRhaukus
  • UnkleRhaukus
yeah you had 2 and 2, so... \[\large \frac{4x^{3} y^{4}}{-8xy^{2}}\\ =\frac{4}{-8} \times x^{(3-1)}y^{(4-2)}\\ =\frac{-1}2x^2y^2\]
Jravenv
  • Jravenv
I don't get that :( I'm sorry I have a horrible instructor
kropot72
  • kropot72
What part don't you get? You have correctly answered the sections that were broken down for you.
Jravenv
  • Jravenv
The formatting makes my brain shut down
Jravenv
  • Jravenv
I don't get it
kropot72
  • kropot72
So the correct choice has -1/2, x^2 and y^2.
anonymous
  • anonymous
4x3y4/8xy2 Reduce the expression by cancelling the common factors. More Steps Factor 4 out of 4x3y4. 4(x3y4)−8xy2 Rewrite the expression. x3y4−2xy2 x3y4−2xy2 Reduce the expression by cancelling the common factors. More Steps Factor x out of x3y4. x(x2y4)−2xy2 Rewrite the expression. x2y4−2y2 x2y4−2y2 Reduce the expression by cancelling the common factors. More Steps Factor y2 out of x2y4. y2(x2y2)−2y2 Rewrite the expression. x2y2/-2 x2y2/-2 Move the negative in front of the fraction. −x2y2/2
Jravenv
  • Jravenv
Thank you all so much. I get it so much more now.
kropot72
  • kropot72
You're welcome :)

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