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rewona [7]
3 years ago
5

Divide 20x^2y^2+4xy^2-8y^2/4x^2y

Mathematics
1 answer:
Ksivusya [100]3 years ago
7 0
I hope this helps you



[20x^2/4x^2y]+[4xy^2/4x^2y]-[8y^2/4x^2y]


[5/y]+[y]-[4y/x^2]
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2 years ago
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This kinda sounds like system of equations to me. Let's use some variables. 'x' would be the fee for the regular selections. 'y' would be the fee for the discounted selections. Let's start making our equations :O. Here's Pat's order: 2x+4y=119.80. Here's Carlos' order: 3x+5y=160.75. Now, there are different ways of solving these variables such as using substitution, but I will use linear combinations method/elimination method to solve this. I will try to make one variable value the opposite of the other equation's variable. I can get 6x+10y=321.50 and -6x-12y=-359.4. I can eliminate the x now and only have variable y. -2y=-37.9. Dividing both sides gets me y=18.95. I can substitute the y value for y in any equation. so 2x+4(18.95)=119.80. 2x+75.8=119.80. 2x=44. 2x/2=44/2. x=22. Therefore, the fee for regular selections is 22 while the fee for discounted selections is 18.95. Hope you enjoyed this session of learning :3

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3 years ago
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