+0  
 
0
229
3
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1: 

 

2: 

 

3: 

Guest Dec 15, 2017
 #1
avatar+7266 
+1

1:

 

(a)  The GCF of these two terms is  9a4b10  ,  so when we factor that out we get

 

36a4b10  –  81a16b20   =   9a4b10( 4 - 9a12b10 )

 

(b)   Let's write the original expression like this...

 

36a4b10  –  81a16b20   =   (6a2b5)2  –  (9a8b10)2

 

Now it is written as a difference of squares, which factors as...

 

(6a2b5)2  –  (9a8b10)2   =   ( 6a2b5  +  9a8b10 )( 6a2b5  –  9a8b10 )

hectictar  Dec 15, 2017
 #2
avatar+7266 
+1

2:     Here's a graph:   https://www.desmos.com/calculator/idxee4c2dz

 

The solution to the system is the part that is shaded by both red and blue.

 

(0, 4)  is a solution to the system.

 

Check the point in both inequalitites to make sure it makes them both true.

 

4  ≥  \(-\frac12(0)+2\frac12\)         ?

 

4  ≥  \(2\frac12\)           True.

 

4  <  \(\frac15(0)+6\)               ?

 

4  <  6              True.

 

 

3:     \(\Large{\frac{\frac{1}{x^2}+\frac{2}{y}}{\frac{5}{x}-\frac{6}{y^2}}}\)

 

 \(\Large{=\,\frac{\frac{y}{x^2y}+\frac{2x^2}{x^2y}}{\frac{5y^2}{xy^2}-\frac{6x}{xy^2}} \\~\\ =\,\frac{\frac{y+2x^2}{x^2y}}{\frac{5y^2-6x}{xy^2}} \\~\\ =\,\frac{y+2x^2}{x^2y}\cdot\frac{xy^2}{5y^2-6x} \\~\\ =\,\frac{y+2x^2}{x}\cdot\frac{y}{5y^2-6x} \\~\\ =\,\frac{y^2+2x^2y}{5xy^2-6x^2}}\)

hectictar  Dec 15, 2017
 #3
avatar+88899 
+2

[    1/x^2  + 2/y  ]   /  [ 5/x  -  6/y^2]

 

Multiply numerator/ denominator by x^2y^2

 

[ y^2  + 2x^2y ]  / [ 5xy^2 - 6x^2]

 

 

cool cool cool

CPhill  Dec 15, 2017

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