Answers

2015-02-21T21:08:46+08:00

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Evaluate (5x² + 3xy) over xy² using the given values of x and y below:

1.) x = -2; y = -2
 \frac{5x^{2}+ 3xy}{xy^{2}}
 \frac{5(-2)^{2} + 3(-2)(-2)}{(-2)(-2)^{2}}
 \frac{5(4)+3(4)}{(-2)(4)}
 \frac{20+12}{-8}
 \frac{32}{-8}
-4

2.) x = 1; y = -1
 \frac{(5 x^{2} + 3xy)}{x y^{2} }
 \frac{5(1)^{2} +3(1)(-1)}{(1)(-1)^{2} }
\frac{5(1) + 3(-1)}{(1)(1)}
\frac{5 - 3}{1} 

 \frac{2}{1}
2

3.) x = -2; y = -1
 \frac{(5x^{2} +3xy)}{xy^{2}}
 \frac{5(-2)^{2} +3(-2)(-1)}{(-2)(-1)^{2}}
 \frac{5(4)+3(2)}{(-2)(1)}
 \frac{20 + 6}{-2}
 \frac{26}{-2} 
-13

4.) x = -1; y = -2
 \frac{5x^{2} + 3xy}{xy^{2} }
 \frac{5(-1)^{2} + 3(-1)(-2)}{(-2)^{2}}
 \frac{5(1) + 3(2)}{(-1)(4)} 
 \frac{5 + 6}{-4}
 \frac{11}{-4}
-2.75

5.) x = 8; y = 7
 \frac{5x^{2} + 3xy}{xy^{2} }
 \frac{5(8)^{2} + (3)(8)(7)}{8(7)^{2}}
 \frac{5(64) + 24(7)}{8(49)}
 \frac{320 + 168}{392}
 \frac{488}{392}
 1\frac{96}{392} or 1 \frac{12}{49}
0
what is the answer on number 2..
The answer in number 2 is 2.
okay
2015-02-22T10:57:23+08:00

This Is a Certified Answer

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Evaluate means like find the value given a certain amount. Ex:
1.x=-2; y=-2
\frac{5x^2+3xy}{xy^2}
Sustitute
\frac{5(-2)^2+3(-2)(-2)}{-2(-2)^2}
\frac{20+12}{-8}
Add like terms
\frac{32}{-8}
Divide:
=-4

2.x=1; y=-1
\frac{5x^2+3xy}{xy^2}
Substitute
\frac{5(1)^2+3(1)(-1)}{1(-1)^2}
\frac{5-3}{1}
Add like terms
=2

3.x=-2; y=-1
\frac{5x^2+3xy}{xy^2}
Substitute
\frac{5(-2)^2+3(-2)(-1)}{-2(-1)^2}
\frac{20+6}{-2}
Add like terms
\frac{26}{-2}
Divide
=-13

4.x=-1; y=-2
\frac{5x^2+3xy}{xy^2}
\frac{5(-1)^2+3(-1)(-2)}{-1(-2)^2}
\frac{5+6}{-4}
Add like terms
\frac{11}{-4}
Divide
=-11/4 or -2.75

5.x=8; y=7
Substitute:
\frac{5(8)^2+3(8)(7)}{7(8)^2}
\frac{320+168}{448}
Add like terms
[tex]\frac{488}{488}{/tex]
Divide
=1

Hope this helps =)
0
sorry the [tex]\frac{488}{488} is 488/488