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Illusion [34]
3 years ago
10

For each ordered pair determine wheatear it is a solution to 2x

Mathematics
1 answer:
stealth61 [152]3 years ago
3 0
All you have to do is plug in the x and y values for each ordered pair into the left hand side of the equation and see if you get 15 = 15. If you do, the ordered pair is a solution; otherwise it is not a solution. For example, let's try (-2,-3); that is x = -2 and y = -3:

2x - 9y = 15
2(-2) - 9(-3) = 15
-4 - (-27) = 15
-4 + 27 = 15
23 = 15

This is obviously not true, so (-2,-3) is not a solution. Now you try it on each of the ordered pairs given in the problem statement.
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Answer:

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Step-by-step explanation:

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nasty-shy [4]
The answer is:  \frac{1}{12}  .
____________________________________________
Explanation:
____________________________________________
We have:   " 6⁻¹  *  (-2)³  * (-4)⁻² = ? " ;
_______________________________________
Note:  "6⁻¹ = \frac{1}{6^1} = \frac{1}{6} "  .
 
Note the following property of exponents:
____________________________________
           a⁻ˣ = \frac{1}{x^a} ;  {x ≠ 0}.
____________________________________________
Note:  (-2)³  = (-2)*(-2)*(-2) = 4*(-2) = -8 .
_______________________________________
Note:  (-4)⁻²  = \frac{1}{(-4)^2}  ;
 
                     = \frac{1}{(-4)*(-4)} ;

                     = \frac{1}{16} ;
____________________________________
So; we have:
_____________________________________
6⁻¹  *  (-2)³  * (-4)⁻²  ;

=  \frac{(1*8*1)}{(6*16)} ;

→ Cancel out the "8" in the "numerator" (to a "1") ; and the "16" in the number (and change to a "2"); since: {"16÷8 = 2" ; and since:  "{8÷8=1"} ;

→and we have:

\frac{(1*1*1)}{(6*2)} ;

 =  \frac{1}{12} .
_____________________________________
 → The answer is:  \frac{1}{12}  .
_____________________________________
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