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photoshop1234 [79]
3 years ago
15

Alexis paid $50 to active her cell phone plus a service fee of $20 per month.She used the phone for 3 months. How much has she p

aid for phone service
Mathematics
2 answers:
TEA [102]3 years ago
7 0

Answer:


Step-by-step explanation:

She paid $50 to activate one time and pays $20/ month

20.00x3=60.00

60.00+50.00=110.00

givi [52]3 years ago
7 0

Answer:

Alexis paid $110 for phone service.

Step-by-step explanation:

Well $50 is like a flat fee; you just pay it and you're done with it.

Meanwhile the $20 is not because it is <em>per month</em>, meaning for every month Alexis using her phone, she has to pay $20. Alexis uses her phone for 3 months.

So she paid $20 three times: 20×3

20×3=60

And then you add the 60 to the flat fee, which is $50.

$60+$50=$110

Alexis paid <u>$110</u> for phone service.

-------

I hope that helps! :)


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oksano4ka [1.4K]

Answer:

These techniques for elimination are preferred for 3rd order systems and higher.  They use "Row-Reduction" techniques/pivoting and many subtle math tricks to reduce a matrix to either a solvable form or perhaps provide an inverse of a matrix (A-1)of linear equation AX=b.  Solving systems of linear equations (n>2) by elimination is a topic unto itself and is the preferred method.  As the system of equations increases, the "condition" of a matrix becomes extremely important.  Some of this may sound completely alien to you.  Don't worry about these topics until Linear Algebra when systems of linear equations (Rank 'n')  become larger than 2.

Step-by-step explanation:

Just to add a bit more information, "Elimination" Can have a variety of other interpretations.  Elimination techniques typically refer to 'row reduction' to achieve 'row echelon form.'  Do not worry if you have not heard of these terms.  They are used in Linear Algebra when referring to "Elimination techniques"

 

Gaussian Elimination

Gauss-Jordan Elimination

LU-Decomposition

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These techniques for elimination are preferred for 3rd order systems and higher.  They use "Row-Reduction" techniques/pivoting and many subtle math tricks to reduce a matrix to either a solvable form or perhaps provide an inverse of a matrix (A-1)of linear equation AX=b.  Solving systems of linear equations (n>2) by elimination is a topic unto itself and is the preferred method.  As the system of equations increases, the "condition" of a matrix becomes extremely important.  Some of this may sound completely alien to you.  Don't worry about these topics until Linear Algebra when systems of linear equations (Rank 'n')  become larger than 2.

 

Substitution is the preferred method for 2 equations in 2 unknowns.  The constants are unimportant other than having a non-zero determinant.  It is always easy to find multiplicative factors using LCMs of one variable or the other to allow substitution into the other equation:

 

Example:

 

4X + 5Y = 9

5X -  4Y = 1

 

Notice that 20 is a LCM of either the X or Y variable.  So multiply the first by 4 and the second by 5 and then adding the two (Y's will drop out allowing for substitution)

 

4(4X + 5Y = 9)

5(5X -  4Y = 1)  

 

Multiplying to produce the LCM factors:

 

16X + 20Y = 36

25X -  20Y = 5

 

Adding the equations

 

41X = 41

X = 1

 

Substitution into either equation yields

Y = 1

 

Elimination techniques are preferred for Rank-n>3

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