Add (2m^2n + 5mn^2 - 7mn) ; (2m^2n - mn^2 + mn

Step-by-step explanation:

(3m+2n)/(m-n2)+(5m+n)/(n-m2)-(2m-3n)/(m2-n)

Final result :

-3m3 - 2m2n + 7m2 - 7mn2 + mn + 2n3 + 2n2

—————————————————————————————————————————

(m - n2) • (n - m2)

Step by step solution :

Step 1 :

2m - 3n

Simplify ———————

m2 - n

Trying to factor as a Difference of Squares :

1.1 Factoring: m2 - n

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : m2 is the square of m1

Check : n1 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares

Equation at the end of step 1 :

(3m+2n) (5m+n) (2m-3n)

(————————+————————)-———————

(m-(n2)) (n-(m2)) m2-n

Step 2 :

5m + n

Simplify ——————

n - m2

Trying to factor as a Difference of Squares :

2.1 Factoring: n - m2

Check : n1 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares

Equation at the end of step 2 :

(3m+2n) (5m+n) (2m-3n)

(————————+——————)-———————

(m-(n2)) n-m2 m2-n

Step 3 :

3m + 2n

Simplify ———————

m - n2

Trying to factor as a Difference of Squares :

3.1 Factoring: m - n2

Check : m1 is not a square !!

Ruling : Binomial can not be factored as the difference of two perfect squares

Equation at the end of step 3 :

(3m + 2n) (5m + n) (2m - 3n)

(————————— + ————————) - —————————

m - n2 n - m2 m2 - n

Step 4 :

Calculating the Least Common Multiple :

4.1 Find the Least Common Multiple

The left denominator is : m-n2

The right denominator is : n-m2

Number of times each Algebraic Factor

appears in the factorization of:

Algebraic

Factor Left

Denominator Right

Denominator L.C.M = Max

{Left,Right}

m-n2 1 0 1

n-m2 0 1 1

Least Common Multiple:

(m-n2) • (n-m2)

Calculating Multipliers :

4.2 Calculate multipliers for the two fractions

Denote the Least Common Multiple by L.C.M

Denote the Left Multiplier by Left_M

Denote the Right Multiplier by Right_M

Denote the Left Deniminator by L_Deno

Denote the Right Multiplier by R_Deno

Left_M = L.C.M / L_Deno = n-m2

Right_M = L.C.M / R_Deno = m-n2

correct answer:

b.(19,-2)

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all