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Snowcat [4.5K]
3 years ago
8

The sum of two polynomials is 8d5 – 3c3d2 + 5c2d3 – 4cd4 + 9. If one addend is 2d5 – c3d2 + 8cd4 + 1, what is the other addend?

Mathematics
2 answers:
9966 [12]3 years ago
8 0
To solve for the the other addend, subtract the given addend from the given sum of the two polynomials,
                (8d^5 - 3c³d² + 5c²d³ - 4cd^4 +9)   -     (2d^5 - c³d² + 8cd^4 + 1)

Subtracting the terms from their like terms of the sum gives and answer of 
                        6d^5 - 2c³d² + 5c²d³ - 12cd^4 +8
Maru [420]3 years ago
8 0

Answer:

6d^5 – 2c^3d^2 + 5c^2d^3 – 12cd^4 + 8

Step-by-step explanation:

To solve for the the other addend, subtract the given addend from the given sum of the two polynomials,

               (8d^5 - 3c³d² + 5c²d³ - 4cd^4 +9)   -     (2d^5 - c³d² + 8cd^4 + 1)

Subtracting the terms from their like terms of the sum gives and answer of 

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Please help me with this one question, and I'll upvote the brainliest answer
ra1l [238]

Answer:   " x = 12 " .

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Step-by-step explanation:

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<u>Note</u>: If 2 (two) triangles are "similar" ;  then their corresponding sides are "proportional".

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Since one of the sides [the "hypotenuse"] of one the triangles given ; shows the measurement containing an expression with a value of "x" — specifically;  " 6x + 28 " ;  

    →  And since we want to solve for the value of "x" ;  we  use this "hypotenuse" of said particular triangle to set up a "proportion" with corresponding sides; so we can solve for "x" :

→  Set up proportions as a "ratio" ; or fraction:

       →  7:28::25:(6x + 28) ;  Solve for "x" ;

Let us write this in the form of a "fraction" :

      →  \frac{7}{28} = \frac{25}{(6x+28)} ;

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<u>Note</u>:  

        →  Rewrite:  " \frac{7}{28} " ; by simplifying to:

                                    " \frac{1}{4} " ;

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         →  Since:  " \frac{7}{28} " ;

              =  " (7÷7) / (28÷7) " ;

              =  " (1 /4) " ;

              =   " \frac{1}{4} " .

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Now, rewrite the "proportion" ; as follows:

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        \frac{1}{4} = \frac{25}{(6x+28)}  ;

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Now:   "Cross multiply" ;  that is:

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→   Given:   " \frac{a}{b} = \frac{c}{d} " ;  

         and:  " b\neq0 " ;  " d\neq0 " ;

→  Then:   " a * d "  =  " b * c " .

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Likewise:   " 1(6x + 28)  =  (4) * (25) "   ;

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SImplify:    " 6x + 28 =  100 " ;  Solve for "x" ;

→ Subtract "28" from each side of the equation:

                  →   " 6x + 28 - 28 = 100 - 28 "  ;

    to get:   →   " 6x  =  72  "  ;

Now, divide each side of the equation by:  " 6 " ;

        to isolate "x" on one side of the equation; & to solve for "x" ; as follows:

   →   "  6x  =  72  "  ;

         →   6x / 6  =  72 / 6 ;

  to get:

         →   " x  = 12 "  ;

        →   which is the answer.

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Hope this helps!

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