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Andrej [43]
2 years ago
14

2.) Simplify: 10x + 6y +4x = x(14) + 6y = 14x +6y

Mathematics
1 answer:
adelina 88 [10]2 years ago
5 0

Answer:

{x,y} = {1,1}

Step-by-step explanation:

[2]    6y = -14x + 20

 [2]    y = -7x/3 + 10/3

// Plug this in for variable  y  in equation [1]

  [1]    -10x - 6•(-7x/3+10/3) = -16

  [1]    4x = 4

// Solve equation [1] for the variable  x

  [1]    4x = 4

  [1]    x = 1

// By now we know this much :

   x = 1

   y = -7x/3+10/3

// Use the  x  value to solve for  y

   y = -(7/3)(1)+10/3 = 1

Solution :

{x,y} = {1,1}

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⌊ -5.2 ⌋ = -6.

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Answer:

5x^2+7x+30+\frac{102}{x-3}

Step-by-step explanation:

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}\\

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}

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zheka24 [161]
<h3>Answer:  D. 5 blinks</h3>

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