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yulyashka [42]
4 years ago
15

The solution of c+d<−6 is c<−8. What is the value of d?

Mathematics
2 answers:
vfiekz [6]4 years ago
7 0
+2 is the answer....
Mashcka [7]4 years ago
4 0

Answer:

The value of D is +2

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General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS
  • Equality Properties

<u>Algebra I</u>

  • Solving systems of equations using substitution/elimination
  • Solving systems of equations by graphing
  • Expanding
  • Finding roots of a quadratic
  • Standard Form: ax² + bx + c = 0
  • Quadratic Formula: x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}

Step-by-step explanation:

<u>Step 1: Define systems</u>

2x - y = 9

4x² + 3y² - 2x + y = 16

<u>Step 2: Rewrite systems</u>

2x - y = 9

  1. Subtract 2x on both sides:                    -y = 9 - 2x
  2. Divide -1 on both sides:                          y = 2x - 9

<u>Step 3: Redefine systems</u>

y = 2x - 9

4x² + 3y² - 2x + y = 16

<u>Step 4: Solve for </u><em><u>x</u></em>

<em>Substitution</em>

  1. Substitute in <em>y</em>:                         4x² + 3(2x - 9)² - 2x + (2x - 9) = 16
  2. Expand:                                    4x² + 3(4x² - 36x + 81) - 2x + (2x - 9) = 16
  3. Distribute 3:                             4x² + 12x² - 108x + 243 - 2x + 2x - 9 = 16
  4. Combine like terms:                16x² - 108x + 234 = 16
  5. Factor GCF:                              2(8x² - 54x + 117) = 16
  6. Divide 2 on both sides:           8x² - 54x + 117 = 8
  7. Subtract 8 on both sides:        8x² - 54x + 109 = 0
  8. Define variables:                      a = 8, b = -54, c = 109
  9. Resubstitute:                            x=\frac{54\pm\sqrt{(-54)^2-4(8)(109)} }{2(8)}
  10. Exponents:                               x=\frac{54\pm\sqrt{2916-4(8)(109)} }{2(8)}
  11. Multiply:                                    x=\frac{54\pm\sqrt{2916-3488} }{16}
  12. Subtract:                                   x=\frac{54\pm\sqrt{-572} }{16}

Here we see that we start to delve into imaginary roots. Since on a real number plane, we do not have imaginary roots, there would be no solution to the systems of equations.

<u>Step 5: Graph systems</u>

<em>We can verify our results.</em>

8 0
3 years ago
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