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sp2606 [1]
3 years ago
12

Solve for K 2(k - 5) + 3k = k + 6

Mathematics
1 answer:
Alexandra [31]3 years ago
3 0

Answer:

k = 4

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtract Property of Equality

<u>Algebra I</u>

  • Terms/Coefficients

Step-by-step explanation:

<u>Step 1: Define</u>

2(k - 5) + 3k = k + 6

<u>Step 2: Solve for </u><em><u>k</u></em>

  1. Distribute 2:                               2k - 10 + 3k = k + 6
  2. Combine like terms:                  5k - 10 = k + 6
  3. Subtract <em>k</em> on both sides:          4k - 10 = 6
  4. Add 10 on both sides:               4k = 16
  5. Divide 4 on both sides:             k = 4

<u>Step 3: Check</u>

<em>Plug in k into the original equation to verify it's a solution.</em>

  1. Substitute in <em>k</em>:                    2(4 - 5) + 3(4) = 4 + 6
  2. (Parenthesis) Subtract:       2(-1) + 3(4) = 4 + 6
  3. Multiply:                               -2 + 12 = 4 + 6
  4. Add:                                     10 = 10

Here we see that 10 does indeed equal 10.

∴ k = 4 is the solution to the equation.

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<u>Step-by-step explanation:</u>

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

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\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

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\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

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