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Andrews [41]
4 years ago
10

How do I solve for x? please explain

Mathematics
2 answers:
PIT_PIT [208]4 years ago
8 0
Divided both sides by 2 the answer would be x = -5
My name is Ann [436]4 years ago
6 0
The aim here is to get all of the x's onto one side and the numbers onto the other side:

We can start by subtracting x from each side

2x+11=1

We can then subtract 11 from both sides

2x=-10

Finally, we can divide both sides by 2 leaving us with:

x = -5
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Help me with these math questions.
Olegator [25]

graph A

no x-intercept

no y-intercept

H.A is y = 5

V. A. is x = 0

*****************************************

Answer: \frac{150}{31}

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

log₆(5x + 6) - log₆(x - 4) = 2

log₆(\frac{5x+6}{x-4}) = 2

(\frac{5x+6}{x-4}) = 6²

(\frac{5x+6}{x-4}) = 36

5x + 6 = 36(x - 4)

5x + 6 = 36x - 144

       6 = 31x - 144

    150 = 31x

     \frac{150}{31} = x

**********************************************************

Answers:

  • a. (-1, -5)
  • b. up
  • c. x = -1
  • d. \frac{-2+\sqrt{15}}{3} and\frac{-2-\sqrt{15}}{3}
  • e. -2
  • f. see attachment
  • g. domain (-∞, ∞)  range [-5, ∞)

<u>Explanation:</u>

f(x) = 3x² + 6x - 2

      a=3  b=6 c=-2

x = \frac{-b}{2a} =\frac{-6}{2(3)} = \frac{-6}{6} = -1

Axis Of Symmetry: x = -1

f(-1) = 3(-1)² + 6(-1) - 2

       = 3 - 6 - 2

       = -5

Vertex: (-1, -5)

x = \frac{-b+/-\sqrt{b^{2}-4ac}}{2a}

  = \frac{-6+/-\sqrt{6^{2}-4(3)(-2)}}{2(3)}

  = \frac{-6+/-\sqrt{36+24}}{6}

  = \frac{-6+/-\sqrt{60}}{6}

  = \frac{-6+/-2\sqrt{15}}{6}

  = \frac{-2+/-\sqrt{15}}{3}

x-intercepts: \frac{-2+/-\sqrt{15}}{3}



6 0
3 years ago
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