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Kamila [148]
3 years ago
13

Find -8+(-13) _______ (-6)°2

Mathematics
2 answers:
Ray Of Light [21]3 years ago
7 0

The answer would be -7/12. ( sorry if wrong ) have a nice night.


olchik [2.2K]3 years ago
4 0

the answer is -8.36
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What are the roots of the polynomial equation? x² + 7x + 12 = 0 Enter your answers in the boxes. x1= x2=
oksian1 [2.3K]
The equation factors as
.. (x +3)(x +4) = 0
By the zero-product rule, the roots are
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Please hurry I need an answer for this quiz
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I think the answer would be C.

Step-by-step explanation:

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What is the slope of a line that passes through the origin<br> and the point (6, 14)?
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Answer:

14/6 or 2 1/3

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Identify as a direct variation, inverse variation or neither.
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inverse

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if you add the number of men, the number of days decrease, making it an inverse variation; as one number increases, the other decreases.

4 0
3 years ago
1:Under what condition will the line px+py+r=0 mat be a normal to the circke x²+y²+2gx+2fy+c=0
ahrayia [7]

Answer:

<h3>#1</h3>

The normal overlaps with the diameter, so it passes through the center.

<u>Let's find the center of the circle:</u>

  • x² + y² + 2gx + 2fy + c = 0
  • (x + g)² + (y + f)² = c + g² + f²

<u>The center is:</u>

  • (-g, -f)

<u>Since the line passes through (-g, -f) the equation of the line becomes:</u>

  • p(-g) + p(-f) + r = 0
  • r = p(g + f)

This is the required condition

<h3>#2</h3>

Rewrite equations and find centers and radius of both circles.

<u>Circle 1</u>

  • x² + y² + 2ax + c² = 0
  • (x + a)² + y² = a² - c²
  • The center is (-a, 0) and radius is √(a² - c²)

<u>Circle 2</u>

  • x² + y² + 2by + c² = 0
  • x² + (y + b)² = b² - c²
  • The center is (0, -b) and radius is √(b² - c²)

<u>The distance between two centers is same as sum of the radius of them:</u>

  • d = √(a² + b²)

<u>Sum of radiuses:</u>

  • √(a² - c²) + √(b² - c²)

<u>Since they are same we have:</u>

  • √(a² + b²) = √(a² - c²) + √(b² - c²)

<u>Square both sides:</u>

  • a² + b² = a² - c² + b² - c² + 2√(a² - c²)(b² - c²)
  • 2c² = 2√(a² - c²)(b² - c²)

<u>Square both sides:</u>

  • c⁴ = (a² - c²)(b² - c²)
  • c⁴ = a²b² - a²c² - b²c² + c⁴
  • a²c² + b²c² = a²b²

<u>Divide both sides by a²b²c²:</u>

  • 1/a² + 1/b² = 1/c²

Proved

6 0
3 years ago
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