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luda_lava [24]
3 years ago
12

Plz help ASAP really need it

Mathematics
2 answers:
Keith_Richards [23]3 years ago
8 0

fjkenfnrfgjvhdfbvourgfn eiurfkjeffujkkf, sufjksfdsf

Kruka [31]3 years ago
3 0

Answer: Both of the ones on the right side of the answers

Hope this helped :)

Please mark me as Brainliest

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You start at (8,0). You move left 6 units. Where do you end?​
tamaranim1 [39]

Answer:

(2, 0)

Step-by-step explanation:

Moving left just subtracts from the x coordinate (example: left 8 subtracts 8 from the x coordinate)

Brₐinliest plz so close to leveling up

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3 years ago
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The missing term in the denominator is
Margaret [11]

\frac{7^{16} }{7^{12} }= \frac{7^{16-34} }{7^{12-34} } =\frac{7^{-18} }{7^{-22} } = 2401

so the missing term in the denominator is 7^(-22)

ok done. Thank to me :>

5 0
3 years ago
How do I graph this?
antoniya [11.8K]

Answer:

Step-by-step explanation:

You would simply change the x value s to its opposite and keep the y values the same.

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Find the x-intercepts y=2x^2 + 5x + 2/x^2-4x+3
max2010maxim [7]

Answer:

{ \tt{y = \frac{2 {x}^{2}  + 5x + 2}{ {x}^{2} - 4x + 3 } }} \\ x - intercept : y = 0 \\ { \tt{ \frac{2 {x}^{2} + 5x + 2 }{ {x}^{2} - 4x + 3  } = 0 }} \\  \\ { \tt{2 {x}^{2}  + 5x + 2 = 0}} \\ x =  \frac{1}{2}  \:  \: and \:  \: x =  - 2

7 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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