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Gnoma [55]
3 years ago
14

Solve x^2 + 5x + 6= 0

Mathematics
2 answers:
Anestetic [448]3 years ago
6 0

x = -3 or x = -2

Step-by-step explanation:

x² + 5x + 6 = 0

(x + 3)(x + 2)

x = -3 \/ x = -2

kodGreya [7K]3 years ago
6 0

Answer:

x₁ = (-2) , & x₂ = (-3).

Step-by-step explanation:

2² + 5x + 6 = 0

/\

2 3

x₂

  • (x + 2)
  • x + 2 = 0
  • x₁ = (-2)

x₁

  • (x + 3)
  • x + 3 = 0
  • x₂ = (-3)
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The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi
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Answer:

The height of cone is decreasing at a rate of 0.085131 inch per second.        

Step-by-step explanation:

We are given the following information in the question:

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525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}

Differentiating with respect to t,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)

Putting all the values, we get,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131

Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.

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