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Alexandra [31]
2 years ago
8

What is the volume of a sphere with a radius of 3 meters?

Mathematics
2 answers:
Pachacha [2.7K]2 years ago
8 0
Hi!
It’s 113.1m³
I hope this helps!!
Fantom [35]2 years ago
6 0

Answer:

v=113.1 m^3

Step-by-step explanation:

v=4/3pi(r)^3

4/3pi(3)^3

v=113.09734m^3

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Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x
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{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is  \dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes arey = k \pm \dfrac{b}{a}(x - h)

1. Calculate h and k. The hyperbola is symmetric about the origin, so  

h = 0 and k = 0

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a = 6/2 = 3  

3. For 'b': The equation for the asymptote with the positive slope is  

y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x

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\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}

4.  The equation of the hyperbola is

\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

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