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Katyanochek1 [597]
3 years ago
5

find an equation of the line that has a y-intercept of -4 that is parallel to the graph of the line y=6

Mathematics
1 answer:
mixas84 [53]3 years ago
3 0

Answer:

y = -4

Step-by-step explanation:

y = 6 is line that is parallel to x-axis

So, the required is also parallel to x -axis

The equation of line parallel to x-axis , y = c, where c is y-intercept

y = - 4

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Cone details:

  • height: h cm
  • radius: r cm

Sphere details:

  • radius: 10 cm

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From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

<u>Using Pythagoras Theorem</u>

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r² = 10² - (h-10)²                                     [change sides]

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===========================

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (\sqrt{20h - h^2})^2  \  ( h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20h - h^2)  (h)

\longrightarrow \sf V = \dfrac{1}{3}  * \pi  * (20 - h) (h) ( h)

\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)

To find maximum/minimum, we have to find first derivative.

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<u>First derivative</u>

\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )

<u>apply chain rule</u>

\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}

<u>Equate the first derivative to zero, that is V'(x) = 0</u>

\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0

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\Longrightarrow \sf  h=0,\:h=\dfrac{40}{3}<u />

<u>maximum volume:</u>                <u>when h = 40/3</u>

\sf \Longrightarrow max=  \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )

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\sf \Longrightarrow min=  \dfrac{1}{3} \pi (0)^2(20-0)

\sf \Longrightarrow minimum=0 \ cm^3

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