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earnstyle [38]
3 years ago
6

Please help... ABCD is a quadrilateral​

Mathematics
1 answer:
IgorC [24]3 years ago
3 0
A ———————————————————
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Which parent function is represented by the graph?
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Linear

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Use greatest common factor and distributive property to write equivalent expressions 5x+20y
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A container has 200 liters and is losing 6 liters per minute. How many will it have after 10 minutes
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140 liters will remain after 10 mins

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A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la
SOVA2 [1]

Answer:

8\pi\text{ square cm}

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

S=2\pi rh

Where,

r = radius,

h = height,

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Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

4^2 = (2r)^2 + h^2   ( see in the below diagram ),

16 = 4r^2 + h^2

16 - 4r^2 = h^2

\implies h=\sqrt{16-4r^2}

Thus, the surface area of the cylinder,

S=2\pi r(\sqrt{16-4r^2})

Differentiating with respect to r,

\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})

=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})

=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})

Again differentiating with respect to r,

\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})

For maximum or minimum,

\frac{dS}{dt}=0

2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0

-8r^2 + 16 = 0

8r^2 = 16

r^2 = 2

\implies r = \sqrt{2}

Since, for r = √2,

\frac{d^2S}{dt^2}=negative

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

S = 2\pi (\sqrt{2})(\sqrt{16-8})

=2\pi (\sqrt{2})(\sqrt{8})

=2\pi \sqrt{16}

=8\pi\text{ square cm}

4 0
3 years ago
¿Cual es el perímetro de un hexágono regular cuyo lado mide 16,5cm?
alukav5142 [94]

Step-by-step explanation:

6(16.5 cm) = 99 cm

6 0
3 years ago
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