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3 years ago

6

Please help and thank you

2 answers:

Sedbober [7]3 years ago

7 0

I am pretty sure that the answer is B.

azamat3 years ago

4 0

Answer:

b

Step-by-step explanation:

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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,

Given,

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,

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

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

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

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3 years ago

Which graph represents the compound inequality? n < –2 or n ≥ 4

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In 25 minutes, Jay can run 10 laps around the track. Write an equation to represent the relationship.

Inessa [10]

Answer:

The equation is $y=\frac{2}{5} x$ .

Step-by-step explanation:

Let $y$ represent the number of laps Jay run, and $x$ the number of minutes it takes him to run those laps.

We know that it it takes Jay 25 minutes to run 10 laps; therefore, the number of laps jay runs per minute is:

$\frac{10\:laps}{25\:minutes} =\frac{2}{5}\:laps/min$

So, if Jay runs for $x$ minutes, the number of laps $y$ he will run be:

$\boxed{y=\frac{2}{5}x}$

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Step-by-step explanation:

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