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Trava [24]
3 years ago
14

Hey guys I need your help on this!!

Mathematics
1 answer:
kogti [31]3 years ago
7 0

Answer:

Domain: (-∞, ∞) or All Real Numbers

Range: (0, ∞)

Asymptote: y = 0

As x ⇒ -∞, f(x) ⇒ 0

As x ⇒ ∞, f(x) ⇒ ∞

Step-by-step explanation:

The domain is talking about the x values, so where is x defined on this graph? That would be from -∞ to ∞, since the graph goes infinitely in both directions.

The range is from 0 to ∞. This where all values of y are defined.

An asymptote is where the graph cannot cross a certain point/invisible line. A y = 0, this is the case because it is infinitely approaching zero, without actually crossing. At first, I thought that x = 2 would also be an asymptote, but it is not, since it is at more of an angle, and if you graphed it further, you could see that it passes through 2.

The last two questions are somewhat easy. It is basically combining the domain and range. However, I like to label the graph the picture attached to help even more.

As x ⇒ -∞, f(x) ⇒ 0

As x ⇒ ∞, f(x) ⇒ ∞

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Colt1911 [192]

Answer:

7/10

51/100

473/1000

27/100

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Step-by-step explanation:

im curious about what the dots are in .16.49 and .5.37, why are there two dots?

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The ubiquitous 12oz aluminum cans used to distribute drinks in this country have a diameter of approximately 2.75 inches and a h
gayaneshka [121]

Answer:

3.5%

Step-by-step explanation:

The volume of a cylinder = \pi r^2h

<em>r</em> = radius of cylinder,

<em>h</em> = height of cylinder

For the non-optimal can,

<em>r</em> = 2.75/2 = 1.375

<em>h</em> = 5.0

V = \pi(1.375^2)\times 5.0 = 9.453125\pi

<em />

For the optimal can,

<em>d</em>/<em>h</em> = 1,

<em>d</em> = <em>h</em>

2<em>r </em>=<em> h</em>

<em>r</em> = h/2

V = \pi\left(\dfrac{h}{2}\right)^2\times h = \pi\left(\dfrac{h^3}{4}\right)

They have the same volume.

<em />\pi\dfrac{h^3}{4} = 9.453125\pi<em />

h^3 = 37.8125

h=3.36 (This is the height of the optimal can)

r = \dfrac{3.36}{2} = 1.68 (This is the radius of the optimal can)

The area of a cylinder is

<em />A = 2\pi r(r+h)<em />

For the non-optimal can,

A = 2\pi\times\dfrac{2.75}{2}\left(\dfrac{2.75}{2}+5.0\right) = 17.53125\pi

For the optimal can,

A = 2\pi\times1.68\left(1.68+3.36\right) = 16.9344\pi

Amount of aluminum saved, as a percentage of the amount used to make the optimal cans = \dfrac{17.53125\pi - 16.9344\pi}{16.9344\pi}\times 100\% = 3.5\%

5 0
3 years ago
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