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Nataly [62]
3 years ago
11

A cylinder and a cone have the same diameter: 8 inches. The height of the cylinder is 3 inches. The height of the cone is 18 inc

hes.
Use π = 3.14.

What is the relationship between the volume of this cylinder and this cone? Explain your answer by determining the volume of each and comparing them. Show all your work.
Mathematics
1 answer:
mina [271]3 years ago
8 0

Vcylinder=\pi r^2 h

Vcone=\frac{1}{3} \pi r^2 h


if we compute the volumes

given that d=8 and height cylinder=3 and height cone=18

note that d/2=r so 8/2=4=r


Vcylidner=[/tex] \pi (4)^2* 3[/tex]=\pi (48)=150.72 cubic inches

Vcone=\frac{1}{3} \pi (4)^2 * 18=\pi (96)=301.44 cubic inches


hmm, if we divide the volume of the cone by thhe volume of the cylinder, we get 301.44/150.72=2


so the relationship is that the cylinder volume is 2 times that of the volume of the cone

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  0, for q ≠ 0 and q ≠ 1

Step-by-step explanation:

Assuming q ≠ 0, you want to find the value of x such that ...

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This is solved using logarithms.

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  x·log(q) = log(1) = 0

The zero product rule tells us this will have two solutions:

  x = 0

  log(q) = 0   ⇒   q = 1

If q is not 0 or 1, then its value is 1 when raised to the 0 power. If q is 1, then its value will be 1 when raised to <em>any</em> power.

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<em>Additional comment</em>

The applicable rule of logarithms is ...

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2 years ago
What is 0.6767 as a simplified fraction
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6 0
3 years ago
D(x)=(x^2-12x+20)/(3x)
krok68 [10]

Answers:

Vertical asymptote: x = 0

Horizontal asymptote: None

Slant asymptote: (1/3)x - 4

<u>Explanation:</u>

d(x) = \frac{x^{2}-12x+20}{3x}

      = \frac{(x-2)(x - 10)}{3x}

Discontinuities: (terms that cancel out from numerator and denominator):

Nothing cancels so there are NO discontinuities.

Vertical asymptote (denominator cannot equal zero):

3x ≠ 0  

<u>÷3</u>   <u>÷3 </u>

x ≠ 0

So asymptote is to be drawn at x = 0

Horizontal asymptote (evaluate degree of numerator and denominator):

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so there is NO horizontal asymptote but slant (oblique) must be calculated.

Slant (Oblique) Asymptote (divide numerator by denominator):

  •        <u>(1/3)x - 4    </u>
  •    3x)    x² - 12x + 20
  •             <u>x²        </u>
  •                  -12x
  •                  <u>-12x         </u>
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So, slant asymptote is to be drawn at (1/3)x - 4



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Answer:

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