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Alborosie
3 years ago
11

When dropped, a super ball will bounce back to 80% of its peak height,continuing on in this way for each bounce.

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
3 0

An expression for the height of the nth bounce is 0.80X^N = Height.

<h3><u>Equations</u></h3>

Since when dropped, a super ball will bounce back to 80% of its peak height, continuing on in this way for each bounce, to determine an expression for the height of the nth bounce the following calculation must be performed:

  • X = Initial value
  • 80% = 0.80
  • N = Number of times the ball bounces
  • 0.80X^N = Height

Therefore, an expression for the height of the nth bounce is 0.80X^N = Height.

Learn more about equations in brainly.com/question/2263981

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Answer:0.25

Step-by-step explanation: 4/16 can be simplified to 1/4. Think about it like quarters. You need 4 quarters to have one dollar. If you have one quarter it is equal to $0.25.

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3x-9 when x=3<br> Solve this
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Answer:

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Step-by-step explanation:

3x - 9 if x = 3

Substitute:

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9 - 9 = 0

This can be written as a function: f(x) = 3x - 9

f(3) = 3x - 9

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3 years ago
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3 years ago
. A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum
Drupady [299]

Answer:

r=1.05\ \text{ft}

h=2.12\ \text{ft}

20.91\ \text{ft}^3

Step-by-step explanation:

r = Radius

h = Height

Volume of cylinder = 7.35\ \text{ft}^3

V=\pi r^2h\\\Rightarrow h=\dfrac{V}{\pi r^2}\\\Rightarrow h=\dfrac{7.35}{\pi r^2}

Surface area is given by

A=2\pi r^2+2\pi rh\\\Rightarrow A=2\pi r^2+2\pi r\dfrac{7.35}{\pi r^2}\\\Rightarrow A=2\pi r^2+\dfrac{14.7}{r}

Differentiating with respect to radius we get

\dfrac{dA}{dr}=4\pi r-\dfrac{14.7}{r^2}

Equating with zero we get

0=4\pi r-\dfrac{14.7}{r^2}\\\Rightarrow 4\pi r=\dfrac{14.7}{r^2}\\\Rightarrow r^3=\dfrac{14.7}{4\pi}\\\Rightarrow r=(\dfrac{14.7}{4\pi})^{\dfrac{1}{3}}\\\Rightarrow r=1.05

\dfrac{d^2A}{dr^2}=4\pi-2\times \dfrac{14.7}{r^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=4\pi+2\times \dfrac{14.7}{1.05^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=37.96>0

So, the value of the function is minimum at r=1.05

h=\dfrac{7.35}{\pi r^2}=\dfrac{7.35}{\pi 1.05^2}\\\Rightarrow h=2.12

So, the radius and height which would minimize the surface area is 1.05 feet and 2.12 feet respectively.

Surface area

A=2\pi r^2+2\pi rh\\\Rightarrow A=2\pi \times 1.05^2+2\pi 1.05\times 2.12\\\Rightarrow A=20.91\ \text{ft}^3

The minimum surface area is 20.91\ \text{ft}^3.

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