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3 years ago

10

a container built in the shape of a cylinder has a circular base with a radius of 1.5 meters. the altitude of the container is 3

.5 meters

2 answers:

HACTEHA [7]3 years ago

8 0

Answer:

im assuming you need the volume of this container:

$Volume=\pi r^{2} h$

$Volume=\pi (1.5^{2} )(3.5)=24.7$

nadezda [96]3 years ago

4 0

Answer: Volume of the container:24.74 cubic meters

Step-by-step explanation:

Hi, since the values given are the radius and the altitude (height) the question is probably asking for the volume of the container.

So, we have to apply the volume formula for a cylinder:

Volume = π radius^2 height

V = π r^2 h

Replacing with the values given and solving for V:

V = π 1.5^2 (3.5) = π (2.25) (3.5) = 24.74 cubic meters .

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HELP! SIMPLIFY!

Vitek1552 [10]

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m⁴ + m³ = m⁷

n⁹ - n² = n⁷

So, our answer for #1 is m⁷n⁷

2) 6a²b³
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Please read rules in #1.

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2 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{e^{-3s}}{s+1}]$

$L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t$

$= e^{-t-3} \left \{ {{1 \ \ \ \ \ t>3} \atop {0 \ \ \ \ \ t$

= $e^{-(t-3)} u (t-3)$

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$y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$

Then

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$y(t) = 7 -7e^{-t} +e^{-t} e^{-3} u (t-3)$

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2 years ago

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spayn [35]

The inequality is solved, and then, the graph is given at the end of this question.

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A similar problem is given at brainly.com/question/24132092

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1 year ago

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dolphi86 [110]

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3 years ago