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3 years ago

The altitude to the hypotenuse of a right triangle is the geometric mean between the segments on the hypotenuse

Mathematics

2 answers:

e-lub [12.9K]3 years ago

5 0

<h2>Answer:</h2>

Option: A is the correct answer.

A. always

<h2>Step-by-step explanation:</h2>

Right Triangle Altitude Theorem--

It states that when an altitude is drawn from the the vertex containing the right angle to the opposite side i.e. hypotenuse then the length of the altitude is the geometric means between the segments on the hypotenuse.

i.e. if we consider a right angled triangle ΔCAB with right angle at A.

Then if a altitude AD is drawn from vertex A then we have:

$AD=\sqrt{CD\cdot DB}$

Maslowich3 years ago

3 0

The answer is always

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3 years ago

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3 years ago

What is the solution to this inequality? <br><br>-13x < 65

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3 years ago

Use the method of cylindrical shells to find the volume v generated by rotating the region bounded by the curves about the given

kondor19780726 [428]

Answer:

The volume is $\frac{16\pi}{e}$

Step-by-step explanation:

* Lets talk about the shell method

- The shell method is to finding the volume by decomposing

a solid of revolution into cylindrical shells

- Consider a region in the plane that is divided into thin vertical

rectangle

- If each vertical rectangle is revolved about the y-axis, we

obtain a cylindrical shell, with the top and bottom removed.

- The resulting volume of the cylindrical shell is the surface area

of the cylinder times the thickness of the cylinder

- The formula for the volume will be: $V=\int\limits^b_a {2\pi xf(x)} \, dx$

where 2πx · f(x) is the surface area of the cylinder shell and dx is its

thickness

* Lets solve the problem

- To find the volume V generated by rotating the region bounded

by the curves y = 4e^x and y = 4e^-x about the y-axis by use

cylindrical shells

- Consider that the height of the cylinder is y = (4e^x - 4e^-x)

- Consider that the radius of the cylinder is x

- The limits are x = 0 and x = 1

- Lets take 2π and 4 as a common factor out the integration

∴ $V=\int\limits^1_0 {2\pi x(4e^{x}-4e^{-x})} \, dx$

∴ $V=2\pi(4)\int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx$

- To integrate $xe^{x}$ and $xe^{-x}$ we will use

integration by parts methods $\int\ {uv'=uv-\int{v}\,u' }\,$

∵ u = x

∴ u' = du/dx = 1 ⇒ differentiation x with respect to x is 1

∵ v' = dv/dx = e^x

- The integration e^x is e^x ÷ differentiation of x (1)

∴ $v=\int\ {e^{x}}\, dx= e^{x}$

∴ $\int\ {xe^{x}} \, dx=xe^{x}-\int\ e^{x}\, dx=xe^{x}-e^{x}$

- Similar we will integrate xe^-x

∵ u = x

∴ u' = du/dx = 1

∵ v' = dv/dx = e^-x

- The integration e^-x is e^x ÷ differentiation of -x (-1)

∴ $v=\int\ {e^{-x}} \, dx=-e^{-x}$

∴ $\int\ {x}e^{-x}\, dx=-xe^{-x}+\int\ {e^{-x}} \, dx=-xe^{-x}-e^{-x}$

∴ V = $8\pi \int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx=8\pi[xe^{x}-e^{x}+xe^{-x}+e^{-x}]$ from 0 to 1

- Lets substitute x = 1 minus x = 0

∴ $V=8\pi[(1)(e^{1})-(e^{1})+(1)(e^{-1})+(e^{-1})-(0)(e^{0})+(e^{0})-(0)(e^{0})-(e^{0})]$

∴ $V=8\pi[e^{1}-e^{1}+e^{-1}+e^{-1}-0+1-0-1]=8\pi[2e^{-1}]=16\pi e^{-1}$

∵ $e^{-1}=\frac{1}{e}$

∴ $V=\frac{16\pi}{e}$

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4 years ago

Select the correct answer.

PilotLPTM [1.2K]

Selecciona la respuesta correcta.