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3 years ago

What is bigger 1/3 or 3/5

Mathematics

1 answer:

raketka [301]3 years ago

6 0

$Method\ 1:$

$\dfrac{1}{3}=\dfrac{1\cdot5}{3\cdot5}=\dfrac{5}{15}\\\\\dfrac{3}{5}=\dfrac{3\cdot3}{5\cdot3}=\dfrac{9}{15}\\\\\dfrac{3}{15} < \dfrac{9}{15}\ therefore\ \dfrac{3}{5}\ is\ bigger$

$Method\ 2:$

$\dfrac{1}{3} < \dfrac{1}{2}=\dfrac{1.5}{3}\\\\\dfrac{3}{5} > \dfrac{1}{2}=\dfrac{2.5}{5}\\\\therefore\ \dfrac{3}{5}\ is\ bigger$

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Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.

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Answer:

The answer is " $cirtical\ points \ (x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$ "

Step-by-step explanation:

Given:

$g(x) = (x-e)^3e^{-(x-e)}$

Find critical points:

$g(x) = (x-e)^3e^{(e-x)}$

differentiate the value with respect of x:

$\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r} \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]$

critical points $g'(x)=0$

$\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3$

So,

The critical points of $(x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$

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3 years ago

Convert and explain 28% to decimal and fraction

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4 years ago

Given: KAL=100, L=25, and OA=21. Find: AK,AL, and KL

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The measure of AK, AL and KL are 19.66, 10.14 and 23.64 respectively

<h3>Circle Geometry</h3>

Given the following parameters

KAL=100, L=25, and OA=21.

Dtermine the measure of m∠AKL

m∠AKL = 180° - 100° - 25° = 55°

m∠AKL = 55 degrees

Since the line KL is a chord, hence the triangle in the circle is isosceles with AO = LO = 12

Given that ∠AOL is a central angle. ∠AKL is an inscribed angle, intersecting the same ark of the circle.

m∠AOL = 2(m∠AKL) = 2 * 55° = 110°

Using the Law of Cosines to determine the value of AL

AL = √(122 + 122 - 2•122•cos(110°) ≅ 19.659649

AL = 19. 66

Similarly for AK

AK = √(122 + 122 - 2•122•cos(50°)

AK ≅ 10.14

We will find the length of LK from ΔLAK.

LK = √(AL2 + AK2 - 2•AL•AK•cos(100°)

LK ≅ 23.64

Hence the measure of AK, AL and KL are 19.66, 10.14 and 23.64 respectively

Learn more on law of cosines here:brainly.com/question/7872492

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Answer with Step-by-step explanation:

We are given that a function is a continuous on R

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Let H be a closed set and function is continuous then R-H is a opens set

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When function is continuous then inverse image of open set is open

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Conversely,

Let inverse image of closed set H is closed

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When inverse image of closed set is closed then R-inverse image of H is opens set

When inverse image of open set is open then the function is continuous.

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