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2 years ago

Which inequality is true? 1.5 greater-than 2 and one-half

Mathematics

1 answer:

Travka [436]2 years ago

7 0

The inequality negative 3 and one-half greater-than negative 4.5 or -3.5 > -4.5 is true.

<h3>What is inequality?</h3>

It is defined as the expression in mathematics in which both sides are not equal they have mathematical signs either less than or greater than known as inequality.

We have inequalities given:

$1.5 > 2\dfrac{1}{2}$

1.5 > 2.5 (false)

1/2 > 0.5

0.5 > 0.5 (false)

-2.5 > -1.5 (false)

$-3 \dfrac{1}{2} > -4.5$

-3.5 > -4.5 (true)

Thus, the inequality negative 3 and one-half greater-than negative 4.5 or -3.5 > -4.5 is true.

Learn more about the inequality here:

brainly.com/question/19491153

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3 years ago

“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

$I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx$

Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

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Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

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Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

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Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

$\implies I=-\dfrac{(t+1)^4}{4t^4}+C$

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