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3 years ago

Long division 236 divided 4

Mathematics

2 answers:

madreJ [45]3 years ago

6 0

The answer would be 59 if you work our your division and do it correctly

vesna_86 [32]3 years ago

3 0

The corret answer is <u><em>59</em></u>

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“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:

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

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

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.)

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

Sustituye:

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

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

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

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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3 years ago

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luda_lava [24]

Answer:

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Step-by-step explanation:

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3 years ago

Name the intersection of AE and CG

Pavel [41]

Is the supposed to be a image

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2 years ago

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s344n2d4d5 [400]

Answer:

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Step-by-step explanation:

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3 years ago

PLEASE HELP ME! FAST

IceJOKER [234]

The quadrants in which all the coordinates given are located is; As explained below.

<h3>How to Identify Quadrants in coordinates?</h3>

1) (2, 4) is located in Quadrant I where both x and y-values are positive.

2) (0, -3) is located in Quadrant II where x - values are positive but y-values are negative.

3) (-1, 1/2) is located in Quadrant IV.

4) (-2 1/2, -7) is located in Quadrant III where x and y values are both negative.

5) (0, 6) is located in Quadrant I where both x and y-values are positive.

6) (-5, 0) is located in Quadrant IV where x is negative but y is positive.

Read more about Quadrant Coordinates at; brainly.com/question/863849

#SPJ1

7 0

2 years ago