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3 years ago

Just tell me the where the x-axis and y-axis goes pls <3

Mathematics

1 answer:

Mazyrski [523]3 years ago

3 0

Answer:

is it to late?

Step-by-step explanation:

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A man jogs from his home to the nearst market 25km away jogs at an average speed 5km 1h. how long does it taken him to reach the

mr_godi [17]

Answer:

1/5

Step-by-step explanation:

first write the formula of average speed

2) then equate the values of given question
3) you will find the solution

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2 years ago

The first 5 terms of a quadratic sequence are

Mashcka [7]

Step-by-step explanation:

an = 2n^2 - 3 · Anonymous. -1, 5, 15, 29, 47, ... an = 2 n^2 - 3.

8 0

2 years ago

I’m super confused new topic not sure what to do for this one

geniusboy [140]

There’s no question here.

8 0

2 years ago

What’s pi? (i actually kno what it is i’m jus bored lol)

Alenkasestr [34]

Answer:

3.14159 and so on

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

“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:

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

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

4 0

3 years ago