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2 years ago

6

20 points five stars a thank you, dont need to awnser first. i will repeat this question every 3 and a half minutes. ive got nin

e hundred points to burn

2 answers:

melomori [17]2 years ago

5 0

Answer:

Thanks, you are so generous

Step-by-step explanation:

klio [65]2 years ago

4 0

Answer:

Thank you so much, you are the best!

Step-by-step explanation:

You might be interested in

To the nearest tenth the value of x that satisfies 2x = −2x + 11 is

olganol [36]

Answer:

2.8 (from 2.75 because of rounding)

Step-by-step explanation:

2x = -2x + 11

Add 2x on both sides

4x = 11

Divide 4 on both sides

x = 11/4

x = 2.75

Round to nearest tenth

x is about 2.8

Please mark me for brainliest :D !! Thanks!

7 0

3 years ago

Read 2 more answers

We are 95% confident that the population proportion in 2000 that supported preferential hiring of women is about 0.0509 and the

alex41 [277]

Answer:

What do Americans think about preferential hiring of women? Has there been a change in the past decade? In 2000(group 1) and 2010(Group 2), the General Social Survey asked participants if they would favor or oppose preferential hiring of women. In 2000, out of 849 respondents, 271 said yes. In 2010, out of 696 respondents, 225 said yes. The 95% confidence interval for p1-p2 is (-0.0509, 0.04273). What would be an appropriate conclusion? O We are 95% confident that the population proportion in 2010 that supported preferential hiring of women is about 0.0509 and the population proportion in 2000 is 0.04723. We are 95% confident that the pulation proportion in 2000 that supported preferential hiring of women is about 0.0509 and the population proportion in 2010 is 0.04723. We are 95% confidence that the population proportion in 2010 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2000. O We are 95% confidence that the population proportion in 2000 that supported preferential hiring of women is between 0.0509 less to 0.04273 more than the population proportion in 2010.

Step-by-step explanation:

5 0

3 years ago

What are the domain and range of each relation? Drag the answer into the box to match each relation. A mapping diagram. Element

ra1l [238]

Answer:

Domain: {−2, −1, 0, 2} Range: {−4, −2, 2}

Step-by-step explanation:

We have been given a mapping diagram as shown below:

X Y

-2 -4

-1 -2

0 2

2

For better view, you can check the attached mapping diagram.

From that diagram we have to find domain and range.

Domain contains only x values so domain will be:

Domain: {-2,-1,0,2}

Range contains only y values so range will be:

Range: {-4,-2,2}

We see that only third choice matches obtained values of domain and range.

Hence final answer is Domain: {−2, −1, 0, 2} Range: {−4, −2, 2}

7 0

3 years ago

Solve<br>4(2x - 3) = 6x + 2

vaieri [72.5K]

｡☆✼★ ━━━━━━━━━━━━━━ ☾

4(2x - 3) = 6x + 2

Expand it:

8x - 12 = 6x + 2

+ 12 to both sides

8x = 6x + 14

- 6x from both sides

2x = 14

Divide both sides by 2:

x = 7

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

｡☆✼★ ━━━━━━━━━━━━━━ ☾

4 0

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:

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

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

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