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2 years ago

15

PLEASE HELP ME ASAP!!! TY :)

1 answer:

Nana76 [90]2 years ago

7 0

The answer is B I had a question similar to that last week

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26. Secondary deviance marks the start of what Erving Goffman called a deviant career, which results in the acquisition of a ___

sleet_krkn [62]

Answer:

d) stigma

Step-by-step explanation:

Secondary deviance marks the start of what Erving Goffman called a deviant career, which results in the acquisition of a stigma.

6 0

3 years ago

One more repair is needed to get the ferris wheel up and running! This one will require a cable to connect chair L to the ground

Elodia [21]

Answer:

<h2>15°</h2>

Step-by-step explanation:

We know that the central angle of a circle is 360°.

In this case, we have the circle divided into 12 equal parts, so each angle would be also the 12th part of 360°.

$\theta = \frac{360\° }{12}= 30\°$

Each circular sector has an angle of 30°, and each arc is equal to 30°.

Now, notice that tha angle formed by the two blue cables is an external angle formed by two secants, that means we can use the following relation

$\alpha = \frac{1}{2}(60\° - 30\°)$

Where $\alpha$ is the angle formed by the two blue cables.

$\alpha = \frac{1}{2}(30\°)=15\°$

Therefore, the angle measures 15°.

4 0

3 years ago

URGENT<br>simplify (sinx+cosx)^2-(sinx-cosx)^2

koban [17]

Answer:

Solution given:

(sinx+cosx)^2-(sinx-cosx)^2
sin²x+2sinxcosx+cos²x-(sin²x-2sinxcosx+cos²)
1+2sinxcosx-1+2sinxcosx
2(2sinxcosx)
2sin2x

Step-by-step explanation:

[note: sin²x+cos²x=1]

3 0

3 years ago

Read 2 more answers

Find the product of 27×6×60 then explain how you would use mental math to find the product of 27×600

Ne4ueva [31]

27x6x60=9,720 *You can use mental math by multiplying 27 by 6 and adding the 2 zeros to the end of the product*

7 0

3 years ago

Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2

NNADVOKAT [17]

$y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}$

Divide both sides by $\dfrac13y^{-2}(x)$ :

$3y^2y'-y^3=xe^x\ln x$

Substitute $v(x)=y(x)^3$ , so that $v'(x)=3y(x)^2y'(x)$ .

$v'-v=xe^x\ln x$

Multiply both sides by $e^{-x}$ :

$e^{-x}v'-e^{-x}v=x\ln x$

The left side can be condensed into the derivative of a product.

$(e^{-x}v)'=x\ln x$

Integrate both sides to get

$e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C$

Solve for $v(x)$ :

$v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

Solve for $y(x)$ :

$y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x$

$\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}$

4 0

3 years ago