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2 years ago

Helpppp pleaseeee helppppp

Mathematics

1 answer:

Brums [2.3K]2 years ago

6 0

Answer:

2x-1=x+7

-x -x

2x-1=7

+1 +1

2x=7

/2 /2

x=3.5

Step-by-step explanation:

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What is36.8+[11.6-(2.5×3)]2

givi [52]

Answer:

Step-by-step explanation:

The answer is 33.4

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3 years ago

What is Szymborska's overall opinion of humanity in "A Contribution to Statistics?" Support your response with at least two deta

aleksandrvk [35]

Answer:

Szymborska's overall opinion of humanity in "A Contribution to Statistics" is overall positive

Step-by-step explanation:

Szymborska is a poet who likes to comment on society with irony and sarcasm.One example of this is when she splits people into categories in her poem “A Contribution to Statistics”. Here Szymborska’s overall opinion of humanity may seem negative due to her bluntness and dry humor. For example, she says in the tenth stanza, “cruel when forced by circumstances -better not to know even ballpark figures.” It’s implied in this line that most people (maybe even all)are cruel or can be. She also says, “taking only things from life -thirty”which comments how lots of people are selfish. However, this seemingly cynical poem takes a sweeter turn near the end with statistics like, “worthy of compassion - ninety-nine” and “mortal- a hundred out of a hundred.” These lines reveal that even though Szymborska believes that humanity is extremely flawed the mass majority of us deserve and are worthy of love. Plus the fact that we’re all mortal may seem dark but the fact remains that both the righteous and the sinners will all end up in the same place. The fact that no one can escape

death gives humans equality in a way. Szymborska’s overall opinion of humanity turned out to be more optimistic after all.

8 0

3 years ago

Read 2 more answers

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$