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Andru [333]
2 years ago
12

HELP!!! Which of the following represents the asymptote of the graph

Mathematics
1 answer:
katen-ka-za [31]2 years ago
7 0
X=2 or y=-1
these are the points that the graph avoids
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How many students tired out for the volleyball team is this a statistical or no
vodomira [7]

That question is not a statistical question since there will be only one answer. If it were to be statistical it would have multiple answers an example of that type of question would be, Why did you decide to try out for the volleyball team? This is statistical because some could say they joined for fun, credits, to do something active, etc.. there would be multiple answers. Your question, How many students tried out for the volleyball team isn't statistical since it will have one answer such as, 26 students, 15 student, 2 students, etc...

Hope this helped!

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3 years ago
The number 1,134 is divisible by all of the following except
Rasek [7]

Answer:

1,134 is not divisible by 12

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2 years ago
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If f(x)=1/9x-2 what is f^-1(x)
Katarina [22]

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3 years ago
Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
babunello [35]

\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k

We want to find f(x,y,z) such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=x^2+y

\dfrac{\partial f}{\partial y}=y^2+x

\dfrac{\partial f}{\partial z}=ze^z

Integrating both sides of the latter equation with respect to z tells us

f(x,y,z)=e^z(z-1)+g(x,y)

and differentiating with respect to x gives

x^2+y=\dfrac{\partial g}{\partial x}

Integrating both sides with respect to x gives

g(x,y)=\dfrac{x^3}3+xy+h(y)

Then

f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)

and differentiating both sides with respect to y gives

y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C

So the scalar potential function is

\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}

By the fundamental theorem of calculus, the work done by \vec F along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it L) in part (a) is

\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}

and \vec F does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them L_1 and L_2) of the given path. Using the fundamental theorem makes this trivial:

\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3

\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4

8 0
2 years ago
Sydney needs to earn $35 so she can buy her father a birthday present. her mom said she can make $6 per hour cleaning around the
grin007 [14]
K = 13The smallest zero or root is x = -10
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3 years ago
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