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Alexus [3.1K]
3 years ago
10

What is the approximate circumference of the circle shown below?

Mathematics
1 answer:
Serggg [28]3 years ago
4 0
Hey, there’s no circle for me to look at unfortunately!
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Which is the correct step below for using the grouping method to factor the following polynomial? 8x - 6 - 12x2 + 9x
IrinaVladis [17]
8x - 6 - 12x^2 + 9x

2(4x-3)-3x(3x-3)

=> (2-3x)(3x-3)

=> 3(x-1)(2-3x)
4 0
3 years ago
Read 2 more answers
How to write 8/15 as a decimal?
Murrr4er [49]

Answer:

0.53

Step-by-step explanation:

Basically just go to your calculator and put in 8 divded by 15

3 0
3 years ago
Please help me with the question
nikdorinn [45]

Answer:

4p,q+r

4p(q+r)

4pq+4pr

3 0
3 years ago
1. There are 3,768 books on the library wall.
guapka [62]

Answer:

157 books in each row

Step-by-step explanation:

24 rows=3768 books

1 row=3768/24 books

=157 books

Therefore,Each row contains 157 books

Mark this answer as brainliest....

3 0
2 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
3 years ago
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