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Nana76 [90]
3 years ago
15

Missing angles in triangle help

Mathematics
2 answers:
Elena L [17]3 years ago
7 0

Answer:

E=35 complete the 90

G=55 complete the 180

F=35 complete the 90

J=15 complete the 90

H = 45 complete the 180

I=45 complete the 90

M=120 complete the 180

K=15 complete the 180

L=75 complete the 90

Q=55 complete the 90

R=45 complete the 180

S=42 complete the 90

T=42 complete the 180

P=21 complete the 90

O=69 complete the 180

N=21 complete the 90

A and B and C = 60

U =69

D=42 complete the 180

hopefully i didnt miss any thing and do it correctly

Dmitry [639]3 years ago
6 0

Answer:

which one?

Step-by-step explanation:

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Hello,
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4 years ago
Find the work required to move an object in the force field F = ex+y <1,1,z> along the straight line from A(0,0,0) to B(-1
storchak [24]

Answer:

Work = e+24

F is not conservative.

Step-by-step explanation:

To find the work required to move an object in the force field  

\large F(x,y,z)=(e^{x+y},e^{x+y},ze^{x+y})

along the straight line from A(0,0,0) to B(-1,2,-5), we have to parameterize this segment.

Given two points P, Q in any euclidean space, you can always parameterize the segment of line that goes from P to Q with

r(t) = tQ + (1-t)P with 0 ≤ t ≤ 1

so  

r(t) = t(-1,2,-5) + (1-t)(0,0,0) = (-t, 2t, -5t)  with 0≤ t ≤ 1

is a parameterization of the segment.

the work W required to move an object in the force field F along the straight line from A to B is the line integral

\large W=\int_{C}Fdr

where C is the segment that goes from A to B.

\large \int_{C}Fdr =\int_{0}^{1}F(r(t))\circ r'(t)dt=\int_{0}^{1}F(-t,2t,-5t)\circ (-1,2,-5)dt=\\\\=\int_{0}^{1}(e^t,e^t,-5te^t)\circ (-1,2,-5)dt=\int_{0}^{1}(-e^t+2e^t+25te^t)dt=\\\\\int_{0}^{1}e^tdt-25\int_{0}^{1}te^tdt=(e-1)+25\int_{0}^{1}te^tdt

Integrating by parts the last integral:

\large \int_{0}^{1}te^tdt=e-\int_{0}^{1}e^tdt=e-(e-1)=1

and  

\large \boxed{W=\int_{C}Fdr=e+24}

To show that F is not conservative, we could find another path D from A to B such that the work to move the particle from A to B along D is different to e+24

Now, let D be the path consisting on the segment that goes from A to (1,0,0) and then the segment from (1,0,0) to B.

The segment that goes from A to (1,0,0) can be parameterized as  

r(t) = (t,0,0) with 0≤ t ≤ 1

so the work required to move the particle from A to (1,0,0) is

\large \int_{0}^{1}(e^t,e^t,0)\circ (1,0,0)dt =\int_{0}^{1}e^tdt=e-1

The segment that goes from (1,0,0) to B can be parameterized as  

r(t) = (1-2t,2t,-5t) with 0≤ t ≤ 1

so the work required to move the particle from (1,0,0) to B is

\large \int_{0}^{1}(e,e,-5et)\circ (-2,2,-5)dt =25e\int_{0}^{1}tdt=\frac{25e}{2}

Hence, the work required to move the particle from A to B along D is

 

e - 1 + (25e)/2 = (27e)/2 -1

since this result differs from e+24, the force field F is not conservative.

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