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marissa [1.9K]
3 years ago
13

2) Given: lines a and b are parallel Prove: m<1 = m<8

Mathematics
1 answer:
monitta3 years ago
6 0

Explanation:

<u><em>* Demonstrate that the related angles are the same.</em></u>

<u><em>* Demonstrate that different interior angles are equivalent.</em></u>

<u><em>* Demonstrate the adjacent interior angles are supplementary.</em></u>

<u><em>* Demonstrate the adjacent external angles are not mutually exclusive.</em></u>

<u><em>* Show that the lines in a plane are perpendicular to the same axis.</em></u>

<u><em></em></u>

<1 and <8 are the same because they are alternate exterior angles that means that they are

Opposite sides of the transversal.

And that they are on the exterior side of the parallel lines.

Now because of that they have the same measures no matter what so would

<2 and <7

<3 and <6

<4 and <5

<u><em></em></u>

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Consider the oriented path which is a straight line segment L running from (0,0) to (16, 16 (a) Calculate the line integral of t
aalyn [17]

This question is missing some parts. Here is the complete question.

Consider the oriented path which is a straight line segment L running from (0,0) to (16,16).

(a) Calculate the line inetrgal of the vector field F = (3x-y)i + xj along line L using the parameterization B(t) = (2t,2t), 0 ≤ t ≤ 8.

Enter an exact answer.

\int\limits_L {F} .\, dr =

(b) Consider the line integral of the vector field F = (3x-y)i + xj along L using the parameterization C(t) = (\frac{t^{2}-256}{48} ,\frac{t^{2}-256}{48} ), 16 ≤ t ≤ 32.

The line integral calculated in (a) is ____________ the line integral of the parameterization given in (b).

Answer: (a) \int\limits_L {F} .\, dr = 384

              (b) the same as

Step-by-step explanation: <u>Line</u> <u>Integral</u> is the integral of a function along a curve. It has many applications in Engineering and Physics.

It is calculated as the following:

\int\limits_C {F}. \, dr = \int\limits^a_b {F(r(t)) . r'(t)} \, dt

in which (.) is the dot product and r(t) is the given line.

In this question:

(a) F = (3x-y)i + xj

r(t) = B(t) = (2t,2t)

interval [0,8] are the limits of the integral

To calculate the line integral, first substitute the values of x and y for 2t and 2t, respectively or

F(B(t)) = 3(2t)-2ti + 2tj

F(B(t)) = 4ti + 2tj

Second, first derivative of B(t):

B'(t) = (2,2)

Then, dot product between F(B(t)) and B'(t):

F(B(t))·B'(t) = 4t(2) + 8t(2)

F(B(t))·B'(t) = 12t

Now, line integral will be:

\int\limits_C {F}. \, dr = \int\limits^8_0 {12t} \, dt

\int\limits_L {F}. \, dr = 6t^{2}

\int\limits_L {F.} \, dr = 6(8)^{2} - 0

\int\limits_L {F}. \, dr = 384

<u>Line integral for the conditions in (a) is 384</u>

<u />

(b) same function but parameterization is C(t) = (\frac{t^{2}-256}{48}, \frac{t^{2}-256}{48} ):

F(C(t)) = \frac{t^{2}-256}{16}-\frac{t^{2}-256}{48}i+ \frac{t^{2}-256}{48}j

F(C(t)) = \frac{2t^{2}-512}{48}i+ \frac{t^{2}-256}{48} j

C'(t) = (\frac{t}{24}, \frac{t}{24} )

\int\limits_L {F}. \, dr = \int\limits {(\frac{t}{24})(\frac{2t^{2}-512}{48})+ (\frac{t}{24} )(\frac{t^{2}-256}{48})  } \, dt

\int\limits_L {F} .\, dr = \int\limits^a_b {\frac{t^{3}}{384}- \frac{768t}{1152} } \, dt

\int\limits_L {F}. \, dr = \frac{t^{4}}{1536} - \frac{768t^{2}}{2304}

Limits are 16 and 32, so line integral will be:

\int\limits_L {F} \, dr = 384

<u>With the same function but different parameterization, line integral is the same.</u>

6 0
3 years ago
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