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3 years ago

Line q passes through the point (4,6)and (0,6).

Mathematics

1 answer:

larisa86 [58]3 years ago

8 0

Answer:

The equation of the line passing through the points (4, 6) and (0, 6) will be:

y = 6

The graph of the equation y = 6 is also attached.

Step-by-step explanation:

The slope-intercept form of the line equation

$y = mx+b$

where

m is the slope

b is the y-intercept

Given the points

(4, 6)

(0, 6)

Finding the slope between (4, 6)and (0, 6).

Using the formula

Slope = m = [y₂ - y₁] / [x₂ - x₁]

= [6 - 6] / [0 - 4]

= 0 / -4

= 0

Thus, the slope of the line = m = 0

The slope m = 0 means the line is horizontal. Because the slope of the horizontal line is 0.

As the y-values are not changing with respect to the x-values.

Therefore,

The equation of the line passing through the points (4, 6) and (0, 6) will be:

y = 6

The graph of the equation y = 6 is also attached.

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Answer:

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Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

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Where P, Q are scalar functions

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$\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy$

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths $\large C_1,C_2,C_3,C_4$ which represents the sides of the rectangle.

$\large C_1$ is the line segment from (0,0) to (5,0)

$\large C_2$ is the line segment from (5,0) to (5,1)

$\large C_3$ is the line segment from (5,1) to (0,1)

$\large C_4$ is the line segment from (0,1) to (0,0)

Then

$\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}$

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r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize $\large C_1$ as

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r'(t) = (5,0) so

$\large \displaystyle\int_{C_1}xydx+x^2dy=0$

Now the second integral. We parametrize $\large C_2$ as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

$\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25$

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r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

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xf(x) = x * 1/1.3 = x/1.3

Integrating x/1.3

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E(x) = 3.730 (approximated)

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First, we'll solve F(c)

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