8+0=0+8 is what property
2 answers:
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Answer:
The equation of required line is:
Step-by-step explanation:
We need to write equation in the form Ax+By+C=0 of the line parallel to and through the point (-2,3)
First we need to find slope and y-intercept of the required line.
Using equation of line to find slope.
Since the given line and required lines are parallel there slope is same.
Writing equation in slope intercept form: where m is slope.
So, the slope m = 5/8
The slope of required line is
Now finding y-intercept using slope and point(-2,3)
So, the equation of required line having slope and y-intercept
So, The equation of required line is:
Answer:
25/2
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]
Where P, Q are scalar functions
We want to compute
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 which represents the sides of the rectangle.
is the line segment from (0,0) to (5,0)
is the line segment from (5,0) to (5,1)
is the line segment from (5,1) to (0,1)
is the line segment from (0,1) to (0,0)
Then
Given 2 points P, Q we can always parametrize the line segment from P to Q with
r(t) = tQ + (1-t)P for 0≤ t≤ 1
Let us compute the first integral. We parametrize as
r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and
r'(t) = (5,0) so
Now the second integral. We parametrize as
r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and
r'(t) = (0,1) so
The third integral. We parametrize as
r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and
r'(t) = (-5,0) so
The fourth integral. We parametrize as
r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and
r'(t) = (0,-1) so
So
Now, let us compute the value using Green's theorem.
According with this theorem
where A is the interior of the rectangle.
so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}
We have
so