8) Solve: X + 250 = 700

When dividing, round your answer to the nearest tenths place (which means you have to find the hundredths place, in order to rou

Naya [18.7K]

Answer:

ok

Step-by-step explanation:

2.1003

5 0

3 years ago

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Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices

Airida [17]

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]

$\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt$

Where P, Q are scalar functions

We want to compute

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

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

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

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 $\large C_1$ as

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

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

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

The third integral. We parametrize $\large C_3$ as

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

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

$\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2$

The fourth integral. We parametrize $\large C_4$ as

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

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

$\large \displaystyle\int_{C_4}xydx+x^2dy=0$

So

$\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2$

Now, let us compute the value using Green's theorem.

According with this theorem

$\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx$

where A is the interior of the rectangle.

so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}

We have

$\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x$

so

$\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2$

3 0

3 years ago

A circle with a radius of 12 inches has an arc that measures 8 pi inches. Find the measure of the central angle determined by th

ivanzaharov [21]

We know that

[length of a circle]=2*pi*r
r=12 in
[length of a circle]=2*pi*12--------------> 24pi in

if 360° (full circle)--------------------> has a length of 24pi in
X-------------------------------------------> 8pi

X=8pi*360/24pi-----------> 120°

the answer is 120°

3 0

3 years ago

5,-15,35,55....<br> What is the common difference

harkovskaia [24]

Answer:

The common difference is -20

Step-by-step explanation:

Simply subtract 5 from -15

-15-5 = -20

3 0

3 years ago

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