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Ghella [55]
3 years ago
14

Evaluate the line integral using the fundamental theorem of line integrals. use a computer algebra system to verify your results

. cos(x) sin(y) dx + sin(x) cos(y) dy c
c.line segment from (0, −π) to 3π 2 , π 2 11
Mathematics
1 answer:
mario62 [17]3 years ago
7 0
\dfrac{\partial f}{\partial x}=\cos x\sin y\implies f(x,y)=\sin x\sin y+g(y)

\dfrac{\partial f}{\partial y}=\sin x\cos y=\sin x\cos y+\dfrac{\mathrm dg}{\mathrm dy}
\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C

f(x,y)=\sin x\sin y+C

\displaystyle\int_{\mathcal C}\cos x\sin y\,\mathrm dx+\sin x\cos y\,\mathrm dy=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f\left(\frac{3\pi}2,\frac\pi2\right)-f(0,-\pi)=-1
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You want to buy a camera. The retail price is $75. The camera is on sale for 15% off . What is the sale price of the camera?
ludmilkaskok [199]

Answer:

$63.75

Step-by-step explanation:

15% off

Get percentage of retail price

1-0.15 = .85

0.85 = 85%

The sale price is 85% of the retail price

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0.85 * 75

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2 years ago
What is the value of 3 in 640.31? Choose ALL the correct answers
grigory [225]

Answer:

3/10 and 0.3

Step-by-step explanation:

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8 0
2 years ago
Determine the volume of the parallelepiped with one vertex at the origin and the three vertices adjacent to it at (2, â1, â1), (
valentinak56 [21]

Answer:

23

Step-by-step explanation:

Here is the complete question

Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at (1, 0, -3), (1, 2, 4), and (5, 1, 0).

Solution

We find the volume of the parallelepiped by making a 3 × 3 column matrix whose columns are the corresponding coordinates of the vertices of the parallelepiped.

So, (1, 0, -3), (1, 2, 4)  and (5, 1, 0)

A = \left[\begin{array}{ccc}1&1&5\\0&2&1\\-3&4&0\end{array}\right]

The determinant of A is the volume of the parallelepiped. So,

detA = 1(2 × 0 - 4 × 1) - 1(0 × 0 - (-3) × 1) + 5(0 × 4 - (-3) × 2)

= 1(0 - 4) - 1(0 + 3) + 5(0 + 6)

= 1(-4) - 1(3) + 5(6)

= -4 - 3 + 30

= 23

So the volume of the parallelepiped is 23

4 0
3 years ago
For each of the following vector fields
olga nikolaevna [1]

(A)

\dfrac{\partial f}{\partial x}=-16x+2y

\implies f(x,y)=-8x^2+2xy+g(y)

\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y

\implies\dfrac{\mathrm dg}{\mathrm dy}=10y

\implies g(y)=5y^2+C

\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}

(B)

\dfrac{\partial f}{\partial x}=-8y

\implies f(x,y)=-8xy+g(y)

\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x

\implies \dfrac{\mathrm dg}{\mathrm dy}=x

But we assume g(y) is a function of y alone, so there is not potential function here.

(C)

\dfrac{\partial f}{\partial x}=-8\sin y

\implies f(x,y)=-8x\sin y+g(x,y)

\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y

\implies\dfrac{\mathrm dg}{\mathrm dy}=4y

\implies g(y)=2y^2+C

\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}

For (A) and (C), we have f(0,0)=0, which makes C=0 for both.

4 0
3 years ago
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