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2 years ago

Are these lines parallel, perpendicular, or neither? 3y=x+4 and 3x+y=1

Mathematics

2 answers:

Goryan [66]2 years ago

6 0

Answer:

These lines are perpendicular

Step-by-step explanation:

3y=x+4

$y = \frac{1}{3}x+\frac{4}{3}$

Slope (m1) = 1/3

3x+y=1

y = -3x + 1

Slope (m2) = - 3

m1 * m2 = $\frac{1}{3}* -3$

= -1

So, these lines are perpendicular.

If product of two slopes is (-1), then they are perpendicular lines.

If both lines have same slope, then they are parallel.

REY [17]2 years ago

5 0

Answer:

perpendicular

Step-by-step explanation:

3y = x + 4 in slope intercept form y = 1/3x +4/3

3x + y = 1 in slope intercept form y = -3x +1

the slopes are negative reciprocals of each other so they are perpendicular

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Which of the following is an example of a variable expense?

olganol [36]

Answer:

Step-by-step explanation:

Rent, wages and car insurance don't vary much. Grocery expenses depend upon the particular groceries purchased each time one goes to the grocery store, and are least likely to be steady / constant from week to week.

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3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$