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3 years ago

Which is the equation of the line with slope 0 passing through the point (-3,6)?

Mathematics

1 answer:

TEA [102]3 years ago

6 0

Answer:

y = 6

Step-by-step explanation:

line with (-3 , 6) with slope = 0 is the line perpendicular to x axis

y = 6

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Why are (1,2) and(2,1) not equal ordered pair?

vladimir1956 [14]

Answer:

The pairs ( 1 , 2 ) and ( 2 , 1 ) are not equal because their respective elements are not equal.

Step-by-step explanation:

The pair of elements which are in specific order is called an ordered pair. The pair ( 1 , 2 ) is not same as the pair ( 2 , 1 ). In the pair ( 1 , 2 ) 1 is in the first position and 2 is in the second position. In the pair ( 2 , 1 ), 2 is in the first position and 1 is in the second position.

Two ordered pairs ( a , b ) and (c , d ) are said to be equal if a = c and b = d. We write ( a , b ) = ( c , d ).

Hope I helped!

Best regards!!

8 0

3 years ago

Ms. Savage’s Art Club is making a mural on a wall in the Annex by reproducing a painting on a different scale. The original pain

lara [203]

Answer:

4 feet

Step-by-step explanation:

i think

5 0

2 years ago

Find the surface Area.

kompoz [17]

Answer: 20

Step-by-step explanation: how i got the answer was to multiply everything by two. then add everything together.

3 0

3 years ago

Meike earned $1565 in tips while working a summer job at a coffee shop. She wants to use this money to take a trip to Europe nex

juin [17]

Do 6.5*9=85.5*1565=91,552.5

5 0

3 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago