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jekas [21]
2 years ago
11

Parallel and perpendicular lines ​

Mathematics
1 answer:
11Alexandr11 [23.1K]2 years ago
3 0

Answer:

A. Slope = -³/2

B. y = -³/2x - 6

C. ³/2x + y = -6

Step-by-step explanation:

A. Slope of the line would be the negative reciprocal of the slope of 2x - 3y = 12, which is written in standard form. Rewrite in slope-intercept form, y = mx + b, where m = slope.

Thus:

2x - 3y = 12

-3y = -2x + 12

y = -2x/-3 + 12/-3

y = ⅔x - 4

The slope of 2x - 3y = 12 is therefore ⅔.

The slope of the line perpendicular to 2x - 3y = 12 would be the negative reciprocal of its slope, ⅔ which is:

-³/2

Therefore, the slope of the line perpendicular to 2x - 3y = 12 is -³/2

B. To find the equation of the line in slope-intercept form, first find the equation in point-slope form using the point given (-6, 3) and slope of the line, -³/2.

Substitute a = -6, b = 3, and m = -³/2 into y - b = m(x - a).

Thus:

y - 3 = -³/2(x - (-6))

y - 3 = -³/2(x + 6)

Rewrite in slope-intercept form, y = mx + b

Multiply both sides by 2

2(y - 3) = -3(x + 6)

2y - 6 = -3x - 18

2y = -3x - 18 + 6

2y = -3x - 12

y = -3x/2 - 12/2

y = -³/2x - 6

C. Rewrite y = -³/2x - 6 in standard form.

y = -³/2x - 6

³/2x + y = -6

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I hope this helps you! :D

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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

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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)}

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3 years ago
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Black_prince [1.1K]

Answer:

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Step-by-step explanation:

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x = 41/-3

x = -13 2/3

or

x = -13,67

make as the brainliest

7 0
3 years ago
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