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Anastaziya [24]
3 years ago
14

How do you use the slope to prove lines are parallel or perpendicular

Mathematics
1 answer:
zaharov [31]3 years ago
3 0

In order to find if two lines are parallel or perpendicular or neither, following rules are used:

  1. If the slope of two lines is the same, the two lines will be parallel
  2. If the product of slopes of two lines is -1, the two lines will be perpendicular to each other.
  3. If none of the above two conditions is being satisfied, this mean lines are neither parallel nor perpendicular.

So the steps that must be followed are:

  • Find the slope of the given lines.
  • Use the values of the slope to check which of the above point is being satisfied and draw the conclusion accordingly.
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Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)
babunello [35]

\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k

We want to find f(x,y,z) such that \nabla f=\vec F. This means

\dfrac{\partial f}{\partial x}=x^2+y

\dfrac{\partial f}{\partial y}=y^2+x

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Integrating both sides of the latter equation with respect to z tells us

f(x,y,z)=e^z(z-1)+g(x,y)

and differentiating with respect to x gives

x^2+y=\dfrac{\partial g}{\partial x}

Integrating both sides with respect to x gives

g(x,y)=\dfrac{x^3}3+xy+h(y)

Then

f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)

and differentiating both sides with respect to y gives

y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C

So the scalar potential function is

\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}

By the fundamental theorem of calculus, the work done by \vec F along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it L) in part (a) is

\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}

and \vec F does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them L_1 and L_2) of the given path. Using the fundamental theorem makes this trivial:

\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3

\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4

8 0
3 years ago
50 POINTS !!<br><br><br> PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.
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Answer:

12

Step-by-step explanation:

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3 years ago
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The office manager of a small office ordered 140 packs of printer paper based on average daily use, she knows that the paper wil
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slope intercept form is y=mx+b

So slope intercept form is The office manager of a small office ordered 140 packs of printer paper based on average daily use, she knows that the paper will last about 80 days

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X axis represents the Number of days paper used

y axis represents the packs of printer paper used

x            y

days      packs of printer paper used

0            0               (0 days , 0 packs used)

80          140           (in 80 days , 140 packs paper used)

The graph is attached below

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y intecept is (0,0)

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m is the slope and b is the y intercept

So slope intercept form of line becomes

y= \frac{7}{4} x

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Total 140 packs of printer paper

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Packs of paper remaining after 30 days = 140- 52.5= 87.5


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Step-by-step explanation:

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{:}\leadsto\sf c={\dfrac {28}{7}}

{:}\leadsto\sf c=4

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