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3 years ago

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:

If the slope of two lines is the same, the two lines will be parallel
If the product of slopes of two lines is -1, the two lines will be perpendicular to each other.
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

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

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

$\dfrac{\partial f}{\partial z}=ze^z$

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:

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

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

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3 years ago

50 POINTS !!<br><br><br> PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.

Artyom0805 [142]

Answer:

Step-by-step explanation:

Use the pythagorean theorem ( $a^2+b^2=c^2$ ). A and b are the two legs of the triangle, and c is ALWAYS the hypotenuse. Plug in the values for a and c

$a^2+b^2=c^2$

$5^2+b^2=13^2$

$25+b^2=169$

$b^2=144$

$b=12$

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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

Mandarinka [93]

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

(A) Lets make a table

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

(B) To find Equation of a line we use points (0,0) and (80,140)

$slope = \frac{y_2-y_1}{x_2-x_1} = \frac{140-0}{80-0} = \frac{7}{4}$

y intecept is (0,0)

so b= 4

Slope intercept form of a line i y=mx + b

m is the slope and b is the y intercept

So slope intercept form of line becomes

$y= \frac{7}{4} x$

Standard form is Ax + By =C

$y= \frac{7}{4} x$

Multiply both sides by 4

4y = 7x

Now subtract 7x on both sides

-7x + 4y =0 is the standard form

$y= \frac{7}{4} x$

$y= \frac{7}{4}(30)= 52.5$

Total 140 packs of printer paper

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alisha [4.7K]

Step-by-step explanation:

${:}\leadsto$ $\sf 59-(2c+3)=4 (c+7)+c$

${:}\leadsto$ $\sf 59-2c-3=4c+28+c$

${:}\leadsto$ $\sf 56-2c=5c+28$

${:}\leadsto$ $\sf 5c+28=56-2c$

${:}\leadsto$ $\sf 5c+2c=56-28$

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

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

${:}\leadsto$ $\sf c=4$

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Sedaia [141]

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s=\cfrac{\theta \pi r}{180}\quad 
\begin{cases}
r=radius\\
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\qquad degrees\\
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