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3 years ago

Please help me with this question :) thanks.

Mathematics

1 answer:

Ostrovityanka [42]3 years ago

4 0

Answer:

( 2 , -1 )

Step-by-step explanation:

If you reflect coordinates across the x-axis, you change the coordinates to match ( x , -y ). If you reflect coordinates across the y-axis, you change the coordinates to match ( -x , y ).

( -2 , 1 ) gets reflected across x-axis ( x , -y ) --- ( -2 , -1 )

( -2 , -1 ) gets reflected across the y-axis ( -x , y ) --- ( 2 , -1 )

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PLEASE HELP FAST ILL GIVE 5 STARS!!!!

MArishka [77]

I believe it's Line y=-x+2 and y=3x+1 intersect the y-axis. From what I've gathered, they are parallel lines, and both are set on the y-axis.

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3 years ago

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Write an equation of the line that passes through (2,-4) and (0,4)

LenKa [72]

Answer:

step 1. y - y1 = m(x - x1). this is the equation given a point and a slope.

step 2. find the slope m. m = (y2 - y1)/(x2 - x1) = (4 - (-4))/(0 - 2) = 8/-2 = -4.

step 3. y - (-4) = -4(x - 2) ; y + 4 = -4x + 8.

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2 years ago

Type the correct answer in the box,

HACTEHA [7]

We'll assume the quadratic equation has real coefficients

Answer:

The other solution is x=1-8i.

Step-by-step explanation:

The Complex Conjugate Root Theorem

if P(x) is a polynomial in x with real coefficients, and a + bi is a root of P(x) with a and b real numbers, then its complex conjugate a − bi is also a root of P(x).

The question does not specify if the quadratic equation has real coefficients, but we will assume that.

Given x=1+8i is one solution of the equation, the complex conjugate root theorem guarantees that the other solution must be x=1-8i.

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2 years ago

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: