In an algebraic expression, what separates the terms

Find the inverse of the function.<br> у= 2х^2-4

Neko [114]

I think 28 (2x16) = 32 -4 + 28

7 0

4 years ago

A line has no thickness, but it does have direction and ___ length.<br> (Fill in the blank)

Mandarinka [93]

A line has no thickness, but it does have direction and infinite length.

A line has length but no width, making it a one-dimensional figure. A line is made up of a collection of points that can be stretched indefinitely in opposing directions.

Given the variety of geometries in modern mathematics, the idea of a line is directly related to how the geometry is described. For instance, a line in the plane is frequently defined in analytic geometry as the collection of points whose coordinates satisfy a given linear equation, but in a more abstract context, such as incidence geometry, a line may be an independent object, distinct from the collection of points that lie on it.

In cartesian plane the general equation of a line is given by :

y=mx+c where the slope is denoted by m and the y-intercept by c.

To learn more about a straight line:

brainly.com/question/17188072

#SPJ9

3 0

2 years ago

Compute the matrix of partial derivatives of the following functions.

s344n2d4d5 [400]

For a vector-valued function

$\mathbf f(\mathbf x)=\mathbf f(x_1,x_2,\ldots,x_n)=(f_1(x_1,x_2,\ldots,x_n),\ldots,f_m(x_1,x_2,\ldots,x_n))$

the matrix of partial derivatives (a.k.a. the Jacobian) is the $m\times n$ matrix in which the $(i,j)$ -th entry is the derivative of $f_i$ with respect to $x_j$ :

$D\mathbf f(\mathbf x)=\begin{bmatrix}\dfrac{\partial f_1}{\partial x_1}&\dfrac{\partial f_1}{\partial x_2}&\cdots&\dfrac{\partial f_1}{\partial x_n}\\\dfrac{\partial f_2}{\partial x_1}&\dfrac{\partial f_2}{\partial x_2}&\cdots&\dfrac{\partial f_2}{\partial x_n}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{\partial f_m}{\partial x_1}&\dfrac{\partial f_m}{\partial x_2}&\cdots&\dfrac{\partial f_n}{\partial x_n}\end{bmatrix}$

So we have

(a)

$D f(x,y)=\begin{bmatrix}\dfrac{\partial(e^x)}{\partial x}&\dfrac{\partial(e^x)}{\partial y}\\\dfrac{\partial(\sin(xy))}{\partial x}&\dfrac{\partial(\sin(xy))}{\partial y}\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}$

(b)

$D f(x,y,z)=\begin{bmatrix}\dfrac{\partial(x-y)}{\partial x}&\dfrac{\partial(x-y)}{\partial y}&\dfrac{\partial(x-y)}{\partial z}\\\dfrac{\partial(y+z)}{\partial x}&\dfrac{\partial(y+z)}{\partial y}&\dfrac{\partial(y+z)}{\partial z}\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}$