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Mathematics

2 answers:

Katyanochek1 [597]3 years ago

3 0

Answer:

it is 3,3 or X<3 you can use either one

D is the answer

Step-by-step explanation:

user100 [1]3 years ago

3 0

Yup what the above me said

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Simplify the ratio of 55/121

cricket20 [7]

Answer:

Thus, 5/11 is the simplified fraction for 55/121 by using the GCD or HCF method. Thus, 5/11 is the simplified fraction for 55/121 by using the prime factorization method.

I hope it's helpful!

3 0

3 years ago

PLEASE HELP ME!!! 5 POINTS!!!

Kamila [148]

Answer:

a:four and ninety one hundredths

Step-by-step explanation:

4 this the whole number the word and is used as a indication of the decimal and then 91 would be in hundredths because of the 1

8 0

4 years ago

Read 2 more answers

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)

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