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Gennadij [26K]
3 years ago
5

Solve 2(x + 3) = x-4 pls <3

Mathematics
1 answer:
Alex Ar [27]3 years ago
4 0

Answer:

Step-by-step explanation:

Step 1  

Having all the  

x

terms on one side of =

Subtract  

2

x

from both sides

2

x

−

2

x

+

3

=

3

x

−

2

x

−

4

0

+

3

=

x

−

4

Swap round

x

−

4

=

3

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Step 2  

Have just  

x

on the left of =

Add  

4

to both sides

x

−

4

+

4

=

3

+

4

x

=

7

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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))

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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}

(c)

Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}

(d)

Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}

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