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3 years ago

What is the solution to this equation?

Mathematics

2 answers:

sattari [20]3 years ago

7 0

Answer:

Step-by-step explanation:

If you 9 times 3 u get 27

mr Goodwill [35]3 years ago

7 0

Answer:

n=27

Step-by-step explanation:

To solve the equation, we want to find out what n is. In order to do this, we have to get n by itself. Perform the opposite of what is being done to the equation. Keep in mind, everything done to one side, has to be done to the other.

n/3=9

n is being divided by 3. The opposite of division is multiplication. Multiply both sides by 3.

3*n/3=9*3

n=9*3

n=27

If we want to check the answer, substitute 27 in for n in the original equation.

n/3=9

27/3=9

9=9

This is true, so we know the answer is n=27

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F=(2xy +z³)i + x³j + 3xz²k find a scalar potential and work done in moving an object in the field from (1,-2,1) to (3,1,4)

Alex73 [517]

Step-by-step explanation:

Given:

$\textbf{F} = (2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}$

This field will have a scalar potential $\varphi$ if it satisfies the condition $\nabla \times \textbf{F}=0$ . While the first x- and y- components of $\nabla \times \textbf{F}$ are satisfied, the z-component doesn't.

$(\nabla \times \textbf{F})_z = \left(\dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y} \right)$

$\:\:\:\:\:\:\:\:\: = 3x^2 - 2x \ne 0$

Therefore the field is nonconservative so it has no scalar potential. We can still calculate the work done by defining the position vector $\vec{\textbf{r}}$ as

$\vec{\textbf{r}} = x \hat{\textbf{i}} + y \hat{\textbf{j}} + z \hat{\textbf{k}}$

and its differential is

$\textbf{d} \vec{\textbf{r}} = dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}}$

The work done then is given by

$\displaystyle \oint_c \vec{\textbf{F}} • \textbf{d} \vec{\textbf{r}} = \int ((2xy + z^3)\hat{\textbf{i}} + x^3\hat{\textbf{j}} + 3xz^2\hat{\textbf{k}}) • (dx \hat{\textbf{i}} + dy \hat{\textbf{j}} + dz \hat{\textbf{k}})$