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3 years ago

What is the slope intercept equation of the line that includes (0,7) and (3,10) ?

Mathematics

2 answers:

nekit [7.7K]3 years ago

7 0

Answer:

y = 1x + 7

simplifies to:

y = x + 7

Step-by-step explanation:

first we need to find the slope or m:

y2 - y1 divided by x2 - x1

10-7 divided by 3-0 =

3/3 = 1

slope is 1

then we need to find the y intercept or b:

Usually you would choose one of the points and find the y intercept but the first point is the y intercept (0,7). 7 is the y intercept

plug it into the equation:

y = mx + b

y = 1x +7

y = x + 7

kompoz [17]3 years ago

6 0

7 is the y intercept, it is given as one of the points to is our b.
The slope can be found by 7-10 over 0-3 which is -3/-3 which is 1. 1 is our m.
so with the slope and intercept we plug it into the formula to get:
y = x + 7

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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

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

$\displaystyle = (x^2y + xz^3) + x^3y + xz^3|_{(1, -2, 1)}^{(3, 1, 4)}$

$= 422$

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Hi There.

Your answer would be

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