A gradient is a vector, and slope is a scalar. Gradients really become meaningful in multivarible functions, where the gradient is a vector of partial derivatives. With single variable functions, the gradient is a one dimensional vector with the slope as its single coordinate (so, not very different to the slope at all). Source(s):

Apr 3, 2015 · This works with slope equal to $\text{rise} / \text{run}$, but would not work if the slope were equal to $\text{run} / \text{rise}$ as you suggest. For a modern example, when you drive down a steep mountain highway, you will notice signs saying something like "Danger: 5% grade ahead. Trucks shift to low gear".

Aug 31, 2023 · Usually, slope (or grade) is determined by the rise in elevation over distance, i.e. $\dfrac {\text {rise}}{\text {run}}$. In your case, $1 \text { in } 20$ means for each $20$ meters you travel, the elevation increases by $1$ meter. If you drew a triangle, the rise would be opposite to the hypotenuse, and the run would be adjacent to the ...

Nov 23, 2016 · My mathematics teacher explained how to find slope using 2 different points but I was never taught how to use the X-Intercept and the Y-Intercept to find the slope. Example question: What is the slope of a line that has an X-Intercept of 8 and a Y-Intercept of 11?

Apr 29, 2015 · Slope and Euclidean geometry (grade 10) ... I had to find slope DF, then perpendicular passing through (7,3).

$\begingroup$ I thought to find slope I would have to use the formula y2-y1/x2-x1. When I used that, I got the slope of -1/2. After converting it into point slope form I rearranged it to slope intercept, which mine was y=-1/2x+67.5. However I am …

The defintion of slope is $$ \text{slope} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$$ where $(x_1,y_1)$ and $(x_2,y_2)$ are any two points on the line. So, if you let $(x_1,y_1) = (0,-3)$, and using the fact that the slope is 3, then any other point on the graph can be obtained by $$ 3 = \frac{y - (- 3)}{x - 0}$$ so $$ y = 3x - 3$$

Feb 12, 2016 · A water main is to be constructed with a $12.5$ % grade in the north direction and a $25$ % grade in the east direction. Determine the angle $\theta$ required in the water main for the turn from north to east.

May 2, 2015 · The symbols d/dx and dy/dx represent derivatives in calculus, indicating rates of change with respect to a variable.

Vertical displacement = % grade x Distance; Treadmill speed = $200$ m/min; Percent Grade = $7.5\% ~\text{or}~ 0.075$ Exercise time = $10$ mins; Total vertical distance traveled $= 200$ m/min $\times 0.075 x 10$ mins = $150 m$ Work = body weight (kg) x total vertical distance traveled = $70 kp \times 150 m = 10,500$ kpm or ~$103$ kilojoules