What is the slope of the line that passes through (1,-19) and (-2,-7)?

Mathematics

1 answer:

Sedbober [7]3 years ago

8 0

The formula for the slope of a line is m = (y2 - y1) / (x2 - x1).

Point 1 = (1, -19)

Point 2 = (-2, -7)

m = (-7 - -19) / (-2 - 1)

m = (-7 + 19) / (-2 - 1)

m = 12 / -3

m = -4

Hope this helps! :)

You might be interested in

What is the surface area of a box if it is scaled up by a factor of 10

xenn [34]

If we are scaling up the surface area by a factor of 10 means that the two dimensions of area are each being increased by 10 times the original size. Therefore, the surface area is increased by a factor of 10 x 10, or by a factor of 100.

7 0

3 years ago

Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integ

liq [111]

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

$\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy$

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

Mx(x,y) = d/dx 8x²y = 16xy
Ly(x,y) = d/dy 7y²x = 14xy

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy$

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2$