First of all we need to find a representation of C, so this is shown in the figure below.
So the integral we need to compute is this:

So, as shown in the figure, C = C1 + C2, so:
Computing first integral:
Applying derivative:

Substituting this value into

Computing second integral:
Applying derivative:

Substituting this differential into


We need to know the limits of our integral, so given that the variable we are using in this integral is x, then the limits are the x coordinates of the extreme points of the straight line C2, so:
![I_{2}= -8\int_{4}^{8}}dx=-8[x]\right|_4 ^{8}=-8(8-4) \rightarrow \boxed{I_{2}=-32}](https://tex.z-dn.net/?f=I_%7B2%7D%3D%20-8%5Cint_%7B4%7D%5E%7B8%7D%7Ddx%3D-8%5Bx%5D%5Cright%7C_4%20%5E%7B8%7D%3D-8%288-4%29%20%5Crightarrow%20%5Cboxed%7BI_%7B2%7D%3D-32%7D)
Finally:
Answer:
y + 4 = -3(x -2)
Step-by-step explanation:
A line parallel to y = -3x + 7 also has a slope of -3.
Use the point-slope formula:
y - k = m(x - h)
Inserting the given info, we get:
y + 4 = -3(x -2)
Simplify what's in brackets first, then move to what's in the parenthesis
Answer:
Please check the explanation.
Step-by-step explanation:
Let the coordinates of the point F be (x, y).
When a point F(x, y) is reflected over the x-axis, the x-coordinate of the point F remains the same, and the y-coordinate of the point reverses the sign.
Thus, the rule of reflection over the x-axis:
F(x, y) → F'(x, -y)
Here,
F'(x, -y) would be coordinates of point F after the reflection over the x-axis.
Let say, the point F(1, 2).
The coordinate of the point F after the reflection over the x-axis would be:
F(1, 2) → F'(1, -2)
Thus, F'(1, -2) would be the coordinates of point F after the reflection over the x-axis.
Answer:
12
Step-by-step explanation:
just find the HCF, which will be
2² × 3 = 12