To determine the centroid, we use the equations:
x⁻ =
1/A (∫ (x dA))
y⁻ = 1/A (∫ (y dA))
First, we evaluate the value of A and dA as follows:
A = ∫dA
A = ∫ydx
A = ∫3x^2 dx
A = 3x^3 / 3 from 0 to 4
A = x^3 from 0 to 4
A = 64
We use the equations for the centroid,
x⁻ = 1/A (∫ (x dA))
x⁻ = 1/64 (∫ (x (3x^2 dx)))
x⁻ = 1/64 (∫ (3x^3 dx)
x⁻ = 1/64 (3 x^4 / 4) from 0 to 4
x⁻ = 1/64 (192) = 3
y⁻ = 1/A (∫ (y dA))
y⁻ = 1/64 (∫ (3x^2 (3x^2 dx)))
y⁻ = 1/64 (∫ (9x^4 dx)
y⁻ = 1/64 (9x^5 / 5) from 0 to 4
y⁻ = 1/64 (9216/5) = 144/5
The centroid of the curve is found at (3, 144/5).
6 acts as the base. 15 acts as the height, because it extends from the base to meet vertex opposite the base at a 90° angle.
General formula for triangle area
a = 1/2 × b × h
plug in the numbers
a = 1/2 × 6 × 15
a = 90/2
a = 45
The area of the triangle is 45 units²
Given that,
Point = (-5,3)
Slope, m = -3/5
To find,
The slope intercept form of the equation of the line.
Solution,
The general form of equation is :
y = mx +b
We have, x = -5, y = 3
So,
The equation is :
Hence, the required equation is .
Because x was *3 you had to y*3 which is
y=6
Number one: if line segments SR & RT are perpendicular line segment Tu and US are perpendicular and angle STR is congruent to angle TSU, then triangle TRS is congruent to triangle SUT
Number two: if line segment AC is congruent to line segment CB and line segment CB bisects line segment AB, then < a is congruent to < B
