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).
Answer:
Seven
Step-by-step explanation:
Let G7, G8, and G9 be the numbers of students in the respective grades.
Then, G7 + G8 + G9 = LCM of G8 and G9
<em>Data:
</em>
G8 = 5
G9 = 3
<em>Calculations:
</em>
<em>Multiples of 5</em>: 1 2 10 15
<em>Multiples of 3</em>: 1 3 6 9 15
The <em>LCM</em> of 5 and 3 is 15.
G7 + 5 + 3 = 15 Combine like terms
G7 + 8 = 15 Subtract 8 from each side
G7 = 7
There are seven students in Grade 7.
Three hundred six thousandths . . . in word form
let the original price be x.
then,
x- 25% of x= 24
x- 25x/100 = 24
x- x/4=24
3x/4=24
3x= 96
x= 32
in short...the original price= 32 dollars
2(-8x-2y=-22)
16x+7y=17
-16x-4y=-44
16x+7y=17
3y=-27
y=-9
16x+7(-9)=17
16x+-63=17
16x=80
X=5
