F(x)=(2/3)x^1.5
The centroid position along the x-axis can be obtained by
integrating the function * x to get the moment about the y-axis,
then divide by the area of the graph,
all between x=0 to x=3.5m.
Expressed mathematically,
x_bar=(∫f(x)*x dx )/(∫ f(x) dx limits are between x=0 and x=3.5m
=15.278 m^3 / 6.1113 m^2
=2.500 m
<span>idk why they want the dratted "k" . it is so simple w/o it !
in the form y = ab^x, a = 4000, b = 4400/4000 = 1.1, x = t/2
N = 4000*1.1^(13/2) = 7432 <------
-----------------
now you won't be satisfied with simplicity, so
4400 = 4000e^2k
e^2k = 1.1
k = ln 1.1/2 = .0477 to 4 dp
N(13) = 4000e^(13*.0477) = 7436 <------
note that apart from a roundabout way, you get a less accurate ans !</span>
A = lw
A = (x + 6)4x
A = (x)4x + (6)4x
A = 4x² + 24x
Answer:
(f-g) (x)=731.............
