L[ y(t) ] = Y(s)
L[ y' ] = sY-y(0) = sY-0 = sY
L[ y'' ] = s^2*Y - sy(0) - y'(0) = s^2*Y - 1
laplace trasform both sides
L[ y'' - 6y' + 9y ] = L[ t ]
= L[ y'' ] - 6 L[ y' ] + 9 L[ y ] = 1/s
= [ s^2*Y - 1 ] - 6[ sY ] + 9Y = 1/s
( s^2 - 6s + 9 ) Y - 1 = 1/s
⇒ ( s^2 - 6s + 9 ) Y = (1/s) + 1
⇒Y = [ 1 / ( s(s^2 - 6s + 9 ) ) ] + [ 1 / ( s^2 - 6s + 9 ) ]
Let inverse laplace trasform , find y(t) :
y(t) = L^(-1)[ Y(s) ] = L^(-1) { [ 1 / ( s(s^2 - 6s + 9 ) ) ] + [ 1 / ( s^2 - 6s + 9 ) ] }
= [ (1/3)*t*e^(3t) - (1/9)*e^(3t) + (1/9) ] + [ t*e^(3t) ]
= (4/3)*t*e^(3t) - (1/9)*e^(3t) + (1/9)
Answer:
115m
Step-by-step explanation:
Note that one of the face of a pyramid is a right angled triangle.
Since the pyramid was about 147m tall, the height of triangle will be 147m. If one of the faces is built with one of its faces at 52° to the incline, we can find the original length of one of its sides using SOH CAH TOA
tan(theta) = opposite/adjacent
Given theta = 52°, opposite = 147m (the side directly opposite the angle)
Substituting we have;
Tan52° = 147/adjacent
Adjacent = 147/tan52°
Adjacent = 114.8m
= 115m
Therefore 115m is the original length of one of its side
r=- 3/8 cos (0)x t maybe is this
Jane has half as much as Elaine at first. The answer is 12, but I honestly just kept using random numbers. After spending 4, Jane has 8 and Elaine has 20.(12 times 2) - 4
Answer:
<h3>b</h3>
Step-by-step explanation:
