What is the slope of a line passing through (-3, -4) and (3, 4)?
1 answer:
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The given displacement at time, t, is
p(t) = 2 sin(t³) + 5 cos(t³)
The initial equilibrium position is
p(0) = 5
To determine future equilibrium postions, define
f(t) = p(t) - 5 = 2 sin(t³) + 5 cos(t³ - 5
The derivative of f(t) is
f'(t) = (3t²)[2 cos(t³) - 5sin(t³)]
Equilibrium is established when f(t) = 0.
To solve this equation numerically, we shall use the Newton-Raphson method, given by.
t(n+1) = t(n) - f[t(n)]/f'[(t(n)], n=0,1,2, ...,
As a guess, let (0) = 1.
The iterative solution for t is shown below.
n t(n)
--- -----------
0 1.0000
1 0.9344
2 0.9147
3 0.9130
4 0.9130
The solution converges rapidly to t = 0.913 s.
The graphical solution (shown below) confirms the numerical solution.
Answer:
The weight first reaches the equilibrium position in 0.913 sec.
p(t) = 2 sin(t³) + 5 cos(t³)
The initial equilibrium position is
p(0) = 5
To determine future equilibrium postions, define
f(t) = p(t) - 5 = 2 sin(t³) + 5 cos(t³ - 5
The derivative of f(t) is
f'(t) = (3t²)[2 cos(t³) - 5sin(t³)]
Equilibrium is established when f(t) = 0.
To solve this equation numerically, we shall use the Newton-Raphson method, given by.
t(n+1) = t(n) - f[t(n)]/f'[(t(n)], n=0,1,2, ...,
As a guess, let (0) = 1.
The iterative solution for t is shown below.
n t(n)
--- -----------
0 1.0000
1 0.9344
2 0.9147
3 0.9130
4 0.9130
The solution converges rapidly to t = 0.913 s.
The graphical solution (shown below) confirms the numerical solution.
Answer:
The weight first reaches the equilibrium position in 0.913 sec.
Answer:
10x.
Step-by-step explanation:
It is given that,
JK = 3x
LM = 12x
JM = 25x
Let as consider the line as shown in the below figure.
From the figure it is clear that,
Therefore, the length of KL is 10x.
