Formatting toolbar because if you look to the right there is the slides and at the top it says: Formatting toolbar
Answer:
Solution: In this case, the system can be modeled as a function with two output signals,
where
where(x(t), y(t)) is the position of the tail in they plane. This model Is Clearly not linear. If the Input torque doubles, for example, the output values will not double. In fact, the output values are constrained to lie on a circle centered at the origin, regardless of the Input. For this reason, the model is BIBO stable. The output is always bounded. Thus, while our previous model was linear and unstable, this one Is nonlinear and stable. Which model is more useful?
Answer:
dax and dear god, also my heart hurts
Explanation:
they are great so is he u should listen to him
Answer:
Sequence of popped values: h,s,f.
State of stack (from top to bottom): m, d
Explanation:
Assuming that stack is initially empty. Suppose that p contains the popped values. The state of the stack is where the top and bottom are pointing to in the stack. The top of the stack is that end of the stack where the new value is entered and existing values is removed. The sequence works as following:
push(d) -> enters d to the Stack
Stack:
d ->top
push(h) -> enters h to the Stack
Stack:
h ->top
d ->bottom
pop() -> removes h from the Stack:
Stack:
d ->top
p: Suppose p contains popped values so first popped value entered to p is h
p = h
push(f) -> enters f to the Stack
Stack:
f ->top
d ->bottom
push(s) -> enters s to the Stack
Stack:
s ->top
f
d ->bottom
pop() -> removes s from the Stack:
Stack:
f ->top
d -> bottom
p = h, s
pop() -> removes f from the Stack:
Stack:
d ->top
p = h, s, f
push(m) -> enters m to the Stack:
Stack:
m ->top
d ->bottom
So looking at p the sequence of popped values is:
h, s, f
the final state of the stack:
m, d
end that is the top of the stack:
m
The question is asking us to swap the values of xp and yp while not changing where they point to. Setting xp equal to yp would not work because then we couldn't change yp since the value for xp was overwritten. We can use a third variable to swap them.
int zp = xp;
xp = yp;
yp= zp;
