Imagine a particle of mass m and energy e in a potential well v (x), sliding frictionlessly back and forth between the classical turning points (call them a and b). classically, the probability of finding the particle in the range dx (if, for example, you took a snapshot at a random time t) is equal to the fraction of the time t it takes to get from a to b that it spends in the interval dx: rho(x)dx = dt t = (dt/dx)dx t = 1 v(x)t dx where v(x) is the speed, and t = z t 0 dt = z b a 1 v(x) dx . thus rho(x) = 1 v(x)t this is perhaps the closest analog to |ψ| 2 . (a) use conservation of energy to express v(x) in terms of e and v (x).

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

idk idk idk idk idk

step-by-step explanation:

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2