The question is:
Check whether the function:
y = [cos(2x)]/x
is a solution of
xy' + y = -2sin(2x)
with the initial condition y(π/4) = 0
Answer:
To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.
Let us do that.
y = [cos(2x)]/x
y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]
Now,
xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x
= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)
= -2sin(2x)
Which is the right hand side of the differential equation.
Hence, y is a solution to the differential equation.
Given:
Women only gym:
60% married women
75% of the married women exercise in the morning
30% of the single women exercise in the morning
60% married women + 40% single women = total women members
60% married
75% exercise in the morning
25% exercise in the afternoon or evening
40% single
30% exercise in the morning
70% exercise in the afternoon or evening
Exercise in the morning
married: 60% x 75% = 45%
single: 40% x 30% = 12%
<span>
B) Yes. P(married and exercise in the morning) = P(married)·P(exercise in the morning) = 45% </span>
Answer:
3/2
Step-by-step explanation:
slope=rise/run=30/20=3/2