7 would probably end up being 6 12/12 if I am doing his correctly.
Remember to include the positive and negative roots
(7x-2)^2=256
sqrt both sides
7x-2=+/-16
add 2 to both sides
7x=2+/-16
divdide both sides by 7
x=(2+/-16)/7)
x=(2+16)/7=18/7=2 and 4/7
x=(2-16)/7=-14/7=-2
x=2 and 4/7 or -2
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.
Answer:
60
Step-by-step explanation:
first you have to find 50% of 40 but multiplying 40 by .5 to get 20. then you add 20 to 40 to get your answer
Answer:
Dont use long words, keep your questions short,
Step-by-step explanation:
