If you get r on the rhs, you would get d/t=r
Hope that helps
Answer: m=-1
Step-by-step explanation: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
m-(-(4+m)+2)=0
1.1 Pull out like factors :
2m + 2 = 2 • (m + 1)
2.1 Solve : 2 = 0
This equation has no solution.
A a non-zero constant never equals zero.
2.2 Solve : m+1 = 0
Subtract 1 from both sides of the equation :
m = -1
If you were to graph the sine and cosine functions on the same set of axes, you'd see that they are 90 degrees, or pi/2 radians, out of sync with one another. cos 0 is 1, whereas sin 0 is 0; sin x does not reach the value 1 until your angle reaches 90 degrees, or pi/2 radians.
Please do some experimentation here. You want to express sin (300t) as a cosine, that is, as cos (300t + [some angle]), where [some angle] is called a "phase shift."
Start with the basic y=sin x. Its graph is usually begun at (0,0). Try simplifying and graphing cos (x-pi/2). Does this produce the same y=sin x, with the same graph? Do you remember that
cos (x-pi/2) = cos x cos pi/2 + sin x sin pi/2?
It happens that cos pi/2 = 0 and that sin pi/2 = 1. Thus,
cos (x-pi/2) = sin x (1) = sin x. So, we have succeeded in obtaining sin x from cos (x-pi/2).
Now, what about obtaining sin 300t from the cosine function?
First: recognize that the standard form of the cosine function with a phase shift is y = a cos (bx + c). What is the period?
Answer: The period is always 2pi/b. So, in the case, the period is 2pi/300, or pi/150.
What is the phase shift?
Answer: the period is always -c/b. So, in this case, the period is -c/b, or
-pi/2 over 300. This simplifies to -pi/150.
Try this: Simplify cos (300t -pi/150) If the end result is sin 300t, you'll know you have this right. If the end result is not sin 300t, experiment with that phase shift.
-2 because the line goes through -2
I would say < because the square root of 2 is 1.414 which is less then 1.8 repeating