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.
Difference means to minus
5 1/4 - 2 3/4 =
[since 1/4 cannot minus 3/4, we need to convert one of the 5 wholes into quarters: 5 1/4 = 4 5/4 ]
5 1/4 - 2 3/4
= 4 5/4 - 2 3/4
= 2 2/4
= 2 1/2
A relation is the relationship between variables. When each input has one output, the relation is a function. When the value of k is 5 or 3, the relation will not be a function.
Given that:
For S, not to be a function, 1 domain element must point to more than one range elements.
This means that:
Collect like terms
Multiply by -1
Remove absolute bracket
Solve for k
Hence, when k is 5 or 3, the relation will not be a function.
Read more about functions and relations at:
brainly.com/question/2253924
Answer:
D 5sqrt(3)
Step-by-step explanation:
We can use the Pythagorean theorem to solve since the altitude is perpendicular to the base
a^2+b^2 = c^2
5^2 + a^2 = 10^2
25+a^2 = 100
a^2 =100-25
a^2 = 75
Taking the square root of each side
a = sqrt(75)
a = sqrt(25)sqrt(3)
a = 5sqrt(3)
D because the question would have to be multiply the numbers in a way that would get you the cost of 2 rubber stamps