Please help!! In radians, what is the period of the function y = .5cos5x

Mathematics

1 answer:

Levart [38]3 years ago

5 0

Answer:

(2pi)/5

Step-by-step explanation:

The cosine function cos(x) has a period of 2*pi.

If we change the argument of the cosine, the new argument will have a period of 2*pi. If the new argument is 5x, we have that the period will be:

5x = 2*pi

x = 2*pi/5

The period of this function will be (2pi)/5, because for every (2pi)/5 change in the value of x, the function will have the same value. The value of 5 multiplying the cosine does not interfere in the period.

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Find a formula for the described function. A rectangle has perimeter 16 m. Express the area A of the rectangle as a function of

Alex17521 [72]

Part A:

Let the length of one of the sides of the rectangle be L, then the length of the other side is obtained as follow.

Let the length of the other side be x, then

$2(L+x)=16 \\ \\ \Rightarrow L+x=8 \\ \\ \Rightarrow x=8-L$

Thus, if the length of one of the side is x, the length of the other side is 8 - L.

Hence, the area of the rectangle in terms of L is given by

$L(8 - L) = 8L-L^2$

Part B:

To find the domain of A

Recall that the domain of a function is the set of values which can be assumed by the independent variable. In this case, the domain is the set of values that L can take.

Notice that the length of a side of a rectangle cannot be negative or 0, thus L cannot be 8 as 8 - 8 = 0 or any number greater than 8.

Hence the domain of the area are the set of values between 0 and 8 not inclusive.

Therefore,

$dom(Area)=0\ \textless \ L\ \textless \ 8 \ or \ (0, 8)$