Answer:
Horizontal asymptote of the graph of the function f(x) = 25/1+4x is y=0
Step-by-step explanation:
I attached the graph of the function. Graphically it can be seen that the horizontal asymptote of the graph of the function is y=0. There is also a <em>vertical</em> asymptote at x=-1/4
When denominator's degree (1) is higher than nominator's degree (0) then the horizontal asymptote is y=0
Let's use an example
Say we had two lines P and Q. If P has slope 2/5, then Q must have slope -5/2 in order for P and Q to be perpendicular.
Note how -5/2 is the negative reciprocal of 2/5. In other words, you flip the fraction and the sign to go from either slope.
Another thing to notice is that the two slopes multiply to -1. This is true for any pair of perpendicular lines as long as neither line is vertical.
Answer: 60 times
Step-by-step explanation:
i took it on edge
This is easily done after writing both complex numbers in exponential form:


Then

In trigonometric form, this is

7/8, 1 is also 8/8 so if you add that to -1/8 (then it automatically subtracts since it’s adding a negative) making 7/8.