Please help me with the problem
1 answer:
Answer:
See the attachment.
Step-by-step explanation:
The approach is to take advantage of the definition of a perpendicular bisector to show the base lengths and right angles are congruent, then use SAS with the second side being CD (congruent to itself). It's mainly a matter of deciding what the various reasons are called.
It seems the goal is to show that AC = BC as marked.
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Step-by-step explanation:
Domain of a rational function is everywhere except where we set vertical asymptotes. or removable discontinues
Here, we have
First, notice we have x in both the numerator and denomiator so we have a removable discounties at x.
Since, we don't want x to be 0,
We have a removable discontinuity at x=0
Now, we have
We don't want the denomiator be zero because we can't divide by zero.
so
So our domain is
All Real Numbers except-2 and 0.
The vertical asymptors is x=-2.
To find the horinzontal asymptote, notice how the numerator and denomator have the same degree. So this mean we will have a horinzontal asymptoe of
The leading coeffixent of the numerator/ the leading coefficent of the denomiator.
So that becomes
So we have a horinzontal asymptofe of 2
it just so happen that √2 is an irrational number, so any product with it as a factor, will yield an irrational number as well.
factoid: Ancient Greeks were scared of irrational numbers, and they steered clear from √2.