Let D = {x| x is a student} be the domain, and let ƒ(x) = “date of birth” be the possible function. Determine if the relation is
1 answer:
Remember that a function is a special type of relation in which each element of the domain is paired with only one element in the range; in other words for each 
value, there is only a 
value.
Since D = {x| x is a student}, the domain of the relation is the set of all students; on the other hand, <span>ƒ(x) = “date of birth”, so the range of the relation is the set of the birth dates of the students. Since in a room of 23 people there is a 50% chance of two people sharing the same date of birth, it is inevitable that in the set of all the students there is at least two students that share the same date of bird; therefore two elements in the domain will be paired with the same element in the range. Given that, We can conclude that the relation is not a function.
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Since D = {x| x is a student}, the domain of the relation is the set of all students; on the other hand, <span>ƒ(x) = “date of birth”, so the range of the relation is the set of the birth dates of the students. Since in a room of 23 people there is a 50% chance of two people sharing the same date of birth, it is inevitable that in the set of all the students there is at least two students that share the same date of bird; therefore two elements in the domain will be paired with the same element in the range. Given that, We can conclude that the relation is not a function.
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Answer(s):
Step-by-step explanation:
<em>OR</em>
You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of <em>sine</em>, then by all means, go for it, but be careful and follow what is explained here. Now, as you can see, the photograph on the right displays the trigonometric graph of in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the <em>co</em><em>sine</em> graph [photograph on the left], accourding to the <u>horisontal shift formula above</u>. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the <em>sine</em> graph [photograph on the right] is shifted
to the right, which means that in order to match the <em>cosine</em> graph [photograph on the left], we need to shift the graph BACKWARD
which means the C-term will be negative, and by perfourming your calculations, you will arrive at
So, the sine graph of the cosine graph, accourding to the horisontal shift, is
Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph hits
from there to
they are obviously
apart, telling you that the period of the graph is
Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the <em>midline</em>. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at
in which each crest is extended <em>one hundred units</em> beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.
**I knew exactly what you were talking about the moment you posted this, so here was what you were looking for. It really does not matter which sinusoid and person you select because you will get the same information either way.
I am delighted to assist you at any time.
Answer:
Step-by-step explanation:
The equation of the line through the point &
can be represented by:
Making m the subject;
∴
we need to carry out the equation of the line through (0,1) and (1,2)
i.e
y - 1 = m(x - 0)
y - 1 = mx
where;
m = 1
Thus;
y - 1 = (1)x
y - 1 = x ---- (1)
The equation of the line through (1,2) & (4,1) is:
y -2 = m (x - 1)
where;
∴
-3(y-2) = x - 1
-3y + 6 = x - 1
x = -3y + 7
Thus: for equation of two lines
x = y - 1
x = -3y + 7
i.e.
y - 1 = -3y + 7
y + 3y = 1 + 7
4y = 8
y = 2
Now, y ranges from 1 → 2 & x ranges from y - 1 to -3y + 7
∴