Part B seems to be missing, but I think I have enough information to be able to answer.
Let's say we had two numbers x and y. Let x be rational and y be irrational.
If x is some nonzero number, then x*y is irrational. The proof for this is a bit lengthy so I'll leave it out.
For instance,
x = 2 is rational, y = sqrt(3) is irrational, x*y = 2*sqrt(3) is irrational.
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If we made x = 0, then
x*y = 0*y = 0
This is true for any value of y that we want. The y value doesnt even have to be irrational. It can be any real number.
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So the distinction is that if x = 0, then x*y = 0 is rational since 0 is rational. Otherwise, x*y is irrational.
Answer:
In mathematics, the domain or set of departure of a function is the set into which all of the input of the function is constrained to fall. It is the set X in the notation f: X → Y, and is alternatively denoted as. . Since a function is defined on its entire domain, its domain coincides with its domain of definition.
Step-by-step explanation:
Answer:
Step-by-step explanation:
We are given that Chantal drives at a constant speed of 55 miles per hour.
If, d represents the total distance in miles, and
h represents number of hours, the following equation can be used to express the given situation:
For every hour, a distance of 55 miles is covered.
Thus, if h = 1,
If h = 2, .
Therefore, , is an ideal equation that represents the situation given in the question above.
this is difficult lol im scared to attempt this so i wont
Answer:
alright
Step-by-step explanation: