Answer:
Solution to determine whether each of these sets is countable or uncountable
Step-by-step explanation:
If A is countable then there exists an injective mapping f : A → Z+ which, for any S ⊆ A gives an injective mapping g : S → Z+ thereby establishing that S is countable. The contrapositive of this is: if a set is not countable then any superset is not countable.
(a) The rational numbers are countable (done in class) and this is a subset of the rational. Hence this set is also countable.
(b) this set is not countable. For contradiction suppose the elements of this set in (0,1) are enumerable. As in the diagonalization argument done in class we construct a number, r, in (0,1) whose decimal representation has as its i th digit (after the decimal) a digit different from the i th digit (after the decimal) of the i th number in the enumeration. Note that r can be constructed so that it does not have a 0 in its representation. Further, by construction r is different from all the other numbers in the enumeration thus yielding a contradiction
Answer:
24
Step-by-step explanation:
6 divided by 1/4 is the same thing as 6 multiplied by 4, so the answer is 24.
Answer:
Part 1) The solution of the system of equations is (2,-5)
Part 2) The solution of the system of equations is (2,4)
Step-by-step explanation:
Part 1) Linear combination
we have
-----> equation A
-----> equation B
Multiply equation B by 2 both sides
-----> equation C
Adds equation A and equation C
Find the value of y
The solution of the system of equations is (2,-5)
Part 2) By graph
-----> equation A
-----> equation B
we know that
The solution of the system of equations is the intersection point both graphs
Using a graphing tool
The intersection point is (2,4)
therefore
The solution of the system of equations is the point (2,4)
see the attached figure
If so then add 6, then divide the 7 off the a
5500(2)^5/6.3
5500*1.7
The answer is 9534
this answer has been rounded.
If you need this to be unrounded, go to https://www.symbolab.com/solver/algebra-calculator
