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:
x | y
10 1 = 1/2 (10) -4=5-5= 1
-2 -5 = 1/2(-2)-4=-1-4= -5
4 -2 = 1/2(4)-4=2-4= -2
-8 -8 = 1/2(-8)-4=-4-4= -8
The points in the graph are (-1,0) and (3,0). I solved the equation using the Factoring method in solving quadratic equations. As I solved for finding x= -1 and 3, I substituted both to the equation in order to find y.
Answer:
Step-by-step explanation:
<u>Coordinates given:</u>
<u>Plotted, these points give a right triangle with hypotenuse DF</u>
- Distance DE = 4 - (-3) = 7
- Distance EF = 2 - (-3) = 5
- Distance DF = √(7²+5²) = √84 = 8.6
<u>Perimeter</u>
