5 brothers want to share the cost of a birthday gift for their mom. The youngest brother is too small to get an allowance so the
1 answer:
Answer:
<em>Total cost=$73.60 </em>
Step-by-step explanation:
<u>Arithmetics To Solve Simple Problems </u>
Sometimes the use of the most basic operations can be handy to solve problems of a variety of complexities without the use of equations or algebra. We'll approach to the solution of this question by pure logic and basic operations.
If the gift was originally meant to be paid for all 5 brothers, each one would have paid one-fifth of the cost. But since the younger brother won't be paying anything, the cost will be divided among the other four brothers, each one of which will pay now one-fourth of the cost. But this new division is $3.68 more than the previous one. If we take the difference between both fractions, we get
That fraction is the portion of the money they are paying in excess, so the total cost should be twenty times that excess, i.e.
Total cost= 20*3.68=$73.60
Let's check. Each brother should have paid $73.60/5=$14,72. But the four brothers really paid $73.6/4=$18.40. The difference is cost is $18.40-$14,72=$3.68 as required
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Answer: First, injective means that for every y, there is only one x such f(x) = y, and surjective means that if f is a function that goes from the set {x} to the set {y}, for every element y in the codomain there is at least one element x in the domain such f(x) = y
(a) f:R----> + R given by f(x) = x2 (i guess you written )
The function is not injective, because f(-2) = f(2), and is surjective, because the codomain is +R, and is always a positive real number.
(b) f:N----> + N given by f(n) = n2 (i guess you written )
As the naturals have no negative numbers, in this case the function is injective, but isn't surjective, because there is no number that when squared is equal to 5, for example.
(c) f: Zx Z → Z given by f(n, k) = n +k:
here the domain is of the form (n,k) and the function is n +k, so the numbers (z,w) and (w,z) return the same value when evaluated in f, then f is not injective. And is easy to see that f is surjective, because in the sum you can reach al the integers on the codomain.