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Alenkinab [10]
3 years ago
5

Jenna used 32 cups of a recipe to make lunch for a party. If Jenna made half a recipe of punch, how many quarts of fruit punch d

id she make?
Mathematics
1 answer:
Allisa [31]3 years ago
7 0

Answer:

She make 4 Quarts of fruit punch.

Step-by-step explanation:

Given:

Jenna used 32 cups of a recipe to make lunch for a party.

Now, to find how many quarts of punch she make if she made half recipe of punch.

Jenna used 32 cups of recipe but she made a half only.

So, <em>half recipe = 1/2 of 32 = 1/2 × 32 =</em><u><em> 16 cups</em></u><em>.</em>

Now,<em> to get the quarts of punch we need to convert the quantities</em>.

<u>1 Quart = 4 cups.</u>

So, if 4 cups = 1 quart.

Then 16 cups = (16 ÷ 4) quarts

                       = 4 quarts.  

Therefore, she make 4 Quarts of fruit punch.

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3. Let A, B, C be sets and let ????: ???? → ???? and ????: ???? → ????be two functions. Prove or find a counterexample to each o
Fiesta28 [93]

Answer / Explanation

The question is incomplete. It can be found in search engines. However, kindly find the complete question below.

Question

(1) Give an example of functions f : A −→ B and g : B −→ C such that g ◦ f is injective but g is not  injective.

(2) Suppose that f : A −→ B and g : B −→ C are functions and that g ◦ f is surjective. Is it true  that f must be surjective? Is it true that g must be surjective? Justify your answers with either a  counterexample or a proof

Answer

(1) There are lots of correct answers. You can set A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. Then g is not  injective (since both 2, 3 7→ 4) but g ◦ f is injective.  Here’s another correct answer using more familiar functions.

Let f : R≥0 −→ R be given by f(x) = √

x. Let g : R −→ R be given by g(x) = x , 2  . Then g is not  injective (since g(1) = g(−1)) but g ◦ f : R≥0 −→ R is injective since it sends x 7→ x.

NOTE: Lots of groups did some variant of the second example. I took off points if they didn’t  specify the domain and codomain though. Note that the codomain of f must equal the domain of

g for g ◦ f to make sense.

(2) Answer

Solution: There are two questions in this problem.

Must f be surjective? The answer is no. Indeed, let A = {1}, B = {2, 3} and C = {4}.  Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. We see that  g ◦ f : {1} −→ {4} is surjective (since 1 7→ 4) but f is certainly not surjective.  Must g be surjective? The answer is yes, here’s the proof. Suppose that c ∈ C is arbitrary (we  must find b ∈ B so that g(b) = c, at which point we will be done). Since g ◦ f is surjective, for the  c we have already fixed, there exists some a ∈ A such that c = (g ◦ f)(a) = g(f(a)). Let b := f(a).

Then g(b) = g(f(a)) = c and we have found our desired b.  Remark: It is good to compare the answer to this problem to the answer to the two problems

on the previous page.  The part of this problem most groups had the most issue with was the second. Everyone should  be comfortable with carefully proving a function is surjective by the time we get to the midterm.

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3 years ago
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The answer is the third one
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2 years ago
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the answer is in the attachment

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At Factory Y, a worker's wages for a 40-hour week is $200. She is paid 10% of her regular weekly wages for every hour that she w
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Answer:

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If she works 40 hours per week and gets $200, subtract $380 - $200 = $180.

If you multiply $180 by 0.1 (which is 10% converted to a decimal), you get 18 which is your final answer.

3 0
3 years ago
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