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Anvisha [2.4K]
3 years ago
10

12,000 g____1.2 kg A greater than B less than C equal to

Mathematics
1 answer:
serious [3.7K]3 years ago
7 0
12000 g would be equal to 12 kg not 1.2 kg so the answer has to be A) greater than.

Answer: A) GREATER THAN
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Determine the two closest integers for each irrational number also 117 not in pic
Grace [21]
1.) 4.358898944
2.) 7.071067812
6 0
3 years ago
I need to know if this Answer is correct. please tell me the answer. if it's wrong. Thanks!!!
yaroslaw [1]
Here, as I can see, you are adding fractions, and making sure you have a common denominator. 

12) is correct
13) is correct

You got a common denominator, and added them up! Great job! <span />
6 0
3 years ago
Read 2 more answers
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.

3 0
3 years ago
What is( 5×100)+(6x10)+(8×1/10)+(9×1/1,000) in standard form
nadya68 [22]

Answer:

6x^10+2504/5+(9*1/1,0)

Step-by-step explanation:

Hope this helps :)

5 0
3 years ago
Kendra is choosing bouquets of flowers for table centerpieces. She decides to buy 40 bouquets for $120 because she thinks that p
Fittoniya [83]

Answer:

Find below:

Step-by-step explanation:

To solve this, we either find the total cos5 of buying 40 bouquets at $2.50 or find the cost of one bouquet at $120.

Cost of one in pack of 40 at $120.

120 divided by 40 = $ 3

Now, we can see that $3>$2.50

This means that Kendra is wrong to buy the bouquet pack of 40 for $120

Hope this helps.

Good Luck

7 0
2 years ago
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