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Mrac [35]
3 years ago
10

43,785 to the nearest ten thousand

Mathematics
2 answers:
denis-greek [22]3 years ago
5 0
40,000
Have a good day
castortr0y [4]3 years ago
4 0

Answer:

40,000

Hope it helps .

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How multiplied whole number with fraction
Fofino [41]

Search it up and you shall find an answer believe me it is there

3 0
3 years ago
Read 2 more answers
45 x 60 / 45 = ______ <br><br> pleasee help
AleksandrR [38]

Answer:

60

Step-by-step explanation:

45 x 60/45 = 2700/45

2700/45 = 60

8 0
3 years ago
Mrs moretti is making welcome bags to give to the 18 students in his homeroom on the first day of school. He puts m mechanical p
julsineya [31]

Answer:

b.) 18(m+2m)

d.) 18m+18(2m)

e.) 54m

Step-by-step explanation:

Given;

total number of Mr Moretti's students = 18

number of mechanical pencils for each student = m

total number of mechanical pencils for all the students = 18m

number of regular pencils for each student = 2m

total number of regular pencils for all the students = 18(2m)

The total number of pencils given to all the 18 students by  Mr Moretti is given by;

T = Total mechanical pencils + total regular pencil

T = 18m + 18(2m)

T = 18(m + 2m)

T = 18(3m)

T = 54m

Therefore, based on the expressions above, the correct options include;

b.) 18(m+2m)

d.) 18m+18(2m)

e.) 54m

3 0
3 years ago
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
How do you Round 96,286 to the nearest thousand.
Varvara68 [4.7K]

Answer:

The correct answer is 96,000

Step-by-step explanation:

To round you look at the thousands place and there is a 9 then you look at the number to the right of it which is 2 and it is lower than 5 since it is lower than five it rounds down so the correct answer is 96,000.

HAVE A GOOD DAY!

8 0
2 years ago
Read 2 more answers
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