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Sonbull [250]
3 years ago
6

Need some help please

Mathematics
2 answers:
Naya [18.7K]3 years ago
4 0
The answer is A, 7 because (23+19)÷6= 7
spayn [35]3 years ago
3 0
It would be A.) 7. hope this helps
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True or false? it's not possible to build a triangle with side lengths of 3 3 and 9
Anit [1.1K]
True

The two shorter lengths do not add up to more than the longest length. 3+3 is less than 9. Therefore, even if the two shorter lengths lay on top of the longer side, the two ends cannot meet to form a closed 3 sided figure
6 0
3 years ago
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Anyone want to ch at?
algol13

Answer:

sure?

Step-by-step explanation:

7 0
2 years ago
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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.

3 0
3 years ago
Is 15 yards to 18 yard increase or decrease
patriot [66]
It is increase, because 15 is less than 18. Other way around, if the question is 18 to 15 yards, then that would be decrease, because it is getting smaller.
8 0
3 years ago
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Creating an Exponential Model
belka [17]

Answer:

P = $300

r = 0.15

n = 12

$544.61  (to the nearest cent)

P(1+r)^t

$524.70  (to the nearest cent)

Step-by-step explanation:

P = principal amount = $300

r = annual interest rate in decimal form = 15% = 15/100 = 0.15

n = number of times interest is compounded per unit t = 12

<u>How much she'll owe in 4 years</u>

P = 300

r = 0.15

n = 12

t = 4

P(1+\frac{r}{n})^{nt}=300(1+\frac{0.15}{12})^{12 \times 4}

= $544.61  (to the nearest cent)

<u>Yearly compounding interest rate</u>

P(1+r)^t

<u>How much she'll owe in 4 years at yearly compounding interest</u>

P(1+r)^t=300(1+0.15)^4

= $524.70  (to the nearest cent)

7 0
2 years ago
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