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Nesterboy [21]
10 months ago
10

Can you solve for me 3ax-n/5= -4

Mathematics
1 answer:
sergeinik [125]10 months ago
7 0

Answer:

If you are solving for n then it would be n=15ax+20

Hope it helps

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Matthew made a scale model of the solar system for a science project. He created the planet Mercury with a diameter of 6 inches.
Ronch [10]

Answer:

3000

Step-by-step explanation:

I'm learning this right now too!

500×6in=3000

6 0
2 years ago
Start at 4, add the next odd number. What is the algebraic expression?
juin [17]

Answer:

It would be 8

Step-by-step explanation:your wellcome

8 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
The lines below are perpendicular. If
RSB [31]

Answer:

Correct answer: s₁ = - 2

Step-by-step explanation:

The relationship between the slope of two lines that are perpendicular is:

s₁ = - 1/s = - 1/(1/2) = - 2

God is with you!!!

5 0
3 years ago
Find the value of x and AC if the midpoint of segment AC is B. AB = 8x and BC = 6x + 24 Draw and label a picture to help you. x
Sindrei [870]

Answer:

x=12

Step-by-step explanation:

plug in 12 to x and results will make sense

7 0
3 years ago
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