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Ray Of Light [21]
3 years ago
14

Which statement is true regarding the functions on the

Mathematics
1 answer:
Leto [7]3 years ago
4 0

Answer:

f(-3) = g(-3)

Step-by-step explanation:

Remember points on the graph get their location from the x-axis (horizontal) and y-axis (vertical).

In function notation, "f(x)" means when "x" is the value in the brackets, what the value of "y" will be.

First option for example: f(-3) = g(4)

<em>Value of "y" for f(x) when x = -3:</em>

On the blue line, when x = -3, y = -4.

<em>Value of "y" for g(x) when x = 4:</em>

On the red line, when x = 4, y = -3.

<u>Therefore</u>:

f(-3) = g(4)

-4 = -3          <= This is false.

In the third option: f(-3) = g(-3)

<em>On f(x) blue, when x = -3: </em>y = -4

<em>On g(x) red, when x = -3: </em>y = -4

<u>Therefore</u>:

f(-3) = g(-3)

-4 = -4                <= This is true.

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The answer to your question is $360
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3 years ago
A rectangle has a perimeter 204 it’s length six feet longer than twice it’s width if L stands for the rectangle and W stands for
andriy [413]

Answer:

W=32

L=70

Step-by-step explanation:

L=2w+6

204=2L+2W

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3 years ago
I WANT TO KNOW<br> WHAT IS THE MULTIPLE OF 6
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6, 12, 18, 24, 30
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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
Factor the trinomial x^2 - 16x +63
Butoxors [25]
Factor the following:
x^2 - 16 x + 63
The factors of 63 that sum to -16 are -7 and -9. So, x^2 - 16 x + 63 = (x - 7) (x - 9):
Answer:  (x - 7) (x - 9)
3 0
3 years ago
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