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denis23 [38]
3 years ago
13

What is the constant of the polynomial 2x3 - 8x2 + 3x - 7

Mathematics
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

-7

Step-by-step explanation:

A constant number is a number that contains no variables like x and y. The only constant in that problem is -7.

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Which is true regarding chords and diameters of circles?
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Answer:

Both chords and diameters have two endpoints on a circle. Diameters must intersect the center of a circle.

Step-by-step explanation:

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2 years ago
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Does (-1, 2), (1, 1),( 1 -1), (2,1), (4,2) represent y as a function
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No. A function maps one input to exactly one output. The given relation maps 1 to both 1 and -1, as indicated by the pairs (1, 1) and (1, -1).

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2 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
For a particular cargo train, the amount of fuel needed for a trip varies directly with the number of miles it plans on travelin
Katyanochek1 [597]
2250 gallons. You have to think of the ratio. 300•750=225000
225000/100=2250
You have to do cross multiplication
4 0
3 years ago
Read 2 more answers
I’ll give brainliest!!!
n200080 [17]

Answer:

A) arithmetic sequence

g(n) = 20 + 3n

Step-by-step explanation:

Each minute, the number of pages increases by 3, which means the common difference is 3 (because it is being added, not multiplied).

If there is a common difference, it's an arithmetic sequence. (Geometric sequences have common ratios.)

If she starts at 20 pages, then you just need to multiply 3 by the number of minutes passed (n) and add it to 20 to find the page she's on.

5 0
3 years ago
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