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jeka94
3 years ago
9

Why is the answer B? Plz explain.

Mathematics
1 answer:
sergeinik [125]3 years ago
4 0
For line l to intersect line m at point (2, 1/2), line m must have the point (2,1/2) on its graph. It is implied that line l already has the point (2, 1/2). If line m does not have it, then there will be no intersection at that specific point.

Try checking every choice to see if it has point (2,1/2) on the graph.
Note that (x,y) = (2,1/2); check by using x=2 and y=1/2.

For A,

   2x = y/2
   2(2) = (1/2) / 2
   4 <span>≠ 1/4
</span>
Cannot be choice A as it results in a false equation; this choice will not go through (2,1/2)

For B:

   2y = 3 - x
   2(1/2) =  3 - 2
   1 = 1

This is a true equation so the point (2,1/2) is on the graph of 2y=3-x. This means that if this is the equation for line m, then line m will have a point at (2,1/2) and therefore intersect with line l. Therefore, B is the answer.

The rest of the choices are false as shown:

For C:
 
   2x + 4y = 8
   2(2) + 4(1/2) = 8
   4 + 2 = 8
   6 ≠ 8

Cannot be choice C as it results in a false equation; this choice will not go through (2,1/2)

For D:

   y = 4 - (5/4)x
   1/2 = 4 - (5/4)(2)
   1/2 = 4 - 5/2
   1/2 = 8/2 - 5/2
   1/2≠ 3/2

Cannot be choice D as it results in a false equation; this choice will not go through (2,1/2)
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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
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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
I need help with this question about similar triangles
Colt1911 [192]

Answer:

ST = 12 units

Step-by-step explanation:

Δ RAS and Δ TES are similar triangles, thus the ratios of corresponding sides are equal, that is, noting that SR = 21 - ST, thus

\frac{ST}{SR} = \frac{TE}{AR} substitute values

\frac{ST}{21-ST} = \frac{8}{6} ( cross- multiply )

6ST = 8(21 - ST)

6ST = 168 - 8ST ( add 8ST to both sides )

14ST = 168 ( divide both sides by 14 )

ST = 12

6 0
3 years ago
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