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lesya [120]
3 years ago
15

If two lines intersect at a right angle, then they are perpendicular

Mathematics
1 answer:
kipiarov [429]3 years ago
4 0

This is TRUE!

The definition for perpendicular only applies when the lines are at a right angle which is 90°

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Help please i rlly need it
jonny [76]
The answer is C.7/3 pi ft lmk if you want more answers
8 0
2 years ago
A ladder leans against the wall and reaches a
stich3 [128]
The base would represent the side adjacent to the angle and the height up the wall would be the opposite side. Using the tangent^-1 (opposite over adjacent) function we are looking for the tan^-1 of 6/24 or 14.04°
5 0
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
In Yellowstone National Park there are 300 species of birds that migrate. This accounts for 2/7 of all the species of birds sigh
Juli2301 [7.4K]

Answer:

\frac{2}{7} x = 300

Step-by-step explanation:

Let x = the number of all the species of birds sighted in the park

“300 species of birds that migrate. This accounts for 2/7 of all the species of birds sighted in the park”

the above sentence means “(2/7)x = 300”

now we can solve it :

(2/7)x = 300

⇔ 2x = 7×300

⇔ 2x = 2100

⇔ x = 2100/2

⇔ x = 1050

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=7.75%20divided%20by%204%201%2F2" id="TexFormula1" title="7.75 divided by 4 1/2" alt="7.75 divi
Andrew [12]

Answer:

2.64516129

Step-by-step explanation:

5 0
3 years ago
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