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Reptile [31]
3 years ago
13

Find the slope of the line that passes through each pair of points. 5. G(–3, 2), H(7, 2)

Mathematics
1 answer:
sladkih [1.3K]3 years ago
5 0

Answer:

y=0x+2 meaning the slope is equal to 0

Step-by-step explanation:

If you think about this question you'll see that our two points have the same y coordinate

This means that the slope is equal to 0

so we have

y=0x+b  

If we plug in -3 for x and set the equation equal to 2 we can find b

0+b=2

b=2

so we have

0x+b=y

You might be interested in
MO bisects angle LMN , angle LMO = 6x-22 and angle NMO = 2x +34. Solve for x and find angle LMN.
Mkey [24]

Answer:

  • x=14
  • \angle{LMN}=124\textdegree

Step-by-step explanation:

<u><em>To Determine:</em></u>

Solve for x and find angle LMN.

<u><em>Fetching Information and Solution Steps:</em></u>

Considering the angle \angle{LMN}

As MO bisects the angle \angle{LMN} into two equal angle parts. These equal angles are:

\angle{LMO}=6x-22

\angle{NMO}=2x+34

As these angles are equal. i.e.

\angle{LMO}=\angle{NMO}

6x-22=2x+34

\mathrm{Add\:}22\mathrm{\:to\:both\:sides}

6x-22+22=2x+34+22

6x=2x+56

\mathrm{Subtract\:}2x\mathrm{\:from\:both\:sides}

6x-2x=2x+56-2x

4x=56

\mathrm{Divide\:both\:sides\:by\:}4

\frac{4x}{4}=\frac{56}{4}

\mathrm{Simplify}

x=14

Hence, x=14

As  \angle{LMN} was cut into two equal parts \angle{LMO} and \angle{NMO}

So,

\angle{LMN} = \angle{LMO} + \angle{NMO}

            = 6x-22+2x+34

            = 8x + 12

            = 8(14) + 12     ∵ x=14

            = 124\textdegree

Therefore, \angle{LMN}=124\textdegree

Keywords: angle bisector, congruent angles

Learn more about angle bisector and congruent angles from brainly.com/question/711370

#learnwithBrainly

3 0
3 years ago
Is 85 prime or composite explain
forsale [732]
85 is a composite number because it has more than two factors.
85 factors are 1, 5, 17, 85

For example 5 is prime because it’s factors are only two of them
5 factors are 1, 5
8 0
3 years ago
Alicia bikes 25% of a 100-mi trip on the first day. she bikes of the remaining distance on the second day. on the third day, she
DochEvi [55]
First day: 25% × 100 = 25 ( miles)
Second day : the remaining distance is 100 - 25 = 75 ( milesl
Third day: 40% × 75 = 30
25 + 75 + 30 = 130 (miles)

5 0
2 years ago
Plz help 100 points!!!
Bad White [126]

(5) Angle ADC is the sum of angles ADB and BDC.

(6) Yes, because lines AG and EC form two pairs of vertical angles, including ADC and EDG, as well as ADE and CDG.

(7) Angle HJK is clearly a right angle, and a right angle has measure 90°.

7 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
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