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kiruha [24]
3 years ago
7

he point-slope form of the equation of a line that passes through points (8, 4) and (0, 2) is y – 4 = (x – 8). What is the slope

-intercept form of the equation for this line? y = x – 12 y = x – 4 y = x + 2 y = x + 6
Mathematics
1 answer:
fgiga [73]3 years ago
5 0

You have y-4 = (x-8)

Add 4 to each side:

Y = (x-8) +4

Remove the parenthesis and combine like terms to get:

y = x-4

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Solve for the x using elimination <br> 3x-2y=13x−2y=1 <br> 2x+2y=42x+2y=4
Nikitich [7]

3x - 2y = 1

2x + 2y = 4

Add the second equation to the first

5x     = 5

2x + 2y = 4

Divide the first equation by 5

x        = 1

2x + 2y = 4

Subtract the first equation from the second

x        = 1

x + 2y = 3

Subtract the first equation from the second again

x        = 1

   2y = 2

Divide the second equation by 2

x        = 1

      y = 1

<h3>So, the solution is  x = 1  and  y = 1  {or: (1, 1)} </h3>
7 0
3 years ago
Which number is equivalent to 7/8
PtichkaEL [24]

Answer:

78 is 7 divided by 8, which equals 0.875. So an equivalent fraction is another fraction that also equals 0.875.

6 0
3 years ago
Wren and Jenni are reading the same book. Wren is on page 12 and reads 7 pages every night. Jenni is on page 8 and reads 11 page
Igoryamba
You should make a table
8 0
3 years ago
A class of students consists of 9 men and 51 women. Write a proper fraction to represent the part of the class that is women. Re
balandron [24]

Answer:

85%

Step-by-step explanation:

9+51 = 60

(51/3) ÷ (60/3) = 17/20

(17/20)% = 85%

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
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