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Harrizon [31]
3 years ago
8

Rewrite the equation by completing the square. x^2 - 4x + 3 = 0

Mathematics
2 answers:
maw [93]3 years ago
8 0

Answer:

After completing the square, we can say:

(x - 2)² = 1

If we solve for x, this tells us that x has a value of both 1 and 3.

Step-by-step explanation:

Given:

x² - 4x + 3 = 0

We can complete the square by adding 1 to both sides:

x² - 4x + 4 = 1

(x - 2)² = 1

We can then solve for x by saying:

(x - 2)² = 1

x - 2 = √1

x = 2 ± 1

x = 1, 3

Degger [83]3 years ago
4 0

Answer:

the answer would be x=3,1

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Answer:

Step 3

Step-by-step explanation:

This is the right one because you are adding two thing in step two which lead to step 3

5 0
3 years ago
The balance after the grace period is $15,000. A payment of $5,000 is made on the
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Answer:

the average daily balance is 500

Step-by-step explanation:

If a payment of 5,000 is made on the tenth day you can divide 5,000 by ten to get the amount payed everyday before hand. Once you do that you will end up with the answer of 500. Therefore the average daily balance is 500

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3 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
Somebody help me solve;
Nadya [2.5K]
P(at least 5 rolls until 1) = P(4 rolls are not 1) = 5/6 x 5/6 x 5/6 x 5/6 = 0.4823 (4sf)

Fewer than 7 rolls to get second 1 after first takes 3 rolls means second occurs on 4th, 5th or 6th roll
The probability of each of these is 1/6, 5/6 x 1/6 and 5/6 x 5/6 x 1/6 respectively.
P(second 1 on 4th, 5th or 6th roll) = 1/6 + 5/36 + 25/216 = 91/216 = 0.4213 (4sf)
3 0
2 years ago
Jolly measured the cake. Then she calculated that it has a circumference of 15cm. what is the cake's diameter? Use 3.14 for pi.
Hatshy [7]

Answer:

4.76cm

Step-by-step explanation:

Given data

circumference= 15cm

The expression for circumference is

C= 2πr

substitute and solve for r

15= 2*3.14*r

r= 15/ 2*3.14

r=15/ 6.28

r= 2.38 cm

Hence the diameter is

D= 2r

D= 2*2.38

D=4.76cm

6 0
2 years ago
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