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mixer [17]
3 years ago
14

Factorise the following fully: 6a⁴b⁶-8a³b⁵+12a²b³​

Mathematics
1 answer:
AlladinOne [14]3 years ago
5 0

Answer:

2a^2 b^3(3a^2 b^3 - 4ab^2 + 6)

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The answer is n=289.
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R/19 &gt; -1<br> please help solve for R
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Answer:

Step-by-step explanation:

\frac{r}{19}>-1

Multiply both sides by 19

\frac{r}{19}*19>-1*19 \\\\r > -19

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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
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
A wire was bent into the shape of a rectangle with width 5 and length 7 . If this wire
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Answer:

B | 6

Step-by-step explanation:

First, find the length of the wire by finding the perimeter of the rectangle

2(5) 2(7) = 24

If a square has 4 sides, put 24 as the numerator and 4 as the denominator.

24/4

Now, divide it and you'll get 6, the answer.

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3 years ago
A triangle has angles measuring 43° and 62°. What is the measure of the triangle's third angle?
jarptica [38.1K]

Answer:

75 degrees

Step-by-step explanation:

I know the measure of the triangle's third angle is 75 degrees because all the angles of a triangle add up to 180 degrees. So, I did 180-62-43 and got 75. So, in order for this triangle to be true, the third triangle would have to be 75 degrees.

To check: 43 degrees+ 62 degrees+ 75 degrees = 180 degrees.

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3 years ago
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