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lesya [120]
3 years ago
10

5. What is the length of side a? Please help

Mathematics
1 answer:
Elena-2011 [213]3 years ago
6 0

Answer:

I a pretty sure it is 10

Step-by-step explanation:

but im not positive

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4x10^4 in standard notation
lyudmila [28]

Answer:

40000

Step-by-step explanation:

4×10^4 = 4×10000 = 40000

6 0
3 years ago
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Consider the following expression and determine which statements are true. m+(5n)(9-p)-6-r^2
Fed [463]

Answer:

(B)The expression (5n)(9-p) is the product

(D)The expression 9-p has exactly two terms

Step-by-step explanation:

In the expression;

m+(5n)(9-p)-6-r^2

The coefficient of m is 1, therefore Option A is not true

The product of 5n and 9-p is (5n)(9-p), therefore Option B is true.

The expression 9-p has exactly two terms,9  and p.

Therefore, Options B and D are true.

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3 years ago
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Please help me quick ASAP!!!!!!!!!!<br> Help me ith all the questions 1-5 PLEASE!!!!!!
elixir [45]
C) Length of 7 and Width of 4. This is because 14 x 2 is 28, and 7 x 4 is 28 as well.

I hope this Helps!
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3 years ago
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Please please answer this with the right answer and thanks for your help???<br><br>14 POINTZ!!<br>​
professor190 [17]
# 5 the answers is A
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3 years ago
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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
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