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FinnZ [79.3K]
3 years ago
6

Which of the following is not an inequality?

Mathematics
2 answers:
emmainna [20.7K]3 years ago
8 0
<h3>Answer: 18x + 9y  (choice C)</h3>

===========================================================

Explanation:

An inequality must have one of the four symbols

<

>

\le

\ge

which are "less than", "greater than", "less than or equal to", and "greater than or equal to" in that exact order. Choices A, B and D have one of these symbols which makes them inequalities. We can cross them off the list. Choice C does not have any of the four symbols mentioned, so it's not an inequality. Instead, choice C is a mathematical expression, or we can say "expression" for shorthand.

kap26 [50]3 years ago
5 0

Answer

C is the correct option

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Prove each of the following statements below using one of the proof techniques and state the proof strategy you use.
pochemuha

Answer:

See below

Step-by-step explanation:

a) Direct proof: Let m be an odd integer and n be an even integer. Then, there exist integers k,j such that m=2k+1 and n=2j. Then mn=(2k+1)(2j)=2r, where r=j(2k+1) is an integer. Thus, mn is even.

b) Proof by counterpositive: Suppose that m is not even and n is not even. Then m is odd and n is odd, that is, m=2k+1 and n=2j+1 for some integers k,j. Thus, mn=4kj+2k+2j+1=2(kj+k+j)+1=2r+1, where r=kj+k+j is an integer. Hence mn is odd, i.e, mn is not even. We have proven the counterpositive.

c) Proof by contradiction: suppose that rp is NOT irrational, then rp=m/n for some integers m,n, n≠. Since r is a non zero rational number, r=a/b for some non-zero integers a,b. Then p=rp/r=rp(b/a)=(m/n)(b/a)=mb/na. Now n,a are non zero integers, thus na is a non zero integer. Additionally, mb is an integer. Therefore p is rational which is contradicts that p is irrational. Hence np is irrational.

d) Proof by cases: We can verify this directly with all the possible orderings for a,b,c. There are six cases:

a≥b≥c, a≥c≥b, b≥a≥c, b≥c≥a, c≥b≥a, c≥a≥b

Writing the details for each one is a bit long. I will give you an example for one case: suppose that c≥b≥a then max(a, max(b,c))=max(a,c)=c. On the other hand, max(max(a, b),c)=max(b,c)=c, hence the statement is true in this case.

e) Direct proof: write a=m/n and b=p/q, with m,q integers and n,q nonnegative integers. Then ab=mp/nq. mp is an integer, and nq is a non negative integer. Hence ab is rational.

f) Direct proof. By part c), √2/n is irrational for all natural numbers n. Furthermore, a is rational, then a+√2/n is irrational. Take n large enough in such a way that b-a>√2/n (b-a>0 so it is possible). Then a+√2/n is between a and b.

g) Direct proof: write m+n=2k and n+p=2j for some integers k,j. Add these equations to get m+2n+p=2k+2j. Then m+p=2k+2j-2n=2(k+j-n)=2s for some integer s=k+j-n. Thus m+p is even.

7 0
3 years ago
QUESTION 9
Ratling [72]
The first one is (v+6)(v-6)
8 0
3 years ago
Write the coordinates of the vertices after a reflection across the line x= -3
anzhelika [568]

(Please vote me Brainliest if this helped!)

  1. C' (-3, -6)
  2. D' (-3, 1)
  3. E' (0, -6)
3 0
3 years ago
What is 5.4 x 2 3/4 show your work
VladimirAG [237]

Answer:

14.85

Step-by-step explanation:

4.4 x 2.75 = 14.84

       2.75

       4.4

      -------

     (Multiply it out)

3 0
3 years ago
Read 2 more answers
Question 7: Which of the following statements is true?
yawa3891 [41]

Problem 7)

The answer is choice B. Only graph 2 contains an Euler circuit.

-----------------

To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.

With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.

================================================

Problem 8)

The answer is choice B) 5

-----------------

Work Shown:

abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10

101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10

101 base 2 = (1*4 + 0*2 + 1*1) base 10

101 base 2 = (4 + 0 + 1) base 10

101 base 2 = 5 base 10


4 0
3 years ago
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