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adoni [48]
3 years ago
10

Please help! I am so confused

Mathematics
1 answer:
levacccp [35]3 years ago
8 0

Answer:

The second option, y + 2x = 10, is the correct answer for this problem.

Step-by-step explanation:

There are many different ways to solve this problem.  I am going to pick a point represented in the table and plug its values into the given equations to find the correct response.

From the table, we can conclude that the point (0,10) must satisfy the equation.  This means that if we plug in 0 for x and 10 for y into the equations below, we should get a true statement.

y - 2x = 14

10 - 2(0) = 14

10 = 14

Since 10 is not equal to 14, we know that the first option is incorrect.

y + 2x = 10

10 + 2(0) = 10

10 = 10

Therefore, the second option may be our answer, but we should make sure the other options are incorrect.

2y + x = 23

2(10) + 0 = 23

20 = 23

Since 20 is not equal to 23, we know that the third option is incorrect.

y + x = 11

10 + 0 = 11

Since 10 is not equal to 11, we know that the fourth option is also incorrect.

Since the second option is the only answer that yielded a true statement when a point from the table was plugged in, we can conclude that the second option (y + 2x = 10) is the answer.  If you wanted to make sure, you could plug in each of the points represented in the table and confirm that they too make the equation true.

Hope this helps!

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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
2 years ago
Who had this before if so can I get help
telo118 [61]

2*3+4=10

5+2x2=9

4-3+2=3

4*3+2=14

20-4x3+6=14

16-30/2+11=12

24/6*5=20

15/5+2=5

2+2*9-5=15

12-3+4*2=1

12/3+3*4=16

9-6/3=7

16-3*4=4

36/4/9=1

The last one is yours

6 0
2 years ago
Can anyone solve this? will give brainliest if correct.
miv72 [106K]

Answer:

  • See below

Step-by-step explanation:

<u>The missing reasons are:</u>

  • 1. k. Given
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  • 7. e/m. Definition of supplementary
  • 8. a/c. Substitution property of congruence
  • 9. i. Subtraction property of congruence
  • 10. f. Alternate interior angles theorem
  • 11. l. Alternate exterior angles theorem
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If they intersect eachother then the answer is true
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Answer:

16

Step-by-step explanation:

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