Put the following statements into order to prove that if 3n+4 is even then n is even. put n next to the 3 statements that should not be used.1. then, 3n+4=3(2k+1)+4=6k+7=2(3k+3)+1.2. by definition of odd, there exists integer k such that n=2k+1.3. thus, if n is odd then 3n+4 is odd or by contradiction, if 3n+4 is even then n is even, .4. therefore, by definition of odd, 3n+4 is odd.5. since k is an integer, t =3k+3 is also an integer.6. thus, there exists an integer t such that 3n+4=2t+1.7. suppose n is odd.8. thus, if n is odd then 3n+4 is odd or by contraposition, if 3n+4 is even then n is even, .9. suppose 3n+4 is even.10. by definition of odd, n=2k+1.

7, 2, 1, 5, 6, 4, 8

Step-by-step explanation:

1. Then, 3n+4=3(2k+1)+4=6k+7=2(3k+3)+1.

2. By definition of odd, there exists integer k such that n=2k+1.

N 3. Thus, if n is odd then 3n+4 is odd or by contradiction, if 3n+4 is even then n is even, .

4. Therefore, by definition of odd, 3n+4 is odd.

5. Since k is an integer, t =3k+3 is also an integer.

6. Thus, there exists an integer t such that 3n+4=2t+1.

7. Suppose n is odd.

8. Thus, if n is odd then 3n+4 is odd or by contraposition, if 3n+4 is even then n is even, .

N 9. Suppose 3n+4 is even.

N 10. By definition of odd, n=2k+1.

Right order:

7. Suppose n is odd.

2. By definition of odd, there exists integer k such that n = 2k+1.

1. Then, 3n+4 = 3(2k+1)+4 = 6k+7 = 2(3k+3)+1.

5. Since k is an integer, t = 3k+3 is also an integer.

6. Thus, there exists an integer t such that 3n+4 = 2t+1.

4. Therefore, by definition of odd, 3n+4 is odd.

8. Thus, if n is odd then 3n+4 is odd or by contraposition, if 3n+4 is even then n is even.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

question: how many people can she serve?

answer: she can serve 16.66666667 people. so she can only give 16 people a complete serving.

question: does she have enough ice cream for her guests?

answer: no. explanation: she only has enough for 16 people to have a full serving. however, she can give a 17th person a partial serving.