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Alexus [3.1K]
3 years ago
8

Show all you're work!! Point's and will mark brainliest

Mathematics
1 answer:
Musya8 [376]3 years ago
4 0

Answer:

d = 7.2

Step-by-step explanation:

Step 1: Write out equation

d/1.2 = 6

Step 2: Multiply both sides by 1.2

d = 7.2

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What is the value of the expression below? -28÷-4
Damm [24]

Answer: +7

Step-by-step explanation: Since a negative divided

by a negative always equals a positive, (-28) ÷ (-4) = +7

7 0
3 years ago
Read 2 more answers
Mitzi has 79 coins she divides the coins equally among herself and her 3 brothers Mitzi will keep any coins left over how many c
Elanso [62]

Answer:

19

Step-by-step explanation:

79÷4=19R3

       ≈ 19

7 0
3 years ago
What is the answer to -5x+7y=11<br> -5x+3y=19
marusya05 [52]
\left\{\begin{array}{ccc}-5x+7y=11\\-5x+3y=19&/\cdot(-1)\end{array}\right\\+\left\{\begin{array}{ccc}-5x+7y=11\\5x-3y=-19\end{array}\right\\-------------\\.\ \ \ \ \ \ \ \ 4y=-8\ \ \ \ |divide\ both\ sides\ by\ 4\\.\ \ \ \ \ \ \ \ \ y=-2\\5x-3(-2)=-19\\5x+6=-19\\5x=-19-6\\5x=-25\ \ \ \ \ |divide\ both\ sides\ by\ 5\\x=-5\\\\Answer:x=-5\ and\ y=-2

3 0
3 years ago
Read 2 more answers
Show that in a group of five people (where any two people are either friends or enemies), there are not necessarily three mutual
sveta [45]

Answer:

Lets call the people A,B,C,D and E. We need to find an example where there are not 3 mutual friends and not three mutual enemies.

Lets start by assuming that A has 2 friends, B and C, and 2 enemies, D and E.

Since we want A,B and C not to be mutually friends, then B and C neccesarily have to be enemies.

Now, since B and C are enemies, they cant be enemies at the same time with D, and they also cant be enemies at the same time with E. Therefore one of them has to be friends with D and one has to be friends with E.

Also, since D and E are enemies with A, they neccesarily need to be friends.

From the fact that D and E are friends, we coclude that they cant have a friend in common, as a consequence, B has to be friends with one of D and E and C has to be friend with the other. We can assume that B is friend with D and C is friend with E. If we use that, we have the following friendship configuration

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

Friends of A: B, C

Enemies of A: D,E

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

Friends of B: A,D

Enemies of B: C, E

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

Friends of C: A,E

Enemies of C: B,D

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

Friends of D: B,E

Enemies of D: A,C

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

Friends of E: C,D

Enemies of E:  A,B

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

As you can check, there is not three mutual friends nor three mutual enemies.

8 0
3 years ago
5 points
Ahat [919]

Answer:

Students

Step-by-step explanation:

Teachers: 13/20= 0.65 = 65%

Students: 37/50= 0.74 = 74%

74% > 65%

Students are more in favor of the new club than teachers

7 0
2 years ago
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