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pantera1 [17]
3 years ago
10

There were twice the number of people

Mathematics
1 answer:
NeX [460]3 years ago
7 0
X times two, x representing what the varible is
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PLEASE HELP, IT'S ARGENT!
ANTONII [103]

Answer:

I'm guessing the right answer should be B

7 0
3 years ago
Find AB<br> Round to the nearest tenth.
zimovet [89]
19.2
To find the hypotenuse the equation would be: a^2 + b^2 = c^2.
12 squared is 144 and 15 squared is 225. Add them and you get 369. Then, square root 369 to isolate c. It'd be about 19.2
6 0
2 years ago
Tiffany is 140 miles away from Maggie. They are traveling towards each other. If Maggie travels 5 mph faster than Tiffany and th
8090 [49]

Answer: Tiffany 15mph, Maggie 20mph

Step-by-step explanation:

Set up the equation 4((x+5) + x) = 140. x+5 represents how many miles Maggie covered in one hour. x represents how much Tiffany traveled in one hour. 140 is the number of miles in total. 4 is the number of hours in total.

Simplify the equation.

(x+5) + x = 35    Divide both sides by 4

2x+5 = 35         Combine like terms

2x = 30              Subtract 5 from both sides

x = 15                 Divide both sides by 2

Tiffany traveled 15mph, while Maggie traveled 15+5=20mph.

3 0
3 years ago
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
I need help please.
erica [24]
Answer is #3
add up all the x’s
5 0
3 years ago
Read 2 more answers
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