1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 are your team numbers
1 - 2
1 - 3
1 - 4
1 - 5
1 - 6
1 - 7
1 - 8
1 - 9
1 - 10
2 - 3
2 - 4
2 - 5
2 - 6
2 - 7
2 - 8
2 - 9
2 - 10
3 - 4
3 - 5
3 - 6
3 - 7
3 - 8
3 - 9
3 - 10
4 - 5
4 - 6
4 - 7
4 - 8
4 - 9
4 - 10
5 - 6
5 - 7
5 - 8
5 - 9
5 - 10
6 - 7
6 - 8
6 - 9
6 - 10
7 - 8
7 - 9
7 - 10
8 - 9
8 - 10
9 - 10

45 matches in all >.< :)

8 0

4 years ago

I’ll give brainliest help ASAP

vazorg [7]

Answer:

your answer will be 6weeks

hope ur help

8 0

3 years ago

Read 2 more answers

Simplify: 2(x+6) show work please

vagabundo [1.1K]

Answer:

2x + 12

Step-by-step explanation:

2(x + 6) =

2(x) + 2(6) =

2x + 12

Hope that helps!

3 0

3 years ago

Read 2 more answers

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

3 0

3 years ago