Given:
1st term = 11
common difference = 6
f(x) = 11 + 6(x - 1)
f(18) = 11 + 6(18-1)
f(18) = 11 + 6(17)
f(18) = 11 + 102
f(18) = 113 number of seats in row 18.
Row
<span>
<span>
</span><span><span>
1 11 11
</span><span>2 11 6 17
</span>
<span>
3 17 6
23
</span>
<span>
4 23 6 29
</span>
<span>
5 29 6 35
</span>
<span>
6 35 6
41
</span>
<span>
7 41 6 47
</span>
<span>
8 47 6 53
</span>
<span>
9 53 6 59
</span>
<span>
10 59 6
65
</span>
<span>
11 65 6 71
</span>
<span>
12 71 6
77
</span>
<span>
13 77 6
83
</span>
<span>
14 83 6
89
</span>
<span>
15 89 6 95
</span>
<span>
16 95 6
101
</span>
<span>
17 101
6
107
</span>
<span>
18 107
6 113
</span></span></span>
Answer:
2. absolute value of p - q
Explanation:
Because when you calculate distance, you always use absolute value.
Can I get a thx and brainliest?
Answer:
christopher columbus
Step-by-step explanation:
A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.
B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4.
(13−10)! 3!
C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =
10! 3!
9!1! 1!2!
10! 3!
8!2! 2!1!
10! 3!
7!3! 3!0!
= 10 · 3 = 30 ways; two women chosen — in C(10,8)C(3,2) =
= 45·3 = 135 ways; three women chosen — in C(10, 7)C(3, 3) =
= 10·9·8 ·1 = 120 ways. Altogether there are 30+135+120 = 285
1·2·3
<span>possible choices.</span><span>
</span>
Answer: 89
Step-by-step explanation: