I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have
C(8,3) = 8!/(8-3)!3!
C(8,3) = 56
There are 56 3-person teams that can be formed from the 8 students.
If you would like to evaluate 6 * [5 * (3 - 9) - 1] + 2 / 7 * (8 - 2) + 4, you can calculate this using the following steps:
6 * [5 * (3 - 9) - 1] + 2 / 7 * (8 - 2) + 4 = 6 * [5 * (-6) - 1] + 2/7 * 6 + 4 = 6 * [- 30 - 1] + 12/7 + 4 = 6 * [- 31] + 12/7 + 4 = - 186 + 12/7 + 4 = - 182 + 12/7 = - 1274/7 + 12/7 = - 1262/7
The correct result would be - 1262/7.
Five times the sum of x and negative two, or five times the difference of x and two
Answer:
A = 40.0 (3 sig. fig)
C = 40
c = 15
Step-by-step explanation:
Using sin law,
sinA/15 = sin100/23
A = 40.0 (3 sig. fig)
Using angle sum of triange,
C = 180 - 40.0 - 100 = 40
Since angleA = angleC, by side equal opp angles, c = 15
