Summation of 3n + 2 from n = 1 to n = 14 = (3(1) + 2) + (3(2) + 2) + (3(3) + 2) + . . . + (3(14) + 2) = 5 + 8 + 11 + ... + 44 ia an arithmetic progression with first term (a) = 5, common difference (d) = 3 and last term (l) = 44 and n = 14
Sn = n/2(a + l) = 14/2(5 + 44) = 7(49) = 343
Therefore, the required summation is 343.
Answer:
i think the answer is 7
Step-by-step explanation:
The answer is choice A.
We're told that the left and right walls of the cube (LMN and PQR) are parallel planes. Any line contained in one of those planes will not meet another line contained in another plane. With choice A, it's possible to have the front and back walls be non-parallel and still meet the initial conditions. If this is the case, then OS won't be paralle to NR. Similarly, LP won't be parallel to MQ.
Table C since it is decreasing at a constant rate of 5.
55 - 5 = 50
50 - 5 = 45
45 - 5 = 40
Answer:
<h2>x = 140, y = 80</h2>
Step-by-step explanation:
