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.
We have been given that
Now, we need to find an ordered pair that must be on the graph of
On substituting x=0 in the above equation, we get
Therefore, the required ordered pair is given by (0,-1)
Answer
Step-by-step explanation:
Answer:
the answer is 1.0893×10 it is now in scientific notation
Answer:the answer is A: (0,80)
Step-by-step explanation:
