Because this is a second degree equation you will have 2 solutions. When you take the square root of a number you have to account for both the positive and negative roots. Since the square root of 289 is 17 then your solutions are +17 and -17.
Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
Answer:
A. 2CO + O2 ---> 2CO2
Step-by-step explanation:
2CO + O2 ---> 2CO2
Answer:
-7
Step-by-step explanation:
6 - 15 = -7
