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.
IF IT EQUALS 0
1. find two number that multiplied to 48 and adds to 14, which are 6 and 8.
2. substitute the new numbers in with x to get x^2 + 6x + 8x + 48.
3. factor out the x and the 8 to get x(x+6)+8(x+6).
4. x = -6, x = -8
IF IT DOES NOT EQUAL 0
then (x+6)*(x+8) is your answer.
Answer:so would i
Step-by-step explanation:
