Answer:
A. Taivon runs 0,285 miles for every mile he rides his bike.
B. Yes
C. No
Step-by-step explanation:
So, Taivon is running 4 miles for every 14 miles he rides his bike. We can identify a ratio of 4:14. However, both numbers have a common multiple and can be reduced to 2:7; saying that taivon runs 4 miles for every 14 miles he rides his bike is the same to say he runs 2 miles for every 7 miles he rides his bike. To find the value of this ratio, we simply divide 2 miles that Taivon runs between 7 miles he rides his bike. The value of the ratio of miles he runs for miles he rides his bike is 0,285.
Once Taivon finished his training the ratio between the of total number of miles he ran to total number of miles he cycled was 80: 280. This is consistent with his training schedule, because if we divide 80 between 280, we obtain the same value of ratio previously calculated: 0,285. This means also that the total number of miles he ran and the miles he runs on one session are multiples; the same applies for the total number of miles he rode and the miles he rides on one session. If we divide 80 between 4, we obtain 20. Furthermore, if we multiply 20 times 14, we obtain 280. We can conclude then that Taivon trained 20 days in preparation to the Duathlon.
In one training session, Taivon ran 4 miles and cycled 7 miles. The ratio of the distance he ran to the distance he cycled in this session changes and for this session is 0,571. This training session does not represent an equivalent ratio of the distance he ran to the distance he cycled, since he actually fell short in the cycling by 7 miles to his usual 14 miles riding the bike.
Number 2 is the answer to you question
The correct answer is A.
lets discard the other options:
For B:
B is always positive, but f(x) is not, so B graph is not the answers
For C:
C is always positive too, so it cannot be the desire graph.
For D:
D is always negative, but we know that x exponent 3 is positive for all positive numbers,
This is the Pythagorean identity, therefore 
.
If i = √-1 ,
then ...
i² = (√-1)² = -1
i³ = - √-1 = -i <== the answer to the question
i⁴ = (i²) · (i²) = (-1) · (-1) = 1
and then the whole thing starts over again, for higher powers of i .
