Let x be the length of a month on the other dimension. Since a year value is 365.242 days
We have 2 × 365.242 = 5 × x
It means that 5x = 730,484
and that x = 730,484 / 5 ≈ 146,1 days rounded to the decimal
This means that a month length in the other dimension is of roughly 146 days.. and If we assume that a month in the other dimension is of 30 days approximately, then one day length would be of 146,1 / 30 ≈ 4.87 days
Answer:
60.32
Step-by-step explanation:
Answer:
Step-by-step explanation:
12x9=108
108x8=864
Step #1 for both: figure out which interval your x-value fits into.
For f(-2), x=-2 and -2 fits with x ≤ -2, the top interval.
For f(3), x=3 and 3 fits into -2 < x ≤ 3, the middle interval.
Step #2 for both, plug in your x-value to the piece of the function that fits with that interval.
For f(-2), we know x≤-2, so we use 2x+8 to evaluate x=-2.
For f(3), we know -2
f(-2) = 2(-2)+8 = -4+8 = 4
f(3) = (3)^2 -3 = 9-3 = 6
