2 AD
3 years. Thats including this year!
1 year
The year 1713 had a total of 365 days, as it was not a leap year. Leap years occur every four years, but the year 1713 is not divisible by 4. Therefore, it followed the standard calendar structure of 365 days.
He was 70 years old when he passed away, which was just this year 2009.
If you are referring to the year 1930 then the year 2005 is the year which is 75 years after the year 1930
As there was no year zero, you would be 312 years old.
ac + 2ad + 2bc + 4bd = a(c + 2d) + 2b(c + 2d) = (a + 2b)(c + 2d) Now expand to confirm your answer: c(a + 2b) + 2d(a + 2b) = ac + 2bc + 2ad + 4bd ≡ ac + 2ad + 2bc + 4bd
All of each of 5BC, 4BC, 3BC, 2BC, 1BC, 1AD, 2AD, 3AD, 4AD, so 9 Or, you might mean 4BC ........... 4AD so 8
ac + 2ad + 2bc + 4bd (ac + 2ad) + (2bc + 4bd) group the figures a(c + 2d) + 2b(c + 2d) remove the common divisors of each set (a + 2b)(c + 2d) take the figures in parentheses as one set, and add the outside figures as the other
About 284 years is one pluto year!
(2a + b)(2c + d)
every 4 years
558 years ago.
vf=vi+at or vf2=vi2+2ad where a=-9.8m/s2
Ptolemy created the Geocentric model in 2AD
4ac + 2ad + 2bc +bd = 2a*(2c + d) + b*(2c + d) = (2c + d)*(2a + b)
To find acceleration using the equation vf^2 = vi^2 + 2ad, you can rearrange the formula to isolate 'a'. First, subtract vi^2 from both sides to get vf^2 - vi^2 = 2ad. Then, divide both sides by 2d to solve for acceleration: a = (vf^2 - vi^2) / (2d).