Answer:
1988 years ago.
Historians have never included a year zero, and it isn't common practice to include it.
Hope this helps,
Azumayay
My father taught me a method that I call, 'throwing some thing over the wall'. There is a sample equation for the first picture that shows you how it works. It's a great way to help isolate the variable and I hope that it works for you. The second picture contains the answers I got through the use of this method. I'm sorry for the messy handwriting; I was on a bus.
Answer:
x=30o+n90on∈Z
Step-by-step explanation:
cos2x=sin(90o−2x)=sin(2x−30o)
Which mean 90o−2x=2x−30o+n360o
or 90o−2x+2x−30o=(2n+1)180o
The later cannot be true.
so x=30o+n90on∈Z
3x - y + z = 5 . . . (1)
x + 3y + 3z = -6 . . . (2)
x + 4y - 2z = 12 . . . (3)
From (2), x = -6 - 3y - 3z . . . (4)
Substituting for x in (1) and (3) gives
3(-6 - 3y - 3z) - y + z = 5 => -18 - 9y - 9z - y + z = 5 => -10y - 8z = 23 . . (5)
-6 - 3y - 3z + 4y - 2z = 12 => y - 5z = 18 . . . (6)
(6) x 10 => 10y - 50z = 180 . . . (7)
(5) + (7) => -58z = 203
z = 203/-58 = -3.5
From (6), y - 5(-3.5) = 18 => y = 18 - 17.5 = 0.5
From (4), x = -6 - 3(0.5) - 3(-3.5) = -6 - 1.5 + 10.5 = 3
x = 3, y = 0.5, z = -3.5
About 67 pairs. one shoe is left over. walking; 26 pairs running: about 41 pairs. sorry if im wrong
