we can always find the x-intercept by simply settting y = 0, and solving for "x".
and we can always find the y-intercept by simply setting x = 0 and solving for "y".
![\bf x-4y=-16\implies \stackrel{x=0}{0-4y=-16}\implies y=\cfrac{-16}{-4}\implies y=4 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (0,4)~\hfill](https://tex.z-dn.net/?f=%20%5Cbf%20x-4y%3D-16%5Cimplies%20%5Cstackrel%7Bx%3D0%7D%7B0-4y%3D-16%7D%5Cimplies%20y%3D%5Ccfrac%7B-16%7D%7B-4%7D%5Cimplies%20y%3D4%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20%280%2C4%29~%5Chfill%20)
You can make two squares with 8, and put a line through one of them making one square two triangles put together.
The answer would probably be 18 degrees below normal because the absolute values of the rest are 8 not 18 which is the absolute value of 18 degrees below normal.
Hello from MrBillDoesMath!
Answer:
a(n) = (-n)^3 where n = 1,2,3,...
Discussion:
The pattern 1,8,27, 64... is immediately recognizable as the the cube of the positive integers. But this question has a minus sign appearing before each entry, suggesting we try this:
- 1 = (-1)^3
-8 = (-2)^3
-27 = (-3)^3
-64 = (-4)^3
That's what the problem statement asked for
. The answer is equivalently
-1 * (n^3)
Thank you,
MrB
