it's 8
The numeral V (5) is symetrical. VI (6), VII (7), VIII (8), and IX (9) are not symetrical.
A regular octagon has 8 line of symmetry.
8/8 - 1 + 3 - 1 + 0 = 1-1+3-1+0 = 0 + 3 - 1 + 0 = 3 - 1 = 2
(8-7)2+(0-0)2 = 1 and the square root of 1 is 1 Answer: 1
it's 8
The numeral V (5) is symetrical. VI (6), VII (7), VIII (8), and IX (9) are not symetrical.
A regular octagon has 8 line of symmetry.
8
8/8 - 1 + 3 - 1 + 0 = 1-1+3-1+0 = 0 + 3 - 1 + 0 = 3 - 1 = 2
x/8 - 1 = 0 (8)(x/8)- (8)(1) = (8)(0) x - 8 = 0 x - 8 + 8 = 0 + 8 x = 8
101011 1 *32 = 32 0 *16 = 0 1 * 8 = 8 0 * 4 = 0 1 * 2 = 2 1 * 1 = 1 -------------- 32+8+2+1=43.
(8-7)2+(0-0)2 = 1 and the square root of 1 is 1 Answer: 1
I think you mean r2 -7r -8 = 0. If so, then factor it: (r+1)(r-8) = 0 Then set r+1 = 0, r+1-1=-1, r = -1 Set r-8 =0, r-8+8 = 8 so r = 8
71 is greater.
yes
(x+8)(x-1)=0 x+8=0 x=-8 x-1=0 x=1 Therefore, the zeroes are -8 and 1.