If r = 3.25, 8r + 6 = 8(3.25) + 6 = 32.
12 + 8r - 6 = 7r - 5 - 3 8r +6 = 7r -8 r = -14
The length of one arch of a cycloid generated by a circle of radius r is 8r.
13r +10 =5r-46 13r-5r=-46-10 8r=-56 8r/8=-56/8 r=-7 use the system of letter one side and numbers one side. also when you move the numeral to the left after the equal sign you need to change the sign if it's positive make it negative, if negative make it positive.
ctrl plus r reloads a webpage.
r + 11 + 8r = 29
8r + 17 = 65 8r = 65 - 17 8r = 48 r = 48 / 8 r = 6
8r + 17 = 65 8r = 48 r = 6
If r = 3.25, 8r + 6 = 8(3.25) + 6 = 32.
It is not correct to say that 8 plus 2 8r-2r 4 3r plus 2 r = 0
Solving for r: 16r2+18r+5=0 (2r+1)(8r+5)=0 2r+1=0 and 8r+5=0 2r=-1 .......... 8r=-5 r= -1/2 ........ r= -5/8
12 + 8r - 6 = 7r - 5 - 3 8r +6 = 7r -8 r = -14
2r + 7r - 6t - r + 7t = 8r + t
-5r + 8r + 5 = 3r +5
-9r
assume that you mean 8r + 112 = 8(r+14)
r=-32