12 + 8r - 6 = 7r - 5 - 3 8r +6 = 7r -8 r = -14
9r-- think of the 'r' as '1r' and add like you normally would.
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.
The length of one arch of a cycloid generated by a circle of radius r is 8r.
4R size is 4" × 6"; 10 cm x 15 cm Other photographic print sizes are 3R (3½" × 5"; 9 cm x 13 cm), 4D (4½" × 6"), the 5R (5" × 7"; 13 cm x 18 cm) and 6R (6" × 8"; double the 4R). The print photo size 8R (8" × 10"; 20 cm x 25 cm) is also well known.
8r + 17 = 65 8r = 48 r = 6
r + 11 + 8r = 29
8r + 17 = 65 8r = 65 - 17 8r = 48 r = 48 / 8 r = 6
12 + 8r - 6 = 7r - 5 - 3 8r +6 = 7r -8 r = -14
-5r + 8r + 5 = 3r +5
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
It is not correct to say that 8 plus 2 8r-2r 4 3r plus 2 r = 0
-6r - s
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2r + 7r - 6t - r + 7t = 8r + t
9r-- think of the 'r' as '1r' and add like you normally would.
if you have 12r and you minus 4r it doesn't equal 48. it would equal 8r to find out what r is i would have to see the whole problem