r=0 is the solution...
a could equal 0 and r could equal 27a could equal 1 and r could equal 23a could equal 2 and r could equal 19a could equal 3 and r could equal 15a could equal 4 and r could equal 11a could equal 5 and r could equal 7a could equal 6 and r could equal 3a could equal 7 and r could equal - 1It could be any of these and many, many more...is it the whole question?
r = 4
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
R2 - 16 = 0 R2 = 16 Take the square root of each side to get rid of the square on R. R = 4
r=0,Tr-r = 0 = r(T-1), since T != 1, then T-1 is non zero so r must be zero.
r=0 is the solution...
pq-r, if p is 3, q is 4, and r is -6 is equal to 3 x 4 - (-6), which is equal to 12 + 6, which is equal to 18.
a could equal 0 and r could equal 27a could equal 1 and r could equal 23a could equal 2 and r could equal 19a could equal 3 and r could equal 15a could equal 4 and r could equal 11a could equal 5 and r could equal 7a could equal 6 and r could equal 3a could equal 7 and r could equal - 1It could be any of these and many, many more...is it the whole question?
451.72=r3 7.6728~=r
300
r2 + r - 20 = 0(r + 5)(r - 4) = 0r + 5 = 0 or r - 4 = 0r = -5 or r = 4
if: r = 5z 15z = 3y then: z = y/5 r = 5(y/5) r = y
r = 4
r+r+r+r+r=10 therefore, r must equal two
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
a = 4, g = 1,e = 9, l = 8, r = 7, o = 2, n = 6, b = 3, d = 5, t = 0