No, they are not.
4 -5 on the r ax = -1
p = 50q/100 = 1/2 q r = 40q/100 = 2/5 q p = (1/2)/(2/5) = (1/2)(5/2) = 5/4 r or 1 1/4 r Thus, p is 125% of r.
distance=rate*time Jill can paint 1 house in 5 hours. 1=r*5 r=1/5 houses/hour Tom can paint 1 house in 4 hours. 1=r*4 r=1/4 houses/hour 1 house = 1/5*x+1/4*x 1=(4x+5x)/20 20=9x x= 2.2 repeating hours (2 hours, 13 minutes, 20 seconds)
5 - 5/3 + 5/9 - 5/27 + ... = 5 + 5(-1/3)¹ + 5(-1/3)² + 5(-1/3)³ + ... The required sum is an infinite GP with initial term a = 5, and common difference r = -1/3 As |r| < 1, the sum can be found from sum = a/(1 - r) → 5 - 5/3 + 5/9 - 5/27 + ... = 5/(1 - (-1/3)) = 5/(1 + 1/3) = 5/(4/3) = 5 × 3/4 = 15/4 = 3¾
Each group of 3 is one octal digit made up of: R = 4 W = 2 X = 1 RWX = 4 + 2 + 1 = 7 R-X = 4 + 1 = 5 R-- = 4 So rwx r-x r-- is 754
r2 + r - 20 = 0(r + 5)(r - 4) = 0r + 5 = 0 or r - 4 = 0r = -5 or r = 4
I Am Weasel - 1997 I-R- Gentlemans 1-4 was released on: USA: 5 August 1997
2r + 4 or 2*(r+2)
1 3 5 2 4 r
1 - E 2 - A 3 - R 4 - I 5 - O
6
Only if you add a +1 onto the 4 Oh well that's not the same problem.How dumb r u ?