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)
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
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
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)
It depends on whether the number order can or cannot be identical. Also whether you can use the same number twice. There are 10 unique pairs possible (i.e. 1 and 5 is the same as 5 and 1), but 20 sequential pairs where the order of the pair makes them different, and 25 if you can use the same number in a pair.10 = 1-2, 1-3, 1-4, 1-5, 2-3, 2-4, 2-5, 3-4, 3-5, 4-520 = (include the reverse of those)25 = (include 1-1, 2-2, 3-3, 4-4, 5-5)---A combination is a selection of a number of items from a group without regard to order (i.e. AB = BA). Since order is not important, then the number of combinations is given by the formula: n!/(n - r)! r!where n = the total number of items. r = the number of items chosen from n(recall that n!, pronounced n factorial, means the continued product of successive natural numbers from 1 to n. So 5!= 5x4x3x2x1, for example)Let the 5 numbers be 1, 2, 3, 4, and 5. Possible combinations are: 1 and 2, 1 and 3, 1 and 4, 1 and 5, 2 and 3, 2 and 4, 2 and 5, 3 and 4, 3 and 5, 4 and 5.(Note how '2 and 1' is not listed because '2 and 1' is the same as '1 and 2' when using combinations).So there are 10 combinations of 2 numbers taken from 5 numbers.Using the formula there are 5!/3!2!=10 ways to pick 2 items from 5 where the order does not matter. We also assume we are not replacing an item after we pick it.When we pick r items from n with replacement, we must use a different method or another formula. We often use a method known as bars and stars. Alternatively, we can use the formula (n+r-1)! / r! (n - 1)!. When we put the items back, we can pick more than n items. So it would make sense to talk about how many ways to pick 7 items from 5, whereas without replacement this is impossible.
6
1 - E 2 - A 3 - R 4 - I 5 - O
1 3 5 2 4 r