2 fraction equivalent to 1/4th is 2/8th and 4/16th
What you get when you add two things together. 1 + 1 = 2. The sum of 1 and 1 is 2.
1 and 4(8th). this can be simplified to 1 and 1/2.
A sum of 4 is more likely. Out of the 36 possible outcomes with a pair of 6-sided dice, only one will have a sum of 2 {1-1}. Three of the possibilities sum to 4. {1-3, 2-2, 3-1}
The question is ambiguous.Does it want the sum of the squares, or the square of the sum ? They're different.Here are both:1). Sum of the squares: . (1)2 + (2)2 + (3)2 + (4)2 + (5)2 = 1 + 4 + 9 + 16 + 25 = 552). Square of the sum: . (1 + 2 + 3 + 4 + 5 )2 = (15)2 = 225
2 fraction equivalent to 1/4th is 2/8th and 4/16th
There are 2 because 2/8 = 1/4
The sum is 50 and 31/40 The difference is 1/40
The sum of 1 and 1 is 2.
Sum is the term for the answer to addition. Any time you add something, you get the sum. 1 + 1 = 2 2 is the sum.
Consider a denominator of r; It has proper fractions: 1/r, 2/r, ...., (r-1)/r Their sum is: (1 + 2 + ... + (r-1))/r The numerator of this sum is 1 + 2 + ... + (r-1) Which is an Arithmetic Progression (AP) with r-1 terms, and sum: sum = number_of_term(first + last)/2 = (r-1)(1 + r-1)/2 = (r-1)r/2 So the sum of the proper fractions with a denominator or r is: sum{r} = ((r-1)r/2)/r = ((r-1)r/2r = (r-1)/2 Now consider the sum of the proper fractions with a denominator r+1: sum{r+1} = (((r+1)-1)/2 = ((r-1)+1)/2 = (r-1)/2 + 1/2 = sum{r) + 1/2 So the sums of the proper fractions of the denominators forms an AP with a common difference of 1/2 The first denominator possible is r = 2 with sum (2-1)/2 = ½; The last denominator required is r = 100 with sum (100-1)/2 = 99/2 = 49½; And there are 100 - 2 + 1 = 99 terms to sum So the required sum is: sum = ½ + 1 + 1½ + ... + 49½ = 99(½ + 49½)/2 = 99 × 50/2 = 2475
The sum of 1/2 and 5/8 is 1 1/8
What you get when you add two things together. 1 + 1 = 2. The sum of 1 and 1 is 2.
1 + 1 = 2 The sum of the digits is therefore 2.
1 and 4(8th). this can be simplified to 1 and 1/2.
A sum of 4 is more likely. Out of the 36 possible outcomes with a pair of 6-sided dice, only one will have a sum of 2 {1-1}. Three of the possibilities sum to 4. {1-3, 2-2, 3-1}
The question is ambiguous.Does it want the sum of the squares, or the square of the sum ? They're different.Here are both:1). Sum of the squares: . (1)2 + (2)2 + (3)2 + (4)2 + (5)2 = 1 + 4 + 9 + 16 + 25 = 552). Square of the sum: . (1 + 2 + 3 + 4 + 5 )2 = (15)2 = 225