one fourth the sum of r and ten is identical to r munus 4
The sum, meaning you add them together. 7+5=13.
sum means addition so the sum of 5 and 2 will be 7
the sum of 10 and the product of 7 and a number 2+5=7 2*5=10
31
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
20
x-7
The sum of 11 and 7 is 18.
The sum of the series a + ar + ar2 + ... is a/(1 - r) for |r| < 1
Sum of r = summer
one fourth the sum of r and ten is identical to r munus 4
the sum of 7 and 23 is 30..
The GCF is 7.
Sum(not div by 7) = Sum(all) - Sum(div by 7) Now the sum of an AP is Sn = n/2 (first + last) where n is the number of terms Sum(All) = 10000/2 (1 + 10000) = 50005000 Sum(div by 7) = (9996/7)/2 (7 + 9996) = 1428/2 (10003) = 7142142 Sum(not div by 7) = Sum(all) - Sum(div by 7) = 50005000 - 7142142 = 42 862 858
Omg u are not for real r u i mean u must be dumber then a rock not to no that BTW 10
The product of 7 and 60 is 420. The sum of 7 and 60 is 67.