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How do you Factor 3r squared equals 13r plus 10?
(3r + 2)(r - 5)
Solve 4r plus 3 equals 3r- 2 for r?
4r +3 = 3r-2 r + 3=-2 r=-5 basically collect terms so that you get one coefficient of r (in this case 4r on one side and 3r on the other. To get the three r over to the other side you subtract 3r on both sides.) Then r+ 3=-2, so that's simple algebraic manipulation.
Multiply r2 plus 7r plus 10 over 3 by 3r-30 over r2-5r-50?
multiply r2+7r+10/3 by 3r-30/r2-5r-50 weegy
What is the sum of all the positive proper fractions with denominators less than or equal to 100?
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
Sum of individual digits of number?
void main() { int num,r,sum=0; clrscr(); printf("enter the number\n"); scanf("%d",&num); while(num!=0) { r=num%10; sum=sum+r; num=num/10 } printf("The sum of individual digit of given number is=%d",sum); getch() }
