Q: What is the sum of r - 10 and 100 - 3r?

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(3r + 2)(r - 5)

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+7r+10/3 by 3r-30/r2-5r-50 weegy

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

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() }

Related questions

1/99 − 1/100 = 1/9900

If: 3r-8 = 2-2r Then: 5r = 10 And: r = 2

(3r + 2)(r - 5)

If: 3r-8 = 2-2r Then: 5r = 10 And: r = 2

3r+10=5r-46 --> add 46 to both sides 3r+10+46=5r-46+46 3r+56=5r --> subtract 3r from both sides 3r+56-3r=5r-3r 56=2r --> divide both sides by 2 56/2=2r/2 28=r

3r-r equals 12 = -9

5

In algebra, "r" typically represents a variable or an unknown quantity. When you see "triple r" in an algebraic expression, it means that you are multiplying the variable "r" by three. So, "triple r" is equivalent to 3r in algebraic terms. This can be written as 3 * r or simply 3r, depending on the context of the expression.

3r-2 for r=65

-5 = 7 + 3r | subtract 7 -12 = 3r | divide by 3 -4 = r

If that's 3r + 12, that factors to 3(r + 4)

r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)r + 2t = -3 . . . . (A)3r - 4t = -9 . . . . (B)2*(A) + (B): 2r + 4t + 3r - 4t = -6 - 95r = - 15 so r = -3substituting in (A), t = 0So the answer is (r, t) = (-3, 0)