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r - 10 + 100 - 3r = -2r + 90
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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.
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
Multiply r2 plus 7r plus 10 over 3 by 3r-30 over r2-5r-50?
r+2
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() }
r=1+3r-3r?
1/99 − 1/100 = 1/9900
What is 3R?
If: 3r-8 = 2-2r Then: 5r = 10 And: r = 2
How do you Factor 3r squared equals 13r plus 10?
(3r + 2)(r - 5)
What is 3r-8 equals2-2r?
If: 3r-8 = 2-2r Then: 5r = 10 And: r = 2
13r plus 10 equals 5r-46?
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
Can r plus 2r - 3r be simplified?
5
What is 3r-r equals 12?
3r-r equals 12 = -9
What is the answer for this math algebra problem 3r-2 for r equals 65?
3r-2 for r=65
What is r in this equation and how do you do it minus 5 equals 7 plus 3r?
-5 = 7 + 3r | subtract 7 -12 = 3r | divide by 3 -4 = r
What is the factor for 3r 12?
If that's 3r + 12, that factors to 3(r + 4)
Which pair of values is the solution to the system of equations below r plus 2t equals -3 and 3r-4t equals -9?
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)
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.
