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R plus 3 plus 2r equals - 21?
r+3+2r=21 to solver this equation you have to first simplify then solve. you have to combine same variables. 3r+3=21 -3 -3 you subtract three to get r by itself (3r=18)/3 you divide by three to get r by itself Answer r=6
Factor the polynomial -18r to the power of 3 plus 242r?
-2r(3r - 11)(3r + 11)
0.3 times 2r- 0.3 times3 equals 0.2r plus 0.9?
0.3(2r) - 0.3(3) = 0.2(r) + 0.9Clear parentheses:0.6r - 0.9 = 0.2r + 0.9Multiply each side by 10:6r - 9 = 2r + 9Add 9 to each side:6r = 2r + 18Subtract 2r from each side:4r = 18Divide each side by 4:r = 18/4 = 4.5
What is the answer to 4x plus 8 equals 2r plus 20?
That depends on what the question is. What you gave is simply a relationship between two variables. You can simplify it like this: 4x + 8 = 2r + 20 (4x + 8) / 2 = (2r + 20) / 2 2x + 4 = r + 10 2x + 4 - 4 = r + 10 - 4 2x = r + 6 You can then solve for x by dividing both sides by two: 2x / 2 = (r + 6) / 2 x = r/2 + 3 You could also solve for r by subtracting six from both sides, and rearranging the equation: 2x = r + 6 2x - 6 = r + 6 - 6 2x - 6 = r r = 2x - 6 r = 2(x - 3)
What is 1 plus 2 plus 2 plus 3 plus 2 plus 4 plus 2 plus 4 plus 3 plus 4 plus 2 plus 6 plus 1 plus 2 plus 3 plus 3 plus 0 plus?
43
What does 12 plus 8r-6 equals 7r-5-3 equal?
12 + 8r - 6 = 7r - 5 - 3 8r +6 = 7r -8 r = -14
Multiply r2 plus 7r plus 10 over 3 by 3r-30 over r2-5r-50?
r+2
R plus 3 plus 2r equals - 21?
r+3+2r=21 to solver this equation you have to first simplify then solve. you have to combine same variables. 3r+3=21 -3 -3 you subtract three to get r by itself (3r=18)/3 you divide by three to get r by itself Answer r=6
Factor the polynomial -18r to the power of 3 plus 242r?
-2r(3r - 11)(3r + 11)
What is 7-2r equals 13?
If: 7 -2r = 13 Then: r = -3
0.3 times 2r- 0.3 times3 equals 0.2r plus 0.9?
0.3(2r) - 0.3(3) = 0.2(r) + 0.9Clear parentheses:0.6r - 0.9 = 0.2r + 0.9Multiply each side by 10:6r - 9 = 2r + 9Add 9 to each side:6r = 2r + 18Subtract 2r from each side:4r = 18Divide each side by 4:r = 18/4 = 4.5
When adding trinomials is the result always a trinomial?
Only if all the variables match up. Example of when they do match: 3x^2 +2y+r plus x^2- 2r+ 3y When they don't match: 3x^4+x+y plus 3x^3-x^2 +r
What is 5r plus 7 equals 13 plus 2r plus 3r?
1)5r+7=13+2r+3r 2)5r+7=13+5r 3)5r's cancel 7=13 4)Subtract thirteen on both sides 7 =13 -13 -13 5)Answer is no solution because -6 doesn't equal to 0. -6=0 NO SOLUTION!!! Have fun with math!
Use the identity below to complete the tasks:a3 + b3 = (a + b)(a2 - ab + b2)Use the identity for the sum of two cubes to factor 8q6r3 + 27s6t3.What is a What is b?
Below ^ denotes power. 8q^6r^3=(2q^2r)^3 denote as a^3 and find a=2q^2r 27s^6t^3=(3s^2t)^3 denote as b^3 and find b=3s^2t Now 8q^6r^3+27s^6t^3=a^3+b^3 = (a+b)(a^2-ab+b^2) | substitute back =(2q^2r+3s^2t)(4q^4r^2-6q^2rs^2t+9s^4t^2). Notice (2q^2r)^2=4q^4r^2 (3s^2t)^2=9s^4t^2. Hence 8q^6r^3+27s^6t^3=(2rq^2+3ts^2)(4r^2q^4-6rts^2t^2+9t^2s^4), a=2q^2r and b=3s^2t.
What is the answer to 4x plus 8 equals 2r plus 20?
That depends on what the question is. What you gave is simply a relationship between two variables. You can simplify it like this: 4x + 8 = 2r + 20 (4x + 8) / 2 = (2r + 20) / 2 2x + 4 = r + 10 2x + 4 - 4 = r + 10 - 4 2x = r + 6 You can then solve for x by dividing both sides by two: 2x / 2 = (r + 6) / 2 x = r/2 + 3 You could also solve for r by subtracting six from both sides, and rearranging the equation: 2x = r + 6 2x - 6 = r + 6 - 6 2x - 6 = r r = 2x - 6 r = 2(x - 3)
What is 1 plus 2 plus 2 plus 3 plus 2 plus 4 plus 2 plus 4 plus 3 plus 4 plus 2 plus 6 plus 1 plus 2 plus 3 plus 3 plus 0 plus?
43
2n plus 3 is an arithmetic sequence how many terms should be taken starting from the first term so that the ratio of the sum of the first third of these terms to the rest of the terms is 7 to 44?
The sum of an arithmetic series of n terms is: S(n) = n(2a1 + (n - 1)d) / 2 where a1 is the first term and d is the difference between terms. The key to solving this is to realise that two sums are required, S1 and S2 such that: S1 = S(r) S2 = S(3r) - S(r). But that this second sum can also be stated as: S2 = sum of 2r terms starting at term ar+1 So that the problem then becomes: Find two sums S1 and S2 of the sequence an = 2n + 3 such that the first (S1) has r terms starting at a1 and the second (S2) has 2r terms starting at ar+1, with the ratio of S1 : S2 = 7 : 44. This gives: S1 = r (2a1 + (r - 1)d) / 2 = r(2(2x1 + 3) + (r - 1)2) / 2 = r(2r + 8) / 2 = r(r + 4) S2 = 2r(2ar+1 + (2r - 1)d) = 2r(2(2(r + 1) + 3) + (2r - 1)2) / 2 = 2r(2(2r + 5) + (2r - 1)2) / 2 = 2r(2r + 5 +2r - 1) = 2r(4r + 4) = 8r(r + 1) As S1 : S2 = 7 : 44, 44 S1 = 7 S2 => 44 r(r + 4) = 7 x 8r(r + 1))0 => 11(r + 4) = 14(r + 1) => 11r + 44 = 14r + 14 => 3r = 30 => r = 10 As r is the number of terms in S1 it is a third of the number of terms required, so 3 x 10 = 30 terms are needed. To check: a1 = 2 x 1 + 3 = 5 S(30) = 30(2 x 5 +(30-1) x 2) / 2 = 30 x 34 = 1020 S(10) = 10(2 x 5 + (10-1) x 2) / 2 = 10 x 14 = 140 Sum of 1st third S1 = S(10) = 140, sum of rest S2 = S(30) - S(10) = 1020 - 140 = 880 Ratio of S1 : S2 = 140 : 880 = 14 : 88 = 7 : 44 Note that the second sum S2 is also 20 terms starting at a11 = 2 x 11 + 3 = 25: S2 = 20(2 x 25 + 19 x 2)/2 = 20 x 44 = 880.
