3s + 1
12r - 15s - r + 3s = 11r -12s
4=4(3s) 4=12s s= 1/3
7
7-2s+4+5s ..... Okay first you combine like terms(add the like terms)..so its going to be 11+3s I did 7+4=11 then -2s+5s=3s So the answer can be 11+3s or 3s+11
3s + 1
135=3s +15 120=3s 40=s
Well, honey, the least common multiple of a monomial like a^3s and s^2 is simply a^3s^2. You just gotta take the highest power of each variable that appears in either monomial, slap 'em together, and there you have it. Math made sassy.
28 3s
-7 = 3s - 1 +1 +1 (add 1 to both sides to get the variable alone) ___ ____ -6 = 3s __ ___ (-6 and 3s both divided by 3) 3 3 -2 = s
4 3s = 4*3 = 12, which is a rational number.
The easiest way is to "flip" the inequality symbol end divide by the negative number:Example:6 < 3 - 3s6 - 3 < 3 - 3s -33 < -3s Method a) Divide by negative coefficient and flip the inequality symbol3/-3 > -3s/-3-1 > s or s< -13 < -3s Method b) Full algorithm, eliminate -3s by adding 3s on both sides3 +3s < -3s + 3s3 + 3s < 03 - 3 + 3s < 0 -33s < -33s/3 < -3/3s < -1 Looks familiar? So basically if you perform the full algorithm (method b) you can understand why we flip the inequality symbol when we have to eliminate a negative coefficient but it is faster just to flip the symbol (method a)
2h2s + so2 - 2h20 + 3s
12r - 15s - r + 3s = 11r -12s
The equation 4R + 3s + 2r = 6r + 3s is an example of the distributive property of addition, where the term 4R is being distributed over the sum of 2r and 6r. To see this more clearly, we can rewrite the equation as: 4R + 3s + 2r = (4R + 6r) + 3s Notice how the terms 4R and 6r are combined and the distributive property allows us to simplify the left-hand side of the equation.
4=4(3s) 4=12s s= 1/3
There are 29 3s because there also a 3 in the how many 3s are in 83333333333333333333333333333