NO
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
A = (s, 2s), B = (3s, 8s) The midpoint of AB is C = [(s + 3s)/2, (2s + 8s)/2] = [4s/2, 10s/2] = (2s, 5s) Gradient of AB = (8s - 2s)/(3s - s) = 6s/2s = 3 Gradient of perpendicular to AB = -1/(slope AB) = -1/3 Now, line through C = (2s, 5s) with gradient -1/3 is y - 5s = -1/3*(x - 2s) = 1/3*(2s - x) or 3y - 15s = 2s - x or x + 3y = 17s
5s-r
Points: (s, 2s) and (3s, 8s) Slope: (8s-2s)/(3s-s) = 6s/2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) => 3y-15s = -1(x-2s) => 3y = -x+17x Perpendicular bisector equation in its general form: x+3y-17s = 0
3s + 4t + 2s + 5s + 6tGroup all of the like 's' terms & 't' terms together:(3s+2s+5s) + (4t + 6t)10s + 10t or 10(s+t)
NO
2s + 16 = 4s - 6 Subtract 2s from both sides: 16 = 2s - 6 Add 6 to both sides: 22 = 2s divide both sides by 2: s = 11
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
A = (s, 2s), B = (3s, 8s) The midpoint of AB is C = [(s + 3s)/2, (2s + 8s)/2] = [4s/2, 10s/2] = (2s, 5s) Gradient of AB = (8s - 2s)/(3s - s) = 6s/2s = 3 Gradient of perpendicular to AB = -1/(slope AB) = -1/3 Now, line through C = (2s, 5s) with gradient -1/3 is y - 5s = -1/3*(x - 2s) = 1/3*(2s - x) or 3y - 15s = 2s - x or x + 3y = 17s
That's a simple example of system of equations. There are quite a number of methods for solving those, but the easiest here would be to calculate one variable from first equation and substitute it into second. 2s + 3t = 28 5s + 6t = 61 Let's calculate s from the first equation: 2s + 3t = 28, 2s = 28 - 3t s = 14 - 3t/2 Substitute s into second equation 5s + 6t = 61, 5(14 - 3t/2) + 6t = 61, 70 - 15t/2 + 6t = 61, 70 - 3t/2 = 61, 3t/2 = 9, t = 6. We can then substitute t back into first equation: 2s + 3t = 28, 2s + 3 * 6 = 28, 2s = 10, s = 5. So the final answer is: s = 5 t = 6
5s-r
Points: (s, 2s) and (3s, 8s) Slope: (8s-2s)/(3s-s) = 6s/2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) => 3y-15s = -1(x-2s) => 3y = -x+17x Perpendicular bisector equation in its general form: x+3y-17s = 0
1 - s = 6 - 6s => 5s = 5 so that s = 1
It is: 5s(5s-3)
5s = 0
7s + 4t