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What is the answer to 5s - 6 equals 2s?
5s - 6 = 2s, ie 5s - 2s = 6, ie 3s = 6, ie s = 2
How do you simplify 7-2s plus 4 plus 5s?
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
How do you determine the equation for the perpendicular bisector of the straight line joining the points s 2s and 3s 8s?
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
What is the perpendicular bisector equation joining the points of s 2s and 3s 8s on the Cartesian plane showing work?
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
What is the perpendicular bisector equation of a line joined by the points of s 2s and 3s 8s showing key stages of work?
It is found as follows:- Points: (s, 2s) and (3s, 8s) Slope: (2s-8s)/(s-3s) = -6s/-2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) Multiply all terms by 3: 3y-15s = -1(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0
What is the answer to 5s - 6 equals 2s?
5s - 6 = 2s, ie 5s - 2s = 6, ie 3s = 6, ie s = 2
How do you simplify 7-2s plus 4 plus 5s?
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
What is 3s plus 4t plus 2s plus 5s plus 6t equals?
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)
How do you determine the equation for the perpendicular bisector of the straight line joining the points s 2s and 3s 8s?
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
What is the perpendicular bisector equation joining the points of s 2s and 3s 8s on the Cartesian plane showing work?
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
What is the perpendicular bisector equation of a line joined by the points of s 2s and 3s 8s showing key stages of work?
It is found as follows:- Points: (s, 2s) and (3s, 8s) Slope: (2s-8s)/(s-3s) = -6s/-2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) Multiply all terms by 3: 3y-15s = -1(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0
What is the perpendicular bisector equation of the line whose end points are at s 2s and 3s 8s on the Cartesian plane?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular equation: y -5s = -1/3(x -2s) => 3y = -x +17s Perpendicular bisector equation in its general form: x +3y -17s = 0
What is the perpendicular bisector equation of the line joined by the points of s 2s and 3s 8s?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular equation: y-5s = -1/3(x-2s) => 3y-15s = -x+2s => 3y = -x+17s Perpendicular bisector equation in its general form: x+3y-17s = 0
What is in its general form the perpendicular bisector equation meeting the line segment of s 2s and 3s 8s at its midpoint?
Points: (s, 2s) and (3s, 8s) Slope: 3 Perpendicular slope: -1/3 Midpoint: (2s, 5s) Equation in its general form: x+3y-17 = 0
What is the perpendicular bisector equation in its general form of the line whose coodinates are at s 2s and 3s 8s on the Cartesian grid showing key stages of work?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular bisector equation: y-5s = -1/3(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0
What is the perpendicular bisector equation of a line with endpoints of s 2s and 3s 8s?
Endpoints: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope of line: 3/1 Slope of perpendicular line: -1/3 Perpendicular bisector equation: y-5s = -1/3(x-2s) => 3y = -x+17s Perpendicular bisector equation in its general form: x+3y-17s = 0
Why is it easier to count by 5s or 10s and 2s instead of counting by 3s or 4s?
It is only because we count in tens - and 5 10 and 2 are factors.
