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What is the distance from the point of 10.5 comma 7.5 that meets the line of 6 comma 1.5 and 0 comma 6 at right angles on the Cartesian plane showing key stages of work?
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∙ 9y ago
Points: (6, 1.5) and (0, 6)
Slope: -3/4
Equation: 4y = -3x+24
Perpendicular slope: 4/3
Perpendicular equation: 3y = 4x-19.5
Both equations intersect ar right angles at (6, 1.5)
Distance from (10.5, 7.5) to (6, 1.5) is 7.5 units
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∙ 9y ago
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What is the perpendicular distance from the point of 2 4 on the Cartesian plane to the straight line equation of y equals 2x plus 10 showing key stages of work?
1 Equation: y = 2x+10 2 Perpendicular equation works out as: 2y = -x+10 3 Point of intersection: (-2, 6) 4 Distance is the square root of: (-2-2)2+(6-4)2 = 2 times sq rt of 5
What is the length of the line and its equation including its perpendicular bisector equation that spans the points of 2 3 and 5 7 on the Cartesian plane showing key stages of work?
Points: (2, 3) and (5, 7)Length: 5 unitsSlope: 4/3Perpendicular slope: -3/4Midpoint: (3.5, 5)Equation: 3y = 4x+1Bisector equation: 4y = -3x+30.5
What is the size of each individual interior angle of a polygon having 464 diagonals showing key stages of work?
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Formula: 0.5*a*b*sinC = area or 0.5*side2*sinC = area Apex angle: 180-75-75 = 30 degrees So: 0.5*side2*sin30 = 100 square cm Side as subject of the formula: side = square root of (200/sin30) = 20 Therefore: each equal sides of the isosceles triangle are 20 cm in length
What is the perpendicular distance from the point of 2 4 on the Cartesian plane to the straight line equation of y equals 2x plus 10 showing key stages of work?
1 Equation: y = 2x+10 2 Perpendicular equation works out as: 2y = -x+10 3 Point of intersection: (-2, 6) 4 Distance is the square root of: (-2-2)2+(6-4)2 = 2 times sq rt of 5
What is the length of the line and its equation including its perpendicular bisector equation that spans the points of 2 3 and 5 7 on the Cartesian plane showing key stages of work?
Points: (2, 3) and (5, 7)Length: 5 unitsSlope: 4/3Perpendicular slope: -3/4Midpoint: (3.5, 5)Equation: 3y = 4x+1Bisector equation: 4y = -3x+30.5
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A boat traveled 8 km east 5 km north then 2km west. How far is the boat from its starting point?
The total EW distance is 8 - 2 = 6 km.The total NS distance is 5 kmTherefore, total distance from the start is sqrt(62 + 52) = sqrt(36 + 25) = sqrt(61) =7.8 km (approx).The answer assumes that the boat is sufficiently far from the poles so that the various stages of its travel is along perpendicular lines.The total EW distance is 8 - 2 = 6 km.The total NS distance is 5 kmTherefore, total distance from the start is sqrt(62 + 52) = sqrt(36 + 25) = sqrt(61) =7.8 km (approx).The answer assumes that the boat is sufficiently far from the poles so that the various stages of its travel is along perpendicular lines.The total EW distance is 8 - 2 = 6 km.The total NS distance is 5 kmTherefore, total distance from the start is sqrt(62 + 52) = sqrt(36 + 25) = sqrt(61) =7.8 km (approx).The answer assumes that the boat is sufficiently far from the poles so that the various stages of its travel is along perpendicular lines.The total EW distance is 8 - 2 = 6 km.The total NS distance is 5 kmTherefore, total distance from the start is sqrt(62 + 52) = sqrt(36 + 25) = sqrt(61) =7.8 km (approx).The answer assumes that the boat is sufficiently far from the poles so that the various stages of its travel is along perpendicular lines.
