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I suspect that the answer is
(x + 6)^2 + (y - 5)^2 = 81
Which transformations could have been used to move the platter to the new location? A. a translation 9 units left and a translation 3 units down B. a reflection across MN and a translation 4 units left C. a reflection across MN and a translation 8 units left D. a rotation 180° clockwise about N and a translation 4 units left
To find the points that are three units to the left of the Y-axis and five units from the origin, we start by noting that moving three units to the left of the Y-axis places us at the X-coordinate of -3. Since we want the distance from the origin to be five units, we can use the distance formula: ( \sqrt{(-3)^2 + y^2} = 5 ). Solving this gives ( y^2 = 16 ), resulting in ( y = 4 ) or ( y = -4 ). Thus, the two points are (-3, 4) and (-3, -4).
The translation is vertical if the added term is outside the main function and horizontal if it is inside it, next to the x. For example, y = x^2 represents a parabola, with its lowest point at (0,0). If we have the equation y = x^2 + 2 then we have translated the parabola up two units -- its lowest point is now x = 0, y = 2. But if we write y = (x + 2)^2, then we are translating two units to the left, and the lowest point is x = -2, y = 0.
Negative five is at five units to the left of the origin while one fourth is a quarter of a unit to the right.
Translation is moving a graph to the left or right, up or down (or both). Given a quadratic equation of the form y = ax^2 + bx + c, if you substitute u = x - p and v = y - q then the graph of v against u will be the same as the x-y graph, translated to the left by p and downwards by q.
Which transformations could have been used to move the platter to the new location? A. a translation 9 units left and a translation 3 units down B. a reflection across MN and a translation 4 units left C. a reflection across MN and a translation 8 units left D. a rotation 180° clockwise about N and a translation 4 units left
(x,y)--(x-4,y+6)
(x1, y1) = (x - 8, y + 9)
translation
To find the points that are three units to the left of the Y-axis and five units from the origin, we start by noting that moving three units to the left of the Y-axis places us at the X-coordinate of -3. Since we want the distance from the origin to be five units, we can use the distance formula: ( \sqrt{(-3)^2 + y^2} = 5 ). Solving this gives ( y^2 = 16 ), resulting in ( y = 4 ) or ( y = -4 ). Thus, the two points are (-3, 4) and (-3, -4).
The translation is vertical if the added term is outside the main function and horizontal if it is inside it, next to the x. For example, y = x^2 represents a parabola, with its lowest point at (0,0). If we have the equation y = x^2 + 2 then we have translated the parabola up two units -- its lowest point is now x = 0, y = 2. But if we write y = (x + 2)^2, then we are translating two units to the left, and the lowest point is x = -2, y = 0.
Yes. For example, if you want to shift the graph 5 units to the right, you must replace every instance of "x" by "x-5".
It's just subtraction. You have five apples (+5). You eat two (-2). What's left?
Negative five is at five units to the left of the origin while one fourth is a quarter of a unit to the right.
Reactants
Reactants at left --> Products at right
The absolute value depends on it's "distance" from zero. So if it's to the right (positive) by 5 units, or to the left (negative) by five units, then it's absolute value is 5