It is a relationship in which changes in one variable are accompanied by changes of a constant amount in the other variable and that the variables are not both zero.In terms of an equation, it requires y = ax + b where a and b are both non-zero.
First assume that the relationship is linear rather than the table. If the table is linear but the relationship is non-linear it is a very different and very difficult task. Suppose, next, that you have two sets of values: inputs (x) and outputs (y). Also, you must have at least two rows of data for different inputs.So, given that there at at least two sets of data, suppose they are as follows:x1 gets mapped to y1andx2 gets mapped to y2Then calculate m = (y2-y1)/(x2-x1)and c = y - mx for any x-y combination from the table.The rule is y = mx + c.First assume that the relationship is linear rather than the table. If the table is linear but the relationship is non-linear it is a very different and very difficult task. Suppose, next, that you have two sets of values: inputs (x) and outputs (y). Also, you must have at least two rows of data for different inputs.So, given that there at at least two sets of data, suppose they are as follows:x1 gets mapped to y1andx2 gets mapped to y2Then calculate m = (y2-y1)/(x2-x1)and c = y - mx for any x-y combination from the table.The rule is y = mx + c.First assume that the relationship is linear rather than the table. If the table is linear but the relationship is non-linear it is a very different and very difficult task. Suppose, next, that you have two sets of values: inputs (x) and outputs (y). Also, you must have at least two rows of data for different inputs.So, given that there at at least two sets of data, suppose they are as follows:x1 gets mapped to y1andx2 gets mapped to y2Then calculate m = (y2-y1)/(x2-x1)and c = y - mx for any x-y combination from the table.The rule is y = mx + c.First assume that the relationship is linear rather than the table. If the table is linear but the relationship is non-linear it is a very different and very difficult task. Suppose, next, that you have two sets of values: inputs (x) and outputs (y). Also, you must have at least two rows of data for different inputs.So, given that there at at least two sets of data, suppose they are as follows:x1 gets mapped to y1andx2 gets mapped to y2Then calculate m = (y2-y1)/(x2-x1)and c = y - mx for any x-y combination from the table.The rule is y = mx + c.
You can write ordered pairs as ratios to determine if two sets of ordered pairs form a linear or non-linear relationship. In a table of x,y values, the ordered pairs are listed as the x value first, then the corresponding y value. Remove from the table and write as a ratio of x over y, (or y over x, if you like). In a linear relationship, all the ratios of x over y, (or y over x) are equivalent.
Not necessarily.
if it passes through (0,0) then it is a direct variation
you put in what x is and solve it for y! thats the answer!
If the ratio between each pair of values is the same then the relationship is proportional. If even one of the ratios is different then it is not proportional.
If the figures in the table are exact and without measurement error then take any two of the points (x1, y1) and (x2, y2) and use these to form the linear relation y - y1 = ((y2 - y1)/(x2 - x1))(x - x1) If, however, you suspect that the values in the table do not exactly follow a linear relationship then use linear regression for which formulae are provided in wikipedia.
For a linear I can see no advantage in the table method.
All the elements are arranged according to atomic number.Number of electrons decide the position of the element.Atomic number is the root main cause of periodic table.
It is a relationship in which changes in one variable are accompanied by changes of a constant amount in the other variable and that the variables are not both zero.In terms of an equation, it requires y = ax + b where a and b are both non-zero.
First assume that the relationship is linear rather than the table. If the table is linear but the relationship is non-linear it is a very different and very difficult task. Suppose, next, that you have two sets of values: inputs (x) and outputs (y). Also, you must have at least two rows of data for different inputs.So, given that there at at least two sets of data, suppose they are as follows:x1 gets mapped to y1andx2 gets mapped to y2Then calculate m = (y2-y1)/(x2-x1)and c = y - mx for any x-y combination from the table.The rule is y = mx + c.First assume that the relationship is linear rather than the table. If the table is linear but the relationship is non-linear it is a very different and very difficult task. Suppose, next, that you have two sets of values: inputs (x) and outputs (y). Also, you must have at least two rows of data for different inputs.So, given that there at at least two sets of data, suppose they are as follows:x1 gets mapped to y1andx2 gets mapped to y2Then calculate m = (y2-y1)/(x2-x1)and c = y - mx for any x-y combination from the table.The rule is y = mx + c.First assume that the relationship is linear rather than the table. If the table is linear but the relationship is non-linear it is a very different and very difficult task. Suppose, next, that you have two sets of values: inputs (x) and outputs (y). Also, you must have at least two rows of data for different inputs.So, given that there at at least two sets of data, suppose they are as follows:x1 gets mapped to y1andx2 gets mapped to y2Then calculate m = (y2-y1)/(x2-x1)and c = y - mx for any x-y combination from the table.The rule is y = mx + c.First assume that the relationship is linear rather than the table. If the table is linear but the relationship is non-linear it is a very different and very difficult task. Suppose, next, that you have two sets of values: inputs (x) and outputs (y). Also, you must have at least two rows of data for different inputs.So, given that there at at least two sets of data, suppose they are as follows:x1 gets mapped to y1andx2 gets mapped to y2Then calculate m = (y2-y1)/(x2-x1)and c = y - mx for any x-y combination from the table.The rule is y = mx + c.
You can write ordered pairs as ratios to determine if two sets of ordered pairs form a linear or non-linear relationship. In a table of x,y values, the ordered pairs are listed as the x value first, then the corresponding y value. Remove from the table and write as a ratio of x over y, (or y over x, if you like). In a linear relationship, all the ratios of x over y, (or y over x) are equivalent.
Not necessarily.
if it passes through (0,0) then it is a direct variation
Linear and Exponetional.
explain vector table?