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Answer:
- linear: slope between any two points is the same
- non-linear: slope between two points may differ
Step-by-step explanation:
A function is a linear function if the slope between any two ordered pairs is the same as it is between any other two. That is, the ordered pairs of a linear function will all fall on the same line when they are graphed.
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A nonlinear function is one that is not linear. That is, its points do not all fall on the same line when graphed. Slopes between different pairs of points may differ.
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<em>Detailed Example</em>
If you are given two tables to compare, it is easiest if the x-values (inputs) are separated by a constant amount. For example, x = -2, 0, 2, 4. These all differ by 2. Then the y-values (outputs) of a linear function will all differ by the same quantity. For example, outputs 10, 15, 20, 25 for the above inputs would correspond to a linear function. This one is y = 2.5x +15.
If the output values for that same set of inputs do not have constant differences, then the function is nonlinear. For example, outputs 10, 15, 21, 27 have differences of 5, 6, 7, so correspond to a nonlinear function. This one happens to be y = (x^2 +22x +120)/8.
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If the input values are not uniformly spaced, then you can check to see if the slope between one pair of points is the same as the slope for the next pair, and so on. The slope formula is ...
m = (y2 -y1)/(x2 -x1)
For our first table, the first pair of points (-2, 10) and (0, 15) are on a line with slope ...
m = (15 -10)/(0 -(-2)) = 5/2
Of course, when the calculations are repetitive, it is convenient to use a spreadsheet or calculator to do them. See the attachment for the calculations for this example.
Answer:
$7,000 at a rate of 7% and $21,000 at a rate of 14%.
Step-by-step explanation:
Let x be amount invested at 7% and y be amount invested at 14%.
We have been given that a women's professional organization made two small-business loans totaling $28,000. We can represent this information in an equation as:
The interest earned at 7% in one year would be and interest earned at 14% in one year would be
.
We are also told that the organization received from these loans was $3,430. We can represent this information in an equation as:
Form equation (1), we will get:
Upon substituting this value in equation (2), we will get:
Therefore, an amount of $21,000 was invested at a rate of 14%.
Therefore, an amount of $7,000 was invested at a rate of 14%.