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klemol [59]
3 years ago
13

If BOTH equations are in slope-intercept form then the--------------? method would be best, but the------------? method would al

so be effective since both y's are already by itself.
If ONE of the equations is solved for x or y and the other equation is not, then the----------? method is best.

If BOTH equations are lined up in standard form & the coefficients of x or y are opposites then the BEST method is definitely the-----------? method.

If BOTH equations are lined up in standard form the elimination method would be best. But if the coefficient of x or y is 1, then the-------------? method is also effective.




PLZ HELP MEEE WITH THIS
Mathematics
1 answer:
nadezda [96]3 years ago
5 0

Answer:

Step-by-step explanation:

If BOTH equations are in slope-intercept form then the-graphing-? method would be best, but the-substitution-? method would also be effective since both y's are already by itself.

If ONE of the equations is solved for x or y and the other equation is not, then the-substitution-? method is best.

If BOTH equations are lined up in standard form & the coefficients of x or y are opposites then the BEST method is definitely the-elimination--? method.

If BOTH equations are lined up in standard form the elimination method would be best. But if the coefficient of x or y is 1, then the-substitution--? method is also effective.

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Answer:

(3, -\frac{1}{6})

Step-by-step explanation:

We can rewrite the equation as

y = \frac{x - 3}{(x - 3)(x - 9)}

Notice that we have x - 3 in both the numerator and the denominator, so it looks like we can divide it out. However, what if x - 3 is 0? Then we would have y = \frac{0}{0 \times (x - 9)} = \frac{0}{0}, which is undefined. So although it looks like the numerator and denominator can be simplified, the resulting function we would get from simplification would not have the same behavior as this one (since such a function would be defined for x = 3, but this one is not).

A point of discontinuity refers to a particular point which is included in the simplified function, but which is not included in the original one. In this case, the point which is not included in the unsimplified function is at x = 3. In the simplified version of the function, if we plug in x = 3, we get

y = \frac{1}{((3) - 9)} = -\frac{1}{6}

So the point (3, -\frac{1}{6}) is our only point of discontinuity.

It's also important to distinguish between specific points of discontinuity and vertical asymptotes. This function also has a vertical asymptote at x = 9 (since it causes the denominator to be 0), but the difference in behavior is that in the case of the asymptote, only the denominator becomes 0 for a specific value of x

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