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hoa [83]
2 years ago
5

Find the slope of the line that passes through (−6, 8)&(−2, 2)

Mathematics
1 answer:
MatroZZZ [7]2 years ago
6 0

Answer

y= 7/4x

Step-by-step explanation:

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Answer:

C.) Solve for the varible

Step-by-step explanation:

As there is currently nothing to do other than one step, the next step would be to solve 81/9 for 9, getting the final answer of a=9.

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Which pairs of quadrilaterals can be shown to be congruent using rigid motions?
Misha Larkins [42]

Answer:

1 and 2 are not congruent; 1 and 3 are congruent; 1 and 4 are congruent; 2 and 3 are not congruent; 2 and 4 are not congruent; 3 and 4 are congruent.

Step-by-step explanation:

From the diagram, we can see that the angle measures and side measures of figures 1, 3 and 4 are the same.  This means that these three figures are congruent.  Figure 2, however, is not congruent to any of the other 3.

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3 years ago
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Combine like terms - y + 9z - 16y - 25z + 4
alexandr402 [8]

- 17y - 16z + 4

Step-by-step explanation:

1) Collect like terms.

( - y - 16y) + (9z - 25z) + 4

2) Simplify.

- 17y - 16z + 4

So, therefor, the answer is -17y - 16z + 4.

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2 years ago
HURRY PLEASE!!!<br> I WILL GIVE YOU BRAINLIEST!!
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WITCHER [35]

Part (a)

The domain is the set of allowed x inputs of a function.

The graph shows that x = 0 is not allowed because of the vertical asymptote located here. It seems like any other x value is fine though.

<h3>Domain: set of all real numbers but x \ne 0</h3>

To write this in interval notation, we can say (-\infty, 0) \cup (0, \infty) which is the result of poking a hole at 0 on the real number line.

--------------

The range deals with the y values. The graph makes it seem like it stretches on forever in both up and down directions. If this is the case, then the range is the set of all real numbers.

<h3>Range: Set of all real numbers</h3>

In interval notation, we would say (-\infty, \infty) which is almost identical to the interval notation of the domain, except this time of course we aren't poking at hole at 0.

=======================================================

Part (b)

<h3>The x intercepts are x = -4 and x = 4</h3>

We can compact that to the notation x = \pm 4

These are the locations where the blue hyperbolic curve crosses the x axis.

=======================================================

Part (c)

<h3>Answer: There aren't any horizontal asymptotes in this graph.</h3>

Reason: The presence of an oblique asymptote cancels out any potential for a horizontal asymptote.

=======================================================

Part (d)

The vertical asymptote is located at x = 0, so the equation of the vertical asymptote is naturally x = 0. Every point on the vertical dashed line has an x coordinate of zero. The y coordinate can be anything you want.

<h3>Answer: x = 0 is the vertical asymptote</h3>

=======================================================

Part (e)

The oblique or slant asymptote is the diagonal dashed line.

It goes through (0,0) and (2,6)

The equation of the line through those points is y = 3x

If you were to zoom out on the graph (if possible), then you should notice the branches of the hyperbola stretch forever upward but they slowly should approach the "fencing" that is y = 3x. The same goes for the vertical asymptote as well of course.

<h3>Answer:  Oblique asymptote is y = 3x</h3>
5 0
2 years ago
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