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arlik [135]
3 years ago
11

Never mind no one will answer it

Mathematics
2 answers:
olga nikolaevna [1]3 years ago
7 0

Answer:

sorry that no one answered your question

Step-by-step explanation:

Vinil7 [7]3 years ago
5 0
What
ask the question I will answer it
go ahead ask
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Below is part of a results table from an experiment.
Sergio [31]
Take all three trial result averages and add them up
Then take that number and divide by three because there was three numbers added.
That number is the average of the three trials
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3 years ago
What is the value of r in: 1/7(105+r)=20
stealth61 [152]
Use the distributive property.

1/7(105 + r) = 20
15 + (1/7)r = 20

Then subtract each side of the equation by 15.

15 + (1/7)r = 20
(1/7)r = 5

Divide both sides by 1/7.

(1/7)r = 5

r = 35

So, the answer to this question is r is equal to 35.
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3 years ago
What is the area of the trapezoid
irga5000 [103]

The equation for a trapezoid is A = a+b all over 2 times h

So, you would simply substitute.

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3 years ago
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MIDDLE SCHOOL MATH <br>help
Sveta_85 [38]
1→3
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Please help I will give Brainliest please!
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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