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krok68 [10]
3 years ago
5

Please help I will give Brainliest please!

Mathematics
1 answer:
WITCHER [35]3 years ago
5 0

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>
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alexdok [17]
The common difference is d = 4 because we add 4 to each term to get the next one.

The starting term is a1 = 3

The nth term of this arithmetic sequence is 
an = a1 + d(n-1)
an = 3 + 4(n-1)
an = 3 + 4n-4
an = 4n - 1

Plug in n = 25 to find the 25th term
an = 4n - 1
a25 = 4*25 - 1
a25 = 100 - 1
a25 = 99

So we're summing the series : 3+7+11+15+...+99

We could write out all the terms and add them all up. That's a lot more work than needed though. Luckily we have a handy formula to make things a lot better
The sum of the first n terms is Sn. The formula for Sn is
Sn = n*(a1+an)/2

Plug in n = 25 to get
Sn = n*(a1+an)/2
S25 = 25*(a1+a25)/2

Then plug in a1 = 3 and a25 = 99. Then compute to simplify

S25 = 25*(a1+a25)/2
S25 = 25*(3+99)/2
S25 = 25*(102)/2
S25 = 2550/2
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The final answer is 1275
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3 years ago
State the slope and the y-intercept for the graph of the equation. y= -2
kotegsom [21]

Answer: 0; -2

Step-by-step explanation:

Data: y-intercept=x and slope=x

Equation=y=-2

Only step, Point out the y-intercept and slope

-2=y-intercept

0=slope

Reason: Since there is no visible slope, the placeholder would be 0 therefore the slope would be 0, -2 is the only number left so that means the only number that can be y-intercept is -2.

That is why the slope is 0 and the y-intercept is -2

I hope this helps!

5 0
3 years ago
Point C is on the graph of the function y = x2 – 3. Its x-coordinate is 4. Which ordered pair gives the location of point C? A.(
Scilla [17]

Answer:

B (4, 2+3)

Step-by-step explanation:

To do this you fill in x with 4 so the equation becomes y = (4)2 - 3

You then solve to y = 8 - 3

then 8 - 3 is 5.

Making the coordinates (4, 5) and in this case with the answers (4, 2+3)

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3 years ago
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Luda [366]
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4 years ago
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