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Allushta [10]
3 years ago
12

Can you help me here?

Mathematics
2 answers:
weqwewe [10]3 years ago
6 0

Answer:

0

Step-by-step explanation:

0 means line is horizontal and no slope means the line is vertical

kati45 [8]3 years ago
3 0

Answer:

no slope

Step-by-step explanation:

the line is strate

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Select the two figures that are similar to the 10 by 4 figure that is shown.
12345 [234]

Answer:

None of the above

Step-by-step explanation:

Since our given figure is 10 by 4, we have an area of 40. The pictures must have a similar area of 40 in order to be similar. Since none of the pictures have an area of 40, we have no pictures that are similar.

8 0
3 years ago
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
3 years ago
2x+11=-3(2x-1) solve for x​
borishaifa [10]

Answer:

x=-1

Step-by-step explanation:

start by distributing the negative 3 to 2x and negative one. after the equation should look like this:

2x+11=-6x + 3

the next step is to combine the X's

8x+11=3

next step:

8x=-8

next step: divide

8/-8 =-1 so x=-1

7 0
3 years ago
Read 2 more answers
Can someone help with this
Darina [25.2K]

T(l) = T = 2\pi \sqrt{\frac{l}{32.2} \\

T^2 = 4\pi^2 \frac{l}{32.2}\\32.2T^2 = 4\pi^2 l\\\frac{32.2}{4\pi^2}T^2 = l\\l (T)= \frac{8.05}{\pi^4} T^2


Then if T = 3s

l(3) = \frac{8.05}{4\pi^2}(3^2)ft

4 0
3 years ago
In a mathematics class, half of the students scored 79 on an achievement test. With the exception of a few students who scored 4
GenaCL600 [577]

Answer:

i went with mean is less than median

Step-by-step explanation:

other website lol

4 0
3 years ago
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