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Marysya12 [62]
2 years ago
13

Select the correct answer.

Mathematics
1 answer:
Phantasy [73]2 years ago
8 0

Answer:

Alia has 11 jelly beans.

Step-by-step explanation:

4^2= 163^3= 2727-16=11

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I not very good with slope and y intercept form. any tips ?
jeyben [28]

Step-by-step explanation:

To find the slope you use the slope formula which is y2-y2/x2-x1

And to find the y intercept you use y=mx +b. Yous substitute the slope as m and you substitute a point from the line to find b. Which is the y intercept

4 0
2 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
2 years ago
HELP!!! MARKING BRAINLIEST
Ilya [14]

Answer:

Not trying to take points but it is too blurry I can’t  see it

Step-by-step explanation:

4 0
2 years ago
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Factor the expression. Use the fundamental identities to simplify, if necessary.
Rasek [7]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identity

sin²x + cos²x = 1 ⇒ sin²x = 1 - cos²x

Given

sin²x + 7cosx + 17

=1 - cos²x + 7cosx + 17

= - cos²x + 7cosx + 18 ← factor out - 1 from each term

= - (cos²x - 7cosx - 18)

Consider the factors of the constant term (- 18) which sum to give the coefficient of the cosx term (- 7)

The factors are - 9 and + 2, thus

= - (cosx - 9)(cosx + 2) ← in factored form

3 0
2 years ago
3/4 (4y+8)= 5+3y<br> solve for y:<br> please help
4vir4ik [10]

Answer:

No Soultion

Step-by-step explanation:

7 0
3 years ago
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