Please help me I dont get this # 25 and 23 D:

If you have a demand function of Qd=48-9P and a supply function of Qs=-12+6P,

jenyasd209 [6]

Answer:

The equilibrium quantity is 26.4

Step-by-step explanation:

Given

$Q_d = 48 - 9P$

$Q_s = 12 + 6P$

Required

Determine the equilibrium quantity

First, we need to determine the equilibrium by equating Qd to Qs

i.e.

$Q_d = Q_s$

This gives:

$48 -9P = 12 + 6P$

Collect Like Terms

$-9P - 6P = 12 - 48$

$-15P = -36$

Solve for P

$P = \frac{-36}{-15}$

$P =2.4$

This is the equilibrium price.

Substitute 2.4 for P in any of the quantity functions to give the equilibrium quantity:

$Q_d = 48 - 9P$

$Q_d = 48 - 9 * 2.4$

$Q_d = 26.4$

<em>Hence, the equilibrium quantity is 26.4</em>

7 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

Given right triangle MNL what is the value of cos(M)?

FromTheMoon [43]

Step $1$