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tester [92]
3 years ago
10

Y=x^2+2;D:-1, 0, 1, 2, 3

Mathematics
1 answer:
IrinaK [193]3 years ago
8 0

For the various values of X the values of Y are given as follows

X=-1; Y=3

X=0; Y=2

X=1; Y=3

X=2; Y=6

X=3; Y=11

Step-by-step explanation:

Step 1:

Let us consider the given expression as

Y= X^{2} + 2

Step 2:

Let us discuss the value of Y for the various values of X

If X = -1, Y = (-1)² + 2 = 3

If X = 0, Y = (0)² + 2 = 2

If X = 1, Y = (1)² + 2 = 3

If X = 2, Y = (2)² + 2 = 6

If X = 3, Y = (3)² + 2 = 11

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You have just opened a new nightclub, Russ' Techno Pitstop, but are unsure of how high to set the cover charge (entrance fee). O
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Answer:

a)q(p)=-15p+ +300

b)R(p)=-15p²+300p

c)C(p)=-30p+1600

d.1)P(p)=-15p²+330p-1600

d.2)p=$11

Step-by-step explanation:

(a).Before getting started, we are going to consider  the cartesian plane, where x-axis corresponds to the cover charge and y-axis corresponds to number of guests per night . There, we will locate the following coordinates according to the previous information:

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Remember that  the slope-intercept form of the  linear equation is y=mx+b where m is the slope and b  is te intercept.

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In this case we will assign  (9,165) as the first coordinate and (10,150) as the second one. The order does not matter. At the end we get the same value for m.

Therefore:

m=\frac{150-165}{10-9} = -15

Now, we have the slope and two poitns. According to this, and taking into account that  point-slope form of the equation of a line is y-y1=m(x-x1),   find a linear demand equation.

Any of two coordintates can be selected. In this case we will select the second one.

y-150=-15(x-10)   equation 1

Solve equation 1 for y

y=-15x+150+150

y=-15x+300

Above we had defined x as cover charge (p) and y as number of guests (q). Thus, rewrite equation 1.

q(p)=-15p+ +300.  

it equation shows the number of guests per night as a function of the cover charge,

(b). The nightly revenue can be calculated multplying the price of the cover charge by  the number of guests. So, we just have to multiply the equation 1 by p.

p*q(p)=p*(-15p +300)

R(p)=-15p²+300p   equation 2

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C(p)=2*(-15p+300)+1000

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P(p)=R(p)-C(p)

P(p)=-15p²+300p - (-30p+1600)

P(p)=-15p²+300p+30p-1600

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Compute the first derivative of P(p)

P'(p)=-30p+330=0

30p=330

p=330/30

p= $11

The entrance fee that we should charge for a maximum profit is p=$11 per guest.

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