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3 years ago

Solve the equation12E − 4 = 0

1 answer:

7 0

This could get complicated. Please try to follow my steps carefully:

Begin by writing the equation to be solved: 12E - 4 = 0

Add 4 to each side of the equation: 12E = 4

Divide each side by 12: E = 4/12 = 1/3 .

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Solve 2x+y=3.5,-x+2y=2.5

padilas [110]

Answer:

y=1.7

x=.9

Step-by-step explanation:

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3 years ago

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A university is trying to determine what price to charge for tickets to football games. At a price of $ per ticket, attendance

Anika [276]

This question is incomplete, the complete question is;

A university is trying to determine what price to charge for tickets to football games. At a price of $24 per ticket, attendance averages 40,000 people per game. Every decrease of $4 adds 10,000 people to the average number. Every person at the game spends an average of $4 on concessions. What price per ticket should be charged in order to maximize revenue? How many people will attend at that price?

Answer:

a) price per ticket = $18

b) 55,000 will attend at that price

Step-by-step explanation:

Given that;

price of ticket = $24

so cost of ticket = 24 - x

every decrease of $4 adds 10,000 people

for $1 adds 2,500 people

Total people = 40000 + 2500x

Revenue function = people × cost + people × 4

R(x) = (40000 + 2500x)(24 - x) + (40000 + 2500x)4

R(x) = 960,000 - 40000x + 60000x - 2500x² + 160,000 + 10000x

R(x) = 1120000 + 30000x - 2500x²

R(x) = -2500x² + 30000x + 1120000

now for maximum revenue; R'(x) is to zero

R'(x) = d/dx( -2500x² + 30000x + 1120000 ) = 0

⇒ -5000x + 30000 + 0 = 0

⇒ -5000x + 30000 = 0

⇒ 5000x = 30000

x = 30000 / 5000

x = 6

R'(X) = -5000 < 0 ⇒ MAXIMUM

∴ Revenue is maximum at x = 6

so cost of ticket ;

price per ticket = (24 - x) = 24 - 6 = $18

so price per ticket = $18

Number of people = 40000 + 2500x

= 40000 + (2500 × 6)

= 40000 + 15,000

= 55,000

Therefore, 55,000 will attend at that price

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2 years ago

What is the greatest common factor (GCF) of 28 and 96?

victus00 [196]

Answer:

Step-by-step explanation:

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3 years ago

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How many lines can be drawn through points J and K? ОО o 1 O 2 O 3

valkas [14]

I'm not so sure but from what I know the answer is option 2

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3 years ago

Simplify

ivolga24 [154]

The correct answer is: [A]: " $\frac{8}{729}$ " .
_________________________________________________________
Explanation:
_________________________________________________________
We are asked to simplify:

  → $(\frac{2}{9})^3$

_________________________________________________________
Note the following property of exponents:
_________________________________________________________

$( \frac{a}{b})^n$   = $\frac{(a^n)}{(b^n)}$ ; { b $\neq$ 0 } .

_________________________________________________________
As such:

→ $( \frac{2}{9})^3$ ;

= $\frac{2^3}{9^3}$ = $\frac{(2*2*2)}{(9*9*9)}$ = $\frac{8}{729}$ .

_____________________________________________________________

The answer is: " $\frac{8}{729}$ " ;

→ which is: "Answer choice: [A]:  " $\frac{8}{729}$ " .
_____________________________________________________________
Hope this answer—and explanation—is of some help to you.
Please comment below or otherwise respond to me if you have any questions or if I can be of further assistance.
Best wishes!

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3 years ago

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