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Nat2105 [25]
3 years ago
10

The owner of a luxury motor yacht that sails among the 4000 Greek islands charges $540 per person per day if exactly 20 people s

ign up for the cruise. However, if more than 20 people (up to the maximum capacity of 90) sign up for the cruise, then each fare is reduced by $7 per day for each additional passenger. Assume at least 20 people sign up for the cruise, and let x denote the number of passengers above 20.
(a) Find a function R giving the revenue per day realized from the charter.
R(x) =
(b) What is the revenue per day if 48 people sign up for the cruise?
$
(c) What is the revenue per day if 78 people sign up for the cruise?
$
Mathematics
1 answer:
lora16 [44]3 years ago
5 0

Answer:

Step-by-step explanation:

Charges up to 20 passengers = $540 per person per day

Total charges = 540 × 20 = $10800

If x passengers above 20 sign up for the cruise then total number of passengers = (20 + x)

Total revenue = $(20+x)

But each fare is reduced by $7 for additional passenger above 20 then revenue generated R(x) = 540(20+x) - 7x

R(x) = 10800 + 540x - 7x

R(x) = 10800 + 533x

a). Revenue per day realized R = 10800 + 533x

b). R(48) = 10800 + 48×533

              = $36,384

C). R(78) = 10800 + 78×543

              = $53,154

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The number on two consecutively numbered gym lockers have a sum of159. what are the locker numbers. Use comma to separate the an
svlad2 [7]
The answer is 79 and 80.

x - the number of the first locker
x + 1 - the number of the consecutive locker
Their sum is 159.

x + x + 1 = 159
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(x+1)(2x-4)(1/x+1) = (x+1)(2x-4)(1-5/2x-4)
Sunny_sXe [5.5K]

Solution:

(x+1)(2x-4)(\frac{1}{x}+1)=(x+1)(2x-4)(1-\frac{5}{2x}-4)\\(x+1)(2x-4)(\frac{1}{x}+1)-(x+1)(2x-4)(1-\frac{5}{2x}-4) = 0\\(x+1)(2x-4)(\frac{1}{x}+1-(1-\frac{5}{2x}-4))=0\\(x+1)(2x-4)(\frac{1}{x}+1-(-3\frac{5}{2x}))=0\\(x+1)(2x-4)(\frac{1}{x}+1+3+\frac{5}{2x})=0\\(x+1)(2x-4)(\frac{1}{x}+4+\frac{5}{2x})=0\\(x+1)(2x-4)(\frac{5x^{2}+8x+2}{2x} )=0\\

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Answer:

x=-1\\x=2\\x=\frac{-4+\sqrt[]{6}}{5}\\x=\frac{-4-\sqrt[]{6}}{5}

<em>Hope this was helpful.</em>

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