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liberstina [14]
3 years ago
15

the population of a city increases by 4000 people each year.in 2025 the population is projected to be 450000 what is an equation

that gives the citys population p in thousands of people x after 2010
Mathematics
1 answer:
Lerok [7]3 years ago
4 0
Assuming this is linear
write everything in thousands smaller

p=mx+b
b=initial amount
m=growth rate

city increases by 4 each year
in 2025, population is 450
find it after 2010, with 2010 being year 0 or x=0

so
2025-2010=15 years
15*4=60
increased by 60
450=result=increase+originalin2010
increase=60
450=60+original
minus 60
390=original

p=4x+390
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If sin tetha=root3/2, what is cos tetha?​
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Answer:

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Step-by-step explanation:

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6 0
3 years ago
PLEASE HELP!!!!<br> I need to solve for a, b, and c
nirvana33 [79]

Answer:

  • 25000 seats in section A
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  • 10900 seats in section C

Step-by-step explanation:

The problem statement tells you half the total number of seats are in section A, so you already know that there are 25000 A seats. The revenue from those seats is

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so the revenue from B and C seats must total

... $1,070,500 - 625,000 = $445,500

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The actual revenue from those seats is $445,500 -375,000 = $70,500 more than that. We know each B seat generates $5 more revenue, so there must be ...

... $70,500/$5 = 14,100 . . . . B seats

Then the balance of the 25000 B and C seats are C seats:

... 25,000 - 14,100 = 10,900 . . . . C seats

_____

<em>Alternate Solution Method</em>

The new Brainly answer format requires the answer be supplied before the working. In order to find the answer quickly so that I can fill in that section, I used a matrix method for solving the problem. The problem equations can be written ...

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  • a - b - c = 0
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so the augmented matrix is ...

\left[\begin{array}{cccc}1&1&1&50000\\1&-1&-1&0\\25&20&15&1070500\end{array}\right]

A graphing calculator can be used to find the solution to this, generally using a function that produces the reduced row-echelon form. The attachment shows the solution using a TI-84 calculator.

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<em>Comment on the Working</em>

Since the number of A seats is equal to the total of B and C seats, the number of A seats must be half the total number of stadium seats. Having figured that out, the problem is reduced to one of finding the mix of B and C seats that will produce the remaining revenue.

As with many mixture problems, it is convenient to look at differences. Start with the assumption that all of the desired revenue comes from the least contributor. Here, that is C seats. Then figure the difference that using a B seat makes ($20 -15 = $5) and the difference of the actual revenue and the amount that you got by assuming all C seats: 445,500 -375,000 = 70,500. Since replacing a C seat by a B seat adds $5 to the revenue, it is easy to figure the number of such replacements required in order to raise the revenue by $70,500.

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... 5b = 445,500 - 375000 . . . . simplify, subtract 375000

... b = 70500/5 = 14100

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Answer:

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Here, slope of the line = \frac{4}{5}

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b = \frac{54}{5}

Equation of the parallel line will be,

y = \frac{4}{5}x+\frac{54}{5}

4 0
3 years ago
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