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inysia [295]
2 years ago
8

"From 1990 to 2000 The population of city A rose from 12,000 to 28,000 and the population of city B rose from 18,000 to 24,000.

If the population of the two cities increased at a constant rate, in what year was the population of both cities the same?"
Mathematics
1 answer:
Virty [35]2 years ago
4 0

Answer:

The year 1996

With population of both 21600

Step-by-step explanation:

From 1990 to 2000 = 10 years

So city A grew from 12000 to 28000 that is city A had an increase of 16000 in 10 years.

While city b grew from 18000 to 24000 , that's an increase of 6000 in 10 years to.

For city A

10 years= 16000

1 year = 16000/10

1 year = 1600

For city B

10 years = 6000

1 year = 6000/10

1 year = 600

So we are to find what year the both cities had same population.

12000 + x1600 = y

18000 + x600 = y

X is the year difference

Y is the population at that year

Eliminating y gives

6000= x1000

X= 6

If x is 6

18000+3600= y

21600= y

So 6 years + 1990 = 1996

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2 years ago
Wendy jogs a 5.5-mile route in the city with her running club once a week. She also jogs a nature trail near her home once a wee
insens350 [35]

Answer:

Wendy jogs 13.75 mile in all each week

Step-by-step explanation:

To determine how many miles Wendy jogs in all each week,

we will sum the distance she jogs in the city with her running club in a week to the distance she jogs along the nature trail near her home in a week.

The distance she jugs with her running club = 5.5 mile.

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Hence, we can write that

x = 1.5 y

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∴ x = 1.5 \times 5.5

x =  8.25 miles

Hence, the nature trail is 8.25 mile

Then, the number of miles Wendy jogs in all each week is

5.5 mile + 8.25 mile = 13.75 mile

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5 0
2 years ago
Which table shows a function that is decreasing over the interval (−2, 0)? A 2-column table with 4 rows. The first column is lab
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Answer:

  A 2-column table with 4 rows. The first column is labeled x with entries negative 3, negative 2, negative 1, 0. The second column is labeled f of x with entries 2, 0, negative 10, negative 24.

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The table shown in the attachment has the function that is decreasing in the interval of interest.

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Answer:

h=7

Step-by-step explanation:

We want to solve the equation:

\frac{1}{h-5}+\frac{2}{h+5}=\frac{16}{h^2-25}

We multiply through by the LCM: h^2-25=(h+5)(h-5)

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