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1 year ago

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"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?"

1 answer:

Virty [35]1 year 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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The function D(t) defines a traveler's distance from home, in miles, as a function of time, in hours. D(t) = StartLayout enlarge

nalin [4]

- At 2 hours, the traveler is 725 miles from home.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE.

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE.

Step-by-step explanation:

The function D(t) is defined as follows:

$D(t) = 300t+125$ for $t< 2.5$

$D(t) = 880$ for $2.5 \leq 3.5$

$D(t) = 75t+612.5$ for $t\leq 6$

Where

t is the time in hours

D(t) is the distance covered, in miles, after t hours

Now let's analyze the different statements:

- The starting distance, at 0 hours, is 300 miles. --> FALSE. In fact, if we substitute t = 0 into the 1st equation, we get

$D(0) = (300)(0)+125 = 125$

So, the distance at t = 0 is 125 miles.

- At 2 hours, the traveler is 725 miles from home. --> TRUE. In fact, if we substitute t = 2 into the 1st equation,

$D(2) = (300)(2)+125 = 725$

- At 2.5 hours, the traveler is 875 miles from home. --> FALSE. In fact, for t=2.5 we have to use the 2nd equation, which states that the distance is:

D(t) = 880

So, not 875 miles.

- At 3 hours, the distance is constant, at 880 miles. --> TRUE. This is clearly visible from the 2nd equation: for t between 2.5 and 3.5 (so, in this case), the distance is

D(t) = 880

- The total distance from home after 6 hours is 1,062.5 miles --> TRUE. In fact, if we replace t = 6 into the last equation,

$D(6)) = 75(6)+612.5=1062.5$

Learn more about functions:

brainly.com/question/3511750

brainly.com/question/8243712

brainly.com/question/8307968

Learn more about distance:

brainly.com/question/3969582

#LearnwithBrainly

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1 year ago

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A tiling company completes two jobs. The first job has $1200 in labor expenses for 40 hours worked, while the second job has $15

daser333 [38]

Answer:

The correct option is;

1560 = 30(52) + b

Step-by-step explanation:

The cost of the first job = $1,200 in labor and expenses

The number of hours worked in the first job = 40

The second job costs $1,560 in labor and expenses

The number of hours worked in the second job = 52

If the cost of labor per hour = L and the expenses = b, we have;

$1,200 = 40×L + b......................(1)

$1,560 = 52×L + b.......................(2)

Subtracting equation (1) from (2), we have;

52×L + b - (40×L + b) = 52×L - 40×L + b - b = $1,560 - $1,200 = $360

12×L = $360

L = $360/12 = $30/hour

From equation (1), we have;

$1,200 = 40×L + X = 40×$30 + b

Therefore, the equation that can be used to calculate the y-intercept of the linear equation is either;

1200 = 40(30) + X or 1560 = 32(52) + b, which gives the correct option as 1560 = 32(52) + b

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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

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(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

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The answer is 20 days.

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(400/460) * 23

= (20 / 23) * 23

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