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klasskru [66]
1 year ago
11

A group of three friends cycle from Llanfyllin to Vyrnwy at 12 km/h.

Mathematics
1 answer:
garik1379 [7]1 year ago
6 0

The average speed for the journey from Llanfyllin to Bala is 14 km/h

Average speed is the ratio of total distance travelled to total time taken. Also, average speed can be given as the average of the different speeds.

Let d₁ represent the distance from Llanfyllin to Vyrnwy and t represent the time taken, since the speed is 12 km/h, hence:

12 = d₁/t

d₁ = 12t

Let d₂ represent the distance from Vyrnwy to bala and t represent the time taken. Since they spent the same time, since the speed is 16 km/h, hence:

16 = d₂/t

d₂ = 16t

Average speed = total distance/total time

Average speed = (d₁ + d₂) / (t + t)

Average speed = (12t + 16t) / 2t

Average speed = 28t / 2t

Average speed = 14 km/h

Hence their average speed for the journey from Llanfyllin to Bala is 14 km/h.

Find out more at: brainly.com/question/23774048

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A rectangular parking lot has a perimeter of 232 feet. The length of the parking lot is 36 feet less than with. Find the length
Ainat [17]

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Length: 80 feet

Width: 36 feet

Step-by-step explanation:

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7 0
2 years ago
the area of a trapezium is 594 sq. cm. The lengths of its parallel sides are in the ratio 4 : 5. The height of the trapezium is
Sophie [7]

Answer:

1st side=44 cm, 2nd side= 55cm

Step-by-step explanation:

let a = '4x' , b = '5x'

h= 12cm

Area of trapezium=1/2*(a+b)h

area=594=1/2*9x*12

594*2/12=9x

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99=9x

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therefore, 4x=4*11=44cm

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5 0
1 year ago
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 (
musickatia [10]
(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).
3 0
2 years ago
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