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1 year ago

Fiona started with 15/ 18 of a piece of pie. then she added another piece which was 13/6. how much did she have in total

Mathematics

1 answer:

PolarNik [594]1 year ago

7 0

First piece of pie = 15/18

Second piece of pie = 13/6.

We need to find the total value.

In order to find the total value, we need to add both fractions.

Total value = First piece of pie + Second piece of pie.

=15/18 + 13/6.

In order to add those fractions, we need to find the common denominators.

We have denominators 18 and 6.

Common denominator of 18 and 6 is 18.

So, we need to multiply second fraction 13/6 by 3 in top and bottom to make the denominator of second fraction equals 18.

Therefore, 15/18 + 13/6 = 15/18 + 13*3/6*3

=15/18 +39/18.

=(15+39)/18.

=54/18.

Therefore, Fiona had 54/18 in total.

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1 year ago

Circle E has a radius of 40 inches with = 324°. Find the exact length of

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2 years ago

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Reika [66]

As per the problem

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Total fraction of money spent cab be given as below

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2 years ago

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musickatia [10]

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

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

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 $n\ge4$ .

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

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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2 years ago

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MatroZZZ [7]

Answer:

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