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2 years ago

Leena walked 2/3 of a mile What is 2/3 written as a sum of unit fractions with the denominator of nine

1 answer:

3 0

Bear in mind that multiplying anything, and anything whatsoever by 1, has a product of the anything itself.

so, let's multiply 2/3 by 1 then and check about,

$\bf \cfrac{same}{same}=1\qquad thus\cfrac{2}{3}\cdot 1\implies \cfrac{2}{3}\cdot \cfrac{3}{3}\implies \cfrac{6}{9}
\\\\\\
\textit{now, we can split that and write it as a sum}\implies \cfrac{1+5}{9}\implies \cfrac{1}{9}+\cfrac{5}{9}$

You might be interested in

A lock has 5 buttons. The lock is opened by pushing two buttons simultaneously and then by pushing one button alone. How many co

Pavel [41]

We first must calculate how many ways 2 oblects can be chosen from 5.

combinations = 5! / 2! * (5-2)!

combinations = 5*4 / 2

combinations = 10

There are 10 ways to choose the 2 buttons and 5 ways to choose the final butto so there are 10 * 5 = 50 different ways.

Source

1728.com/combinat.htm

7 0

1 year ago

If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

Hence, they encircle the planet $3.76\times 10^{11}$ times.

8 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

1 year ago

Erick, Mia, and Isabelle golfed 9 holes. Erick scored 10 more than Mia, and Isabelle scored 16 less than twice Mia's score. Use

klasskru [66]

X + x + 10 + 2x - 16

x = mia's score
x + 10 = erick's score
2x - 16 = isabelle's score

the entire expression is the sum of all 3 scores <==

3 0

1 year ago

Jordan prepares 200 name tags to use at a meeting. The number for each color of the name tag is described below.

nadya68 [22]

Answer:

The answer would be A. 55.

Step-by-step explanation:

For the 35% you would take your 200 and divide it by 2 and have two 100's. Since 35% is out of 100% you could take 35 from each 100 and add them together to get 70.

For the 3/8 you would divide and get .375 then you would multiply that by 200 which would give you 75 yellow tags. Then you would add 75 and 70 and get 145. Then subtract 145 from 200 to get 55. Therefore there would be 55 red tags.

5 0

1 year ago