Consider the following recurrence relation: if n = 0 Hn) In. Hin - 1) + 1 if n > 0. Prove that H(n) = n!(1/1! + 1/2 + 1/3! + ... + 1/n!) for all n 2 1. (Induction on n.) Let f(n) = n!(1/1! + 1/2! + 1/3! + ... + 1/n!). Base Case: If n = 1, the recurrence relation says that H(1) = 1 . H(0) + 1 = 1.0 + 1 = 1, and the formula says that f(1) = 1!(1/1!) = 1, so they match. Inductive Hypothesis: Suppose as inductive hypothesis that H(k-1) = ! + 1/2 + 1/3! + ... + 1/(k - 1)!) for some k > 1. Inductive Step: Using the recurrence relation, H(K) = k· H(k-1) + 1, by the second part of the recurrence relation (1/1! + 1/2 + 1/3! + ... + 1/(k − 1)!) + 1, by inductive hypothesis (1/1! + 1/2! + 1/3! + ... + 1/(k-1)!) + k!/k! (1/11 + 1/2! + 1) (1/1! + 1/2 + 1/3! + ... + 1/(k-1)! + 1/k!) so, by induction, H(n) = f(n) for all n 2 1.

The function equality that represents the point where f(x) = g(x) is; A: f(0) = g(0) and f(2) = g(2)

From the question, the curved red line represents g(x) while the straight blue line represents f(x).

Now, the equality of functions f(x) = g(x)  is represented as a common function between their curves. Thus, we will just need to find a common point for both. This is;

f(x) has points (0, 4) and (2, 0).

g(x) has points (0,4) and (2,0).

It is pertinent to note that both functions have the same y-value for x=0 and x = 2. Thus, we can say that;

f(2) = g(2) and f(0) = g(0)

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The simplified form is [tex]6x^{3}[/tex].

What are indices in maths?

The equation is -

[tex]\sqrt{\frac{2160x^{3} }{60x^{2} } }[/tex]

We can rewrite this using property of multiplication to obtain

[tex]\sqrt{\frac{2160}{60} * \frac{x^{8} }{x^{2} } }[/tex]

We simplify variable expression in the radicand using

[tex]\frac{a^{m} }{a^{n} } = a^{m - n}[/tex]

[tex]\sqrt{36 * x^{8 - 2} }[/tex]

[tex]\sqrt{36 * x^{6} }[/tex]

[tex]\sqrt{36 * (x^{3})^{2} }[/tex]

[tex]\sqrt{36} * \sqrt{(x^{3})^{2} }[/tex]

[tex]6 * x^{3}[/tex]

Therefore,  the simplified form is [tex]6x^{3}[/tex]

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Answer:

Step-by-step explanation:

Area = h/2 (b_1+b_2)

[tex]\sf\cfrac{7}{2}\:(22+34)[/tex]

Add 22 and 34 = 56

[tex]\sf \cfrac{7}{2} \times 56[/tex]

Let's express 7/2 * 56 as a single fraction:-

[tex]\sf \cfrac{7\times 56}{2}[/tex]

Multiply 7 and 56 = 392

[tex]\sf \cfrac{392}{2}[/tex]

Divide

[tex]\sf 196[/tex]

Therefore, the area of the trapezoid is 196cm².

_______________________

The function for the given graph is f(x) = (x - 4)(x + 2)(x + 4).

In the given graph,

The points that lie on the x-axis are (-4, 0), (-2, 0), and (4, 0) where y-coordinate is 0 and the x coordinates are -4, -2, and 4.

On solving the given functions,

Option A:

f(x) = (x - 4)(x + 2)(x + 4)

Since y = 0 consider f(x) becomes 0

So,

(x - 4)(x + 2)(x + 4) = 0

According to the zero product rule,

x - 4 = 0, x + 2 = 0, and x + 4 = 0

⇒ x = 4, x = -2 and x = -4.

Thus, option A represents the function of the given graph.

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Question:

Select the correct answer. which equation could represent the graphed function (given in the figure below)?

The tripling time of population is 6 days and the population will be doubled in 3.5 days.

Given that the population of a bacteria is tripling in 6 days.

We have to find the tripling time of population and the time when the population will be doubled.

Suppose the population of bacteria in beginning be x.

The population of bacteria seems like arithmetic progression in which a =x, nth term =3x and n=6.

Aritmetic progression is a sequence which is having common difference.

nth term=a+(n-1)d

3x=x+(6-1)*d

3x-x=5d

2x=5d

d=2x/5

d=0.4x

So now we have to find the value of n where population will be 2x.

nth term=a+(n-1)d

put nth term=2x,  a=x,d=0.4x

2x=x+(n-1)*0.4x

2x-x=(n-1)*0.4x

x/0.4x=n-1

2.5=n-1

n=2.5+1

n=3.5

Hence the population of bacteria willbe doubled after 3.5 days.

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Answer:

[tex]\frac{FH}{RX} =\frac{HG}{XS} =\frac{FG}{RS}[/tex]

Step-by-step explanation:

The order of the letters matters in any similarity theorem.

When triangle FHG is similar to triangle RXS, the lengths of FH must be similar to RX.

When 2 similar lines are divided, that number is the factor by which one triangle can be dilated to match the other.

That being said, each of these proportion statements can be read as individual division expressions that equal the next division expression.

This means [tex]FH\div RX = HG\div XS = FG\div RS[/tex].

If you look at it this way, you can see that the sides you need to fill in must be the corresponding similar sides, so that you end up with the factor for the dilation of the 2 triangles.

Question 2

1) [tex]\overline{PR} \perp\overline{QS}, \overline{PQ} \cong \overline{PS}[/tex] (given)

2) [tex]\overline{PR} \cong \overline{PR}[/tex] (reflexive property)

3) [tex]\angle PRQ, \angle PRS[/tex] are right angles (perpendicular lines form right angles)

4) [tex]\triangle PRQ, \triangle PRS[/tex] are right triangles (a triangle with a right angle is a right triangle)

5) [tex]\triangle PRQ \cong \triangle PRS[/tex] (HL)

Question 3

1) [tex]\angle C[/tex] is a right angle, [tex]\overline{AC} \cong \overline{AE}, \overline{DE} \perp \overline{AB}[/tex] (given)

2) [tex]\angle DEA[/tex] is a right angle (perpendicular lines form right angles)

3) [tex]\triangle ACD, \triangle DAE[/tex] are right triangles (a triangle with a right triangle is a right angle)

4) [tex]\overline{AD} \cong \overline{AD}[/tex] (reflexive property)

5) [tex]\triangle ACD \cong \triangle AED[/tex] (HL)

6) [tex]\angle CAD \cong \angle DAE[/tex] (CPCTC)

7) [tex]\overline{AD}[/tex] bisects [tex]\angle BAC[/tex] (if a segment splits an angle into two congruent parts, it is an angle bisector)

Answer:

6y^2 + 18y

Step-by-step explanation:

2y (3y + 9)=

6y^2 + 18y

[2y (3y becomes 6y^2]

[2x(      + 9) becomes 18y]

Add them to obtain 6y^2 + 18y

The inequality of the graph that is given is: y < 2x - 3.

First find the slope (m) = rise / run of the line.

Rise = 2 units

Run = 1 unit

Slope (m) = 2/1 = 2

Y-intercept (b) = -3.

Since the shaded side is at the left, we are going to use the ">" sign. Substitute m = 2 and b = -3 into y < mx + b.

y < 2x - 3

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196 36 correct answer is 232

Answer:

728 cm²

Step-by-step explanation:

the surface area is the area of the 6 faces.

the opposite rectangular faces are congruent , then

SA = front/ back + sides + top/ bottom

     = 2(14 × 14) + 2(6 × 14) + 2(14 × 6)

     = 2 × 196 + 2 × 84 + 2 × 84

     = 392 + 168 + 168

     = 728 cm²

The value is 52. 674

We have;

4(3) / 4(8) +.9(7.8 x 7.45)

From BODMAS, we have that we must must open up the bracket

= [tex]\frac{12}{32} + 0. 9 (58. 11)[/tex]

Divide the fraction to get decimals

= [tex]0. 375 + 52. 299[/tex]

Add the decimals

= [tex]52. 674[/tex]

Thus, the value is 52. 674

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Answer:

555 meters

Step-by-step explanation:

toymaker = height/shadow

= 0.9/0.2

CN tower = height/shadow

= x/123.3

solve for x:

;[tex]\frac{0.9}{0.2} = \frac{x}{123.3} \\\\0.2(x) = (0.9)(123.3)\\0.2x = 110.97\\(\frac{10}{2} )\frac{2}{10}x = 110.97(\frac{10}{2})\\ x = 554.85[/tex]

They are all in meters, so we don't need to convert anything so rounded to the nearest meter, the answer is 555 meters

Answer: 555 m

Step-by-step explanation:

       We will set up a proportion to solve.

[tex]\displaystyle \frac{\text{vertical side}}{\text{horizontal side}} \;\;\;\;\;\;\;\frac{0.9\;m}{0.2\;m} =\frac{x\;m}{123.3 \;m}[/tex]

    Now we will cross multiply.

0.9 * 123.3 = 0.2 * x

110.97 = 0.2x

554.85 = x

x = 554.85

    Lastly, we will round the nearest m.

554.85 ➜ 555 m

The correct options about parabola are:

The focus is located at (0,–3).

The parabola can be represented by the equation x^2 = –12y.

According to the statement

we have given that the vertex is at the origin and directrix at y = 3.

and from these given information and we have to find the all abot the parabola like its focus point etc.

So,

We know that the equation of parabola is

(x-h)^2 = 4p(y-k)

Here The vertex is (h,k). and the focus is at (h,k+p). and the directrix is   y(k - p.)

So, From the given information

Vertex at the origin means that h=0 and k=0

Directrix at y = 3 means that p=-3

Directrix at the y-axis means the parabola opens upwards.

Thus, the focus is: (0,-3)

And The p-value becomes

: 4(-3) = -12.

And from all these the equation of the parabola is becomes

(x)^2 = -12y

So, The correct options about parabola are:

The focus is located at (0,–3).

The parabola can be represented by the equation x^2 = –12y.

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Question:

A parabola, with its vertex at the origin, has a directrix at y = 3. Which statements about the parabola are true? Select two options.

The focus is located at (0,–3).

The parabola opens to the left.

The p value can be determined by computing 4(3).

The parabola can be represented by the equation x2 = –12y.

The parabola can be represented by the equation y2 = 12x.

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Answer:

Step-by-step explanation:

[tex]\sf 1.1)\dfrac{Sin \ \theta-Cos \ \theta}{Sin \ \theta + Cos \ \theta}=\dfrac{(Sin \ \theta-Cos \ \theta)}{((Sin \ \theta + Cos \ \theta)}*\dfrac{(Sin \ \theta-Cos \ \theta)}{(Sin \ \theta-Cos \ \theta)}[/tex]

                           [tex]\sf = \dfrac{(Sin \ \theta-Cos \ \theta)^2}{Sin^2 \ \theta-Cos^2 \ \theta}\\\\=\dfrac{Sin^2 \ \theta+Cos^2 \ \theta-2Sin \ \theta \ Cos\ \theta}{(Sin \ \theta-Cos \ \theta)}\\\\\\\bf Identity: \ (a +b)^2= a^2 + b^2 - 2ab\\\\\\=\dfrac{1-2Sin \ \theta \ Cos \ \theta}{(Sin \ \theta-Cos \ \theta)} = LHS[/tex]

[tex]\sf 1.2) LHS = tan^2 \ x - Sin^2 \ x = \dfrac{Sin^2 \ x}{Cos^2 \ x}-Sin^2 \ x[/tex]

                                        [tex]\sf =\dfrac{Sin^2 \ x}{Cos^2 \ x}-\dfrac{Sin^2 \ x*Cos^2 \ x}{1*Cos^2 \ x}\\\\\\ = \dfrac{Sin^2 \ x - Sin^2 \ x*Cos^2 \ x}{Cos^2 \ x}\\\\\\= \dfrac{Sin^2 \ x *(1 -Cos^2 \ x)}{Cos^2 \ x}\\\\=\dfrac{Sin^2 \ x*Sin^2 \ x}{Cos^2 \ x} \\\\ \bf 1 - Cos^2 \ x = Sin^2 \ x\\\\= \dfrac{Sin^2 \ x}{Cos^2 \ x}*Sin^2 \ x\\\\=tan^2 \ x * Sin^2 \ x = RHS[/tex]

[tex]\sf 1.3) LHS = \dfrac{1-Cos \ x}{Sin \ x}-\dfrac{Sin \ x}{1+Cos \ x} =\dfrac{(1-Cos \ x)(1+Cos \ x)}{Sin \ x*(1+Cos \ x)}-\dfrac{Sin \ x*Sin \ x}{(1+Cos \ x)*Sin \ x}\\[/tex]

              [tex]\sf =\dfrac{1 - Cos^2 \ x}{Sin \ x*(1+Cos \ x)}-\dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)}\\\\=\dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)} - \dfrac{Sin^2 \ x}{Sin \ x*(1+Cos \ x)}\\\\=\dfrac{Sin^2 x - Sin^2 \ x}{Sin \ x*(1+Cos \ x)} \\\\= 0 = RHS[/tex]

[tex]\sf 1.4) LHS = Sin x - \dfrac{1}{Sin \ x + Cos \ x}+Cos \ x \\[/tex]

              [tex]\sf = \dfrac{Sin \ x *(Sin \ x + Cos \ x) - 1 + Cos \ x * (Sin \ x + Cos \ x)}{Sin \ x + Cos \ x }\\\\\\= \dfrac{Sin \ x * Sin \ x + Sin \ x*Cos \ x -1 + Cos \ x*Sin \ x + Cos \ x*Cos \ x}{Sin \ x + Cos \ x}\\\\\\=\dfrac{Sin^2 \x + Sin \ x \ Cos \ x - 1 + Cos \ x \ Sin \ x + Cos^2 \x}{Sin \ x + Cos \ x}\\\\=\dfrac{Sin^2 \ x + Cos^2 \ x - 1 + Sin \ xCos \x +Sin \ x Cos \ x}{Sin \ x + Cos \ x}\\\\= \dfrac{1 - 1 +2Sin \ x Cos \ x}{Sin \ x + Cos \ x}\\\\= \dfrac{2Sin \ x Cos \ x}{Sin \ x + Cos \ x}[/tex]

Answer:

24 combinations

Step-by-step explanation:

To get to this answer, I multiplied 6 x 4, since there are 6 shirts and 4 pairs of jeans, there will be a total of 24 combinations. For combinations of two different things, you can just multiply the amount of each item by the other item. hope this helps :)

The sum of areas of circle C and circle D exists 98 square units.

Given: Radius of circle = 7 units

Area of circle [tex]=\pi r^2[/tex] ..............(1)

Where, r be the radius of circle

The radius of circle C, r = 7 units

(1) becomes

Area of circle, C [tex]= \pi 7^2[/tex] square units

[tex]= 49\pi[/tex] units²

Area of circle, D [tex]= \pi 7^2[/tex] square units

[tex]= 49\pi[/tex] units²

Sum of areas of circles C and D,

= Area of circle C + Area of circle D

= 49 + 49

= 98 units²

The sum of areas of circles C and D exist 98 units².

Therefore, the correct answer is option d) 98 units².

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The probability of randomly getting first a red marble and then a blue marble is:

P = 0.083

There are:

For a total of 9 marbles.

The probability of getting a red marble is equal to the quotient between the number of red marbles and the total, so:

P(red) = 2/9

Then the probability of getting a blue marble is equal to the quotient between the number of blue marbles and the total, but because we already took one marble, now the total is 8.

P(blue) = 3/8

The joint probability is given by the product:

P = (2/9)*(3/8) = 0.083

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Answer:

2625 different numbers

Step-by-step explanation:

This type of problem is classified as a "Combinatorics" problem:  Counting up the number of ways something can happen.

Many times when attempting to count a large number of items, patterns are recognized and shortcuts to count are developed so that one doesn't need to physically count all of the items.  Perhaps the trickiest part of counting with these shortcuts is to ensure that one does not over-count (counting a scenario more than once), although it is equally important that we don't under-count (fail to count a situation that applies).

It should be noted that for a number to be a 4 digit number, it must have 4 digits, and thus the first digit must not be zero (despite that that would be an even digit, the number itself would not be a 4 digit number).

There are 2 main scenarios:

1.  All 4 digits are even

2. Exactly 3 digits are even (meaning that exactly one digit is odd).

There are two sub-cases to Scenario 2:

2.1. The first digit is odd, and all three of the other digits are even.

2.2. The first digit is even, and exactly two of the other digits are even (meaning that exactly one of the last three digits is odd).

None of scenarios 1, 2.1, and 2.2 overlap, so we're not over-counting:  

Further, this is all of the possibilities for a 4 digit number with least three even digits, so we're not under-counting.

Scenario 1 -- All 4 digits are even.

If all 4 digits are even, then the first digit has fours choices (2,4,6,8), and the next 3 digits each have 5 choices (2,4,6,8,0).

[tex]4*5*5*5=500\text{ choices}[/tex]

Scenario 2.1 -- The first digit is odd, and all three of the other digits are even

If the first digit is odd, there are 5 choices for the first digit (1,3,5,7,9), and the next 3 digits each have 5 choices (2,4,6,8,0).

[tex]5*5*5*5=625\text{ choices}[/tex]

Scenario 2.2 -- The first digit is even, and exactly two of the other digits are even

If the first digit is even, there are 4 choices for the first digit (2,4,6,8), and if exactly two of the next 3 digits are even, then there are 5 choices for each of the two even digits (2,4,6,8,0), and 5 choices for the odd digit (1,3,5,7,9), and there are "3 permuted by 2" ways of ordering those three digits.

[tex]4* \left [(5*5*5) * {}_3 \! P_2 \right ]=\\\\=4*\left [5*5*5 * \dfrac{3!}{2!} \right ]\\\\=4*\left [ 5*5*5 * \dfrac{3*2*1}{2*1} \right ] \\\\=4*\left [5*5*5 * 3 \right ] \\\\=1500\text{ choices}[/tex]

Scenario 2.2 broken down (calculated without using the permutation operation) -- The first digit is even, and exactly two of the other digits are even

Then either the second digit is odd (scenario 2.2.1), the third digit is odd (scenario 2.2.2), or the fourth digit is odd (scenario 2.2.3).

Scenario 2.2.1.  The second digit is odd.

The first digit is even: 4 choices for the first digit (2,4,6,8)

The second digit is odd:  5 choices for the odd digit (1,3,5,7,9)

The third and fourth digits are even: 5 choices for each even digit (2,4,6,8,0).

[tex]4*5*5*5=500\text{ choices}[/tex]

Scenario 2.2.2.  The third digit is odd.

The first digit is even: 4 choices for the first digit (2,4,6,8)

The second and fourth digits are even:  5 choices for each even digit (2,4,6,8,0).

The third digit is odd:  5 choices for the odd digit (1,3,5,7,9)

[tex]4*5*5*5=500\text{ choices}[/tex]

Scenario 2.2.3.  The last digit is odd.

The first digit is even: 4 choices for the first digit (2,4,6,8)

The second and third digits are even:  5 choices for each even digit (2,4,6,8,0).

The last digit is odd:  5 choices for the odd digit (1,3,5,7,9)

[tex]4*5*5*5=500\text{ choices}[/tex]

Scenarios 2.2.1, 2.2.2, and 2.2.3 comprise all of Scenario 2.2, so [tex]500+500+500=1500\text{ choices}[/tex]

How many ways can a 4 digit number be formed where at least 4 of the digits are even?

This is the sum of the choices from Scenario 1, Scenario 2.1, and Scenario 2.2, so    [tex]500+625+1500=2625\text{ choices}[/tex].

Answer:

88

Step-by-step explanation: make me brainliest if it is correct

There are 24 ways in which 5 guys can sit if arranged from oldest to youngest.

We have,

Five guys.

Now,

We know that,

Total number of ways to arranged around a table (n) = (n-1)!

So,

For n = 5,

I.e.

(n - 1)! = (5 - 1)! = 4!

So,

Total number of ways to arranged Five guys rom oldest to youngest (n) = (n-1)!

i.e.

= (5 - 1)! = 4!

We get,

= 4 × 3 × 2 × 1

i.e.

Total number of ways to arranged Five guys rom oldest to youngest (n) = 24.

Hence we can say that there are 24 ways in which 5 guys can sit if arranged from oldest to youngest.

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No, this not plagiarism .You can use generally accepted facts without citing them.

What defines self-plagiarism?

Self-plagiarism is defined as a type of plagiarism in which the writer republishes a work in its entirety or reuses portions of a previously written text while authoring a new work.

What is considered general knowledge in plagiarism?

How to Avoid plagiarism common knowledge?

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Answer:

0.42

Step-by-step explanation:

Answer:

width = 9 inches

Step-by-step explanation:

let width be w then length is w + 12

the perimeter ( P) is calculated as

P = 2( width + length )

here P = 60 , then

2(w + w + 12) = 60 ( divide both sides by 2 )

2w + 12 = 30 ( subtract 12 from both sides )

2w = 18 ( divide both sides by 2 )

w = 9

that is the width of the rectangle is 9 inches

Based on the length and width of the rectangular lawn, and the percentage the man has mowed, the strip width would be 40.5 m.

first, find the area of the lawn:

= 100 x 200

= 20,000 ft²

the area mowed is:

= 40.5% x 20,000

= 8,100 ft²

the width is therefore:

= 8,100 / 200

= 40.5 m

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The system of linear equations can be used to find the number of nickels, n, and the number of quarters, q, yolanda used is n + q = 28

n + q = 280.05n + 0.25q = 4

n + q = 28

0.05n + 0.25q = 4

n = 28 - q

0.05n + 0.25q = 4

0.05(28 - q) + 0.25q = 4

1.4 - 0.05q + 0.25q = 4

- 0.05q + 0.25q = 4 - 1.4

0.20q = 2.6

q = 2.6 / 0.20

q = 13

Substitute into

n + q = 28

n + 13 = 28

n = 28 - 13

n = 15

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Since beta is in the first quadrant, the final answer will be positive.

To find cos(beta) so we can use the half angle identity, we can substitute into the Pythagorean identity. Doing so gives us that

[tex] \sin( \beta ) = \frac{ \sqrt{17} }{5} [/tex]

So, this means that

[tex] \sin( \frac{ \beta }{2} ) = \sqrt{ \frac{1 - \frac{ \sqrt{17} }{2} }{2} } = \sqrt{ \frac{2 - \sqrt{17} }{4} } = \frac{ \sqrt{2 - \sqrt{17} } }{2} [/tex]

The lowest score you can earn and still be eligible for employment is 85.6925.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean  and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

Top 5%.

At least the 100-5 = 95th percentile.

The 95th percentile is X when Z has a pvalue of 0.95. So X when Z = 1.645.

x-75 = 1.645*6.5

x = 85.692

Thus the lowest score you can earn and still be eligible for employment is 85.6925.

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The solution to the expression [tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex] is [tex]\frac{216x^6 - 108x^4 + 18x^2 - 1}{27x^3}[/tex]

The expression is given as:

[tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex]

Take the LCM of the expression

[tex]\frac{8x^3 * 27x^3 - 4x * 27x^3 + 2 * 9x^2 - 1}{27x^3}[/tex]

Evaluate the products

[tex]\frac{216x^6 - 108x^4 + 18x^2 - 1}{27x^3}[/tex]

Hence, the solution to the expression [tex]8x^3-4x+\frac{2}{3x}-\frac{1}{27x^3}[/tex] is [tex]\frac{216x^6 - 108x^4 + 18x^2 - 1}{27x^3}[/tex]

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Answer:

Step-by-step explanation:

Given the 4th order linear ODE

[tex]u^{(4)} - u = 0[/tex]

we substitute

[tex]x_1 = u[/tex]

[tex]x_2 = {x_1}' = u'[/tex]

[tex]x_3 = {x_2}' = {x_1}'' = u''[/tex]

[tex]x_4 = {x_3}' = {x_2}'' = {x_1}''' = u'''[/tex]

Then the given equation transforms to

[tex]{x_4}' - x_1 = 0[/tex]

but we also need to relate this to the other derivative substitutions. This gives the system of differential equations

[tex]\begin{cases} {x_1}' = x_2 \\ {x_2}' = x_3 \\ {x_3}' = x_4 \\ {x_4}' = x_1 \end{cases}[/tex]

In matrix form,

[tex]\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}' = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}[/tex]