HELP ME!!!!!!!!!!!!!!!!!!!

Answer: 10*100*100

10 times 100 times 100 = 100,000

hope this helped =)

According to the information, the unit price of the products is: $0.28 for each can of soft drink, $0.022 for each ounce of detergent, $0.18 for each ounce of peanuts, $0.0033 for each gram of cooked chicken.

To calculate the unit price of the products we must perform the following procedure:

We must divide the value of each product into the amount of product as shown below:

Accordingly, the unit price of these products is the result of these divisions.

Note: This question is incomplete. Here is the complete information:

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The domain of the given relation is:

D: {-3, -2, -1, 0, 1, 3}

For any relation that maps elements from one set into elements of another set, the domain is just the set of the inputs, and usually the domain is the se of the values "x"

Here the domain is the listof numbers in the left,the domain is:

D: {-3, -2, -1, 0, 1, 3}

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Proportionately, the number of sundaes that will be vanilla is 10 sundaes.

Proportion refers to the part or portion of a whole.

Proportions show the relative size (ratio) of a value compared to another.

The difference between proportion and ratio is that proportions are two ratios equated to each other.

The total number of sundaes Maggie is making = 30

The types of sundaes = mint, chocolate, and vanilla ice cream.

Mint ice cream = 1/3 = 10 sundaes (1/3 x 30)

The remaining sundaes = 2/3 (1 - 1/3)

= 20 (30 - 10) or (30 x 2/3)

Chocolate ice cream = 1/2 of 20 = 10 sundaes

Vanilla sundaes = 10 (30 - 10 - 10) or (1/2 of 20)

Thus, using the proportional or fractional relationships among the different sundaes, the number of vanilla sundaes that Maggie will make is 10.

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The required values are,

x = 3

y = 5.

An equation is said to be linear equation in two variables if it is written in the form of ax + by + c=0, where a, b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero

Now in the given question,

we have two equations,

3x - 2y = -1                                 ......(1)

3x + y = 14                                 .......(2)

subtract (2) from (1), we get

-3y = -15

divide both sides by -3, we get

y = 5

now put this value of y in (2),

3x + 5 = 14

subtarct both sides by 5, we get

3x = 9

divide both sides by 3, we get

x = 3

Hence, the required values are,

x = 3

y = 5

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As research have shown that GDP has many known biases that make it a poor absolute measure of the welfare of a society. A reason why GDP will still be a good measure of changes in welfare will likely be because biases may be relatively constant over time. The Option B is correct.

GDP is a poor indicator of a society's standard of living because it does not directly account for leisure, environmental quality, levels of health and education, activities conducted outside the market, changes in income inequality, increases in variety, increases in technology, and so on.

So, GDP does not directly measure the things that make life worthwhile, but it does measure our ability to obtain many of the inputs that make life worthwhile. GDP, however, is not a perfect measure of happiness. Some factors that contribute to a happy life are excluded from GDP.

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The graph of the given inequality of x > 23 is attached.

The relationship between two expressions or values that are not equal to each other is called inequality.

A number line can be used to represent numbers placed on regular intervals. A number line can be used to represent an inequality.

Given that the inequality of x > 23

We are asked to plot the given inequality on a number line.

x > 23

The above inequality says that, the value of x is equal to or greater than 23.

Hence, the graph of the given inequality is attached.

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Answer: We can use trigonometry to solve this problem. Let's call the angle of elevation of the lift from the lodge area to the top of Giffin's Gallop "θ". Then we can break down the lift into two parts: the horizontal distance from the lodge area to the top of Giffin's Gallop, and the vertical distance from the lodge area to the top of Giffin's Gallop.

Horizontal distance:

The horizontal distance from the lodge area to the top of Giffin's Haven is 1200 m * cos(43°) + 1500 m * cos(53°) + 30 m = 1759.29 m

The horizontal distance from the top of Giffin's Haven to the top of Giffin's Gallop is 30 m.

So, the total horizontal distance from the lodge area to the top of Giffin's Gallop is 1759.29 m + 30 m = 1789.29 m

Vertical distance:

The vertical distance from the lodge area to the top of Giffin's Haven is 1200 m * sin(43°) + 1500 m * sin(53°) + 40 m = 1174.70 m

The vertical distance from the top of Giffin's Haven to the top of Giffin's Gallop is 40 m.

So, the total vertical distance from the lodge area to the top of Giffin's Gallop is 1174.70 m + 40 m = 1214.70 m

Finally, using the tangent function, we can find the angle of elevation of the lift from the lodge area to the top of Giffin's Gallop:

θ = tan^-1(vertical distance / horizontal distance) = tan^-1(1214.70 m / 1789.29 m) = tan^-1(0.67967)

θ = 37.90° (approximately)

So, the angle of elevation of the lift from the lodge area to the top of Giffin's Gallop is approximately 37.90°.

Step-by-step explanation:

The balance (future value) after 6 years of investing $100 at 9% compounded semi-annually is B. $169.58.

The future value represents the present value or investment compounded at an interest rate for a period.

Compounding involves computing interest on accumulated interest, not only on the principal like simple interest.

The future value can be determined using an online finance calculator as follows:

N (# of periods) = 12 semi-annual periods (6 x 12)

I/Y (Interest per year) = 9%

PV (Present Value) = $100

PMT (Periodic Payment) = $0

Results:

Future Value (FV) = $169.58

Total Interest = $69.58

Thus, the investment is expected to be worth B. $169.58 in 6 years.

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Answer:

[tex]\frac{1}{4}[/tex].

Step-by-step explanation:

[tex]\frac{4}{16}[/tex] is not in simplest form, so that will be our first step.

[tex]4[/tex] and [tex]16[/tex] are both have factors/are divisible by [tex]2[/tex] and [tex]4[/tex].

In order to get the fraction in simplest form fastest, divide by the GCF. (Greatest Common Factor)

Therefore, divide the Numerator, (top number) and Denominator, (bottom number) by [tex]4[/tex].

[tex]4[/tex] ÷ [tex]4[/tex] [tex]= 1[/tex].

[tex]16[/tex] ÷ [tex]4[/tex] [tex]= 4[/tex].

Therefore, the answer is [tex]\frac{1}{4}[/tex].

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The graph of the system of inequalities is in the image at the end.

Here we have a system of inequalities, first we want to solve them:

-x + 3y ≤ 3

-2x + y ≥ -2

Isolating y in both of these we will get:

y  ≥ -2 + 2x

y ≤ (3 + x)/3

So we just needto graph the two lines, on the first one we will shade the region above the line and on the second one we will shade the region below the line.

The graph of the system is on the image below.

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The missing statement from step 4 is that the angles OAKRV, OASVT, OARQP, and OAPQR are congruent.

Congruent is a term used in mathematics to describe two figures or objects that have the same shape and size. This means that the two objects have identical angles and sides, so they can be perfectly superimposed on each other.

This can be concluded using the Angle-Angle-Side (AAS) Theorem, which states that if two triangles have two angles and a side in common, the triangles are similar. Since the angles OAKRV, OASVT, OARQP, and OAPQR are the corresponding angles of the similar triangles AVKR and APQR, the reason that completes step 4 is the converse of the Isosceles Triangle Theorem, which states if two sides of a triangle are congruent, then the angles opposite those sides are congruent.

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Answer: Suppose there exists a positive rational number r such that r^2 < 2. Then we have 2 - r^2 > 0. Let h = (2 - r^2)/4. Then h > 0 because r^2 < 2.

Consider the number rh = r + h. We have:

(rh)^2 = (r + h)^2 = r^2 + 2rh + h^2 = r^2 + 2(2 - r^2)/2 + (2 - r^2)/16

= r^2 + 2 + (2 - r^2)/16

< 2 + 2 + (2 - 2)/16 = 2.

Thus, for any positive rational number r such that r^2 < 2, there exists a larger positive rational number rh = r + h such that (rh)^2 < 2.

Step-by-step explanation:

The 90% confidence interval is [64.21, 73.79].

To find the margin of error and confidence interval,

we can use the following formula:

Margin of error [tex]= z* (sigma / \sqrt{(n))[/tex]

where z* is the z-score corresponding to the desired confidence level, sigma is the population standard deviation, n is the sample size, and sqrt represents the square root.

Since we want a 90% confidence interval, the corresponding z-score can be found using a standard normal distribution table or calculator.

The z-score for a 90% confidence level is 1.645.

Plugging in the values we have:

Margin of error = 1.645 * (5 / sqrt(7))

≈ 4.05

So the margin of error is approximately 4.05 inches.

To find the confidence interval, we need to add and subtract the margin of error from the sample mean:

Lower bound = x¯ - margin of error

= (68+63+70+69+75+63+72) / 7 - 4.05

≈ 64.21

Upper bound = x¯ + margin of error

= (68+63+70+69+75+63+72) / 7 + 4.05

≈ 73.79

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She has completed 3/4 of her walk, she has walked 3/5 miles.

We are given that Sonia takes 4/5 mile walk everyday.

A proportion is a fraction of a total amount, the relations between variables, could be direct or inverse proportional, can be built to find the desired measures in the problem.

Consider 4/5 represent the 100 percent.

Using the proportion we can find which part of her walk she completed once she has walked 3/5 miles.

Find out what percentage represent 3/5

Let x be the percentage that represent 3/5

Then convert to fraction number

4/500 = 3/5x

x = 75

Therefore, the value of x in percentage is 75 percent.

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The complete question is

Sonia takes a 4/5 mile walk every day. What part of her walk has she completed once she has walked 3/5 miles

The solution of the expression 2 x 2[tex]\frac{1}{5}[/tex] is 4.4.

Multiplication is a type of mathematical operation. The repetition of the same expression types is another aspect of the practice.

For instance, the expression 2 x 3 indicates that 3 has been multiplied by two.

Given:

Two fractions are 2 and 2[tex]\frac{1}{5}[/tex].

To find the product of two fractions:

Applying multiplication operation,

we get,

2 x 2[tex]\frac{1}{5}[/tex]

To simplify further;

Converting mixed fractions to improper fractions,

we get,

2 x 11/5

= 22/5

= 4.4

Therefore, 2 x 2[tex]\frac{1}{5}[/tex] = 4.4.

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Answer:

I believe the answer would be A- $69. I just do 230-30% and I got the answer of 69.

I hope this helped!

Answer:

Option 1. A

Step-by-step explanation:

= (30/100)*230

= (30*230)/100

= 6900/100 = 69.

Thus, your answer is, A.

( There is nothing else I could explain to you! )

Ascending order is a term that applies to an ordered set of observations from smallest to largest.

This means that the values are arranged in order of increasing magnitude, starting from the lowest value. The cumulative relative frequency is the sum of the relative frequencies for all values that are less than or equal to the given value. This is calculated by adding the relative frequencies of each value from the lowest value up to the given value, and then dividing the sum by the total number of observations. This helps us to determine the percentage of observations that lie below or equal to the given value. For example, given a set of five values, the cumulative relative frequency of the third value would be calculated by adding the relative frequencies of the first two values and then dividing by 5.

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The system of equation that does not belong to the other two is the third system of equations; y = -4·x - 3, y + 4·x = -5, This is so because, the third system has no solutions.

A linear system of equations consists of two or more linear equations that consists of common variables.

The possible equations are;

y = x + 6, y = -x + 2

3·x + y = -1, y = 4·x + 6

y = -4·x - 3, y + 4·x = 5

Evaluation of the system of equations, we get;

First system of equations;

y = x + 6, y = -x + 2

x + 6 = -x + 2

x + x = 2 - 6 = -4

2·x = -4

x = -4/2 = -2

x = -2

y = x + 6

y = -2 + 6 = 4

y = 4

The solution is; x = -2, y = 4

Second system of equation;

3·x + y = -1, y = 4·x + 6

3·x + 4·x + 6 = -1

7·x + 6 = -1

7·x  = -1 - 6 = -7

x = -7/7 = -1

x = -1

y = 4·x + 6

y = 4 × (-1) + 6 = 2

y = 2

The solution to the second system of equation is; x = -1, y = 2

Third system of equation;

y = -4·x - 3, y + 4·x = 5

y + 4·x = 5

-4·x - 3 + 4·x = 5

-4·x + 4·x - 3 = 5

0 - 3 = 5

-3 = 5

The third system of equation has no solution

The system of equations that does not belong with the other two is the third system of equation; y = -4·x - 3, y + 4·x = 5, that has no solution.

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The size of sample are both large enough 40 and 35. The mean and standard deviation of given data is 0.05 and 0.105 respectively.

The shape of the sampling distribution of [tex]\hat{p}_j - \hat{p}_s[/tex]is approximately normal, according to the Central Limit Theorem. This is because the sample sizes are both large enough (40 and 35, respectively) and the population proportions are unknown but assumed to be independent.

The mean of the sampling distribution is the difference in the population proportions of females, which is 0.54 - 0.49 = 0.05.

The standard deviation of the sampling distribution can be calculated as:

[tex]$\sqrt{\frac{\hat{p}_j(1 - \hat{p}_j)}{n_j} + \frac{\hat{p}_s(1 - \hat{p}_s)}{n_s}}$[/tex]

where [tex]\hat{p}_j = 0.54$, $n_j = 40$, $\hat{p}_s = 0.49$, and $n_s = 35$.[/tex]Plugging in these values, we get:

[tex]$\sqrt{\frac{0.54(1 - 0.54)}{40} + \frac{0.49(1 - 0.49)}{35}} \approx 0.105$[/tex]

Interpretation: The standard deviation of the sampling distribution tells us how much we can expect the sample proportion difference to vary across different random samples. In this case, we can expect the difference between the sample proportions of females in the junior and senior classes to vary by about 0.105 on average across different samples.

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The number of quarters is 2 and mass of quarter is 0.006 Kg.

With scientific notation, one can express extremely big or extremely small values. If a number between 1 and 10 is multiplied by a power of 10, the result is written in scientific notation.

Given:

The number of quarters minted in 2018 was about 2 x 10°

and, The mass of a quarter is about 6 X [tex]10^{-3[/tex] kg.

As, in standard form the decimal is place after one significant and the rest raised to 10 to the powers and if the number are in power then it can be written in form of decimal.

Here, number of quarters minted is already in standard form 2 x 10° or 2 x 1

and,  mass of a quarter can be written as

= 6 X [tex]10^{-3[/tex] k

= 6/ 10³

= 6/ 1000

= 0.006

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(a) For no solution, we need the two equations to be inconsistent, which means that they cannot be satisfied simultaneously. We can achieve this by making the first equation a multiple of the second equation:

4(x1 + hx2) = 8

4x1 + 4hx2 = 8

4x1 + 8x2 = k

Now, we can see that the second equation is not compatible with the first equation since they imply contradictory statements:

4x1 + 8x2 = k and 4x1 + 4hx2 = 8

(b) For a unique solution, we need the two equations to be independent, which means that they are not multiples of each other. We can achieve this by choosing different coefficients for x1 and x2 in the two equations.

x1 + hx2 = 2 and 4x1 + 8x2 = k

To find the values of h and k that give a unique solution, we can solve the system by elimination or substitution. For example, we can multiply the first equation by 4 and subtract it from the second equation:

4x1 + 8x2 = k

-4x1 - 4hx2 = -8

Simplifying and dividing by -4, we get:

x2 = (2 + h)/2

x1 = (k - 4x2)/4

Since x1 and x2 are expressed in terms of h and k, we can choose any values of h and k that satisfy these equations, and the system will have a unique solution.

(c) For many solutions, we need the two equations to be dependent, which means that they are multiples of each other or one is a linear combination of the other. We can achieve this by making the second equation a multiple of the first equation:

x1 + hx2 = 2

4(x1 + hx2) = 8 + 4hkx2

4x1 + (4h - k)x2 = 8

Now, we can see that the second equation is a linear combination of the first equation, so the system has infinitely many solutions. To find the solutions, we can choose any value of x2 and solve for x1 in terms of x2:

x1 = (8 - (4h - k)x2)/4

Since x1 and x2 are expressed in terms of h and k, we can choose any values of h and k that satisfy the equation 4h - k = 0, and the system will have many solutions.

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The correct students in their approach to solving the problem are Aditi, Erick, and Raj. Nico and Sandra's approaches are incorrect.

The following are the correct approaches to solving the equation:

Aditi: two-fifths (5 + x) = eight. five-hundredths (5) + five-hundredths x = (five-halves) 8

Erick: two-fifths (5 + x) equals eight. two-fifths (5 + x) minus 5 equals eight minus five.

Raj: two-fifths (5 + x) = eight. two-fifths (5) + two-fifths x = eight.

As a result, Aditi, Erick, and Raj are the correct students in their approach to solving the problem. The approaches of Nico and Sandra are incorrect.

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The probability of the continuous random variable X taking on a value greater than one is e⁻¹, which is approximately 0.368.

In probability theory, random variables are used to represent uncertain events. A continuous random variable is a variable that can take any value within a given range of values. The probability density function (PDF) of a continuous random variable is a function that describes the likelihood of the variable taking on a certain value. In this question, we are given the PDF of a continuous random variable and asked to find the probability of the variable taking on a value greater than one.

The given PDF is

=> f(x) = e⁻ˣ, 0 < x < ∞.

To find P{X > 1}, we need to integrate the PDF from 1 to infinity.

P{X > 1} = ∫(1 to ∞) e⁻ˣ dx

Using integration by parts, we get:

P{X > 1} = [-e⁻ˣ](1 to ∞)

= lim t → ∞ [-e⁻ˣ + e⁻¹)]

= e⁻¹

In conclusion, the probability density function of a continuous random variable is used to describe the likelihood of the variable taking on a certain value. In this question, we used the PDF to find the probability of the continuous random variable taking on a value greater than one. By integrating the PDF, we obtained the probability to be e⁻¹, which is approximately 0.368.

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Complete Question:

A continuous random variable X has a probability density function f ( x ) = e⁻ˣ , 0 < x < ∞ . Then P{X > 1} is

The probability that a data point falls within one standard deviation of its mean value is 0.6827 or 68.27%.

The normal distribution is a continuous probability distribution that describes the distribution of a set of data around its mean. It is often represented by a bell-shaped curve, where the mean is located at the center and the standard deviation determines the width of the curve.

To find the probability that a data point falls within one standard deviation of the mean, we need to calculate the area under the normal distribution curve between the mean minus one standard deviation and the mean plus one standard deviation. This area represents the proportion of the data that falls within one standard deviation of the mean.

The probability of a data point falling within one standard deviation of the mean can be calculated using the following formula:

P(x - s < X < x + s) = 0.6827

Where P represents the probability, X represents the variable we are measuring, x represents the mean value, and s represents the standard deviation.

The value 0.6827 represents the proportion of the data that falls within one standard deviation of the mean in a normal distribution.

This means that if we have a data set with a mean value of 50 and a standard deviation of 5, then there is a 68.27% chance that a randomly chosen data point will fall between 45 and 55.

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The equation of tangent to the curve y = 8√x  at the point (4,16) is  y = 2x + 8 .

We have to find equation of tangent line to the curve y = 8√x at the point (4, 16), we first find the slope ;

So , slope of the tangent line is  derivative of function y = 8√x at point (4, 16).

which means :  y' = 4[tex]x^{-\frac{1}{2} }[/tex]  ;

At the point (4, 16), the value of x is 4.

So , y' = 4 × [tex]4^{-\frac{1}{2} }[/tex] = 2 .

By Using the point slope form, the equation of the tangent line is ;

⇒ y - 16 = (2)(x - 4)  ;

Simplifying this equation, we get:

⇒ y - 16 = 2x - 8

⇒ y = 2x + (16 - 8)

⇒ y = 2x + 8 .

Therefore, the equation of the tangent line to the curve is y = 2x + 8 .

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The given question is incomplete , the complete question is

Find an equation for the tangent to the curve y = 8√x  at the point (4,16) .

The object will hit the ground in 2 seconds.

Expression in maths is defined as the collection of numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

The distance d an object falls after t seconds is given by d = 16t²

To detemine the height of an object launched upward from ground level at a rate of 32 feet per second, use the expression 32t - 16t^2, where t is the time in seconds.

Therefore, put h = 0 in the equation;

0 = 32t - 16t²

16t² = 32t

16t = 32

t = 2

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(a) The rate of change of the brightness after t days is dB/dt = (2π/4.4) * 0.45 * cos(2πt/4.4).

(b) The rate of increase after one day is  0.22 radians/day.

For the given case the equation for the brightness of cepheid joe is  B(t)=4.2 +0.45sin(2pit/4.4). This equation tells us that the brightness of the star is determined by the sine of the time multiplied by a constant. Since the sine of a number is always changing, the brightness of the star is always changing too.

Therefore, the rate of change of the brightness is given by the derivative of the equation, which is   2π/4.4)*0.45*cos(2πt/4.4), when t is  1 day, we can plug this value into the equation to get the rate of increase after one day, which is dB/dt = (2π/4.4) * 0.45 * cos(2π/4.4) = 0.22 radians/day.

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a3 is greater than or equal to a4, then the subsequence a1, a2, a4 would form a monotonically increasing 3-chain. Hence, a3 must be less than a4.  If a1 < a5 < a4, then the subsequence a1, a4, a5 would form a monotonically increasing 3-chain any value of a5 results in a 3-chain.any sequence of five distinct integers must contain a 3-chain.

(a) Assume that a1 < a2 and there is no 3-chain in our sequence. Then, a3 cannot be greater than or equal to a2 (otherwise, the subsequence a1, a2, a3 would form a monotonically increasing 3-chain). Similarly, a3 cannot be less than or equal to a2 (otherwise, the subsequence a3, a2, a1 would form a monotonically decreasing 3-chain). Therefore, a3 must be strictly between a1 and a2. Now, if a3 is greater than or equal to a4, then the subsequence a1, a2, a4 would form a monotonically increasing 3-chain. Hence, a3 must be less than a4.

(b) From part (a), we know that a3 < a1. Also, since there is no 3-chain, a3 < a4 < a2. Combining these inequalities, we get a3 < a4 < a2 and a3 < a1. Hence, a3 < a4 < a2 < a1.

(c) Assume that a1 < a2 and a3 < a4 < a2. If a5 is less than a4, then the subsequence a3, a4, a5 would form a monotonically decreasing 3-chain. If a5 is greater than a2, then the subsequence a2, a5, a4 would form a monotonically decreasing 3-chain. If a4 < a5 < a2, then the subsequence a3, a4, a5 would form a monotonically increasing 3-chain. If a1 < a5 < a4, then the subsequence a1, a4, a5 would form a monotonically increasing 3-chain. Therefore, any value of a5 results in a 3-chain.

(d) Assume that there is a sequence of five distinct integers with no 3-chain. Without loss of generality, we can assume that a1 < a2. From part (a), we know that a3 < a1. From part (b), we know that a3 < a4 < a2 < a1. From part (c), we know that any value of a5 results in a 3-chain. Therefore, we have a contradiction and our assumption is false. Hence, any sequence of five distinct integers must contain a 3-chain.

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dydt = k(y-5)(y-0) where k is a positive constant. This equation has equilibrium solutions of y=0 and y=5, and y' is positive for 0<y<5 and negative for y<0 or y>5.

We can solve this problem by rearranging the equation to separate the y terms from the y' terms. We can do this by factoring the y terms on the left side, and then writing the equation as y' = k(y-5)(y-0). This equation has an equilibrium solution at y=0 and y=5, and the sign of y' depends on the sign of k. Since k is a positive constant, y' is positive when 0<y<5, and y' is negative when y<0 or y>5.  Thus, the equation dydt = k(y-5)(y-0) meets all of the given criteria. Therefore, dydt = k(y-5)(y-0) where k is a positive constant. This equation has equilibrium solutions of y=0 and y=5, and y' is positive for 0<y<5 and negative for y<0 or y>5.

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