A triangle is placed in a semicircle with a radius of 5 mm, as shown below. Find the area of the shaded region. Use the 3.14 value for pi, and do not round your answer. Be sure to include the correct unit in your answer.

The value of particular solution to the nonhomogeneous differential equation is,

⇒ y (p) = 2x + 1/2 e⁻ˣ - 8/5

We have to given that;

The nonhomogeneous differential equation is,

⇒ y'' + 4y' + 5y = 10x + e⁻ˣ  . (i)

To find homogeneous solution,

D² + 4D + 5 = 0

(D + 2)² = - 1

D + 2 = ±i

D = 2 ± i

Hence, We get;

y = e⁻²ˣ (c₁ cos x + c₂ sin x)  .. (ii)

To find the particular solution,

y (p) = A + Bx + Ce⁻ˣ

y' (p) = B - Ce⁻ˣ

y'' (p) = Ce⁻ˣ

Substitute all the values in (i);

⇒ y'' + 4y' + 5y = 10x + e⁻ˣ

⇒ Ce⁻ˣ + 4(B - Ce⁻ˣ) + 5(A + Bx + Ce⁻ˣ) = 10x + e⁻ˣ

Equating the coefficient;

A = 2

B = - 8/5

C = 1/2

So, We get;

⇒ y (p) = 2x + 1/2 e⁻ˣ - 8/5

The value of particular solution to the nonhomogeneous differential equation is,

⇒ y (p) = 2x + 1/2 e⁻ˣ - 8/5

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The required two way frequency table is shown below.

We know that a two-way frequency table is nohting but the way to display frequencies for two different categories collected from a single group of people.

While making the two-way frequency tables first we need to identify the two variables of interest. Then we need to determine the possible values of each variable. Select a variable to be represented by the rows and the other to be represented by the columns. And then complete the table with frequencies.

Here we have two variables as:

Time ( practiced for atleast 10,000 hours and did not practiced for atleast 10,000 hours) and the second (won a prize and do not won a major)

Based on the information provided, we can obtain a two way frequency table as shown below:

                                   practiced for atleast          did not practiced for

                                       10,000 hours               atleast 10,000 hours

have won a major                  9                                      1

haven't won a major              1                                       2

Thus the required two-way frequency table is shown in attached figure.

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Find the complete question below.

The patterns of the sum and difference of two cubes can be used to factorize polynomial expressions. To utilize these patterns, we need to identify if the polynomial expression we want to factorize can be written in the form of a sum or difference of two cubes, and then apply the corresponding pattern to factorize it.

The sum and difference of two cubes are useful patterns that can be used to factorize polynomial expressions. To utilize these patterns, we need to identify if the polynomial expression we want to factorize can be written in the form of a sum or difference of two cubes. The sum of two cubes can be expressed as:

a³ + b³ = (a + b)(a² - ab + b²)

And the difference of two cubes can be expressed as:

a³ - b³ = (a - b)(a² + ab + b²)

To use these patterns, we need to look for polynomials in the form of a³ + b³ or a³ - b³, where a and b are integers or algebraic expressions. If we find such expressions, we can factorize them using the corresponding pattern.

For example, let's consider the polynomial expression x³ + 8. This can be written in the form of a sum of two cubes, where a = x and b = 2:

x³ + 8 = x³ + 2³

Now we can use the sum of two cubes pattern to factorize the expression:

x³ + 2³ = (x + 2)(x² - 2x + 4)

Similarly, if we have an expression in the form of a³ - b³, we can use the difference of two cubes pattern to factorize it. For example, let's consider the expression y³ - 27:

y³ - 27 = y³ - 3³

We can use the difference of two cubes pattern to factorize this expression:

y³ - 3³ = (y - 3)(y² + 3y + 9)

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How will you utilize the patterns in the sum and difference of two cubes in any case?

The answer is true. A random variable is a variable whose value is subject to variations due to chance. The variance of a random variable measures how spread out its values are.

It is a measure of the average distance between the values of the variable and its expected value. In this case, V(cX) represents the variance of a new random variable obtained by multiplying X by a constant c. According to the properties of variance, V(cX) = c^2 V(X). Therefore, the equation V(cX) = 4V(X) can be rewritten as c^2 V(X) = 4V(X).
Dividing both sides of the equation by V(X), we get c^2 = 4. Taking the square root of both sides, we obtain c = 2 or c = -2. However, since c represents a scaling factor, we can disregard the negative solution. Therefore, c must be 2.
In conclusion, if V(cX) = 4V(X), then c must be 2. This result shows that multiplying a random variable by a constant affects its variance by the square of that constant.

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We need to use about 10 terms to estimate the value of the series with an error at most 0.000017.

To estimate the value of the convergent series 9 + 15 + ... with an error at most 0.000017, we need to use the formula for the error bound of a convergent series:

|En| ≤ (Mn+1/2) * r^n

where En is the error bound, Mn is the maximum value of the remainder term for the first n terms of the series, r is the common ratio, and n is the number of terms used to estimate the series.

In this case, the series has a common ratio of 5/3 (since each term is 5/3 times the previous term), and the remainder term for the first n terms is:

Rn = (5/3)^n * 9/(3n+3)

To find Mn, we need to find the maximum value of Rn for n terms. This can be done by taking the derivative of Rn with respect to n, setting it equal to zero, and solving for n. However, since we only need an estimate of the number of terms, we can use trial and error to find the smallest n such that Rn ≤ 0.000017:

n = 10: R10 ≈ 0.000013
n = 11: R11 ≈ 0.000021


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We would expect about 979 babies to weigh between 5 and 9 pounds.

We would expect about 1063 babies to weigh less than 8 pounds.

We would expect about 720 babies to weigh more than 7 pounds.

We would expect about 692 babies to weigh between 6.9 and 10 pounds.

We have,

We can use the normal distribution to answer these questions.

1)

To find the number of babies expected to weigh between 5 and 9 pounds, we need to find the area under the normal curve between these two values.

The z-scores for the lower and upper bounds are:

z1 = (5 - 6.9) / 2 = -0.95

z2 = (9 - 6.9) / 2 = 1.05

The area between these z-scores is approximately 0.653.

Now,

= 0.653 x 1500

= 979.5

So we would expect about 979 babies to weigh between 5 and 9 pounds.

2)

To find the number of babies expected to weigh less than 8 pounds, we need to find the area under the normal curve to the left of this value.

The z-score for 8 pounds.

z = (8 - 6.9) / 2 = 0.55

The area to the left of this z-score is approximately 0.7088.

To get the actual number of babies, we need to multiply this proportion by the total number of babies:

= 0.7088 x 1500

= 1063.2

So we would expect about 1063 babies to weigh less than 8 pounds.

3)

To find the number of babies expected to weigh more than 7 pounds, we need to find the area under the normal curve to the right of this value.

The z-score for 7 pounds.

z = (7 - 6.9) / 2 = 0.05

The area to the right of this z-score is approximately 0.4801.

So,

= 0.4801 x 1500

= 720.15

So we would expect about 720 babies to weigh more than 7 pounds.

4)

To find the number of babies expected to weigh between 6.9 and 10 pounds, we can use the z-scores for these values:

z1 = (6.9 - 6.9) / 2 = 0

z2 = (10 - 6.9) / 2 = 1.55

The area between these z-scores is approximately 0.4616.

So,

0.4616 x 1500 = 692.4

So we would expect about 692 babies to weigh between 6.9 and 10 pounds.

Thus,

We would expect about 979 babies to weigh between 5 and 9 pounds.

We would expect about 1063 babies to weigh less than 8 pounds.

We would expect about 720 babies to weigh more than 7 pounds.

We would expect about 692 babies to weigh between 6.9 and 10 pounds.

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It will take about 3 years for the bird population to reach 7,200, assuming exponential growth. Assuming exponential growth, we can use the formula N = N0 x (1+r)^t, where N is the final population, N0 is the initial population, r is the annual growth rate, and t is the time in years.

In this case, we know that N0 = 900 and N = 7,200. We can find the annual growth rate, r, by using the fact that the population doubled in one year.
If the population doubles in one year, then the growth rate is 100%. So r = 1.
Now we can plug in the values we know and solve for t:
7,200 = 900 x (1+1)^t
Dividing both sides by 900:
8 = 2^t
Taking the logarithm of both sides:
log(8) = t x log(2)
Solving for t:
t = log(8) / log(2)
t ≈ 3
So it will take about 3 years for the bird population to reach 7,200, assuming exponential growth.

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The limit of (t) as t approaches infinity is (infinity, undefined, 0). The limit of (t) cannot be evaluated for all values of t.

To find the limit of (t), we need to evaluate it as t approaches some value. Let's first simplify the expression inside the parentheses:
14 (t*cos(3), t4 t cost t + tan(3t)'1- Vt+1 Voti) = (14t*cos(3), t^5 cos(t) + t^4 tan(3t), sqrt(t+1) - sqrt(t))

Now, we can evaluate the limit as t approaches some value. Let's evaluate it as t approaches infinity:

lim (t) as t approaches infinity = (lim 14t*cos(3) as t approaches infinity, lim t^5 cos(t) + t^4 tan(3t) as t approaches infinity, lim sqrt(t+1) - sqrt(t) as t approaches infinity)

Since cosine function oscillates between -1 and 1, and t is growing to infinity, the second term in the limit above will become infinitely large and oscillatory. Therefore, it does not have a limit as t approaches infinity.

The first and third terms, however, can be evaluated. As t approaches infinity, t*cos(3) approaches infinity as well. And since the difference between sqrt(t+1) and sqrt(t) is infinitesimal compared to t, we can approximate it as 1/2sqrt(t), which approaches 0 as t approaches infinity.

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The initial value problem. y'(t) = 1 + e^t, y(0) = 20 The specific solution is y(t)= t + e^t + 19.

Let's go step-by-step:
1. Identify the problem: We are given a differential equation y'(t) = 1 + e^t and an initial value y(0) = 20.

2. Integrate the differential equation: To find y(t), we need to integrate the given equation with respect to t.
  ∫(y'(t) dt) = ∫(1 + e^t dt)

3. Perform the integration: After integrating, we obtain the general solution of the problem:
  y(t) = t + e^t + C, where C is the constant of integration.

4. Apply the initial value: We are given y(0) = 20, so we can plug this into the general solution to find the specific solution.
  20 = 0 + e^0 + C
  20 = 1 + C

5. Solve for the constant of integration C: From the above equation, we find the value of C.
  C = 19

6. Write the specific solution: Now that we have the value of C, we can write the specific solution for y(t).
  y(t) = t + e^t + 19

So, the specific solution for this initial value problem is y(t) = t + e^t + 19.

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The cutoff score for the top 10% of students is approximately 644.

We have,

To find the cutoff score for the top 10%, we need to calculate the z-score that corresponds to the top 10% of the distribution.

Using a standard normal distribution table or a calculator, we can find that the z-score corresponding to the top 10% is approximately 1.28.

We can use the formula for the z-score:

z = (x - μ) / σ

where z is the z-score, x is the score we want to find, μ is the mean, and σ is the standard deviation.

Substituting the value.

1.28 = (x - 503) / 109

Multiplying both sides by 109.

140.52 = x - 503

Adding 503 to both sides.

x = 643.52

Therefore,

The cutoff score for the top 10% of students is approximately 644.

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There are a total of (10 choose 5) possible ways to randomly group the 10 kids into two teams of 5. The probability that Alex and Jose end up on the same team is 7/31 or approximately 0.2258 (rounded to 4 decimal places). This is because we are choosing 5 kids out of 10 for one team, and the remaining 5 kids automatically make up the other team.

To calculate the probability of Alex and Jose ending up on the same team, we can think of it as choosing 3 more kids to be on their team out of the remaining 8 kids. There are (8 choose 3) ways to do this. Therefore, the probability of Alex and Jose ending up on the same team is:
(8 choose 3) / (10 choose 5) = 0.357 or approximately 35.7%
So there is a 35.7% chance that Alex and Jose will end up on the same team when the 10 kids are randomly grouped into an A team and a B team.
Since the 10 kids are randomly grouped into two teams, we can use combinations to determine the possible groupings. The total number of ways to divide the kids into two groups of 5 is given by the combination formula:
Total groupings = C(10, 5) = 10! / (5! * 5!) = 252
Now, let's consider the groupings where Alex and Jose are on the same team. There are 8 other kids left, and we need to select 3 of them to complete the team of 5. So, the number of groupings with Alex and Jose together is given by:
Groupings with Alex and Jose together = C(8, 3) = 8! / (3! * 5!) = 56
Finally, we can find the probability of Alex and Jose being on the same team by dividing the number of groupings with them together by the total groupings:
Probability = (Groupings with Alex and Jose together) / (Total groupings) = 56 / 252 = 7/31
So, the probability that Alex and Jose end up on the same team is 7/31 or approximately 0.2258 (rounded to 4 decimal places).

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Both methods give us the same answer, which is -1/2. In terms of which order is superior, it really depends on the integrand and the region of integration.

To evaluate the integral ∫ r xcos(xy) da where r = [0,π] ×[1,2], we can use either the order dxdy or dydx. Using the order dxdy, we have:
∫ r xcos(xy) da = ∫π0 ∫21 xcos(xy)dydx
Integrating with respect to y first, we have:
∫ r xcos(xy) da = ∫π0 [sin(2x)-sin(x)]dx
Using the order dydx, we have:
∫ r xcos(xy) da = ∫21 ∫π0 xcos(xy)dxdy
Integrating with respect to x first, we have:
∫ r xcos(xy) da = ∫21 [-cos(2y)+cos(y)]dy
Sometimes one order may be easier to work with than the other. However, Fubini's Theorem tells us that the answer should not depend on the order of integration as long as the integral is well-defined. The moral of the story is to always check both orders of integration and use the one that is easier or more convenient for the given problem.

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The standard deviation of wait time is 13.8564.

The length of time patients must wait to see a doctor in a local clinic is uniformly distributed between 25 minutes and 73 minutes. We have to find the standard deviation of the wait time.

The square root of the variance of a random variable, sample, statistical population, data collection, or probability distribution is its standard deviation.

The standard deviation in statistics is a measure of the degree of variation or dispersion in a set of values.

A low standard deviation implies that the values are close to the set's mean, whereas a high standard deviation shows that the values are spread out over a larger range.

S² = (73 - 25)²/12

S² = (48)²/12

S² = 192

S = √192

S = 13.8564

Hence, The standard deviation of wait time is 13.8564.

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complete question:

the length of time patients must wait to see a doctor in a local clinic is uniformly distributed between 25 minutes and 73 minutes. what is the standard deviation of wait time? group of answer choices

Answer:

I believe 0.143

Step-by-step explanation:

Well the chance is 2 out of 14 so 2/14 then you reduce that and get 1/7 equals 0.143. I may have done that wrong

Both sections cover important mathematical concepts with challenging aspects but also offer numerous applications for various professional fields and everyday life.

In Section 1, computing partial derivatives algebraically can be tricky and frustrating because it involves using the chain rule, product rule, and quotient rule in complex functions. However, it can also be exciting to see how these rules can be applied to find rates of change in multivariable functions.

Two things that could be useful as a scientist, engineer, mathematician or in your personal life from this section are:
1. Understanding partial derivatives can help in optimizing systems in engineering and science.
2. Partial derivatives can also be used in finance to calculate sensitivity analysis in portfolio management.

In Section 2, local linearity and the differential can be difficult to grasp because it involves understanding the tangent plane of a surface and how it approximates the surface near a point. However, it can be exciting to see how this concept can be applied to approximating solutions to nonlinear equations.

Two things that could be useful as a scientist, engineer, mathematician or in your personal life from this section are:
1. Local linearity can be used in computer graphics to render 3D objects.
2. The differential can be used in physics to calculate small changes in variables in differential equations.

Section 1: Computing Partial Derivatives Algebraically
1. Tricky aspects:
  a. Differentiating with respect to one variable while treating other variables as constants can be challenging, especially in functions with multiple variables.
  b. Applying the chain rule for partial derivatives may be confusing for some due to the interplay of different variables.

2. Applications:
  a. Scientists and engineers use partial derivatives to model and understand how different parameters affect complex systems.
  b. Mathematicians use partial derivatives in optimization problems to find the maxima or minima of multivariable functions.

Section 2: Local Linearity & The Differential
1. Tricky aspects:
  a. Understanding the concept of local linearity and how it connects to differentiability can be challenging for some learners.
  b. Applying differentials to approximate changes in functions can be tricky due to the need to find the right balance between accuracy and simplicity.

2. Applications:
  a. Engineers use the concept of local linearity to analyze how systems behave under small changes and make approximations that simplify their calculations.
  b. In personal life, differentials can be used to estimate how small changes in one aspect, like the price of a product, might affect the overall cost.

In summary, both sections cover important mathematical concepts with challenging aspects but also offer numerous applications for various professional fields and everyday life.

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We can say with 95% confidence that the true proportion of people who have seen the video is between 0.841 and 0.897.

To create a 95% confidence interval for the proportion of people who have seen the video, we can use the following formula:

[tex]CI = \hat{p} \pm z*√((\hat{p}(1-\hat{p}))/n)[/tex]

where:

[tex]\hat{p}[/tex]  = sample proportion (352/405)

z = z-score for the desired confidence level (1.96 for 95% confidence interval)

n = sample size (405).

Plugging in the values, we get:

CI = 0.869 ± 1.96*√((0.869(1-0.869))/405)

CI = 0.869 ± 0.028

Rounding to three decimal places, we get:

CI = (0.841, 0.897).

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The sale price of the item after a 50% discount is $133.75.

To calculate a sale price after a 50% discount;

Find the cost after a 7% increase. To do this, we multiply the original cost by 1 + the percentage increase. In this case, the original cost is $250 and the percentage increase is 7%, so the cost after the increase is;

Cost after increase = $250 + 7% of $250

= $250 + 0.07 × $250

= $250 + $17.50

= $267.50

Find the sale price after a 50% discount. To do this, we multiply the cost after the increase by (1 - 50%), which is equivalent to multiplying by 0.5. So the sale price is;

Sale price = Cost after increase × (1 - 50%)

= $267.50 × 0.5

= $133.75

Therefore, the sale price of the item after a 50% discount is $133.75.

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The situation that is best modeled by a linear function is option D: the amount of interest earned for a year on a savings account earning 5.5% simple interest.

A linear function has a constant rate of change, which means that the output (dependent variable) changes by a constant amount for every unit change in the input (independent variable). In option D, the amount of interest earned on a savings account earning 5.5% simple interest is a linear function of the principal amount of the account. The rate of change is constant and equal to the interest rate, so the interest earned increases linearly with the principal amount.

In contrast, options A, B, and C all involve exponential growth or decay, which cannot be modeled by a linear function. Option A involves a decreasing value of a new automobile that depreciates 20% each year, which follows an exponential decay model. Option B involves the size of a culture of yeast that doubles in size every 20 minutes, which follows an exponential growth model. Option C involves an investment that earns compound interest, which also follows an exponential growth model.

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If the population is half women, then we can expect that half of the 408 people chosen would also be women. Therefore, we can expect 204 women to be in the sample if it was randomly drawn from the population.

To determine how likely it is that the sample has only 184 women or fewer, we need to use a statistical test. We can use a binomial distribution with n=408 and p=0.5 (since half the population is women). We want to find the probability of getting 184 women or fewer in the sample if it was randomly drawn from the population. Using a binomial calculator, we find that the probability of getting 184 women or fewer in the sample if it was randomly drawn from the population is 0.0036, or 0.36%. This means that if the sample truly was randomly drawn from the population, it would be very unlikely to get a sample with only 184 women or fewer. However, if the sample did have only 184 women or fewer, it could suggest that the sample was not truly randomly chosen and that there may be bias against women in the selection process. Further investigation would be needed to confirm this suspicion.

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The coefficient of determination, or r-squared, tells us the proportion of variance in the dependent variable that is explained by the independent variable(s). In this case, the value of r-squared being 0.61 means that 61% of the variance in the proportion of voters in the eastern and southern wards who vote for the Republican candidate for state congress can be explained by the relationship between the two variables.

In other words, there is a moderate-to-strong positive correlation between the proportion of Republican voters in the eastern and southern wards. However, it's important to note that correlation does not necessarily imply causation, and there may be other variables at play that influence voter preferences. Additionally, a coefficient of determination of 0.61 leaves 39% of the variance unexplained, so there may be other factors that contribute to voter preferences that are not captured in this particular relationship.

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The sum of the given arithmetic series is 6727.

What is arithmetic series?

The arithmetic series is the sequence of terms where the common difference remains constant between any two successive terms.  A sequence is a collection of numbers which follow a definite pattern. For example, the sequence 1, 5, 9, 13, … is an arithmetic sequence because here is a pattern where each number is obtained by adding 4 to its previous term.

a)

17+20+23+26+ ... + 200

This is in arithmetic progression.

First term (a₁)= 17

common difference (d)= 3

Let the nth term be aₙ

aₙ= a₁ + (n-1)×d

200= 17 + (n-1)×3

61= n-1

n= 62

Let the sum is Sₙ = n/2(2a+(n-1)×d)

                             = 62/2 ( 34+ 61×3)

                            = 31×217

                           = 6727.

Hence, the sum of the given arithmetic series is 6727.

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Therefore, one relief employee can cover 2 regular employees in a week by probability.

Assuming that each regular employee works for 5 days a week, and one relief employee is available to cover the remaining two days, we can calculate the number of regular employees that can be covered by one relief employee as follows:

One relief employee covers 2 days/week.

So, the number of regular employee days that one relief employee can cover in a week is:

2 days/week × 1 week = 2 days

Therefore, the number of regular employees that one relief employee can cover in a week is:

5 days/week ÷ 2 days = 2.5 regular employees

However, since we cannot have half of an employee, we round down to the nearest whole number.

Therefore, one relief employee can cover 2 regular employees in a week.

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The factorization of 81x² - 4 is (9x + 2)(9x - 2).

Factorization, also known as factoring, is the process of expressing a number or an algebraic expression as a product of two or more factors that are smaller than the original number or expression.

We can factorize 81x² - 4 by recognizing that it is a difference of squares, which can be factored as:

a² - b² = (a + b)(a - b)

In this case, a = 9x and b = 2, so we can write:

81x²- 4 = (9x + 2)(9x - 2)

Therefore, the factorization of 81x² - 4 is (9x + 2)(9x - 2).

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The equation given models the distance from the ground of a person riding on a ferris wheel. it takes 80 seconds for the Ferris wheel to make one revolution.

To determine how long it will take for the ferris wheel to make one revolution, we need to find the period of the function. The period is the amount of time it takes for the function to complete one full cycle.
In this case, the function is d = 30sin(pi/40t) + 20, where t is measured in seconds. The period of the function can be found using the formula T = (2pi)/b, where b is the coefficient of t in the argument of the sine function. In this case, b = pi/40, so T = (2pi)/(pi/40) = 80 seconds.
Therefore, it will take 80 seconds for the ferris wheel to make one full revolution. The answer is option C, 80 seconds.
The time it takes for a Ferris wheel to make one revolution can be determined using the given equation: d = 30 * sin((π/40) * t) + 20. In this equation, d represents the distance (in feet) of the person above the ground, and t represents the time in seconds.
A full revolution occurs when the angle inside the sine function completes a cycle of 2π radians. To find the time it takes for this to happen, we need to equate the angle (π/40) * t to 2π:
(π/40) * t = 2π
To solve for t, we can divide both sides of the equation by (π/40):
t = 2π * (40/π)
The π in both the numerator and denominator cancels out:
t = 2 * 40
t = 80 seconds
Therefore, it takes 80 seconds for the Ferris wheel to make one revolution.

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The solution is 5 divided by 1/4 is 20.

We have,

A mathematical arithmetic operation is a multiplication. Moreover, it is the practice of repeatedly adding the same expression kinds.

Example: 2 + 3 means that 2 is multiplied by 3 or that 3 is multiplied by 2 times.

Given:

A phrase: 5 divided by 1/4.

To solve a fraction division problem using reciprocals:

Let n be the required value of the quotient.

n = 5 ÷ 1/4

n = 5/ 1/4

To convert the division to multiplication:

Reverse the number in the denominator,

n = 5 x 4/1

n = 20

Therefore, the value of the quotient is 20.

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Since, function. f(t)= e3t, 0
Therefore, the Laplace transformation of f(t) = e^(3t) is F(s) = -1/(s-3).


To find the Laplace transformation of the function f(t) = e^(3t), we'll use the definition of the Laplace transform, which is:

L{f(t)} = F(s) = ∫(0 to infinity) e^(-st) * f(t) dt

Now, let's substitute f(t) = e^(3t) into the definition:

F(s) = ∫(0 to infinity) e^(-st) * e^(3t) dt

To simplify, combine the exponentials:

F(s) = ∫(0 to infinity) e^((3-s)t) dt

Now, we'll integrate with respect to t:

F(s) = (-1/(s-3)) * e^((3-s)t) | evaluated from 0 to infinity

When we evaluate the limit as t approaches infinity, we get:

lim (t→infinity) (-1/(s-3)) * e^((3-s)t) = 0, as long as s > 3 (since the exponent will be negative and the exponential term will go to 0)
Simplifying the expression inside the integral, we get:

F(s) = ∫0^∞ e^[(3-s)t] dt

Using the formula for integration of exponential functions, we get:

F(s) = [e^[(3-s)t]] / (3-s)  [evaluated from 0 to infinity]

Since e^(-∞) is equal to zero, the lower limit of the integral does not affect the value of F(s), so we get:

F(s) = [0 - 1/(3-s)] = -1/(s-3)
Now, let's evaluate the lower limit at t=0:

(-1/(s-3)) * e^((3-s)*0) = (-1/(s-3)) * e^0 = -1/(s-3)

So, the Laplace transform of f(t) = e^(3t) is:

F(s) = -1/(s-3), for s > 3

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The total weight of the liquid in the tank is 126295 pounds.

To calculate the total weight of the liquid in the tank, we need to first calculate the volume of the tank and then multiply it by the density of the liquid.

Given; Radius of the hemisphere (r) = 4 feet

Density of the liquid = 95 pounds per cubic foot

The formula for volume of a hemisphere is:

Volume = (2/3) × π × r³

Plugging in the given value of the radius (r):

Volume = (2/3) × π × (4 feet)³

Volume = (2/3) × π × 64 cubic feet

Next, we can multiply the volume by the density of the liquid to get the total weight of the liquid in the tank;

Total weight = Volume × Density

Plugging in the given value of the density:

Total weight = [(2/3) × π × 64 cubic feet] × 95 pounds per cubic foot

Total weight = 120160/3 × π pounds

Using the value of π as approximately 3.14 and rounding to the nearest full pound;

Total weight = 120160/3 × 3.14 pounds

Total weight = 126295.45 pounds

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a) The area of the first circle is approximately 254.34 square inches

b) The area of the second circle is approximately 70650 square inches.

a) The area of a circle can be calculated using the formula A = πr², where π (pi) is a mathematical constant approximately equal to 3.14, and r is the radius of the circle.

For the first circle with a diameter of 18 inches, we can find the radius by dividing the diameter by 2:

r = 18/2 = 9 inches

Now we can calculate the area using the formula:

A = πr² = 3.14 x 9² = 254.34 square inches

Therefore, the area of the first circle is approximately 254.34 square inches.

b) For the second circle with a diameter of 25 feet, we need to convert the diameter to inches, since our formula uses radius in inches:

25 feet = 25 x 12 inches = 300 inches

Then we can find the radius by dividing by 2:

r = 300/2 = 150 inches

Now we can calculate the area using the formula:

A = πr² = 3.14 x 150² = 70650 square inches

Therefore, the area of the second circle is approximately 70650 square inches.

Note that the units for the second calculation are in square inches, not square feet, because we used the formula that requires radius in inches.

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The area of the remaining wooded area is 2070 square meters.

To solve this problem, we need to first find the area of the entire trapezoid and then subtract the area of the promenade to get the remaining wooded area.

The formula for the area of a trapezoid is:

Area = (b1 + b2) * h / 2

where b1 and b2 are the lengths of the bases and h is the height (or width) of the trapezoid.

In this case, we are given that the bases measure 132 m and 96 m, and the width of the zone (which is the height of the trapezoid) is 30 m. So we can plug these values into the formula:

Area of trapezoid = (132 + 96) * 30 / 2 = 2280 square meters

Next, we need to find the area of the promenade, which is a rectangle with a width of 7 m and a length equal to the height of the trapezoid (30 m). So the area of the promenade is:

Area of promenade = 7 * 30 = 210 square meters

Finally, we can find the area of the remaining wooded area by subtracting the area of the promenade from the area of the trapezoid:

Area of remaining wooded area = 2280 - 210 = 2070 square meters

Therefore, the area of the remaining wooded area is 2070 square meters.

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Translated Question: A wooded area has the shape of a trapezoid, whose bases measure 132 m and 96 m. The width of the zone measures 30 m. A 7 m wide promenade is built perpendicular to the two bases. Calculate the area of ​​the remaining wooded area.​: