How to turn a fraction into a decimal or percent (percent has to written in fraction parts). A decimal into a fraction percent (still in fractions part). And Percents (still in fraction parts) into a decimal or fraction

If two methods agree perfectly in a method comparison study, the slope equals 1.0 and the y-intercept equals 0.0. Therefore, option (b) is the correct answer.

In a method comparison study, the goal is to compare the agreement between two different measurement methods or instruments. The relationship between the measurements obtained from the two methods can be described by a linear equation of the form y = mx + b, where y represents the measurements from one method, x represents the measurements from the other method, m represents the slope, and b represents the y-intercept.

When the two methods agree perfectly, it means that there is a one-to-one relationship between the measurements obtained from each method. In other words, for every x value, the corresponding y value is the same. This indicates that the slope of the line connecting the measurements is 1.0, reflecting a direct proportional relationship.

Additionally, when the two methods agree perfectly, there is no systematic difference or offset between the measurements. This means that the line connecting the measurements intersects the y-axis at 0.0, indicating that the y-intercept is 0.0.

Therefore, in a perfect agreement scenario, the slope equals 1.0 and the y-intercept equals 0.0, which corresponds to option (b).

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The solution to the initial value problem is:

[tex]y(t) = (-1/15)e^{(-t)} + (2/5)e^{(2t) }+ (2/15)e^{(-t/2)[/tex]

To solve this initial value problem using Laplace transform, we need to take the Laplace transform of both sides of the differential equation, apply initial conditions, and then solve for the Laplace transform of y. Once we have the Laplace transform of y, we can take its inverse Laplace transform to get the solution y(t).

Taking the Laplace transform of both sides of the differential equation yields:

2L{y'''} + 3L{y''} - 3L{y'} - 2L{y} = L{e^{-t}}

Applying the Laplace transform formulas for derivatives and using the initial conditions, we get:

[tex]2(s^3 Y(s) - s^2 y(0) - sy'(0) - y''(0)) + 3(s^2 Y(s) - sy(0) - y'(0)) - 3(sY(s) - y(0)) - 2Y(s) = 1/(s+1)[/tex]

Substituting y(0) = 0, y'(0) = 0, y''(0) = 1, and simplifying, we get:

[tex](2s^3 + 3s^2 - 3s - 2)Y(s) = 1/(s+1) + 2s[/tex]

Solving for Y(s), we get:

[tex]Y(s) = [1/(s+1) + 2s] / (2s^3 + 3s^2 - 3s - 2)[/tex]

We can now use partial fraction decomposition to express Y(s) in terms of simpler fractions:

Y(s) = [A/(s+1)] + [B/(2s-1)] + [C/(s-2)]

Multiplying both sides by the denominator and solving for A, B, and C, we get:

A = -1/15, B = 4/15, C = 2/5

Therefore, we have:

Y(s) = [-1/(15(s+1))] + [4/(15(2s-1))] + [2/(5(s-2))]

Taking the inverse Laplace transform of Y(s), we get the solution to the initial value problem:

[tex]y(t) = (-1/15)e^{(-t) }+ (2/5)e^{(2t) }+ (2/15)e^{(-t/2)[/tex]

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To solve this initial-value problem using the Laplace transform, we first take the Laplace transform of both sides of the equation. Applying the linearity and derivative properties of the Laplace transform, we get:

2L{y'''} + 3L{y''} - 3L{y'} - 2L{y} = L{e^(-t)}
Using the initial-value  conditions given, we can simplify this expression further:

2s^3Y(s) - 2s^2 - 3s - 2 = 1/(s+1)

Solving for Y(s), we get:

Y(s) = (1/(2s^3 - 3s^2 + 3s - 3)) * (1/(s+1))
Using partial fraction decomposition, we can rewrite this expression as:

Y(s) = (1/3) * (1/s) - (1/2) * (1/(s-1)) + (1/6) * (1/(s+1))
Taking the inverse Laplace transform of this expression, we get:

y(t) = (1/3) - (1/2)e^(t) + (1/6)e^(-t)

Therefore, the solution to the initial-value problem using the Laplace transform is y(t) = (1/3) - (1/2)e^(t) + (1/6)e^(-t).
To solve the given initial-value problem using Laplace transform, follow these steps:

1. Take the Laplace transform of both sides of the differential equation: L{2y'''+3y''-3y'-2y} = L{e^(-t)}.
2. Apply Laplace transform properties to the left side: 2(s^3Y(s)-s^2y(0)-sy'(0)-y''(0))+3(s^2Y(s)-sy(0)-y'(0))-3(sY(s)-y(0))-2Y(s).
3. Substitute initial values (y(0)=0, y'(0)=0, y''(0)=1) and find the Laplace transform of e^(-t) (1/(s+1)).
4. Simplify and solve for Y(s): Y(s) = (2s^2+3s+2)/(s^4+4s^3+6s^2+4s).
5. Find the inverse Laplace transform: y(t) = L^(-1){Y(s)}.
By following these steps, you will find the solution to the given initial-value problem.

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The distribution that X follows in this scenario is A. X ~ Expo(1/16), which means that the waiting time until you see Peter the Anteater follows an exponential distribution with a rate parameter of 1/16.

This can be determined by considering the characteristics of an exponential distribution, which models the waiting time for an event to occur given a constant rate. In this case, the event is seeing Peter the Anteater, and the rate is the average time it takes for him to appear, which is given as 16 minutes.

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The double intergral value is 288 units

By using symmetry, we can simplify the double integral to only consider the region where x is positive. Therefore, we can rewrite the integral as 2 times the integral of 9xyx⁴ over the region 0 ≤ x ≤ 2, 0 ≤ y. Evaluating this integral gives us 288.

Symmetry allows us to take advantage of the fact that the function 9xyx⁴ is an odd function in y, meaning that it flips signs when y is negated. Therefore, we can split the region of integration into two halves, one where y is positive and one where y is negative.

Because the integrand changes sign in the negative y half, we can ignore it and simply double the integral of the positive y half to get the total value. This simplifies the computation and reduces the possibility of errors.

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False.

The 85th percentile of a distribution cannot be less than zero. The percentile is a measure that indicates the percentage of data points below a given value. Therefore, the 85th percentile refers to the point in the distribution where 85% of the data falls below that point. Since zero is the lowest possible value in any distribution, it is impossible for the 85th percentile to be less than zero. It is important to note that percentiles are relative measures and can only be interpreted in the context of the distribution they are derived from.
The statement "the 85th percentile of a distribution can sometimes be less than zero" is a. true. In a distribution, the percentile represents the value below which a given percentage of the data falls. In this case, the 85th percentile indicates the value below which 85% of the data points lie. If the distribution is negatively skewed, with most of its data points concentrated on the left side and towards negative values, the 85th percentile can indeed be less than zero. It ultimately depends on the specific distribution and the range of values it contains.

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The asymptotes of the hyperbola with equation[tex](x - 2)^2/81 (y + 2)^2/4 = 1[/tex]are [tex]y = +(x - 2) + 2.[/tex]

The given hyperbola has a horizontal transverse axis and its center is at (2, -2). The standard form of a hyperbola with a horizontal transverse axis is[tex](x - h)^2/a^2 - (y - k)^2/b^2 = 1[/tex] , where (h, k) is the center of the hyperbola, a is the distance from the center to each vertex along the transverse axis, and b is the distance from the center to each vertex along the conjugate axis.

Comparing the given equation to the standard form, we can see that

[tex]a^2[/tex]= 81, so a = 9, and [tex]b^2[/tex] = 4, so b = 2. Therefore, the distance between the center and each vertex along the transverse axis is 9, and the distance between the center and each vertex along the conjugate axis is 2.

The asymptotes of a hyperbola with a horizontal transverse axis have equations y = +/- (b/a)(x - h) + k. Substituting the values of a, b, h, and k, we get:

y = +(2/9)(x - 2) - 2 and y = -(2/9)(x - 2) - 2

Therefore, the equations of the asymptotes for the given hyperbola are

y = +(x - 2)/9 - 2 and y = -(x - 2)/9 - 2.

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Assessing the regression model on data other than the sample data that was used to generate the model is known as cross-validation.

Cross-validation is a technique used in machine learning and statistics to evaluate the performance of a predictive model. It involves dividing the available data into multiple subsets, using one subset as a training set to build the model and the remaining subsets as validation sets to assess its performance.

The purpose of cross-validation is to estimate how well the model would generalize to unseen data. By evaluating the model on different subsets of the data, it provides a more robust measure of its performance and helps detect potential issues such as overfitting. Cross-validation allows researchers and practitioners to make more informed decisions about the model's predictive power and suitability for real-world applications.

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The curl of the vector field

curl f = (-8y sin(z)/z)i - (5ex sin(z) - 4x tan^-1(x/z)/z)j + (5exy cos(z) + 4y/x)k and the the divergence of the vector field div f = 5y sin(z) + 4/x for the given vector field. f(x, y, z) = 5exy sin(z)j 4y tan−1(x/z)k.

To find the curl of the vector field f(x, y, z) = 5exy sin(z)j + 4y tan−1(x/z)k, we use the formula:

curl f = ∇ × f

where ∇ is the del operator.

Using the del operator, we have:

∇ = i(∂/∂x) + j(∂/∂y) + k(∂/∂z)

Taking the curl of the vector field f, we have:

curl f = ∇ × f

= i(det |j k| ∂/∂y ∂/∂z + |k i| ∂/∂z ∂/∂x + |i j| ∂/∂x ∂/∂y) (5exy sin(z)j + 4y tan−1(x/z)k)

= i((-4y sin(z)/z) - (4y sin(z)/z)) - j((5ex sin(z)) - (4x tan^-1(x/z)/z)) + k((5exy cos(z)) + (4y/x))

Therefore, the curl of the vector field is:

curl f = (-8y sin(z)/z)i - (5ex sin(z) - 4x tan^-1(x/z)/z)j + (5exy cos(z) + 4y/x)k

To find the divergence of the vector field f(x, y, z) = 5exy sin(z)j + 4y tan−1(x/z)k, we use the formula:

div f = ∇ · f

where ∇ is the del operator.

Using the del operator, we have:

∇ = i(∂/∂x) + j(∂/∂y) + k(∂/∂z)

Taking the divergence of the vector field f, we have:

div f = ∇ · f

= (∂/∂x)(5exy sin(z)) + (∂/∂y)(4y tan−1(x/z)) + (∂/∂z)(0)

= (5y sin(z)) + (4/x) + 0

= 5y sin(z) + 4/x

Therefore, the divergence of the vector field is:

div f = 5y sin(z) + 4/x

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The interval of convergence for the power series is -20 < x < 30. The interval of convergence is (-infinity, infinity).

To find the interval of convergence for the power series ∑n=2[infinity] n(x−5)^n/(5^2n), we can use the ratio test:

|[(n+1)(x-5)^(n+1)/(5^2(n+1))]/[n(x-5)^n/(5^2n)]| = |(n+1)(x-5)/(25)|

Taking the limit as n approaches infinity, we get:

lim |(n+1)(x-5)/(25)| = |x-5| lim (n+1)/25

Since lim (n+1)/25 = infinity, the series diverges if |x-5| > 25, and the series converges if |x-5| < 25. We need to test the endpoints of the interval to determine if they converge:

When x = 5 - 25 = -20, we have:

∑n=2[infinity] n(-20-5)^n/(5^2n) = ∑n=2[infinity] (-1)^n n(25/400)^n

Using the ratio test, we have:

lim |[(n+1)(25/400)^n+1]/[n(25/400)^n]| = lim |(n+1)/25| = 0

Therefore, the series converges when x = -20.

When x = 5 + 25 = 30, we have:

∑n=2[infinity] n(30-5)^n/(5^2n) = ∑n=2[infinity] n(25/625)^nUsing the ratio test, we have:

lim |[(n+1)(25/625)^n+1]/[n(25/625)^n]| = lim |(n+1)/25| = infinity

Therefore, the series diverges when x = 30.

Therefore, the interval of convergence for the power series is -20 < x < 30.

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Step-by-step explanation:

1, x = -4 and y = 0

2, 3a =6 and 2b = -8

a =2 and b = -4 by dividing both side of equate equations respectivily.

3, quadrant-IV

4, on x-axis

5, i, 7

ii, 5

6 i) -3

ii) -2

The above given figures can be name in two different ways as follows:

13.)line WRS or SRW

14.) line XHQ or QHX

15.) line LA or AL

16.) Line UJC or CJU

17.) Line LK or KL

18.) line PXL or LXP

The names of a figure are gotten from the points on the figure. For example in figure 13, The names of the figure are WRS and SRW.

There are three points on the given figure, and these points are: point W, point R and point S, where Point R is between W and S.

This means that, when naming the figure, alphabet R must be at the middle while alphabets W and S can be at either sides of R.

Figure 13.)The possible names of the figure are: WRS and SRW.

Figure 14.)The possible names of the figure are: XHQ or QHX

Figure 15.)The possible names of the figure are:LA or AL

Figure 16.)The possible names of the figure are:UJC or CJU

Figure 17.)The possible names of the figure are:LK or KL

Figure 18.)The possible names of the figure are:PXL or LXP.

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The eigenvector corresponding to the eigenvalue λ = 4 is: v = [-3, -1, 1]

To determine if λ = 4 is an eigenvalue of the matrix

2 2 -4

3 -1 4

0 1 5

we need to check if there exists a non-zero vector v such that Av = λv, where A is the given matrix.

We have the equation:

A - λI = 0

where I is the identity matrix and 0 is the zero matrix. Let's substitute the values:

A - 4I =

2 2 -4

3 -1 4

0 1 5

4 0 0

0 4 0

0 0 4

Performing the subtraction, we get:

-2 2 -4

3 -5 4

0 1 1

Now, we set this resulting matrix equal to the zero matrix:

-2v₁ + 2v₂ - 4v₃ = 0

3v₁ - 5v₂ + 4v₃ = 0

v₂ + v₃ = 0

Simplifying the system of equations, we have:

-2v₁ + 2v₂ - 4v₃ = 0

3v₁ - 5v₂ + 4v₃ = 0

v₂ = -v₃

We can choose v₃ as a free variable and set v₃ = 1, which gives us v₂ = -1. Then, substituting these values back into the equations, we find:

-2v₁ + 2(-1) - 4(1) = 0

3v₁ - 5(-1) + 4(1) = 0

Simplifying these equations, we get:

-2v₁ - 6 = 0

3v₁ + 9 = 0

Solving these equations, we find v₁ = -3 and v₂ = -1.

Therefore, the eigenvector corresponding to the eigenvalue λ = 4 is:

v = [-3, -1, 1]

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The rate of the river's current is 3 mph.

Now, Let the rate of the river's current "c".

To solve the problem, we can use the formula:

distance = rate x time

Let's start by finding the time it takes the boat to travel 72 miles upstream in the river.

Let's call this time "t". Going upstream, the effective speed of the boat will be

21 - c

since the current is working against the boat.

So we can write:

72 = (21 - c) t

Now let's find the time it takes the boat to travel 72 miles downstream in the river.

According to the problem, this time is 3/4 of the time it takes to travel the same distance upstream.

So the time it takes to travel downstream is:

(3/4) t

Going downstream, the effective speed of the boat will be

21 + c

since the current is now helping the boat. So we can write:

72 = (21 + c) (3/4) t

Now we have two equations:

72 = (21 - c) * t 72 = (21 + c) (3/4) t

We can solve for "t" in the first equation:

t = 72 / (21 - c)

Now we can substitute this value of "t" into the second equation:

72 = (21 + c) (3/4) (72 / (21 - c))

Simplifying: 72 = (21 + c) * (54 / (21 - c))

Multiplying both sides by

(21 - c): 72 (21 - c) = (21 + c) 54

1512 - 72c = 1134 + 54c

Solving for "c":

126c = 378 c = 3

Therefore, the rate of the river's current is 3 mph.

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The volume of the cylinder is [tex]5725.28\ cm^3[/tex].

According to the question:

The diameter of the cylinder [tex]d = 18\ cm[/tex]

Therefore radius [tex]r[/tex] = [tex]9\ cm[/tex]

Height [tex]h = 2.5\times 9 = 22.5\ cm[/tex]

To find:

The volume of the cylinder

We know that the volume of a cylinder is:

[tex]V = \pi r^2h[/tex]

substitute the given values into this equation and take [tex]\pi = 3.1415[/tex], we get:

[tex]V = 3.1415\times 9^2\times 22.5\ cm^3[/tex]

[tex]V = 5725.28\ cm^3[/tex]

Therefore, The volume of the cylinder is [tex]5725.28\ cm^3[/tex].

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The area of the figure is 197 cm².

We have,

From the figure,

We can make three shapes.

Rectangle 1:

Area = 9 x 3 = 27 cm²

Rectangle 2:

Area = (9 + 3) x (17 - 6) = 12 x 11 = 132 cm²

Trapezium:

Area = 1/2 x (parallel sides sum) x height

= 1/2 x (12 + 7) x (15 + 6 - 17)

= 1/2 x 19 x 4

= 19 x 2

= 38 cm²

Now,

The area of the figure.

= 27 + 132 + 38

= 197 cm²

Thus,

The area of the figure is 197 cm².

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The eigenvalues of the given matrix are λ₁ = 1/10 and λ₂ = 21/10, and their associated eigenvectors are [3, 1] and [1, -2], respectively.

To find the eigenvalues and eigenvectors of the matrix, we need to solve the equation (A - λI)v = 0, where A is the given matrix, λ is the eigenvalue, I is the identity matrix, and v is the eigenvector.

For the given matrix |2 3 ||0 9/10 |, subtracting λI gives the matrix |2 - λ 3 ||0 9/10 - λ |. Setting this matrix equal to zero and solving the system of equations yields the eigenvalues.

By solving (2 - λ)(9/10 - λ) - 3*0 = 0, we obtain the eigenvalues λ₁ = 1/10 and λ₂ = 21/10.

To find the eigenvectors, we substitute each eigenvalue back into the equation (A - λI)v = 0 and solve for v.

For λ₁ = 1/10, solving (2 - (1/10))x + 3y = 0 and 3x + ((9/10) - (1/10))y = 0 gives the eigenvector [3, 1].

Similarly, for λ₂ = 21/10, solving (2 - (21/10))x + 3y = 0 and 3x + ((9/10) - (21/10))y = 0 gives the eigenvector [1, -2].

In summary, the eigenvalues of the given matrix are λ₁ = 1/10 and λ₂ = 21/10, and their associated eigenvectors are [3, 1] and [1, -2], respectively

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Answer:

First choice

Step-by-step explanation:

This is much like slicing a stick of butter....you want the cut face to have the same dimensions as the current face  5.5 x 4  inches

The indefinite integral of (2x-3)/(2x^2-6x+3)^2 dx is -(1/(2x^2-6x+3)) + C, where C is the constant of integration.

To evaluate the indefinite integral, we can use the substitution method or partial fractions. Let's proceed with the substitution method for this problem.

Step 1: Perform the substitution:

Let u = 2x^2-6x+3. Taking the derivative of u with respect to x, we have du = (4x-6) dx.

Step 2: Rewrite the integral:

We can rewrite the integral as ∫(2x-3)/(2x^2-6x+3)^2 dx = ∫(1/u^2) du.

Step 3: Evaluate the integral:

Now we can integrate ∫(1/u^2) du. Applying the power rule of integration, the result is -(1/u) + C, where C is the constant of integration. Substituting back u = 2x^2-6x+3, we get -(1/(2x^2-6x+3)) + C as the final answer.

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The absolute magnitude of the reduction in the variation of y when x is introduced into the regression model represents the amount by which the variability of y decreases due to the inclusion of x.

The absolute magnitude of the reduction in the variation of y when x is introduced into the regression model can be determined by calculating the difference in the variability of y before and after the inclusion of x. Here are the steps to explain it:

Calculate the variation of y (also known as the total sum of squares, SST) before introducing x into the regression model.

Fit a regression model with both y and x as variables and calculate the residuals (the differences between the observed y values and the predicted y values).

Calculate the sum of squares of the residuals (also known as the residual sum of squares, SSE) after introducing x into the model.

Calculate the absolute magnitude of the reduction in the variation of y by subtracting SSE from SST.

Reduction in variation = SST - SSE

This value represents the amount by which the variability of y decreases when x is introduced into the model. It indicates how much of the total variation in y can be explained by the inclusion of x in the regression model.

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It is indeed possible to find non-empty families f and g such that the intersection of f and g, denoted as (f ∩ g), is not equal to the intersection of f and the intersection of g, denoted as (f) ∩ (g).

Let's consider the following example to prove this statement. Assume we have two families of sets: f = {{1, 2, 3}, {2, 3, 4}} and g = {{3, 4, 5}, {4, 5, 6}}. In this case, the intersection of f and g is f ∩ g = {{3}}.

Now, let's find the intersection of f and the intersection of g. The intersection of g, denoted as (g), is {3, 4, 5, 6}. Therefore, (f) ∩ (g) = {{1, 2, 3}, {2, 3, 4}} ∩ {3, 4, 5, 6} = {}.

As we can see, f ∩ g = {{3}} is not equal to (f) ∩ (g) = {}, which confirms that there exist non-empty families f and g for which the intersection of f and g is not equal to the intersection of f and the intersection of g.

This example illustrates that the intersections of families of sets do not necessarily distribute over each other, leading to distinct results in different orderings of intersections.

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True, the slope measures the rate of change in the y-values with respect to the x-values. In other words, it indicates how much the y-value changes when the x-value changes by a certain amount. When referring to a linear equation in the form y = mx + b, the slope is represented by the coefficient m.

If the slope is positive, it means that as the x-value increases by a certain amount, the y-value also increases. Conversely, if the slope is negative, it means that as the x-value increases, the y-value decreases. The slope can be calculated using the formula (change in y) / (change in x), which can be written as (y2 - y1) / (x2 - x1) for any two points (x1, y1) and (x2, y2) on the line.In your specific question, the slope would indeed represent how much the y-value changes when the x-value changes by 2 units of the variable being measured. Therefore, if you know the slope and you have a starting point, you can calculate the corresponding y-value after the x-value has changed by 2 units. This concept is important in various fields, such as mathematics, physics, and economics, where the relationships between variables are often represented using linear equations with slopes.

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A vertical line drawn through a normal distribution at z = –0.75 will separate the distribution into two sections then the proportion in larger section, separated by the vertical line at z = -0.75 is 0.7266.

In a standard normal distribution, the area to the left of a particular z-score represents the proportion of values below that z-score. The area to the right represents the proportion of values above that z-score.

Since the proportion in the smaller section is given as 0.2734, it corresponds to the area to the left of z = -0.75.

To find the proportion in the larger section, we subtract the given proportion from 1 since the total area under the curve is 1.

Proportion in larger section = 1 - 0.2734 = 0.7266

Therefore, the proportion in the larger section, separated by the vertical line at z = -0.75, is 0.7266.

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The equation in vertex form. y=a(x-h)^2+k is y = 0.25(x + 4)^2 - 3.

We are given that;

The points and form y=a(x-h)^2+k

Now,

for the first set of points, we get:

a = -0.5 h = 3 k = 4.5

So the equation of the parabola in vertex form is:

y = -0.5(x - 3)^2 + 4.5

By repeating this process for the other sets of points to find their equations;

y = 0.5(x + 6)^2 + 1

y = -0.75(x - 3)^2 + 6

y = 0.25(x + 4)^2 - 3

Therefore, by equations the answer will be y = 0.25(x + 4)^2 - 3.

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1) The statement of Lin is correct.

2) (a) When scale factor, k = 9, area of the new triangle = 8100 square inches

(b) When k = 3/4, area of the new triangle = 56.25 square inches.

Given that,

A triangle has area 100 square inches.

It's dilated by a factor of k = 0.25.

When the triangle is dilated by a scale factor of k, then, each of the base and height is dilated by the scale factor of k.

So new area of the triangle after the dilation with the original triangle having base = b and height = h is,

Area of new triangle = 1/2 (kb)(kh) = k² (1/2 bh) = k² × Area of original triangle

Here original area = 100 square inches.

k = 0.25

New area = (0.25)² 100 = 6.25 square inches

So the correct statement is that of Lin.

Mai may found the new area by just multiplying the scale factor with 100, instead of taking the square of the scale factor. That is why she got 25 square inches as the new area.

2) (a) When k = 9,

Area of the new triangle = 9² (100) = 8100 square inches

(b) When k = 3/4

Area of the new triangle = (3/4)² (100) = 56.25 square inches

Hence the areas are found.

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The domain of f(x) is all real numbers because there are no restrictions on x in the given function. So, the domain of f(x) is: (-∞, ∞)

The function f(x) is not differentiable at x = 1 and x = -4 because of the corners in the absolute value function. However, it is differentiable for all other values of x. Therefore, the interval notation for the values of x where f is differentiable is:

(-∞, -4) U (-4, 1) U (1, ∞)

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Finding the derivatives of x and y with respect to t

We need to find the values of t for which the given parametric equations for x and y intersect.

We need to find the values of t for which the given parametric equations for x and y intersect.

To do that, we first find the derivatives of x and y with respect to t.

dx/dt = 2t - 90t^4

dy/dt = (2t^3 - 28t^2)/7

Setting the derivatives equal to zero and solving for t

Next, we set each derivative equal to zero and solve for t.

2t - 90t^4 = 0

t(2 - 90t^3) = 0

t = 0 or t = (2/90)^(1/3) ≈ 0.382

(2t^3 - 28t^2)/7 = 0

t(2t - 28)/7 = 0

t = 0 or t = 14/2 = 7

Therefore, the points on the curve that have horizontal or vertical tangent lines are (0,0), (7,49/2), and approximately (1.176,-9.724).

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The parcel of land described as ne¼, nw¼, se¼, sec.24, t2n, r7e, 6th p.m. would be a total of 40 acres.

This is because each ¼ section is equal to 40 acres, and this description includes 4 ¼ sections.

In the Public Land Survey System (PLSS), land is divided into 6-mile-square townships. Each township is then divided into 36 sections, each section being a square mile or 640 acres.

Each section can be further divided into quarters, and each quarter section is equal to 160 acres.

Therefore, a description of ne¼, nw¼, se¼, sec.24, t2n, r7e, 6th p.m. refers to the northeast quarter of the northwest quarter of the southeast quarter of section 24, township 2 north, range 7 east, 6th principal meridian. Since this description includes 4 quarter sections, the total acreage would be 40 acres.

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To find the critical points in MATLAB, you can use the 'solve' function in combination with symbolic variables.


1. First, define the symbolic variables and the function. For example, let's find the critical points of the function f(x) = x^3 - 6x^2 + 9x.
```MATLAB
syms x;
f = x^3 - 6*x^2 + 9*x;
```
2. Compute the first derivative of the function using the 'diff' function.
```MATLAB
f_derivative = diff(f, x);
```
3. Solve the first derivative for x using the 'solve' function.
```MATLAB
critical_points = solve(f_derivative, x);
```
4. Display the critical points.
```MATLAB
disp(critical_points);
```

By following these steps, you can find the critical points of a function in MATLAB using symbolic variables and the 'solve' function. Remember to define the symbolic variable, the function, compute the first derivative, and then solve the first derivative for x to obtain the critical points.

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