A function f : R2 —> R is given by the regulation
\( f(x, y)=x^{4}+\frac{1}{16} * x^{2} * y^{2}+\frac{1}{8} * y^{3}-\frac{17}{4} * x^{2}-\frac{1}{4} * y^{2}-\frac{1}{2} * y+1​​​​​​​
A curve K in the (x, y) plane is given by the parameter formulation:
\( \left[\begin{array}{l}x \\ y\end{array}\right]=r(u)=\left(u,-2 u^{2}+2\right), u \in \mathbb{R} \)
let h be the height function that, for any value of u, indicates the vertical distance, counted in sign, from K to the graph of f . Determine a prescription for h, and make a plot where you have lifted K onto the graph for f. Determine the values ​​of u in which the differential quotient of h is respectively 0, negative and positive
a) state the value of the height function h(-2) =
----------------------------------------------------------------------------------------------------------
It is stated that f has two stationary points. In the first, which we call Q, f actually has
local extremum. In the second, which we call R , the Hessian matrix has the eigenvalue 0.
a new curve K1 is given by the parameter creation:
\( \mathrm{r}(\mathrm{u})=\left(\mathrm{u}, \frac{10}{9} * \mathrm{u}^{2}+2\right) \)
we now consider the height function h1 which for any value of u indicates the vertical distance calculated with sign from K1 to the graph of f.
b) determine a prescription for h and determine whether h1 has: local maximum, local minimum or no local extrema in u=0

The solution is:  standard deviation. So, correct option is C.

Here, we have,

Explanation:

However, note that the mean of a sample from a normal population (that is, a type of "standard deviation" known as "standard deviation from a sample mean", is an "unbiased estimator" of the "population mean".

We shall represent the "sample mean" as:  X  ;

              (pronounced: "x-bar"; that is; "ex-bar"); this is the usual symbol                                                                                                              used;

We shall represent the sample size as "n" ;

                                                (This is the usual variable used; note that

                                                                               "n" is a numeric value).

The standard deviant from the sample mean:

(n *  X)  / (n + 1) is a "biased estimator of the mean" that becomes "unbiased" as the sample size increases; since as "n" increases in values; the value of the entire entire expression becomes smaller (i.e the standard deviation becomes smaller.

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Complete question:

Which biased estimator will have a reduced bias based on an increased sample size? mean median standard deviation range

Answer:

70

Step-by-step explanation:

-70 / -1

= 70 / 1

= 70

(3w/55c6w2)*(11c4/6w2) = 33c4/330c6w4.

To multiply fractions, we first need to express the fractions in the same denominator. To do this, we need to find the least common denominator (LCD) of the two fractions. In this case, the LCD is 55c6w2. Once we have the LCD, we can then rewrite the fractions with that as the denominator.

So the first fraction will be (3w/55c6w2) and the second fraction will be (11c4/6w2). Now that the fractions have the same denominator, we can multiply the numerators of the fractions and keep the same denominator.

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Answer: 240 more cards than Alex!

Step-by-step explanation:

After 8 months, Filip will have more trading cards than Alex. This is because over 8 months, Filip will have bought 8 boxes of 30 cards and Alex will have bought 8 boxes of 22 cards, meaning that Filip will have a total of 8*30 = 240 more cards than Alex.

Answer:

240 more

Step-by-step explanation:

Yes, there is sufficient evidence to conclude that display at point of purchase helps in increasing the sales of the product. This is because the mean difference in sales after display (300) is greater than the standard deviation (314.8), which indicates that there is a significant difference between the two groups.

To further support this conclusion, we can use a 99% confidence interval (CI) for the true difference. The formula for a 99% CI is:

CI = mean difference ± (t-value)(SD/sqrt(n))

Where n is the sample size, t-value is the critical value for a 99% CI, and SD is the standard deviation. For a sample size of 11 and a 99% CI, the t-value is 3.106. Plugging in the values, we get:

CI = 300 ± (3.106)(314.8/sqrt(11))

CI = 300 ± 294.6

CI = (5.4, 594.6)

Since the 99% CI does not include 0, we can conclude that there is sufficient evidence to support the claim that display at point of purchase helps in increasing the sales of the product. In other words, we can be 99% confident that the true difference in sales between after display and before display is between 5.4 and 594.6, which supports the claim that display at point of purchase helps in increasing sales.

True difference in sales lies between -162 and 662

Yes, there is sufficient evidence to conclude that display at point of purchase helps in increasing the sales of the product. The mean of the differences in sales between after display and before display was found to be 300 with a standard deviation of 314.8. This indicates that, on average, the display helped to increase sales by 300.

Furthermore, with a 99% confidence interval, the true difference in sales lies between -162 and 662, which is statistically significant. Therefore, the display at point of purchase has a statistically significant effect on sales and should be implemented.

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Cylinders X and Y have the same lateral surface area in square inches.

The y-intercept is 20 and, in the situation, it represents the initial pledge plan.

In Mathematics, the y-intercept is sometimes referred to as an initial value or vertical intercept and the y-intercept of any graph such as a linear function, generally occur at the point where the value of "x" is equal to zero (x = 0).

Based on the information provided, an equation that describes the relationship between the money ($) received and the distance (meters) walked is given by;

M = 20 - 3d

M = 20 - 3(0)

M = 20

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a. The variable is the type of communication preference: postal service or email.

b. The population is the adult population in Puerto Rico.

c. The parameter being estimated is the percentage of adults who prefer postal mail to electronic mail.

d. The parameter being estimated is the percentage of adults who prefer postal mail to electronic mail.

e. The sample is the 600 adults that the graduate student surveyed.

a. The variable in this research project is the preferred method of communication for adults in Puerto Rico (postal mail or email).

b. The population in this research project is the adult population in Puerto Rico.

c. The sample in this research project is the 600 adults that the graduate student surveyed.

d. The parameter in this research project is the percentage of adults in Puerto Rico who prefer postal mail to email.

e. The statistic in this research project is the 250 adults who indicated their preference for the postal service out of the 600 adults surveyed. This statistic was used to estimate the parameter of 48% of adults in Puerto Rico preferring the postal service to email.

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z3 = 6 - 2xz3x + 4 in the form a + b i, where a,b in R.

=> z3 + (2z3 + i)x = 4 - 6i

Expand the bracket:

z3 + 2xz3x + ix = 4 - 6i

Subtract 4 from both sides:

z3 + 2xz3x + ix - 4 = - 6i

Rearrange and set ix = -1:

z3 + 2xz3x - 4 = 6

Subtract 2xz3x from both sides:

z3 - 4 = 6 - 2xz3x

Solve for z3:

z3 = 6 - 2xz3x + 4

Therefore, z3 = 6 - 2xz3x + 4 in the form a + b i, where a,b in R.

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The system performance is not affected by the thermodynamic losses of 1.5 °C

In a brine circulation MSF system, thermodynamic losses occur when heat is lost from the system, resulting in a decrease in the efficiency of the system. To compare the system performance if the thermodynamic losses are equal to 1.5 °C, we need to calculate the performance ratio (PR) of the system with and without the thermodynamic losses.

Without thermodynamic losses:
PR = (Production capacity) / (Heating steam flow rate)
= (1 kg/s) / ((116 °C - 40 °C) / (106 °C - 30 °C))
= 1 / (76 / 76)
= 1

With thermodynamic losses of 1.5 °C:
PR = (Production capacity) / (Heating steam flow rate)
= (1 kg/s) / ((116 °C - 40 °C - 1.5 °C) / (106 °C - 30 °C - 1.5 °C))
= 1 / (74.5 / 74.5)
= 1

The performance ratio of the system remains the same with and without the thermodynamic losses of 1.5 °C. This means that the system performance is not affected by the thermodynamic losses of 1.5 °C. However, it is important to note that thermodynamic losses can have a significant impact on the system performance if they are larger.

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The statement, "Only datasets having a linear relationship between variables can be assessed using regression analyses" is False, because Non-Linear relationships also can be assessed, the correct option is (b)False.

The Regression Analysis can be used to assess the relationship between variables, including non-linear relationships.

There are various types of regression models that can capture non-linear relationships, such as polynomial regression, exponential regression, and logarithmic regression.

The interpretation of the regression coefficients and other statistics may be different in non-linear regression models compared to linear regression models.

Therefore, the given statement is (b)False.

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Answer:

Step-by-step explanation:

For an expression 2(3a+2b−7), an equivalent expression is 6a + 4b - 14

The correct answer is an option (D)

Consider an expression 2(3a+2b−7)

We simplify this expression.

2(3a+2b−7)

To simplfy this expression, multiply each term of (3a+2b−7) by 2

2(3a+2b−7)

= 2 × (3a + 2b - 7)

= (2 × 3a) + (2 × 2b) - (2 × 7)

= 6a + 4b - 14

So, we get 2(3a+2b−7) = 6a + 4b - 14

Therefore, 6a + 4b - 14 is an equivalent expression to 2(3a+2b−7)

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The complete question is:

Which expression is equivalent to the given expression? 2(3a+2b−7) ​

A 5a+4b−5

B 6a+2b−7

C 6a+4b−7

D 6a+4b−14

For the function a(t) = -1/2t + 19, the graph is plotted using the x and y intercepts.

What is a function?

In mathematics, a function is a unique arrangement of the inputs (also referred to as the domain) and their outputs (sometimes referred to as the codomain), where each input has exactly one output and the output can be linked to its input.

To graph the function a(t) = -1/2t + 19, we can follow these steps -

Choose a range of values for t.

Since the function represents the amount of snow remaining after a certain number of hours, we should choose a range that makes sense for the context.

Let's choose t values from 0 to 38, since it's unlikely that there would be much snow left after 38 hours on a warm day.

Substitute each t value into the function to find the corresponding value of a(t).

For example, when t = 0, we have -

a(0) = -1/2(0) + 19 = 19

When t = 10, we have -

a(10) = -1/2(10) + 19 = 14

And so on, for each value of t in our range.

Plot the (t, a(t)) points on a coordinate plane.

For example, the first point is (0, 19), and the second point is (10, 14). Continue plotting points for each value of t.

Draw a smooth curve through the plotted points to represent the function.

The curve should be a straight line with a negative slope, since the function is linear with a negative coefficient on t.

The y-intercept is 19, which means that there is 19 units of snow remaining when t = 0.

The x-intercept can be found by setting a(t) = 0 and solving for t.

0 = -1/2t + 19

1/2t = 19

t = 38

Therefore, the graph for the function is plotted.

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The equation for the exponential function is f(x) = (4.5)((0.5/4.5)-1/2)x. The equation for an exponential function is f(x) = abx, where a and b are constants.

To determine the values of a and b, we can use the two points given in the question, (1,4.5) and (-1,0.5).

Let's substitute the point (1,4.5) into the equation.

f(1) = a*b1
4.5 = a*b

Now let's substitute the point (-1,0.5) into the equation.

f(-1) = a*b-1
0.5 = a*b-1

We can now solve for a and b.

a = 4.5 / b
b-1 = 0.5 / a
b-1 = 0.5 / (4.5/b)
b-1 = 0.5b/4.5
b-2 = 0.5/4.5
b = (0.5/4.5)-1/2

Thus, the equation for the exponential function is f(x) = (4.5)((0.5/4.5)-1/2)x

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The polynomial function has zeros of x=-2 with a multiplicity of 2, x=1 with a multiplicity of 1 , and a y-intercept of 2 is y = (x+2)^2(x-1) + 2.

To find out which polynomial function has zeros of x=-2 with a multiplicity of 2, x=1 with a multiplicity of 1, and a y-intercept of 2, we can use the factored form of a polynomial function.

This is given by:

f(x) = a(x - r₁)^n₁(x - r₂)^n₂ ... (x - rₖ)^nₖ where a is a constant, r₁, r₂, ..., rₖ are the zeros of the function, and n₁, n₂, ..., nₖ are their respective multiplicities.

Using the given zeros and multiplicities, we can write the factored form of the polynomial as:

f(x) = a(x + 2)²(x - 1) where the y-intercept is 2. To find the value of a, we can substitute the y-intercept, (0, 2), into the function:

f(0) = a(0 + 2)²(0 - 1) = -4a

Since the y-intercept is 2, we have:

f(0) = 2-4a = 2 => -4a = 0 => a = 0

Therefore, the polynomial function is:

f(x) = a(x + 2)²(x - 1) = 0(x + 2)²(x - 1) = 0

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Answer:

Step-by-step explanation:

x + 4 + 2x + 23 = 90

3x + 27 = 90

3x = 63

x = 21

x + 4 = 21 + 4 = 25 for <1

2(21) + 23  = 42 + 23 = 65 for <2

The values of x that satisfy the equation are (l + sqrt(11)l)/10 and (l - sqrt(11)l)/10.

To find the values of x such that the angle between the vectors < 2, 1, -1 > and < l,x,0 > is 45 degrees, we can use the formula for the dot product of two vectors:

= |u||v|cos(theta)

Where  is the dot product of the vectors u and v, |u| and |v| are the magnitudes of the vectors, and theta is the angle between them. Plugging in the values from the question, we get:

< 2, 1, -1 > . < l,x,0 > = |< 2, 1, -1 >||< l,x,0 >|cos(45)

Simplifying the dot product, we get:

2l + x = sqrt(6)sqrt(l^2 + x^2)/sqrt(2)

Squaring both sides and rearranging, we get:

4l^2 + 4lx + x^2 = 6l^2 + 6x^2

2l^2 + 2lx - 5x^2 = 0

Using the quadratic formula, we can solve for x:

x = (-2l +/- sqrt(4l^2 - 4(2l^2)(-5)))/(2(-5))

x = (-l +/- sqrt(l^2 + 10l^2))/(-10)

x = (-l +/- sqrt(11)l)/(-10)

x = (l +/- sqrt(11)l)/10

Therefore, the values of x that satisfy the equation are (l + sqrt(11)l)/10 and (l - sqrt(11)l)/10.

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The correct option is A. Yes, λ=4 is an eigenvalue of the matrix.

To find the corresponding eigenvector, we need to solve the equation (A - λI)x = 0, where A is the given matrix, λ is the eigenvalue, I is the identity matrix, and x is the eigenvector.
First, we subtract λI from A:
A - λI = 3-4 -4 0 0 3-4 3 2-4 -2 6
= -1 -4 0 0 -1 3 2 -2 2
Next, we set the equation (A - λI)x = 0 and solve for x:
(-1 -4 0) (x1) = 0
(0 -1 3) (x2) = 0
(2 -2 2) (x3) = 0
Simplifying the equations gives us:
-x1 - 4x2 = 0
-x2 + 3x3 = 0
2x1 - 2x2 + 2x3 = 0
We can solve this system of equations to find the eigenvector. One possible solution is x1 = 2, x2 = 1, x3 = 1/3. Therefore, one corresponding eigenvector is (2, 1, 1/3).
So the correct answer is A. Yes, λ=4 is an eigenvalue of the matrix. One corresponding eigenvector is (2, 1, 1/3).

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Answer: Triangle 1: -6x + 105      Triangle 2: 20x - 30        Triangle 1 would have a greater perimeter if x = 5

Step-by-step explanation:

The perimeter of a shape is the sum of all the sides. (Add them together)

Triangle 1:

17 + 6x + 4(-3x + 22)

Simplify:

17 + 6x -12x + 88

Perimeter = -6x + 105

Triangle 2:

24 + 5x + 3(5x - 18)

Simplify:

24 + 5x + 15x - 54

Perimeter = 20x - 30

If x = 5: (Plug in 5 for x)

Triangle 1:

If x = 5: (Plug in 5 for x)

-6(5) + 105

-30 + 105 = 75

Triangle 2:

20(5) - 30

100 - 30 = 70

75 > 70

Triangle 1 would have a greater perimeter if x = 5

Hope this helps!

The next time when a train and a bus leave the station together is 4:30pm.

The next time when a train and a bus leave the station together will be the least common multiple (LCM) of the two intervals, 8 minutes and 10 minutes.

To find the LCM of 8 and 10, we can list the multiples of each number until we find a common multiple:
8: 8, 16, 24, 32, 40
10: 10, 20, 30, 40

The LCM of 8 and 10 is 40. This means that a train and a bus will leave the station together every 40 minutes.

Since the train and bus both leave the station at 3:50pm, the next time they will leave together will be 40 minutes later, at 4:30pm.

Therefore, the next time when a train and a bus leave the station together is 4:30pm.

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Answer:

There are two situations involving rational exponents or radicals that will never result in a negative real solution:

Even-indexed roots: If we take the square root, fourth root, sixth root, etc. of a non-negative real number, the result will always be non-negative. For example, the square root of 9 is 3, and the fourth root of 16 is 2, both of which are non-negative. This is because even-indexed roots always produce a non-negative result, regardless of the sign of the original number.

Exponents with even denominators: If we raise a non-negative real number to an exponent with an even denominator, the result will always be non-negative. For example, (4^2/4) is equal to 4, which is non-negative. This is because any negative base raised to an even power results in a positive number, and any positive base raised to an even power also results in a positive number. Therefore, any exponent with an even denominator will always produce a non-negative result, regardless of the sign of the original number.

If G is a non-Abelian group, it has an automorphism that is not the identity. This is proven by constructing a map φ that is a bijective homomorphism but not the identity, using an element a that is not in the center of G.

An automorphism is a bijective homomorphism of a group onto itself. If G is a non-Abelian group, we need to prove that there is an automorphism that is not the identity.

To prove this, we can use the following steps:

1. Let G be a non-Abelian group.
2. Choose an element a ∈ G that is not in the center of G. This means that there exists an element b ∈ G such that ab ≠ ba.
3. Define a map φ: G → G by φ(x) = axa⁻¹ for all x ∈ G.
4. We can prove that φ is a homomorphism by showing that φ(xy) = φ(x)φ(y) for all x, y ∈ G.
5. We can also prove that φ is bijective by showing that it has an inverse map φ⁻¹: G → G defined by φ⁻¹(x) = a⁻¹xa for all x ∈ G.
6. Since φ is a bijective homomorphism, it is an automorphism of G.
7. However, φ is not the identity, because φ(b) = aba⁻¹ ≠ b.
8. Therefore, G has an automorphism that is not the identity.

In conclusion, if G is a non-Abelian group, it has an automorphism that is not the identity. This is proven by constructing a map φ that is a bijective homomorphism but not the identity, using an element a that is not in the center of G.

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ln(144) can be written in terms of ln(3) and ln(4) as:

ln(144) = 4 ln(2) + 2 ln(3) ≈ 4 ln(1.386) + 2 ln(3) ≈ 2.7726 + 2 ln(3)

We can use the logarithmic property that states that ln(a * b) = ln(a) + ln(b) to rewrite ln(144) in terms of ln(3) and ln(4).

First, we can find the prime factorization of 144:

[tex]144 = 2^4 * 3^2[/tex]

Using the property above, we can rewrite ln(144) as:

[tex]ln(144) = ln(2^4 * 3^2) = ln(2^4) + ln(3^2)[/tex]

Now, we can use another logarithmic property that states that ln(aᵇ) = b * ln(a) to simplify ln(2⁴) and ln(3²):

ln(144) = 4 ln(2) + 2 ln(3)

Therefore, ln(144) can be written in terms of ln(3) and ln(4) as:

ln(144) = 4 ln(2) + 2 ln(3) ≈ 4 ln(1.386) + 2 ln(3) ≈ 2.7726 + 2 ln(3)

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The quotient for the associated division is -0.5x^(3)-x^(2)+x+1 and the remainder is -9.

Synthetic division is a method for dividing polynomials by monomials. It is a simplified form of the long division of polynomials, and is useful when the divisor is a monomial. The method involves arranging the coefficients of the dividend in a row, and then dividing each term by the divisor.

To find P(-2) for P(x)=x^(4)+2x^(3)-2x-7, we simply need to substitute -2 for x and evaluate the expression.

P(-2)=(-2)^(4)+2(-2)^(3)-2(-2)-7

P(-2)=16+2(-8)-2(-2)-7

P(-2)=16-16+4-7

P(-2)=-3

Therefore, the value of P(-2) is -3.

The associated division would be (x^(4)+2x^(3)-2x-7)/(-2), which can be simplified using polynomial long division. The quotient is -0.5x^(3)-x^(2)+x+1 and the remainder is -9.

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Using the angle addition postulate, we found that the measure of the angle, ∠LMN is 135°.

What is the angle addition postulate?

The measure of the angle created by the non-common sides of two adjacent angles is equal to the total of the measures of the two adjacent angles. The angle addition postulate in geometry asserts that if we position two or more angles side by side, with a shared vertex and an arm between each pair of angles, the sum of those angles will be equal to the sum of the resulting angle. Adjacent angles are those two angles that are connected by a common ray. Any pair of neighbouring angles in mathematics can be applied to this postulate.

The figure is given below.

We can solve this using the angle addition postulate.

Given,

m∠FMN = 99°

m∠LMF = 36°

We are asked to find the measure of angle ∠LMN.

According to the angle addition postulate,

m∠FMN + m∠LMF =  m∠LMN

99 + 36 = m∠LMN

m∠LMN = 135°

Therefore using the angle addition postulate, we found that the measure of the angle, ∠LMN is 135°.

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The linear transformation T is defined as:

[tex]\( T(x) = \left[\begin{array}{c}-6x_{1}+x_{2}+180 \\ 24x_{1}-4x_{2}-720\end{array}\right] \)[/tex]

The linear transformation T is defined as T(x) = Ax+b, where A is a matrix and b is a vector. In this case, we have

[tex]\( A=\left[\begin{array}{cc}-6 & 1 \\ 24 & -4\end{array}\right] \)[/tex]  and  [tex]\( b=\left[\begin{array}{c}180 \\ -720\end{array}\right] \).[/tex]

So, for any vector   [tex]\( x=\left[\begin{array}{c}x_{1} \\ x_{2}\end{array}\right] \)[/tex] , we have:

[tex]\[ T(x) = \left[\begin{array}{cc}-6 & 1 \\ 24 & -4\end{array}\right]\left[\begin{array}{c}x_{1} \\ x_{2}\end{array}\right] + \left[\begin{array}{c}180 \\ -720\end{array}\right] \][/tex]

[tex]\[ T(x) = \left[\begin{array}{c}-6x_{1}+x_{2} \\ 24x_{1}-4x_{2}\end{array}\right] + \left[\begin{array}{c}180 \\ -720\end{array}\right] \][/tex]

[tex]\[ T(x) = \left[\begin{array}{c}-6x_{1}+x_{2}+180 \\ 24x_{1}-4x_{2}-720\end{array}\right] \][/tex]

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Using the scale factors obtained the values are -

The value of k is 8, since g(−2) = 8.

The value of k is 8, since g(−2) = 8.

The value of k is 8, since g(−2) = 1.

The value of k is 4, since g(−2) = 2.

What is scale factor?

The scale factor of a shape refers to the amount by which it is increased or shrunk. It is applied when a 2D shape, such as a circle, triangle, square, or rectangle, needs to be made larger.

For each part, we can use the given table to determine how the transformation affects the function values -

g(x) = f(2x)

This transformation is a horizontal compression by a factor of 2.

To see this, notice that when we evaluate g at x = -1, we get the same value as f evaluated at x = -2.

Similarly, when we evaluate g at x = 0, we get the same value as f evaluated at x = 0.

And so on. In other words, the function values of g(x) are the same as the function values of f(x) at every other point (starting with x = -2).

So, g(x) is a compressed version of f(x) horizontally by a factor of 2.

Therefore, the value of k is 8, since g(−2) = 8 = f(2×(-2)).

g(x) = 2f(x)

This transformation is a vertical stretch by a factor of 2.

To see this, notice that every value of g(x) is twice the corresponding value of f(x).

So, g(x) is a stretched version of f(x) vertically by a factor of 2.

Therefore, the value of k is 8, since g(−2) = 2f(−2) = 2×4 = 8.

g(x) = f(x/2)

This transformation is a horizontal stretch by a factor of 2.

To see this, notice that when we evaluate g at x = -2, we get the same value as f evaluated at x = -4.

Similarly, when we evaluate g at x = -1, we get the same value as f evaluated at x = -2.

And so on. In other words, the function values of g(x) are the same as the function values of f(x) at every other point (starting with x = -4).

So, g(x) is a stretched version of f(x) horizontally by a factor of 2.

Therefore, the value of k is 8, since g(−2) = f(−1) = 1.

g(x) = (1/2)f(x)

This transformation is a vertical compression by a factor of 2.

To see this, notice that every value of g(x) is half the corresponding value of f(x).

So, g(x) is a compressed version of f(x) vertically by a factor of 2.

Therefore, the value of k is 4, since g(−2) = (1/2)f(−2) = (1/2)×4 = 2.

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15000 rational numbers from (-1,1) interval using black

Answer: Code for the given mathematical function# Python code for plotting sine and cosine functions import matplotlib.py plot as plt import numpy as npa = [0.25, 1, 3] # given set of valuesx = np.linspace(-15, 1, 15000) # from (-1, 1) interval and 15000 rational numbersb = 2.5c = 1.6fig, ax = plt.subplots()# Plotting the graph with different colorsplt.plot(x, a[0]*np.cos(b*x)+ (1-a[0])*np.sin(c*x), color='black', label='a=0.25')plt.plot(x, a[1]*np.cos(b*x)+ (1-a[1])*np.sin(c*x), color='red', label='a=1')plt.plot(x, a[2]*np.cos(b*x)+ (1-a[2])*np.sin(c*x), color='blue', label='a=3')# Adding labels and titlesplt.xlabel('x-axis')plt.ylabel('y-axis')plt.title('Plot of given mathematical function')plt.legend()# Displaying the plotplt.show()Graphical output for the given mathematical function:Here, the above code gives a line plot of the given mathematical function for all a € (0.25k: 1 ks 3,k e Z+), and x € {x € R:-15x 1), where b=2.5, C = 1.6 by creating 15000 rational numbers from (-1,1) interval using black, red, and blue colors for respective k values.

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