Ex2. Prime Numbers ( 40 points) You will implement in this exercise an ancient Greek algorithm for finding the prime numbers less than a given number. (ask your instructor about the name of the algorithm after class!) Reminder: A prime number is a positive integer greater than 1 that is divisible only by itself and by 1 . Here is how the algorithm works assuming we would like to find the prime numbers ≪=20 : 1. Initially, assume that all the numbers are prime by marking them with 1 (0 means not prime). 2. For each number that is marked as prime, starting at 2, mark all of its multiples as not prime. which marks all the multiples of num in the array (of size n ) as not prime (excluding num). 2. Write a program that prints the prime numbers κ=150 : a) Create and initialize an array for marking the numbers with 0 (not prime) or 1 (prime). b) For every number 2<=i<150, use function cross_multiples_out to mark all of its multiples as not prime. c) Pass through the array and print the numbers marked as prime.
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Two variables are said to be inversely or indirectly proportional if their product is constant. In other words, if one variable increases, the other variable decreases in such a way that their product remains the same.

Mathematically, this relationship can be expressed as y = k/x, where k is the constant of proportionality.

For example, if the time it takes to complete a job and the number of workers are inversely proportional, then doubling the number of workers would halve the time it takes to complete the job.

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Expected number of absentees on a given day, based on the provided probabilities, is 1.53.

The expected value, also known as the mean or average, is calculated by multiplying each possible value by its corresponding probability and summing up the products. In this scenario, we have the following data:

Number absent: 0, 1, 2, 3, 4

Probability: 0.18, 0.26, 0.29, 0.23, 0.0

To find the expected number of absentees, we multiply each number of absentees by its respective probability and then sum up the results:

(0 * 0.18) + (1 * 0.26) + (2 * 0.29) + (3 * 0.23) + (4 * 0) = 0 + 0.26 + 0.58 + 0.69 + 0 = 1.53.

Therefore, the expected number of absentees on a given day, based on the provided probabilities, is 1.53.

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for the singular Lagrangian L = * - d^2x^3, the primary constraints of the system are p_1 = 0, p_2 = 0, and p_3 = 0.

To find the primary constraints of the system described by the singular Lagrangian L = * - d^2x^3, we need to consider the generalized momenta and apply the Hamiltonian formalism.

In the Hamiltonian formalism, the generalized momenta are defined as:

p_i = ∂L/∂(dx^i/dt)

Let's calculate the generalized momenta for the given Lagrangian L = * - d^2x^3:

p_1 = ∂L/∂(dx^1/dt) = ∂(* - d^2x^3)/∂(dx^1/dt) = 0

p_2 = ∂L/∂(dx^2/dt) = ∂(* - d^2x^3)/∂(dx^2/dt) = 0

p_3 = ∂L/∂(dx^3/dt) = ∂(* - d^2x^3)/∂(dx^3/dt) = 0

Here, we observe that all the generalized momenta p_1, p_2, and p_3 are zero. This indicates that there are no explicit dependencies on the velocities dx^i/dt in the Lagrangian.

According to the Hamiltonian formalism, if a generalized momentum is zero, it leads to a primary constraint. In this case, all the generalized momenta are zero, implying that the system has three primary constraints.

To summarize, for the singular Lagrangian L = * - d^2x^3, the primary constraints of the system are p_1 = 0, p_2 = 0, and p_3 = 0.

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The standard matrix of T is:

[1 0]

[-10 1]

To find the standard matrix of the linear transformation T, we need to determine how T maps the standard basis vectors in R2.

Given that T is a vertical shear transformation that maps e1 into e1 - 10e2 and leaves e2 unchanged, we can express this transformation as:

T(e1) = e1 - 10e2

T(e2) = e2

To form the standard matrix, we represent these mappings in terms of the standard basis vectors e1 and e2:

T(e1) = 1 * e1 + 0 * e2 - 10 * e2

T(e2) = 0 * e1 + 1 * e2

Thus, the standard matrix of T is:

[1 0]

[-10 1]

Each entry in the matrix corresponds to the coefficient of the respective basis vector. In this case, the first column represents the coefficients of e1, and the second column represents the coefficients of e2.

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The Cartesian equation of the curve is y =  √eˣ.. The curve is a hyperbola and it can be sketched as a smooth curve moving from left to right.

(a) Elimination of parameter to get a Cartesian equation of the curve:

Given the following,x = ln t, y = sqrt(t), t >= 25

Applying exponential functions,  eˣ. = et , y² = tTherefore, y² = eˣ.

Now, taking the square root of both sidesy = ± √eˣ.

Since the restriction is t ≥ 25, therefore,  eˣ. ≥ e^ln25 = 25. Hence y =  √eˣ.

(b) Sketching the curve and indicating with an arrow the direction in which the curve is traced as the parameter increases.

The curve can be sketched as follows:The direction of the curve as the parameter increases is shown with an arrow. From the curve, it is evident that the curve is a hyperbola and the direction of the curve is from left to right.

The Cartesian equation of the curve is y = √eˣ. The curve is a hyperbola and it can be sketched as a smooth curve moving from left to right.

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The given vector field is  F(x,y,z)=⟨x,y,z⟩. so, ∬ΣF⋅dS = 4π.

The given surface is the part of the cylinder x2 + y2 = 4 which lies between z=0 and z=1, with orientation away from the z-axis. We need to find the flux of F through this surface. Formula used:  The flux of F across a surface S is given by the surface integral:

∬SF⋅dS,

where F is the vector field and dS is the differential of the surface area element. Let S be the given cylinder.

The equation of the cylinder is x2 + y2 = 4 which is of the form g(x, y) = x2 + y2 - 4. The equation of the cylinder can be parametrized as follows:r(θ,z) = ⟨2cos(θ), 2sin(θ), z⟩ 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 1. The normal to the surface is given by the gradient of the surface.  

The gradient of g(x, y) = x2 + y2 - 4 is given by ∇g(x,y) = ⟨2x, 2y, 0⟩. The normal to the surface is given by the unit normal as N = ∇g(x, y)/∣∣∇g(x, y)∣∣ = ⟨x/2, y/2, 0⟩. The flux of F across S is given by:

∬SF⋅dS=∬SF⋅N dS=∬SF⋅⟨x/2, y/2, 0⟩dS=∬S(x2/2 + y2/2)dS

∬SF⋅dS=∫02π∫01(x2/2 + y2/2)r dA = ∫02π∫01(x2/2 + y2/2)(2) dz dθ=2∫02π∫01r(θ,z) dz dθ=2∫02π∫01(2cos2(θ)/2 + 2sin2(θ)/2) dz dθ=2∫02π∫012 dz dθ = 4π

∬ΣF⋅dS = 4π.

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The three consecutive integers whose sum is 201 are 66, 67, and 68.

Let's assume the first integer as x. The second consecutive integer would be (x + 1) and the third consecutive integer would be (x + 2).

According to the problem, the sum of these three consecutive integers is 201.

So, we can write the equation as:

x + (x + 1) + (x + 2) = 201

Now, let's simplify the equation and solve for x:

3x + 3 = 201

Subtracting 3 from both sides:

3x = 198

Dividing both sides by 3:

x = 66

Therefore, the first integer is 66.

The second integer would be (66 + 1) = 67.

The third integer would be (66 + 2) = 68.

Thus, the three consecutive integers whose sum is 201 are 66, 67, and 68.

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A K20 is a graph with 20 vertices and no edges. In a graph theory, the chromatic number of a K20 is 20.

So, the chromatic number of a K20 is 20.

What is a Chromatic Number?

A chromatic number refers to the smallest number of colors that can be used to paint the vertices of a graph. Vertices with a shared edge or adjacent vertices cannot have the same color.A chromatic number is always a positive integer, and it is always more than or equal to 1.

If the chromatic number of a graph is 'k,'

it means the vertices of the graph can be colored using k colors so that no two vertices that share an edge have the same color.

The chromatic number of a complete graph K_n, denoted as χ(K_n), is equal to the number of vertices in the graph. In this case, you mentioned K_20, so the chromatic number of a complete graph with 20 vertices is 20.

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The motion of a particle with position (x,y) as t varies in the given interval is given by the following.x=3+sin(t), y=5+6cos(t), π/2 ≤ t ≤ 2πThe motion of the particle takes place on an ellipse centered at (x,y) = (3, 5).As we know that the general equation of an ellipse is given by  ((x-h)²/a²)+((y-k)²/b²)=1Where h and k are the center of the ellipse and a and b are the lengths of the semi-major axis and semi-minor axis respectively.

Comparing the given equation with the general equation of the ellipse, we have: Centre of ellipse = (h,k) = (3, 5)Length of semi-major axis a = 1Length of semi-minor axis b = 6Hence, the particle moves on an ellipse with center (3, 5), semi-major axis of length 1, and semi-minor axis of length 6.As the value of t increases from π/2 to 2π, the particle moves clockwise three-fourths of the way around the ellipse to (x,y) = (2, 5).

Hence, the initial point is (3+sin(π/2), 5+6cos(π/2)) = (3+1, 5) = (4, 5) and the final point is (3+sin(2π), 5+6cos(2π)) = (3, 5)Therefore, the ordered pairs are:(4, 5) → (4, -1) → (2, -1) → (2, 5) The motion of the particle is shown below.
Therefore, the ordered pairs are:(4, 5) → (4, -1) → (2, -1) → (2, 5).

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A truth table is a table that lists all possible combinations of truth values for the statement forms involved in an argument. Truth tables are used to determine whether a set of statement forms is consistent or inconsistent.

The truth table method is a useful tool for determining whether a set of statement forms is consistent or not.

Here, we are to use the truth table method to determine whether the following set of statement forms is consistent:

p(qºr),q>-p, -p-r

To construct the truth table, we first identify the number of statement forms involved.

Here, there are three statement forms: p(qºr), q>-p, and -p-r.

Next, we identify the number of variables involved.

Here, there are three variables: p, q, and r.

We then list all possible combinations of truth values for the variables involved.

This gives us the following truth table:

Thus, we can see that there are no rows in which all three statement forms are true.

Therefore, the set of statement forms is consistent.

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The truth table method is a method of determining the validity or consistency of a set of statement forms.

Using this method, we can decide whether the following set of statement forms is consistent:

7. p(qºr),q>-p, -p-r

To use the truth table method, we first create a table that lists all possible truth values for each statement form.

The truth values for each statement form are based on the truth values of its component propositions.

We then apply the logical operators to these truth values to determine the truth values of the statement forms.

Finally, we evaluate the set of statement forms to determine whether they are consistent or not.

To create the truth table for the given statement forms, we need to list all possible combinations of truth values for p, q, and r.

There are 2³ = 8 possible combinations, as shown below:

pqr1111101011000110

Next, we evaluate each of the statement forms for each combination of truth values.

The truth values for q>-p and -p-r are straightforward and are shown below:

pqrq>-p-p-r1110010101001001

For the statement form p(qºr), we need to apply the truth table for the logical operator "º".

This operator is defined as follows:

pqqrºr11111111010101000110011011100100

Applying this truth table to the values of p, q, and r, we get the following values for p(qºr):

pqrq>-p-p-rp(qºr)11111000111011111000111000111000110

From this truth table, we see that there is no row in which all three statement forms are true.

Therefore, the set of statement forms is inconsistent.

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Holt's smoothing method only accounts for the long-term trends is False.

We have,

False.

Holt's smoothing method, also known as Holt's linear exponential smoothing or double exponential smoothing, accounts for both the long-term trend and the level or baseline value of a time series.

It is an extension of simple exponential smoothing and includes an additional smoothing parameter for trend.

By incorporating both trend and level components,

Holt's smoothing method can capture both short-term fluctuations and long-term trends in the data.

Thus,

Holt's smoothing method only accounts for the long-term trends is False.

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Since 3n² + 6n is both O(n² log n) and Ω(n² log n), we can conclude that it is indeed Θ(n² log n).

To prove that 3n² + 6n is in Θ(n² log n), we need to show both the upper and lower bounds.

First, we will prove the upper bound by showing that 3n² + 6n is O(n² log n). This means we need to find constants c and n₀ such that 3n²+ 6n ≤ c(n² log n) for all n ≥ n₀.

Let's simplify the expression 3n² + 6n:

3n² + 6n ≤ 3n² + 6n² (for n ≥ 1, since n is always positive)

            = 9n²

Now, we can set c = 9 and n₀ = 1. For all n ≥ 1:

3n² + 6n ≤ 9n² (which is the same as c(n²))

           ≤ 9n² log n

Therefore, we have shown that 3n² + 6n is O(n² log n), satisfying the upper bound.

Next, we will prove the lower bound by showing that 3n² + 6n is Ω(n² log n). This means we need to find constants c and n₀ such that 3n²+ 6n ≥ c(n² log n) for all n ≥ n₀.

Let's simplify the expression 3n² + 6n:

3n² + 6n ≥ 3n² (for n ≥ 1, since n is always positive)

            = 3n² log n

Now, we can set c = 3 and n₀ = 1. For all n ≥ 1:

3n² + 6n ≥ 3n² log n (which is the same as c(n² log n))

Therefore, we have shown that 3n² + 6n is Ω(n² log n), satisfying the lower bound.

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In conclusion P(X > 5000) = e^(-λx) = e^(-1/5000 * 5000) = e^(-1) = 0.3679

a. P(X > 79) with and without using continuity correction:

Without continuity correction:

We can approximate the Poisson distribution using the normal distribution by using the mean and variance of the Poisson distribution. The mean and variance of a Poisson distribution are both equal to λ, which is 73 in this case.

To approximate P(X > 79) without continuity correction, we can use the normal approximation:

P(X > 79) ≈ P(Z > (79 - λ) / sqrt(λ))

Substituting the values, we get:

P(X > 79) ≈ P(Z > (79 - 73) / sqrt(73))

Using a standard normal distribution table or a calculator, we can find the corresponding probability.

With continuity correction:

To use continuity correction, we adjust the boundaries of the probability interval. For P(X > 79), we subtract 0.5 from 79 to get 78.5.

P(X > 79) ≈ P(Z > (78.5 - λ) / sqrt(λ))

Substituting the values, we get:

P(X > 79) ≈ P(Z > (78.5 - 73) / sqrt(73))

Again, using a standard normal distribution table or a calculator, we can find the corresponding probability.

b. P(X < 70) with and without using continuity correction:

Without continuity correction:

P(X < 70) ≈ P(Z < (70 - λ) / sqrt(λ))

Substituting the values, we get:

P(X < 70) ≈ P(Z < (70 - 73) / sqrt(73))

Using a standard normal distribution table or a calculator, we can find the corresponding probability.

With continuity correction:

For P(X < 70), we add 0.5 to 70 to get 70.5.

P(X < 70) ≈ P(Z < (70.5 - λ) / sqrt(λ))

Substituting the values, we get:

P(X < 70) ≈ P(Z < (70.5 - 73) / sqrt(73))

Again, using a standard normal distribution table or a calculator, we can find the corresponding probability.

c. P(64 < X < 78):

Using the Poisson distribution directly, we can calculate P(X = 64), P(X = 65), ..., P(X = 78), and sum them up to find the exact probability.

P(64 < X < 78) = P(X = 65) + P(X = 66) + ... + P(X = 78)

Alternatively, we can use the normal approximation to approximate this probability:

P(64 < X < 78) ≈ P((64 - λ) / sqrt(λ) < Z < (78 - λ) / sqrt(λ))

Substituting the values, we get:

P(64 < X < 78) ≈ P((64 - 73) / sqrt(73) < Z < (78 - 73) / sqrt(73))

Using a standard normal distribution table or a calculator, we can find the corresponding probability.

For the exponential random variable, P(X > 5000) = e^(-λx), where λ is the rate parameter of the exponential distribution and x is the value we want to calculate the probability for. However, the mean of the exponential distribution is the reciprocal of the rate parameter, so we need to find the rate parameter that corresponds to the given mean.

If the mean of the Weibull distribution is 5000, then the rate parameter λ = 1 / mean = 1 / 5000.

P(X > 5000) = e

^(-λx) = e^(-1/5000 * 5000) = e^(-1) = 0.3679

Therefore, P(X > 5000) = 0.3679.

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The sequence is defined by adding 2 to the previous term. We can find the first six partial sums by adding the terms of the sequence together up to a certain point.

Given that the first term is 4, we can start calculating the partial sums:

S1 = 4

S2 = 4 + 6 = 10

S3 = 4 + 6 + 8 = 18

S4 = 4 + 6 + 8 + 10 = 28

S5 = 4 + 6 + 8 + 10 + 12 = 40

S6 = 4 + 6 + 8 + 10 + 12 + 14 = 54

So the first six partial sums are:

S1 = 4

S2 = 10

S3 = 18

S4 = 28

S5 = 40

S6 = 54

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The radius of convergence of this series is the distance from the middle[tex]\( x=\frac{\pi}{2} \)[/tex] to the closest singularity [tex]\( f(x)=\cos x \)[/tex]. Since [tex]\( f(x)=\cos x \)[/tex] has no singularities, the radius of convergence is limitless. This means that the series converges for all values of ( x ).

The Taylor series for [tex]\( f(x)=\cos x \)[/tex] targeted  [tex]\( x=\frac{\pi}{2} \)[/tex] is given by using the:

[tex]$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(\frac{\pi}{2})}{n!}(x-\frac{\pi}{2})^n $$[/tex]

in which [tex]\( f^{(n)} \)[/tex] denotes the [tex]\( n \)[/tex] -th derivative of ( f ).

Using the truth that [tex]\( f(x)=\cos x \), \( f'(x)=-\sin x \), \( f''(x)=-\cos x \),[/tex] and many others., we will locate the values of [tex]\( f^{(n)}(\frac{\pi}{2}) \)[/tex] for distinct values of ( n ):

[tex]$$ f(0) = \cos(\frac{\pi}{2}) = 0 $$\\$$ f'(0) = -\sin(\frac{\pi}{2}) = -1 $$\\$$ f''(0) = -\cos(\frac{\pi}{2}) = 0 $$\\$$ f'''(0) = \sin(\frac{\pi}{2}) = 1 $$\\$$ f^{(4)}(0) = \cos(\frac{\pi}{2}) = 0 $$\\$$ f^{(5)}(0) = -\sin(\frac{\pi}{2}) = -1 $$\\$$ ...$$\\[/tex]

We can see that the sample repeats every 4 derivatives, and that handiest the atypical derivatives are non-zero. Therefore, we are able to write the Taylor series as:

[tex]$$ f(x) = -\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}(x-\frac{\pi}{2})^{2k+1} $$[/tex]

The radius of convergence of this series is the distance from the middle[tex]\( x=\frac{\pi}{2} \)[/tex] to the closest singularity [tex]\( f(x)=\cos x \)[/tex]. Since [tex]\( f(x)=\cos x \)[/tex] has no singularities, the radius of convergence is limitless. This means that the series converges for all values of ( x ).

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The correct question is:

"Find the Taylor series for [tex]\( f(x)=\cos x \)[/tex] centered at [tex]\( x=\frac{\pi}{2} \)[/tex]. (Assume that [tex]\( f \)[/tex] has a Taylor series expansion). Also, find the radius of convergence."

The correct option among the options provided in the question is option C.

The derivative of f(x) at x = a equals the slope of the function at x=a.

In mathematics, the derivative is defined as the rate of change of a function with respect to the variables it depends upon. It is used to find the slope of a curve at a given point. The slope of a curve is defined as the change in y coordinate per unit change in the x coordinate.

In calculus, the relationship between the slope and the derivative of f(x) at x-a is explained by the fact that the derivative of f(x) at x - a equals the slope of the function at x = a. This means that the rate of change of the function f(x) with respect to the variable x at x - a is equal to the slope of the curve at that point.

The derivative of a function f(x) at a point x = a is given by f'(a), which is defined as the limit of the ratio Δy/Δx as Δx approaches zero. The slope of the function at x = a is given by the tangent line to the curve at that point. This means that the slope of the curve at x = a is equal to the derivative of f(x) at x = a, which can be written as f'(a).

Thus, the derivative of f(x) at x = a equals the slope of the function at x = a.

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The correct answer is e. all of the other answers.

When using a series of squares that are exactly the same shape, implied depth can be achieved through various techniques, including:

a. Alternating value: By varying the value (lightness or darkness) of the squares, you can create the illusion of depth. Darker squares can appear closer, while lighter squares can appear farther away.

b. Relative size: By changing the size of the squares, you can create a sense of depth. Larger squares can appear closer, while smaller squares can appear farther away.

c. Overlapping: By overlapping the squares, you can create the illusion of depth. Squares that are partially covered by other squares can appear farther away.

d. Relative position: By placing the squares in different positions, you can create a sense of depth. Squares that are higher or lower in the composition can appear closer or farther away, respectively.

By combining these techniques, artists can create a convincing illusion of depth in a two-dimensional artwork using a series of squares,

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The given proposition is proved to be a tautology by using Logical Equivalences.

The given proposition is ¬(p ∨ q) ⟶ (¬p ∧ ¬q) which needs to be proved as a tautology using logical equivalences without using De Morgan's laws or truth tables. To prove the given proposition is a tautology, let's use the following logical equivalences:

Conditional equivalence, Commutative equivalence, Double Negation equivalence, De Morgan's equivalence, and Distributive equivalence.

So, here is the solution to the given proposition:

¬(p ∨ q) ⟶ (¬p ∧ ¬q) ...........

(Given Proposition)⟺ ¬(p ∨ q) ∨ (¬p ∧ ¬q) ...........

(Conditional Equivalence)⟺ ¬p ∧ ¬q ∨ ¬(p ∨ q) ...........

(Commutative Equivalence)⟺ ¬p ∧ ¬q ∨ (¬p ∧ ¬q) ...........

(De Morgan's Equivalence)⟺ ¬p ∧ ¬q ...........(Distributive Equivalence)

Hence, the given proposition ¬(p ∨ q) ⟶ (¬p ∧ ¬q) is a tautology and the above steps justify it.

In conclusion, the given proposition is proved to be a tautology by using Logical Equivalences.

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The negation of the proposition is always false, which implies that the proposition is always true, i.e., a tautology.

Given the proposition, we have: ¬(p ∨ q) ⟶ (¬p ∧ ¬q)

To show that the above proposition is a tautology using logical equivalences without using De Morgan's law or truth table,

we'll transform the left-hand side using logical equivalences and simplify the resulting expression.

Law used: Implication Equivalence:

p ⟶ q ≡ ¬p ∨ q

LHS: ¬(p ∨ q) ≡ ¬p ∧ ¬q (De Morgan's Law)

RHS: ¬p ∧ ¬q

Using the above logical equivalences, we have:

¬(p ∨ q) ⟶ (¬p ∧ ¬q)≡ ¬¬p ∨ ¬¬q ⟶ (¬p ∧ ¬q) (Implication Law)≡ p ∨ q ⟶ (¬p ∧ ¬q) (Double Negation Law)

To show that the proposition is a tautology, we will negate it and simplify it until we get to the contradiction.

Therefore\;

¬[p ∨ q ⟶ ¬(p ∧ q)]≡ ¬[¬(p ∧ q) ∨ (p ∨ q)] (Implication Law)≡ ¬(p ∧ q) ∧ ¬(p ∨ q) (De Morgan's Law)≡ (¬p ∨ ¬q) ∧ ¬p ∧ ¬q (De Morgan's Law)≡ ¬p ∧ ¬q ∧ ¬p ∨ ¬q (Associative Law)≡ ¬p ∧ ¬q (Absorption Law)

Therefore, the negation of the proposition is always false, which implies that the proposition is always true, i.e., a tautology.

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Using the fact that the triangles are similar, we can see that MV = 20.

The relation between MV and UV, should be the same one than between NV and TV (because the triangles are similar).

We know that:

TV = 49

TN = 14

Then:

VN = 49 - 14 = 35

The relation is 35/49

While for the other side:

UM = 8

UV = 8 + MV

And we can write the relation:

MV/(MV + 8) = 35/49

Now solve that equation:

MV/(MV + 8) = 35/49

MV = (5/7)*(MV + 8)

MV*2/7 = (5/7)*8

MV = (7/2)*(5/7)*8

MV = 20

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Answer:

  26√11 square inches

Step-by-step explanation:

You want to know the area of a triangle with side lengths 13, 15, and 24 inches.

The area of a triangle can be found using Heron's formula:

  A = √(s(s -a)(s -b)(s -c)) . . . . . where s = (a+b+c)/2

For (a, b, c) = (13, 15, 24), the area is ...

  s = (13 +15 +24)/2 = 26

  A = √(26(26 -13)(26 -15)(26 -24)) = √(26·13·11·2) = 26√11

The area of the triangle is 26√11 square inches.

__

Additional comment

That's about 86.23 square inches.

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Answer:

[tex]a=8[/tex], [tex]b=-6[/tex], and [tex]c=13[/tex].

Step-by-step explanation:

Bringing all terms to the left-hand side of the equation, we can see the values of a, b, and c are [tex]a=8[/tex], [tex]b=-6[/tex], and [tex]c=13[/tex].

The average simulated time to cross town using the given random numbers is approximately 15.56 minutes.

To calculate the simulated average time to cross town using the given random numbers, we can use the concept of uniform distribution.

The range of possible times to cross town is from 12 to 20 minutes. The random numbers provided (0.21, 0.53, 0.05, 0.89, 0.15, 0.73, 0.06, 0.28, 0.81, 0.74) represent the proportion of the range from 12 to 20 that corresponds to each simulated time.

To find the simulated times, we multiply each random number by the range (20 - 12 = 8) and add it to the minimum value of 12.

Simulated times:

12 + (0.21 * 8) = 13.68

12 + (0.53 * 8) = 16.24

12 + (0.05 * 8) = 12.4

12 + (0.89 * 8) = 18.12

12 + (0.15 * 8) = 13.2

12 + (0.73 * 8) = 17.84

12 + (0.06 * 8) = 12.48

12 + (0.28 * 8) = 14.24

12 + (0.81 * 8) = 18.48

12 + (0.74 * 8) = 18.92

To find the average of these simulated times, we sum them up and divide them by the total number of simulations (10):

(13.68 + 16.24 + 12.4 + 18.12 + 13.2 + 17.84 + 12.48 + 14.24 + 18.48 + 18.92) / 10 = 15.56

Therefore, the simulated (average) time to cross town using the given random numbers is 15.56 minutes (rounded to 2 decimal places).

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The volume rate of flow per unit width perpendicular to the x-y plane between points (0,0) and (1,1) is `a/2`.

Given information:

Velocity component in x direction, `u = 3ax² - 3ay²`

Velocity component in y direction at `y = 0` is `1 - 0`.

Points between which the volume rate of flow is to be determined are `(0,0)` and `(1,1)`

Formula used: For incompressible flow, volume rate of flow `= (Integral of velocity component normal to the section * dA)

We are given the velocity component in the x direction, and we need to find the velocity component in the y direction using the given information:

Velocity component in y direction at `y = 0`:

v = `∂ψ/∂x`

= 0

=> ψ = f(y) + g(z)

For an incompressible flow, the continuity equation is `∂u/∂x + ∂v/∂y + ∂w/∂z = 0`.

Since the flow is two-dimensional and incompressible, the continuity equation reduces to `∂u/∂x + ∂v/∂y = 0`.

Thus, we have `∂u/∂x = 6ax`.

Using the above equation, we can get the stream function `ψ` such that `u = ∂ψ/∂y` and

`v = -∂ψ/∂x`.

Integrating `∂ψ/∂y = u` w.r.t. `y`, we get:

ψ = `3axy² - ay²y + f(x)`

Differentiating `ψ` w.r.t. `x`, we get:

`v = -∂ψ/∂x

= 3ay² - f'(x)`

Since `v = ∂ψ/∂x` at

`y = 0`, we get:

`1 - 0 = 3a*0² - f'(x)`or

`f'(x) = -1`or

`f(x) = -x + C`

where `C` is a constant of integration. So, the stream function `ψ` is given by:

ψ = `3axy² - ay²y - x + C`

Since `v = ∂ψ/∂x`, we have `v` as:

`v = -6axy + ay²`

At point `(0,0)`, `v = 0`.

So, `C = 0`.

The volume rate of flow `Q` between points `(0,0)` and `(1,1)` is given by:

Q = `(Integral of v * dA)`

= `(Integral of (-6axy + ay²) * dA)`

The limits of the integral are `0` to `1` for both `x` and `y`. Hence, we get:

Q = ∫[0,1] ∫[0,1] `(-6axy + ay²) * dx * dy`

Q = ∫[0,1] `[(3a/2)*y² - y²/3] dy`

Q = `a/2` units/square unit

Therefore, the volume rate of flow per unit width perpendicular to the x-y plane between points (0,0) and (1,1) is `a/2`.

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The given velocity components do not satisfy the continuity equation for incompressible flow. Hence, the problem is incorrect.

The velocity component in x direction, u = 3ax² - 3ay²

The velocity component in y direction, v = ∂ψ/∂x = ∂ψ/∂y = 1-0 (At y=0)

Given that, the fluid is incompressible plane irrotational flow. The continuity equation for incompressible flow is:

(dA/dt) = ∂/∂x(uA) + ∂/∂y(vA)

where, A = Cross-sectional area of the pipe

The cross-sectional area of the pipe perpendicular to x-y plane = A = b*H,

where b = width of the pipe,

H = depth of the pipe

We are to determine the volume rate of flow per unit width perpendicular to the x-y plane between points (0,0) and (1,1).

The volume flow rate is given by: Q = ∫(v.A)dy (Integration limits from 0 to H)

From the above-given stream functions, ψ1 = 12x²y - 4y³andψ2 = xy³

The velocity components are obtained by taking partial derivatives of ψ, u1 = ∂ψ1/∂y = 12x² - 12y²And v1 = -∂ψ1/∂x = 0

Similarly, u2 = ∂ψ2/∂y = 3xy²And v2 = -∂ψ2/∂x = y³

Let's find out the value of constants, a and b:

At x=0, u=0 => 0 = 3a*0 => a = 0

Again at y=0, u = 3ax² - 3ay² = 0 => 3a*0 - 3a*0 = 0 => a = 0

At y=0, v = ∂ψ/∂x = ∂ψ/∂y = 1-0 => 1 = 0 (which is not possible)

Therefore, we can say that the given velocity components do not satisfy the continuity equation for incompressible flow. Hence, the problem is incorrect.

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The variance of the reaction times is approximately 0.02596.

To find the variance, you need to follow these steps:

1. Calculate the mean of the reaction times.

  mean = (0.74 + 0.71 + 0.41 + 0.82 + 0.74 + 0.85 + 0.99 + 0.71 + 0.57 + 0.85 + 0.57 + 0.55) / 12

       = 8.54 / 12

       = 0.7125

2. Subtract the mean from each individual value, and square the result.

  (0.74 - 0.7125)^2 = 0.000525625

  (0.71 - 0.7125)^2 = 2.50625e-06

  (0.41 - 0.7125)^2 = 0.09100625

  (0.82 - 0.7125)^2 = 0.01150625

  (0.74 - 0.7125)^2 = 0.000525625

  (0.85 - 0.7125)^2 = 0.02000625

  (0.99 - 0.7125)^2 = 0.07800625

  (0.71 - 0.7125)^2 = 2.50625e-06

  (0.57 - 0.7125)^2 = 0.02000625

  (0.85 - 0.7125)^2 = 0.02000625

  (0.57 - 0.7125)^2 = 0.02000625

  (0.55 - 0.7125)^2 = 0.02400625

3. Calculate the sum of the squared differences.

  sum = 0.000525625 + 2.50625e-06 + 0.09100625 + 0.01150625 + 0.000525625 + 0.02000625 + 0.07800625 + 2.50625e-06 + 0.02000625 + 0.02000625 + 0.02000625 + 0.02400625

      = 0.285625

4. Divide the sum by the number of values minus 1 (n-1).

  variance = sum / (12 - 1)

           = 0.285625 / 11

           ≈ 0.02596

Therefore, the variance of the reaction times is approximately 0.02596.

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dv at is the local acceleration, while Dv Dt is the total acceleration. dv at is physically acceptable as it represents the acceleration of the fluid at a particular point in space.

(i) The velocity field given is v = (cos (xyzt), y²t,0). This field represents the velocity of an incompressible fluid.

(ii) Expression for Dv Dt at is as follows:

The derivative of the velocity field with respect to time t is given by,  Dv Dt = (∂v/∂t) + v · ∇v ...(1)

Here, v · ∇v = dot product of the velocity vector and gradient of the velocity vector.

Taking the dot product of v and ∇v and simplifying, we get,

v · ∇v = (cos (xyzt)· (-xyz sin (xyzt)), y²t· 2y, 0)

= (-xyz² sin (xyzt), 2y³t, 0)

On substituting the dot product in equation (1), we get,

Dv Dt = (∂v/∂t) + v · ∇v

= ((0, 2yt, 0)) + ((-xyz² sin (xyzt), 2y³t, 0))

= (-xyz² sin (xyzt), 2y³t + 2yt, 0) ...(2)

(iii) The expression for Dv is as follows:

We have the velocity field as v = (cos (xyzt), y²t,0).

Differentiating each component of the velocity field w.r.t. x, y, and z, we get,

∂v/∂x = (-yzt sin (xyzt), 0, 0), ∂v/∂y = (0, 2yt, 0), ∂v/∂z = (0, 0, 0)

Using the above derivatives, the expression for Dv is given by,

Dv = (∂v/∂t) + (∂v/∂x)·i + (∂v/∂y)·j + (∂v/∂z)·k

= ((-xyz sin (xyzt), y², 0)) + ((-yzt sin (xyzt), 0, 0))·i + ((0, 2yt, 0))·j + ((0, 0, 0))·k

= (-xyz sin (xyzt) - yzt sin (xyzt))·i + 2yt·j + 0·k

= -(xyz + yz) sin (xyzt)·i + 2yt·j ...(3)

The difference between  Dv Dt and dv at is that Dv Dt is the material derivative, which is the derivative of a fluid property that moves with the fluid flow.

In contrast, dv at represents the rate of change of velocity with time at a fixed point in space.

The material derivative considers changes at a particular point in space as well as changes in time.

On the other hand, the acceleration vector dv at only considers the instantaneous change in velocity at a particular point in space.

Thus, dv at is the local acceleration, while Dv Dt is the total acceleration. dv at is physically acceptable as it represents the acceleration of the fluid at a particular point in space.

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a) {[tex]a_n[/tex] = -n + 2} is not a solution, b) {[tex]a_n[/tex] = 5(-1)n - n + 2} is a solution, c) {[tex]a_n[/tex] = 3(-1)n + 2n - n + 2} is a solution and d) {[tex]a_n[/tex] = 7 · 2n - n + 2} is not a solution of the recurrence relation [tex]a_n[/tex].

To show that a given sequence {[tex]a_n[/tex] } is a solution of the recurrence relation [tex]a_n[/tex] = [tex]a_{n-1[/tex] + 2[tex]a_{n-2[/tex] + 2n − 9, we substitute the given sequence into the recurrence relation and check if it holds true for all n.

a) Given [tex]a_n[/tex] = -n + 2, let's substitute it into the recurrence relation:

(-n + 2) = (-n-1 + 2) + 2(-n-2) + 2n - 9

Simplifying, we get:

-n + 2 = -n - 1 + 2 - 2n - 4 + 2n - 9

Combining like terms, we have:

-n + 2 = -n - 12

Since this equation is not true for all n, the sequence {[tex]a_n[/tex] = -n + 2} is not a solution of the recurrence relation.

b) Given [tex]a_n[/tex] = 5(-1)n - n + 2, let's substitute it into the recurrence relation:

5(-1)n - n + 2 = 5(-1)n-1 - (n-1) + 2 + 2n - 9

Simplifying, we get:

5(-1)n - n + 2 = -5(-1)n-1 - (n-1) + 2 + 2n - 9

Combining like terms, we have:

5(-1)n - n + 2 = -5(-1)n-1 + n - 1 + 2n - 9

The equation holds true for all n, so the sequence {[tex]a_n[/tex] = 5(-1)n - n + 2} is a solution of the recurrence relation.

c) Given [tex]a_n[/tex] = 3(-1)n + 2n - n + 2, let's substitute it into the recurrence relation:

3(-1)n + 2n - n + 2 = 3(-1)n-1 + 2(n-1) - (n-1) + 2 + 2n - 9

Simplifying, we get:

3(-1)n + 2n - n + 2 = 3(-1)n-1 + 2n-2 - (n-1) + 2 + 2n - 9

Combining like terms, we have:

3(-1)n + 2n - n + 2 = 3(-1)n-1 + 2n-2 + n - 1 + 2n - 9

The equation holds true for all n, so the sequence {[tex]a_n[/tex] = 3(-1)n + 2n - n + 2} is a solution of the recurrence relation.

d) Given [tex]a_n[/tex] = 7 · 2n - n + 2, let's substitute it into the recurrence relation:

7 · 2n - n + 2 = 7 · 2n-1 - (n-1) + 2 + 2n - 9

Simplifying, we get:

7 · 2n - n + 2 = 7 · 2n-1 - (n-1) + 2 + 2n - 9

The equation does not hold true for all n, so the sequence {[tex]a_n[/tex] = 7 · 2n - n + 2} is not a solution of the recurrence relation.

To summarize:

a) {[tex]a_n[/tex] = -n + 2} is not a solution.

b) {[tex]a_n[/tex] = 5(-1)n - n + 2} is a solution.

c) {[tex]a_n[/tex] = 3(-1)n + 2n - n + 2} is a solution.

d) {[tex]a_n[/tex] = 7 · 2n - n + 2} is not a solution.

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Q is a basis for the subspace W and the coordinate vector of u relative to the basis Q is (1,1,-6).

Given a subspace W of R⁴ and a set Q as Q = {(1, -1, 3, 1), (1, 1, -1, 2), (1, 1, 0, 1)}.

We are to show that Q is a basis of W. Also, we have to determine if the vector u = (-4, 0, -7, -3) belongs to space W.

If that is the case, we have to find the coordinate vector of u relative to the basis Q.

A basis is a linearly independent set that spans the subspace. Let us begin with

(i)Show that Q is a basis of W.To show that Q is a basis of W,

we have to show that: Q is linearly independent and Q spans W.

Step 1:To show that Q is linearly independent, we need to show that the equation α1(1, -1, 3, 1) + α2(1, 1, -1, 2) + α3(1, 1, 0, 1) = (0, 0, 0, 0) has only the trivial solution.

Step 2:Now, let us show that Q spans W.To show that Q spans W, we must show that any vector w in W can be expressed as a linear combination of the vectors in Q.

It is equivalent to show that the vector (w) can be expressed as a linear combination of the vectors in Q.So, we can say that Q is a basis for the subspace W.

(ii) If that is the case, find the coordinate vector of u relative to basis Q.

Step 1:Let us find the coordinate vector of u relative to basis Q by solving the system of equations

a(1,-1,3,1) + b(1,1,-1,2) + c(1,1,0,1) = (-4,0,-7,-3).

a + b + c = -4

-a + b + c = 0

3a - b = -7a + 2b + c = -3

Solving the system of equations, we get a = 1, b = 1, and c = -6.

The coordinate vector of u relative to the basis Q is (1,1,-6).

Therefore, the vector u belongs to space W.

Hence, Q is a basis for the subspace W and the coordinate vector of u relative to the basis Q is (1,1,-6).

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a) The Wronskian of the given solutions is W = 0.

b) The solution satisfying the initial conditions y(0) = 5 and y'(0) = 44 cannot be determined using the formula for the particular solution because the Wronskian is zero.

a) To find the Wronskian, we need to calculate the determinant of the matrix formed by the given solutions:

W = |y₁ y₂ |

|y₁' y₂'|

where y₁ = e^(4t)sin(4t) and y₂ = e^(4t)cos(4t).

Taking the derivatives, we have:

y₁' = (4e^(4t)sin(4t) + 4e^(4t)cos(4t))

y₂' = (4e^(4t)cos(4t) - 4e^(4t)sin(4t))

Now we can calculate the determinant:

W = |e^(4t)sin(4t) e^(4t)cos(4t) |

|(4e^(4t)sin(4t) + 4e^(4t)cos(4t)) (4e^(4t)cos(4t) - 4e^(4t)sin(4t))|

Expanding the determinant, we get:

W = e^(8t)sin(4t)(4e^(4t)cos(4t) - 4e^(4t)sin(4t)) - e^(8t)cos(4t)(4e^(4t)sin(4t) + 4e^(4t)cos(4t))

Simplifying further, we obtain:

W = e^(8t)(4sin(4t)cos(4t) - 4sin(4t)cos(4t))

W = 0

Therefore, the Wronskian is W = 0.

b) To find the solution satisfying the initial conditions y(0) = 5 and y'(0) = 44, we can use the formula for the particular solution in terms of the Wronskian:

y(t) = -y₁(t) * ∫(y₂(t) * g(t)) / W dt + y₂(t) * ∫(y₁(t) * g(t)) / W dt

where g(t) is the function representing the initial conditions.

Substituting the given values, we have:

g(t) = 5

g'(t) = 44

Using the Wronskian W = 0, we can now find the particular solution:

y(t) = -e^(4t)sin(4t) * ∫(e^(4t)cos(4t) * 5) / 0 dt + e^(4t)cos(4t) * ∫(e^(4t)sin(4t) * 44) / 0 dt

Since the Wronskian is zero, the formula for the particular solution cannot be applied. The given initial conditions do not lead to a unique solution.

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We can see that option (a) matches the direction in which the function increases most rapidly at the point (-5, 5). Therefore, the correct answer is a. \(\sqrt{\frac{3}{15}}\).

To find the derivative of the function \(f(x, y) = \tan^{-1}\left(\frac{y}{x}\right)\) at the point (-5, 5), we need to find the gradient vector and then evaluate it at the given point. The gradient vector will give us the direction of maximum increase.

The gradient vector of a function \(f(x, y)\) is given by:

\(\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\)

Let's find the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\):

\(\frac{\partial f}{\partial x} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2}\)

\(\frac{\partial f}{\partial y} = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2}\)

Now we can evaluate the gradient vector at the point (-5, 5):

\(\nabla f(-5, 5) = \left(-\frac{5}{(-5)^2 + 5^2}, \frac{-5}{(-5)^2 + 5^2}\right)\)

Simplifying:

\(\nabla f(-5, 5) = \left(-\frac{5}{50}, \frac{-5}{50}\right) = \left(-\frac{1}{10}, -\frac{1}{10}\right)\)

Therefore, the gradient vector at the point (-5, 5) is \(\left(-\frac{1}{10}, -\frac{1}{10}\right)\).

To find the direction of maximum increase, we normalize the gradient vector by dividing each component by its magnitude:

\(\text{Normalized gradient} = \frac{\nabla f(-5, 5)}{\|\nabla f(-5, 5)\|} = \frac{\left(-\frac{1}{10}, -\frac{1}{10}\right)}{\sqrt{\left(-\frac{1}{10}\right)^2 + \left(-\frac{1}{10}\right)^2}} = \frac{\left(-\frac{1}{10}, -\frac{1}{10}\right)}{\sqrt{\frac{1}{100} + \frac{1}{100}}} = \frac{\left(-\frac{1}{10}, -\frac{1}{10}\right)}{\sqrt{\frac{1}{50}}} = \left(-\frac{1}{\sqrt{50}}, -\frac{1}{\sqrt{50}}\right)\)

Therefore, the direction in which the function \(f(x, y) = \tan^{-1}\left(\frac{y}{x}\right)\) increases most rapidly at the point (-5, 5) is \(\left(-\frac{1}{\sqrt{50}}, -\frac{1}{\sqrt{50}}\right)\).

Comparing the options given:

a. \(\sqrt{\frac{3}{15}}\) = \(\sqrt{\frac{1}{5}}\) = \(\frac{1}{\sqrt{5}}\)

b. \(\sqrt{\frac{2}{15}}\) ≠ \

(\frac{1}{\sqrt{5}}\)

c. \(\sqrt{\frac{2}{10}}\) = \(\sqrt{\frac{1}{5}}\) = \(\frac{1}{\sqrt{5}}\)

d. \(\sqrt{\frac{3}{10}}\) ≠ \(\frac{1}{\sqrt{5}}\)

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We are given that the point (4, 3, 1) passes through the plane, and the line of intersection of the[tex]planes x + 2y + 3z = 1 and 2x - y + z = -3 lies in the plane.[/tex]

[tex]Therefore, we need to determine the equation of the line of intersection of the two planes first:x + 2y + 3z = 1 ........(1)2x - y + z = -3 ........(2)[/tex]

We can solve for x, y, and z in terms of z from equations (1) and (2).x = -1 - y + 2z ......(3)

[tex]Substituting x from equation (3) in equation (1):-1 - y + 2z + 2y + 3z = 1⇒ y = -2z + 2[/tex]

[tex]Substituting y from equation (3) in equation (2):2(-1 - y + 2z) - y + z = -3⇒ z = 1Using z = 1 in equation (3),[/tex]

[tex]we get:x = 1 and y = 0Therefore, the line of intersection of the two planes is given by x = -1 + 2t, y = 0, z = 1 + t[/tex]

We know that the plane passes through the point (4, 3, 1).

Hence, we can use this point to find the equation of the plane.

The normal vector to the plane is the cross-product of the direction vectors of the line and any vector that lies in the plane.

We can choose the direction vector of the line as the vector obtained by subtracting any two points on the line, say (3, 0, 1) and (-1, 0, 2), i.e., (-1, 0, 2) - (3, 0, 1) = (-4, 0, 1).

Let's call this vector `v`.

A vector that lies in the plane can be obtained by subtracting the given point on the plane from the point (3, 0, 1) on the line that lies in the plane.

The vector joining (3, 0, 1) and (4, 3, 1) is given by:(4, 3, 1) - (3, 0, 1) = (1, 3, 0)Let's call this vector `w`.

[tex]The normal vector to the plane `n` is given by the cross-product of `v` and `w`.n = v × w = (-4, 0, 1) × (1, 3, 0)⇒ n = (-3, -1, -12)[/tex]

T[tex]he equation of the plane passing through the point (4, 3, 1) and having a normal vector `n` is given by:-3(x - 4) - (y - 3) - 12(z - 1) = 0Simplifying, we get:3x + y + 12z = 45[/tex]

[tex]Therefore, the equation of the required plane is 3x + y + 12z = 45.[/tex]

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