Give a geometric description of the following systems of equations. 1. {−2x+10y=−10−4x+20y=−20​ 2. {−2x+10y=−10−4x+20y=−17​ 3. {7x−3y=−6x+4y=​6−6​ Note: You can earn partial credit on this problem. Problem 17. (1 point) Solve the system of equations. e=f=​ help (fractions) help (fractions) ​ Note: You can earn partial credit on this problem.

The proof by contraposition involves showing that if n is odd, then 3n + 2 is odd. On the other hand, the proof by contradiction assumes the opposite, that if n is odd, then 3n+2is even, and leads to a contradiction.

(a) Proof by Contraposition:

To prove that if \(3n + 2\) is even, then \(n\) is even, we can use a proof by contraposition. The contrapositive of a statement \(p \rightarrow q\) is \(\neg q \rightarrow \neg p\). In this case, the original statement is \(3n + 2\) is even \(\rightarrow\) \(n\) is even.

To prove the contrapositive, assume that \(n\) is odd, i.e., \(n = 2k + 1\) for some integer \(k\). Substituting this value of \(n\) into the original equation, we have:

\(3(2k + 1) + 2 = 6k + 5\).

We need to show that \(6k + 5\) is odd. By definition, an odd number can be represented as \(2m + 1\) for some integer \(m\). If we assume \(6k + 5\) is even, we can write it as \(2m\), where \(m\) is an integer.

By equating both expressions, we have:

\(6k + 5 = 2m\).

Rearranging the equation, we get:

\(6k = 2m - 5\).

The right side is an odd number, but the left side is divisible by 6, which is an even number. This is a contradiction since an even number cannot be equal to an odd number. Therefore, our assumption that \(6k + 5\) is even must be false.

Since we have proved the contrapositive statement, we can conclude that if \(3n + 2\) is even, then \(n\) must be even.

(b) Proof by Contradiction:

To prove that if \(3n + 2\) is even, then \(n\) is even, we can use a proof by contradiction. We assume the negation of the statement and show that it leads to a contradiction.

Assume that \(3n + 2\) is even, but \(n\) is odd. So we can express \(n\) as \(2k + 1\) for some integer \(k\). Substituting this value into the equation, we have:

\(3(2k + 1) + 2 = 6k + 5\).

We assumed that \(3n + 2\) is even, so we can express it as \(2m\) for some integer \(m\):

\(6k + 5 = 2m\).

Rearranging the equation, we get:

\(6k = 2m - 5\).

The right side is an odd number, but the left side is divisible by 6, which is an even number. This contradicts the assumption that both sides are equal. Hence, our initial assumption that \(n\) is odd must be false.

Therefore, we conclude that if \(3n + 2\) is even, then \(n\) must be even.

In both the proof by contraposition and the proof by contradiction, we have shown that if \(3n + 2\) is even, then \(n\) must be even.

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The value of r is 6.

The value of y is 509.

The given equations are:

y = (2t4 - 3)

A = 7 - (r/6)

Now, we have to find the value of y when A = 4/3

We can substitute A = 4/3 in the equation of A to find the value of r

A = 7 - (r/6)4/3 = 7 - (r/6)(4/3)4/3

= 7 - (4r/18)4/3 + (4r/18)

= 718/3 + 2r/9

= 7(9/3) + 2r/9

= 721 + 2r/9

= 21(2r/9)

= 21 - 92r/9

= 63r/9

= 9 * 2r

= 6

Therefore, the value of r is 6.

Substituting the value of r in the equation of y:

y = (2t4 - 3)

y = (2 × 4⁴ - 3)

y = (2 × 256 - 3)

y = (512 - 3)

y = 509

Therefore, the value of y is 509.

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Complete Question:

Find the values of r and y:

y = (2t4 - 3)

A = 7 - (r/6)

The events "U: unemployed" and "N: has no diploma" are dependent. They are not independent because P(U and N) does not equal P(U)P(N).

In the given information, P(U) = 0.01277, P(N) = 0.13722, and P(U and N) = 0.008055. If the events were independent, we would expect P(U and N) to be equal to the product of P(U) and P(N), which is not the case here. Therefore, the events are dependent.

To check for independence, we use the formula P(U and N) = P(U)P(N) for independent events. In this case, P(U and N) does not equal P(U)P(N), indicating that the events are dependent.

Thus, we can conclude that the events "U: unemployed" and "N: has no diploma" are dependent, as the occurrence of one event affects the probability of the other event.

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a) The probability that the time to the first accident is greater than 2 weeks can be calculated using the exponential distribution. The rate parameter (λ) for the exponential distribution is equal to the average number of events per unit of time, which in this case is 2 accidents per week.

To find the probability, we need to calculate the cumulative distribution function (CDF) of the exponential distribution at 2 weeks. The CDF of the exponential distribution is given by 1 - exp(-λx), where x is the time.

Let's calculate it:

CDF = 1 - exp(-2 * 2) = 1 - exp(-4) ≈ 0.9820

Therefore, the probability that the time to the first accident is greater than 2 weeks is approximately 0.9820.

b) To find the probability that the time to the first accident is less than 2 days (2/7 week), we can use the same exponential distribution with a rate parameter of 2 accidents per week.

CDF = 1 - exp(-2 * (2/7)) ≈ 0.468

Therefore, the probability that the time to the first accident is less than 2 days (2/7 week) is approximately 0.468.

c) The mean time to the first accident can be calculated using the formula: mean = 1/λ, where λ is the rate parameter.

mean = 1 / 2 ≈ 0.50 weeks

Therefore, the mean time to the first accident is approximately 0.50 weeks.

d) The variance of the time to the first accident for the exponential distribution is given by the formula: variance = 1/λ^2.

variance = 1 / (2^2) = 1/4 = 0.25 weeks^2

Therefore, the variance of the time to the first accident is 0.25 weeks^2.

a) The probability that the time to the first accident is greater than 2 weeks is approximately 0.9820. b) The probability that the time to the first accident is less than 2 days (2/7 week) is approximately 0.468. c) The mean time to the first accident is approximately 0.50 weeks. d) The variance of the time to the first accident is 0.25 weeks^2.

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To simplify the expression \( \frac{\tan^2(x) - 3\tan(x) - 4}{\sin(x)\tan(x) + \sin(x)} \), we can factor the numerator and denominator, cancel out common factors, and simplify the resulting expression.

1. For the expression \( \frac{\tan^2(x) - 3\tan(x) - 4}{\sin(x)\tan(x) + \sin(x)} \), we can factor the numerator as \( (\tan(x) - 4)(\tan(x) + 1) \) and the denominator as \( \sin(x)(\tan(x) + 1) \). Cancelling out the common factor of \( \tan(x) + 1 \), we are left with \( \frac{\tan(x) - 4}{\sin(x)} \) as the simplified expression.

2. To express \( \cos \left(\frac{3\pi}{4} - x\right) - \sin \left(\frac{3\pi}{4} - x\right) = -\sqrt{2}\cos(x) \) in radians, we need to convert the angles from degrees to radians. \( \frac{3\pi}{4} \) in radians is equivalent to \( 135^\circ \), and subtracting \( x \) gives us \( \frac{3\pi}{4} - x \) in radians. By applying trigonometric identities, we know that \( \cos \left(\frac{3\pi}{4} - x\right) = \cos(x)\sin\left(\frac{\pi}{4}\right) - \sin(x)\cos\left(\frac{\pi}{4}\right) \), which simplifies to \( \cos(x)\frac{\sqrt{2}}{2} - \sin(x)\frac{\sqrt{2}}{2} \). Similarly, \( \sin \left(\frac{3\pi}{4} - x\right) = \sin(x)\frac{\sqrt{2}}{2} + \cos(x)\frac{\sqrt{2}}{2} \). Combining these results, we get \( \cos(x)\frac{\sqrt{2}}{2} - \sin(x)\frac{\sqrt{2}}{2} - \sin(x)\frac{\sqrt{2}}{2} - \cos(x)\frac{\sqrt{2}}{2} = -\sqrt{2}\cos(x) \), which matches the right side of the equation.

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To simplify the expression \( \frac{\tan^2(x) - 3\tan(x) - 4}{\sin(x)\tan(x) + \sin(x)} \), we can factor the numerator and denominator, cancel out common factors, and simplify  we are left with ( \frac{\tan(x) - 4}{\sin(x)} \) as the simplified expression..

1. For the expression ( \frac{\tan^2(x) - 3\tan(x) - 4}{\sin(x)\tan(x) + sin(x)} \), we can factor the numerator as ( (\tan(x) - 4)(\tan(x) + 1) \) and the denominator as ( \sin(x)(\tan(x) + 1) \). Cancelling out the common factor of ( \tan(x) + 1 \), we are left with ( \frac{\tan(x) - 4}{\sin(x)} \) as the simplified expression.

2. To express ( \cos \left(\frac{3\pi}{4} - x\right) - sin \left(\frac{3\pi}{4} - x\right) = -sqrt{2}\cos(x) \) in radians, we need to convert the angles from degrees to radians. ( \frac{3\pi}{4} \) in radians is equivalent to ( 135^\circ \), and subtracting ( x \) gives us ( \frac{3\pi}{4} - x \) in radians. By applying trigonometric identities, we know that ( \cos \left(\frac{3\pi}{4} - x\right) = \cos(x)\sin\left(\frac{\pi}{4}\right) - sin(x)\cos\left(\frac{\pi}{4}\right) \), which simplifies to ( \cos(x)\frac{\sqrt{2}}{2} - \sin(x)\frac{\sqrt{2}}{2} \).

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The equation for the line in point-(1, 1) is y = -0.5x + 0.5.

Given that the slope of the line below is -0.5. We are to enter the equation for the line in point-(1, 1).The equation for the slope-intercept form of the line is y = mx + c where m is the slope and c is the y-intercept.

Now, the slope of the line is given as -0.5.Therefore, the equation for the slope-intercept form of the line is y = -0.5x + c. Now we need to find the value of c for the equation of the line.

To find the value of c, substitute the values of x and y in the equation of the slope-intercept form of the line.

Given that the point is (-1,1), x=-1 and y=1y = -0.5x + c⇒ 1 = (-0.5) (-1) + c⇒ 1 = 0.5 + c⇒ c = 1 - 0.5⇒ c = 0.5

Therefore, the equation for the line in point-(1, 1) is y = -0.5x + 0.5.The slope of a line refers to how steep the line is and is used to describe its direction. Slope is defined as the vertical change between two points divided by the horizontal change between them.A positive slope moves up and to the right, while a negative slope moves down and to the right. If a line has a slope of zero, it is said to be a horizontal line.

The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line and b is the y-intercept, or the point at which the line crosses the y-axis. To find the equation of a line with a given slope and a point, we can use the point-slope form of a linear equation.

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(a) The component form of vector v can be found by subtracting the coordinates of the initial point from the coordinates of the terminal point. Therefore, v = (7 - 2, 3 - (-3)) = (5, 6).

(b) To write v as a linear combination of the standard unit vectors i and j, we can express v = 5i + 6j.

(a) The component form of a vector represents the change in coordinates from the initial point to the terminal point. We subtract the coordinates of the initial point (2, -3) from the coordinates of the terminal point (7, 3) to get the change in x and y coordinates, which gives us (5, 6).

(b) The standard unit vectors i and j represent the horizontal and vertical directions, respectively. We can express vector v as a linear combination of these unit vectors by multiplying the corresponding components of v by the unit vectors. In this case, we have 5 times i and 6 times j, so we can write v as 5i + 6j.

(c) To sketch vector v with its initial point at the origin, we start at the origin (0, 0) and draw an arrow from the origin to the terminal point (7, 3). The length of the arrow represents the magnitude of the vector, and the direction of the arrow represents the direction of the vector. In this case, the arrow would start at (0, 0) and end at (7, 3), indicating the magnitude and direction of vector v.

In conclusion, the vector v can be represented in component form as (5, 6) and as a linear combination of the standard unit vectors i and j as 5i + 6j. It can be sketched by drawing an arrow from the origin to the terminal point (7, 3).

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The series Σ √√k diverges. It can be determined using the Divergence Test, where the limit as k approaches infinity of √√k is infinity.

The series Σ √√k converges or diverges, we can apply the Divergence Test. According to the Divergence Test, if the limit of the nth term of a series does not approach zero as n approaches infinity, then the series diverges.

In this case, the nth term of the series is √√k. To find the limit as k approaches infinity, we can simplify the expression by taking the square root of both sides, which gives us √k. Taking the limit as k approaches infinity, we have lim(k→∞) √k = ∞.

Since the limit of the nth term is not zero, but rather approaches infinity, the series diverges. Therefore, the correct choice is (C) The series diverges by the Divergence Test. The value of the limit as k approaches infinity, lim(k→∞) √√k, is infinity.

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The main answer is that the first investment would earn $417,612 more than the second investment. Here is an explanation of how to get to that answer:For the first investment, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)where:A = the future value of the investmentP = the principal (initial investment)r = the annual interest raten = the number of times interest is compounded per yeart = the number of yearsFor this investment, we have:P = $20,000r = 0.12n = 1 (compounded annually)t = 30 yearsPlugging these values into the formula, we get:

A = $20,000(1 + 0.12/1)^(1*30)A = $20,000(1.12)^30A

= $20,000(17.446)A = $348,916.45

Rounding this to the nearest dollar gives us $348,916.For the second investment, we use the same formula, but with different values of n and r since the interest is compounded monthly:P = $20,000r = 0.06n = 12 (compounded monthly)t = 30 years Plugging these values into the formula, we get

:A = $20,000(1 + 0.06/12)^(12*30)

A = $20,000(1.005)^360

A = $20,000(2.214)

A = $44,280.90

Rounding this to the nearest dollar gives us $44,281.To find the difference between the two investments, we subtract the second investment's future value from the first investment's future value:$348,916 - $44,281 = $304,635Rounding this to the nearest dollar gives us $304,635.The first investment would earn $417,612 more than the second investment ($304,635 rounded to the nearest dollar).

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Using the angle addition formula, simplify the equation to sin(-5x)=-0.3. Taking the inverse sine, the smallest positive solution is approximately x=0.06.



To solve the equation \( \sin (5x) \cos (10x) - \cos (5x) \sin (10x) = -0.3 \) for the smallest positive solution, we can rewrite it using the angle addition formula for sine:

\[ \sin (a - b) = \sin a \cos b - \cos a \sin b. \]

Comparing this with the given equation, we can see that \( a = 5x \) and \( b = 10x \). Therefore, we can rewrite the equation as:

\[ \sin (5x - 10x) = -0.3. \]

Simplifying further, we have:

\[ \sin (-5x) = -0.3. \]

Now, we can solve for \( x \) by taking the inverse sine of both sides:

\[ -5x = \sin^{-1}(-0.3). \]

To find the smallest positive solution, we need to consider the principal value of the inverse sine function. In this case, the range of the inverse sine function is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

Therefore, the smallest positive solution for \( x \) is:

\[ x = -\frac{1}{5} \sin^{-1}(-0.3). \]

Evaluating this expression numerically, we have:

\[ x \approx 0.064.\]

Hence, the smallest positive solution for \( x \) is approximately 0.06 (accurate to two decimal places).

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(I) The characteristic polynomial of matrix A is p(λ) = 2λ² - 2λ. (II) Two eigenvalues: λ = 0 and λ = 1 (III) The eigenspace corresponding to λ = 0 is the zero vector. The eigenspace corresponding to λ = 1 is spanned by the vector [2, 0]. (IV) The algebraic multiplicity is 2 and the geometric multiplicity is 0. The algebraic multiplicity is also 2 and the geometric multiplicity is 1.

(V) The matrix A is not diagonalizable. (VI) There is need to calculate A¹⁰ using a different approach. (VII) Aᵏ = Aᵏ ᵐᵒᵈ ⁵ for all non-negative integers k. (VIII) A¹⁰ × b = [-2, 2]. (IX) A is similar to B, and there is an invertible matrix P such that P⁻¹ × A × P = B.

(I) To find the characteristic polynomial of matrix A, we need to calculate the determinant of the matrix (A - λI), where λ is the eigenvalue and I is the identity matrix.

A - λI =

[1 - λ]

[1 - λ]

[-1 - λ]

[1 - λ]

det(A - λI) = (1 - λ)(1 - λ) - (1 - λ)(-1 - λ)

= (1 - λ)² - (-1 - λ)(1 - λ)

= (1 - λ)² - (λ + 1)(1 - λ)

= (1 - λ)² - (1 - λ²)

= (1 - λ)² - 1 + λ²

= (1 - 2λ + λ²) - 1 + λ²

= 2λ² - 2λ

Therefore, the characteristic polynomial of matrix A is p(λ) = 2λ² - 2λ.

(II) To find the eigenvalues of matrix A, we set the characteristic polynomial equal to zero and solve for λ:

2λ² - 2λ = 0

Factorizing the equation, we have:

2λ(λ - 1) = 0

Setting each factor equal to zero, we find two eigenvalues:

λ = 0 and λ = 1

(III) To find a basis for the eigenspaces of matrix A, we need to find the eigenvectors corresponding to each eigenvalue.

For λ = 0:

(A - 0I)v = 0, where v is the eigenvector.

Simplifying the equation, we have:

A × v = 0

Substituting the values of A and v, we get:

[1 0] [v1] = [0]

[1 -1] [v2] [0]

This gives us the system of equations:

v1 = 0

v1 - v2 = 0

Solving these equations, we find v1 = 0 and v2 = 0.

Therefore, the eigenspace corresponding to λ = 0 is the zero vector.

For λ = 1:

(A - I)v = 0

Substituting the values of A and v, we get:

[0 0] [v1] = [0]

[1 -2] [v2] [0]

This gives us the system of equations:

v2 = 0

v1 - 2v2 = 0

Solving these equations, we find v1 = 2 and v2 = 0.

Therefore, the eigenspace corresponding to λ = 1 is spanned by the vector [2, 0].

(IV) The algebraic multiplicity of an eigenvalue is the power of its factor in the characteristic polynomial. The geometric multiplicity is the dimension of its eigenspace.

For λ = 0, the algebraic multiplicity is 2 (since (λ - 0)² appears in the characteristic polynomial), and the geometric multiplicity is 0.

For λ = 1, the algebraic multiplicity is also 2 (since (λ - 1)² appears in the characteristic polynomial), and the geometric multiplicity is 1.

(V) To show that the matrix is diagonalizable, we need to check if the algebraic and geometric multiplicities are equal for each eigenvalue.

For λ = 0, the algebraic multiplicity is 2, but the geometric multiplicity is 0. Since they are not equal, the matrix is not diagonal

izable for λ = 0.

For λ = 1, the algebraic multiplicity is 2, and the geometric multiplicity is 1. Since they are not equal, the matrix is not diagonalizable for λ = 1.

Therefore, the matrix A is not diagonalizable.

(VI) To find A¹⁰ × b, we can write b as a linear combination of eigenvectors of A and use the fact that Aᵏ × v = λᵏ × v, where v is an eigenvector corresponding to eigenvalue λ.

We have two eigenvectors corresponding to the eigenvalue λ = 1: [2, 0]. Let's denote it as v1.

b = [-2, 2] = (-2/2) × [2, 0] = -1 × v1

Using the fact mentioned above, we can calculate A¹⁰ × b:

A¹⁰ × b = A¹⁰ × (-1 × v1)

= (-1)¹⁰ × A¹⁰ × v1

= 1 × A¹⁰ × v1

= A¹⁰ × v1

Since A is not diagonalizable, we need to calculate A¹⁰ using a different approach.

(VII) To find a formula for Aᵏ for all non-negative integers k, we can use the Jordan canonical form of matrix A. However, without knowing the Jordan canonical form, we can still find Aᵏ by performing repeated matrix multiplications.

A² = A × A =

[1 0] [1 0] = [1 0]

[1 -1] [1 -1] [1 -2]

A³ = A² × A =

[1 0] [1 0] = [1 0]

[1 -2] [1 -1] [-1 2]

A⁴ = A³ × A =

[1 0] [1 0] = [1 0]

[-1 2] [-1 2] [-2 2]

A⁵ = A⁴ × A =

[1 0] [1 0] = [1 0]

[-2 2] [-1 2] [0 0]

A⁶ = A⁵ × A =

[1 0] [1 0] = [1 0]

[0 0] [0 0] [0 0]

As we can see, starting from A⁵, the matrix Aⁿ becomes the zero matrix for n ≥ 5.

Therefore, Aᵏ = Aᵏ ᵐᵒᵈ ⁵ for all non-negative integers k.

(VIII) Using the formula from (VII), we can find A¹⁰ × b:

A^10 * b = A¹⁰ ᵐᵒᵈ ⁵ × b

= A⁰ × b

= I × b

= b

We previously found that b = [-2, 2].

Therefore, A¹⁰ × b = [-2, 2].

(IX) To determine if A is similar to B, we need to check if there exists an invertible matrix P such that P⁻¹ × A × P = B.

Let's calculate P⁻¹ × A × P and check if it equals B:

P = [v1 v2] = [2 0]

[0 0]

P⁻¹ = [1/2 0]

[ 0 1]

P⁻¹ × A × P =

[1/2 0] [1 0] [2 0] = [0 0]

[ 0 1] [1 -1] [0 0] [0 0]

The result is the zero matrix, which is equal to B.

Therefore, A is similar to B, and we found an invertible matrix P such that P⁻¹ × A × P = B. In this case, P = [2 0; 0 0].

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The derivatives of the given functions are as follows: For the function y = x * e^(-x), its derivative is given by dy/dx = e^(-x) - x * e^(-x). For the function F(t) = ln(t-1) / (t^(1/4)), its derivative is given by dF/dt = [(1/(t-1)) - (1/4) * (t^(-3/4))] / sqrt[4]{t}.

To find the derivative of y = x * e^(-x), we can apply the product rule. The derivative of x with respect to x is 1, and the derivative of e^(-x) with respect to x is -e^(-x). Therefore, using the product rule, we get dy/dx = (x * -e^(-x)) + (1 * e^(-x)) = e^(-x) - x * e^(-x).

To find the derivative of F(t) = ln(t-1) / (t^(1/4)), we can use the quotient rule. The derivative of ln(t-1) with respect to t is (1/(t-1)), and the derivative of (t^(1/4)) with respect to t is (1/4) * (t^(-3/4)). Applying the quotient rule, we have dF/dt = [(1/(t-1)) * (t^(1/4))] - [ln(t-1) * (1/4) * (t^(-3/4))] / (t^(1/4))^2 = [(1/(t-1)) - (1/4) * (t^(-3/4))] / sqrt[4]{t}.

These derivatives represent the rates of change of the given functions with respect to their independent variables.

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Under the given conditions, we can find the exact value of each expression. For sin(x + β), the exact value is sqrt(61)/61, and for cos(x + 3), the exact value is 1/2

Given conditions:

cos θ = -6√61 in Quadrant II

sin α = 0° with 0° < α < 17.0°

(a) To find sin(x + β), we need to determine the value of sin β. Since sin α = 0, we know that α = 0°. Therefore, sin β = sin(α + β - α) = sin(β - α) = sin(0° - α) = -sin α = 0.

Thus, sin(x + β) = sin x cos β + cos x sin β = sin x (1) + cos x (0) = sin x.

(b) To find cos(x + 3), we can use the angle addition formula for cosine:

cos(x + 3) = cos x cos 3 - sin x sin 3.

Since sin α = 0°, we know that α = 0°, so sin α = 0. Therefore, sin 3 = sin(α + 3) = sin 3°.

Using a calculator, we find that sin 3° = 0.052336. So, cos(x + 3) = cos x cos 3 - sin x sin 3 = cos x (1) - sin x (0.052336) = cos x.

Therefore, under the given conditions, sin(x + β) simplifies to sin x, and cos(x + 3) simplifies to cos x.

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PDF using the method of transformation for the given functions are:fY(y)=-1/2 e*-Y/2, for Y=-2in(X)and fY(y)=1/|y|×1/√(2π)× e-(lny)²/2, for Y=exand fY(y)=1/2√(2πy)× e-y/2, for Y=x².

Probability density function (PDF) is a function that describes the relative likelihood of a continuous random variable X taking on a particular value x. In order to derive the PDF using the method of transformation, the formula given below is used:

fY(y)=fX(x)|d/dyG-1(y)|

where X and Y are two random variables, and G is a function used for transforming X to Y.

Here, d/dyG-1(y) denotes the derivative of the inverse function of G with respect to y.

Now, we have to find the PDF of Y=-2in(X). Given, X is Uniform(0,1).We know that, X~Uniform(0,1).

Using the transformation formula, fY(y)=fX(x)|d/dyG-1(y)|

where G(x)= -2in(x) and G-1(y)= e*-y/2, we get

Y=G(X)= -2in(X).So, G-1(Y)= e*-Y/2.

To calculate the PDF of Y=-2in(X), we will first find the PDF of X.

PDF of X:fX(x)=1, for 0≤x≤1; otherwise 0.

From the transformation formula,

fY(y)=fX(x)|d/dyG-1(y)

By differentiating G-1(Y)= e*-Y/2

d/dyG-1(Y)=-1/2 e*-Y/2

Using the above formula,

fY(y)=fX(x)|d/dyG-1(y)|=-1/2 e*-Y/2×1=-1/2 e*-Y/2

The PDF of Y=-2in(X) is given by,

fY(y)=-1/2 e*-Y/2, where y>0.

Now, we have to find the PDF of Y=ex.

Given, X~N(0,1).

We know that, X~N(0,1).

Using the transformation formula,

fY(y)=fX(x)|d/dyG-1(y)

where G(x)=ex and G-1(y)= ln(y),

we get

Y=G(X)= ex.So, G-1(Y)= ln(Y).

To calculate the PDF of Y=ex, we will first find the PDF of X. PDF of X:fX(x)=1/√(2π)× e-x²/2

From the transformation formula,

fY(y)=fX(x)|d/dyG-1(y)|

By differentiating G-1(Y)= ln(Y), we get

d/dyG-1(Y)= 1/Y

Using the above formula,

fY(y)=fX(x)|d/dyG-1(y)|

=1/Y×1/√(2π)× e-(ln(y))²/2

=1/Y×1/√(2π)× e-(lny)²/2

=1/|y|×1/√(2π)× e-(lny)²/2

The PDF of Y=ex is given by,

fY(y)=1/|y|×1/√(2π)× e-(lny)²/2, where y≠0.

Now, we have to find the PDF of Y=x².

Given, X(x)= (x+1)/2.

We know that, X(x) is a function of X.

Let Y=x².

So, we need to find the PDF of Y. PDF of

Y:fY(y)=fX(x)|d/dyG-1(y)|where G(x)=x²

G-1(y)=√y.

Since, X(x)= (x+1)/2.

We can write X as

x=2X-1.

Now, G(X)=X², so

G(X(x))=[2X(x)-1]² = 4X²(x)-4X(x)+1

To calculate the PDF of Y=x², we will first find the PDF of X.

PDF of X:

fX(x)=1/√(2π)× e-x²/2

From the transformation formula

fY(y)=fX(x)|d/dyG-1(y)|

By differentiating G-1(Y)= √Y, we get

d/dyG-1(Y)= 1/2√y

Using the above formula,

fY(y)=fX(x)|d/dyG-1(y)|

=1/2√y×1/√(2π)× e-(√y)²/2

=1/2√(2πy)× e-y/2

The PDF of Y=x² is given by,

fY(y)=1/2√(2πy)× e-y/2, where y≥0.

Therefore, the PDF using the method of transformation for the given functions are:

fY(y)=-1/2 e*-Y/2,

for Y=-2in(X)and

fY(y)=1/|y|×1/√(2π)× e-(lny)²/2, for

Y=exand

fY(y)=1/2√(2πy)× e-y/2, for Y=x².

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The angle between vectors \(v\) and \(w\) can be found using the dot product and vector magnitudes. The angle between \(v\) and \(w\) is approximately 111.9 degrees.

To find the angle between vectors \(v\) and \(w\), we can use the formula:

\[

\theta = \cos^{-1}\left(\frac{{v \cdot w}}{{\lVert v \rVert \lVert w \rVert}}\right)

\]

where \(v \cdot w\) represents the dot product of vectors \(v\) and \(w\), and \(\lVert v \rVert\) and \(\lVert w \rVert\) represent the magnitudes of \(v\) and \(w\) respectively.

Given \(v = 3i + 7j\) and \(w = -4i - j\), we can calculate the dot product as follows:

\(v \cdot w = (3 \cdot -4) + (7 \cdot -1) = -12 - 7 = -19\)

Next, we determine the magnitudes of \(v\) and \(w\):

\(\lVert v \rVert = \sqrt{(3^2) + (7^2)} = \sqrt{9 + 49} = \sqrt{58}\)

\(\lVert w \rVert = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}\)

Substituting these values into the formula, we have:

\(\theta = \cos^{-1}\left(\frac{{-19}}{{\sqrt{58} \cdot \sqrt{17}}}\right)\)

Using a calculator, we can evaluate this expression and find that the angle \(\theta\) is approximately 111.9 degrees.

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[tex]Given the homogeneous differential equation is y′′−16y=0[/tex]

[tex]The given functions y1=e4x and y2=e−4x[/tex][tex]are both solutions for the homogeneous differential equation y′′−16y=0.[/tex]

[tex]To find the general solution of the nonhomogeneous differential equation y′′−16y=32x−16,[/tex] we will have to use [tex]the method of undetermined coefficients where the solution is assumed to be of the form y=Ax+B.[/tex]

First, we find the complementary solution by solving the homogeneous[tex]differential equationy′′−16y=0[/tex]
[tex]Auxiliary equation: m² - 16 = 0[/tex]
[tex]m² = 16m = ±√16m1 = 4, m2 = -4[/tex]

The complementary solution is
[tex]y_c = c1e^(4x) + c2e^(-4x)where c1 and c2 are arbitrary constants.[/tex][tex]Now, we find the particular solution of the nonhomogeneous differential equation y′′−16y=32x−16[/tex]by the method of undetermined coefficients:[tex]Let y = Ax + Bdy/dx = ASecond derivative of y, d²y/dx² = 0[/tex]

Substituting these values in the differential equation, we getA = 2Comparing coefficients, [tex]we get the particular solution asy_p = 2x - 1[/tex]The general solution of the nonhomogeneous [tex]differential equation y′′−16y=32x−16 is given byy = y_c + y_p[/tex]
[tex]y = c1e^(4x) + c2e^(-4x) + 2x - 1So, the correct option is:y=c1e^(4x)+c2e^(-4x)+2x-1.[/tex]

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The newspaper article presents data from a study comparing virus leak rates in vinyl and latex laboratory gloves.

The null and alternative hypotheses for testing whether vinyl gloves have a greater virus leak rate than latex gloves are provided as options. In hypothesis testing, the null hypothesis (H0) represents the claim being tested, while the alternative hypothesis (H1) represents the opposing claim. In this case, the claim being tested is whether vinyl gloves have a greater virus leak rate than latex gloves.

To determine the appropriate null and alternative hypotheses, we need to consider the direction of the claim. The claim states that vinyl gloves have a greater virus leak rate, suggesting a one-sided comparison. Therefore, the null hypothesis should assume no difference or equality, and the alternative hypothesis should indicate a difference or inequality.

Looking at the options provided:

A. H0: p1 = p2 (null hypothesis assumes no difference)

  H1: p1 ≠ p2 (alternative hypothesis assumes a difference)

B. H0: p1 = p2 (null hypothesis assumes no difference)

  H1: p1 > p2 (alternative hypothesis assumes a greater virus leak rate for vinyl gloves)

C. H0: p1 > p2 (null hypothesis assumes a greater virus leak rate for vinyl gloves)

  H1: p1 = p2 (alternative hypothesis assumes no difference)

D. H0: p1 = p2 (null hypothesis assumes no difference)

  H1: p1 (missing alternative hypothesis)

The correct option is B, which states H0: p1 = p2 and H1: p1 > p2. This aligns with the claim that vinyl gloves have a greater virus leak rate than latex gloves. In conclusion, the null and alternative hypotheses for testing whether vinyl gloves have a greater virus leak rate than latex gloves are H0: p1 = p2 and H1: p1 > p2, respectively.

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To verify if the given differential equation is exact, we check if the partial derivative of the coefficient of dx with respect to y is equal to the partial derivative of the coefficient of dy with respect to x.

[Question 1]

a) To verify the exactness of the given differential equation, we compute the partial derivatives of the coefficients: ∂/∂y (-y^8 sin(x) + 7x^6y^3) = -8y^7 sin(x) + 21x^6y^2 and ∂/∂x (8y cos(x) + 3x^7y^2) = 8y cos(x) + 21x^6y^2. Since these partial derivatives are equal, the equation is exact.

b) To find the general solution, we integrate the coefficient of dx with respect to x while treating y as a constant, which gives -y^8 sin(x) + 7/2x^7y^3 + C(y). Then, we integrate the coefficient of dy with respect to y while treating x as a constant, resulting in 4y^2 cos(x) + x^7y^3 + D(x). Combining these results, the general solution is given by -y^8 sin(x) + 7/2x^7y^3 + C(y) + 4y^2 cos(x) + x^7y^3 + D(x) = C(y) + D(x).

[Question 2]

a) By performing row operations on the augmented matrix, we find the values of x, y, and z as x = -8/11, y = -11/22, and z = -1/22.

b) The system is consistent as it has a solution. Since it has a unique solution for x, y, and z, the solution type is unique.

[Question 3]

a) To make the system homogeneous, we set the right side of each equation equal to zero. By equating -3x - 6y to zero, we obtain k^2 + 3k - 18 = 0. Solving this quadratic equation gives us the values for k as k = 3 and k = -6.

b) By substituting k = 0 into the equations and applying Gaussian elimination, we can find the solution to the homogeneous system. The solution may be trivial (all variables are zero) or nontrivial (at least one variable is nonzero). The uniqueness of the solution depends on the results obtained from the Gaussian elimination process.

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In this problem, we applied this formula to convert 3.8 radians to decimal degrees. We found that 3.8 radians is equivalent to 217.555 degrees in decimal form.

In order to convert 3.8 radians to decimal degrees, we use the formula:

Radians = (π/180) x Degrees where π/180 is a conversion factor to convert from radians to degrees.

Now, we can substitute 3.8 radians into the formula to find the equivalent decimal degrees:

3.8 radians = (π/180) x Degrees

Multiplying both sides by 180/π, we get:

3.8 radians x (180/π) = Degrees

Simplifying this expression gives us:

3.8 radians x (180/π) = 217.555 degrees

Therefore, 3.8 radians is equivalent to 217.555 degrees in decimal form. In this problem, we are asked to convert 3.8 radians to decimal degrees.

To do this, we use the formula Radians = (π/180) x Degrees, where π/180 is a conversion factor to convert from radians to degrees. We start by substituting 3.8 radians into the formula to find the equivalent number of degrees.

This gives us 3.8 radians = (π/180) x Degrees, which we can simplify by multiplying both sides by 180/π.

Doing this gives us 3.8 radians x (180/π) = Degrees.

Simplifying this expression yields 217.555 degrees, which is the final answer.

To convert from radians to degrees, we use the formula

Radians = (π/180) x Degrees, where π/180 is a conversion factor to convert from radians to degrees. In this problem, we applied this formula to convert 3.8 radians to decimal degrees.

We found that 3.8 radians is equivalent to 217.555 degrees in decimal form.

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The points A and B are approximately 101.39 feet apart. To find the distance between points A and B, we can use the Law of Cosines.

The Law of Cosines states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of the two sides and the cosine of the included angle.

In this case, we have side AC = 53 feet, side BC = 73 feet, and the included angle ACB = 46 degrees. Let's denote the distance between points A and B as x.

Applying the Law of Cosines, we have:

x^2 = 53^2 + 73^2 - 2(53)(73)cos(46)

Calculating this expression, we find that x^2 is approximately 10278.39.

Taking the square root of both sides, we get:

x ≈ √10278.39 ≈ 101.39 feet

Therefore, points A and B are approximately 101.39 feet apart.

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Given the point P(-2, -3) on a circle with equation x^2 + y^2 = r^2 and also on the terminal side of an angle θ in standard position, we are asked to find the value of cot(θ) given the expression cot(θ) = 3/2 √(10/13) / 2/√13.

To find the value of cot(θ), we need to determine the ratio of the adjacent side to the opposite side of the right triangle formed by the point P(-2, -3) and the origin (0, 0).

First, we calculate the length of the hypotenuse using the distance formula. The distance from the origin to P is √((-2 - 0)^2 + (-3 - 0)^2) = √(4 + 9) = √13.

Next, we determine the lengths of the adjacent and opposite sides of the triangle. The adjacent side is the x-coordinate of P, which is -2. The opposite side is the y-coordinate of P, which is -3.

Now, we can calculate the value of cot(θ). Since cot(θ) = adjacent/opposite, we have cot(θ) = -2/-3 = 2/3.

Comparing this with the given expression cot(θ) = 3/2 √(10/13) / 2/√13, we can simplify it to cot(θ) = (3/2) * (2/√13) / (2/√13) = 3/2.

Therefore, cot(θ) = 3/2.

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The exact value of sin(17π/36)cos(13π/36) + cos(17π/36)sin(13π/36) is 1/2.

This can be found using the sum-to-product formula, which states that sin

A cos B + cos A sin B = sin (A + B).

In this case, A = 17π/36 and B

= 13π/36, so A + B

= 30π/36

= 5π/12.

Therefore, sin(17π/36)cos(13π/36) + cos(17π/36)sin(13π/36)

= sin(5π/12)

= 1/2.

The sum-to-product formula is a trigonometric identity that relates the product of two sines or cosines to the sum or difference of two sines or cosines. It is a useful tool for simplifying trigonometric expressions and finding exact values. In this case, by applying the sum-to-product formula, we were able to simplify the expression and determine that its exact value is 1/2.

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The rate of increase in the direction (-6, -6) is 6[tex]\sqrt{(2)}[/tex]. The direction of the fastest decrease is (6, 6), and the rate of decrease in that direction is 6[tex]\sqrt{(2)}[/tex].

To find the direction in which the function f(x, y) = [tex]x^3[/tex] - x[tex]y^2[/tex] is increasing the fastest at points (1, 3), we need to calculate the gradient vector and evaluate it at the given point. The gradient vector represents the direction of the steepest ascent.

First, we compute the partial derivatives of f with respect to x and y:

∂f/∂x = 3[tex]x^2[/tex] - [tex]y^2[/tex]

∂f/∂y = -2xy

Next, we substitute the coordinates (1, 3) into the partial derivatives:

∂f/∂x = 3[tex](1)^2[/tex] - [tex](3)^2[/tex]= 3 - 9 = -6

∂f/∂y = -2(1)(3) = -6

Therefore, the gradient vector at (1, 3) is (-6, -6). This vector points in the direction of the fastest increase of the function at that point.

To determine the rate of increase in that direction, we calculate the magnitude of the gradient vector:

|∇f| = [tex]\sqrt{((-6)^2 + (-6)^2)}[/tex] = [tex]\sqrt{(36 + 36)}[/tex] = [tex]\sqrt{(72)}[/tex] = 6[tex]\sqrt{(2)}[/tex]

Hence, the rate of increase in the direction (-6, -6) is 6[tex]\sqrt{(2)}[/tex].

To find the direction of the fastest decrease at points (1, 3), we consider the opposite of the gradient vector, which is (6, 6). This vector points in the direction of the steepest descent or fastest decrease of the function at that point.

Similarly, the rate of decrease in that direction is the magnitude of the gradient vector:

|∇f| = [tex]\sqrt{((6)^2 + (6)^2)}[/tex] = [tex]\sqrt{(36 + 36)}[/tex] = [tex]\sqrt{(72)}[/tex] = 6[tex]\sqrt{(2)}[/tex]

Therefore, the direction of the fastest decrease is (6, 6), and the rate of decrease in that direction is 6[tex]\sqrt{(2)}[/tex].

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The exact ordered pair associated with the angle 150 degrees on the unit circle is (-√3 / 2, 1 / 2).

To determine the point (x, y) on the unit circle associated with the angle 150 degrees, we can use the trigonometric functions sine and cosine.

Let's convert the angle from degrees to radians since trigonometric functions typically work with radians. We know that π radians is equivalent to 180 degrees. Therefore, we can use the conversion factor:

150 degrees ×(π radians / 180 degrees) = (5π / 6) radians

The angle (5π / 6) radians lies in the second quadrant of the unit circle. In this quadrant, the x-coordinate is negative, and the y-coordinate is positive.

Now, we can calculate the values of x and y using sine and cosine:

x = cos(5π / 6) = -√3 / 2

y = sin(5π / 6) = 1 / 2

Therefore, the exact ordered pair associated with the angle 150 degrees on the unit circle is (-√3 / 2, 1 / 2).

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If each consumer demands one item of groceries and faces a travel cost of $12 per kilometer. BetaMarket chooses the price of $33.33 in equilibrium.

AlphaMart sells groceries at the west end of Main Street, a street that is one kilometer long. AlphaMart competes with BetaMarket, which is located at the east end of the street. AlphaMart and BetaMarket sell groceries that are identical in every respect, apart from the locations of the two stores. The marginal cost of an item of groceries is $3 for both retailers.

Main Street is home to 200 consumers; the consumers are evenly spaced along the street. Each consumer demands one item of groceries and faces a travel cost of $12 per kilometer. To calculate the equilibrium price of BetaMarket, we first need to find out the quantity demanded at each price point.

The quantity demanded for each price point can be found by subtracting the number of consumers who are closer to AlphaMart than to BetaMarket from the total number of consumers. Let the price charged by BetaMarket be P. If BetaMarket charges P, then the demand for BetaMarket's groceries is given by:

QB = 200/2 - 1/2 (P + 12) = 100 - 1/2 (P + 12)

QB = 100 - 1/2P - 6

We can now write down BetaMarket's profit function as:

πB = QB(P - 3) = (100 - 1/2P - 6)(P - 3)

πB = 100P - 3/2P² - 309

From this, we can find the first-order condition for profit maximization by differentiating the profit function with respect to P and setting it equal to zero:

∂πB/∂P = 100 - 3P = 0P = 100/3

Thus, BetaMarket chooses to set the price at $33.33 in equilibrium.

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That is cos θ = adjacent / hypotenuse cos θ = 1 / 2Hence, the exact value of the arc tan (√3) is π/3.

What is a circle? A circle is a set of points that are all equidistant from a given point known as the center of the circle.

A circle can also be defined as a closed curve that has the same distance between any two points on the curve, and the curve is traced by a point that moves in such a way that its distance from a fixed point is constant.

What is the Unit Circle? A unit circle is a circle with a radius of 1 and a center at the origin (0, 0) of a coordinate plane.

The unit circle is used to determine the exact values of trigonometric functions.

What is arc tan (√3)?Arc tan (√3) is a function of trigonometry that relates to finding the angle whose tangent is √3.

This value is approximately 60 degrees or π/3 in radians.

Solution Using the unit circle, we can determine the exact value of the arc tan (√3).

The arc tan (√3) can be simplified by considering the right triangle whose opposite side is √3, adjacent side is 1, and the hypotenuse is √4.

This is because tan θ = Opposite / Adjacent Therefore, we have tan θ = √3 / 1 = √3Now, we know that θ is in the first quadrant, and the hypotenuse of the triangle is equal to 2.

The sine of the angle is equal to the opposite side divided by the hypotenuse of the triangle.

That is Sin θ = opposite / hypotenuse Sin θ = √3 / 2We can also find the cosine of the angle, which is adjacent to the hypotenuse. That is cos θ = adjacent / hypotenuse cos θ = 1 / 2

Hence, the exact value of the arc tan (√3) is π/3.

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The relationship between (a, b) and (b, r) is that they have a common divisor.

If d is a common divisor of b and r, then d is a common divisor of a and b. It means that if d divides b and r, then d should divide the linear combination of b and r, which is a.

Therefore, d is a common divisor of a and b. This shows that (a, b) and (b, r) have a common divisor d, namely d, where d ∈ Z.2) If d is a common divisor of a and b, then d is a common divisor of b and r. It means that d divides a and b, so d must divide a-bq = r.

Therefore, d is a common divisor of b and r. This shows that (a, b) and (b, r) have a common divisor d, namely d, where d ∈ Z.3) It can be concluded that (a, b) = (b, r) because if (a, b) = d, and (b, r) = e, then by applying parts 1 and 2, we can say that d = e.

Therefore, (a, b) = (b, r).The relationship between (a, b) and (b, r) is that they have a common divisor.

This common divisor can be found by applying part 1 and part 2, as shown above.

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The largest composite less than 1000 with a GPF of 3 is 996. To prove that it is the largest, we can note that any larger multiple of 3 would either be a prime or have a larger prime factor than 3.

a) To determine the sequence of GPFs for the composites from 60 to 65, we can list the prime factors of each number and take the largest:

- 60 = 2 x 2 x 3 x 5, so the GPF is 5

- 61 is prime

- 62 = 2 x 31, so the GPF is 31

- 63 = 3 x 3 x 7, so the GPF is 7

- 64 = 2 x 2 x 2 x 2 x 2 x 2, so the GPF is 2

- 65 = 5 x 13, so the GPF is 13

Therefore, the sequence of GPFs for the composites from 60 to 65 is 5, prime, 31, 7, 2, 13.

b) The given sequence of GPFs is 41, 19, 79. All of these numbers are prime, so any successive composites that would give this sequence of GPFs would have to be divisible by each of these primes. The product of 41, 19, and 79 is 62,999, which is a four-digit number. Therefore, any composite that would give the sequence of GPFs 41, 19, 79 would have to have at least four digits.

c) To find the smallest composites that give the sequence of GPFs 17, 73, 2, 19, we can start with 17 x 73 x 2 x 19 = 45634, which is a five-digit number. The next composite with these GPFs would be obtained by adding the product of these primes to 45634. This gives 3215678, which is a seven-digit number. Therefore, the smallest successive composites that give the sequence of GPFs 17, 73, 2, 19 are 45634 and 3215678.

d) To find the largest composite less than 1000 with a GPF of 3, we can list the multiples of 3 less than 1000 and eliminate the primes by inspection:

- 3 x 1 = 3

- 3 x 2 = 6

- 3 x 3 = 9 (prime)

- 3 x 4 = 12

- 3 x 5 = 15

- 3 x 6 = 18

- 3 x 7 = 21 (prime)

- 3 x 332 = 996

- 3 x 333 = 999 (prime)

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The exact points of intersection are (π/6, √3/2) and (5π/6, -√3/2) for the two functions f(x) = g(x) means cos x = sin 2x.

To find the exact values where f(x) = g(x),

we have to equate the two functions. f(x) = g(x) means cos x = sin 2x.

Rewriting sin 2x in terms of cos x:

sin 2x = 2 sin x cos x

Hence, cos x = 2 sin x cos x

Dividing both sides by cos x:

1 = 2 sin xor sin x = 1/2

Since sin x = 1/2 has two solutions in the interval (0, 47),

we can find them by solving sin x = 1/2 for x in this interval.

The solutions of sin x = 1/2 in the interval (0, 47) are given by x = π/6 and x = 5π/6.

The x-coordinates of the points of intersection are π/6 and 5π/6.

To find the exact points (x, y) of the intersections, we can substitute these values of x in either of the functions f(x) = cos x or g(x) = sin 2x.

So the exact points of intersection are (π/6, √3/2) and (5π/6, -√3/2).

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Given the position vector c(t)=(5cos(t),5sin(t),t^2) for a particle moving on a helix.

Find the speed s(t0) of the particle at time t0=13π.

At t = t0 = 13π, the position vector is c(t0) = (5cos(13π), 5sin(13π), (13π)²) = (-5, 0, 169π²).

The velocity vector is given by the derivative of the position vector w.r.t t as v(t) = c'(t) = (-5sin(t), 5cos(t), 2t).

At t = t0 = 13π,

we have v(t0) = (-5sin(13π), 5cos(13π), 2(13π)) = (0, -5, 26π).

Hence, the speed is given by s(t0) = |v(t0)| = √(0² + (-5)² + (26π)²) = 5√(1 + 676π²).

The parametric equation of the tangent line to the helix at t = t0 is given by the equation r(t) = c(t0) + t.v(t0),

where c(t0) is the position vector of the helix at t0, and v(t0) is the velocity vector of the helix at t0.

Hence, we have r(t) = (-5, 0, 169π²) + t(0, -5, 26π) = (-5t, -5t, 169π² + 26πt).

The line will intersect the xy-plane when z = 0, i.e., at the point (x, y, 0),

where -5t = -5t = 0 and 169π² + 26πt = 0.

Hence, t = -169π²/26 and the point of intersection is (5t, 5t, 0) = (-845π, -845π, 0).

Therefore, l(t) = (-5t, -5t, 169π² + 26πt) = (845πt, 845πt, -2197π²).

The line will intersect the xy-plane at (845πt, 845πt, 0) = (-845π, -845π, 0).

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