determine the matrix of the linear transformation t : r 4 → r 3 defined by t(x1, x2, x3, x4) = (2x1 3x2 x4, 5x1 9x3 − x4, 4x1 2x2 − x3 7x4).

a) Cylindrical coordinates for point (-5, 5, 5) are (r, θ, z) = (√50, 3π/4, 5).
b) Cylindrical coordinates for point (-3, 3√3, 1) are (r, θ, z) = (6, 5π/6, 1).

We will use the following equations:

1. r = √(x² + y²)
2. θ = arctan(y/x) (note: make sure to take the quadrant into account)
3. z = z (z-coordinate remains the same)

(a) For the point (-5, 5, 5):

1. r = √((-5)² + 5²) = √(25 + 25) = √50
2. θ = arctan(5/-5) = arctan(-1) = 3π/4 (in the 2nd quadrant)
3. z = 5

So, the cylindrical coordinates for point (-5, 5, 5) are (r, θ, z) = (√50, 3π/4, 5).

(b) For the point (-3, 3√3, 1):

1. r = √((-3)² + (3√3)²) = √(9 + 27) = √36 = 6
2. θ = arctan((3√3)/-3) = arctan(-√3) = 5π/6 (in the 2nd quadrant)
3. z = 1

So, the cylindrical coordinates for point (-3, 3√3, 1) are (r, θ, z) = (6, 5π/6, 1).

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In this sample, 154 college graduates and 132 non-college graduates support drilling for oil and natural gas off the coast of California. Therefore, the percentage of college graduates who support drilling is (154/438) x 100 = 35.16%, while the percentage of non-college graduates who support drilling is (132/389) x 100 = 33.95%.

It is worth noting that college graduates have a larger proportion of support than non-college graduates, although the difference is not statistically significant. The percentages of those who oppose and those who are unsure, on the other hand, differ dramatically between the two categories. In this sample, 41.1% of college graduates were opposed to drilling, compared to 32.4% of non-college graduates, and 23.7% were uncertain, compared to 33.6% of non-college graduates.

Overall, the evidence reveals that, while there is some difference in beliefs between college graduates and non-college graduates, the differences are not statistically significant. In both categories, the percentages of support, opposition, and undecided are quite identical. It is worth noting, however, that a sizable proportion of both groups (about one-third) are undecided, implying that there is still substantial disagreement and confusion around this subject.

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The tangential component of the acceleration vector is (4t / (1 + 4t² + 9)[tex]^{1/2}[/tex])i + (8t²/ (1 + 4t² + 9)[tex]^{1/2}[/tex])j + (12t / (1 + 4t² + 9)[tex]^{1/2}[/tex])k, and the normal component of the acceleration vector is -4t / (1 + 4t² + 9)[tex]^{1/2}[/tex] * i + (2 - 8t² / (1 + 4t² + 9)[tex]^{1/2}[/tex])j - 12t / (1 + 4t² + 9)[tex]^{1/2}[/tex] * k.

To find the tangential and normal components of the acceleration vector, we first need to find the acceleration vector itself by taking the second derivative of the position vector r(t):

r(t) = ti + [tex]t^{2j}[/tex] + 3tk

v(t) = dr/dt = i + 2tj + 3k

a(t) = dv/dt = 2j

The acceleration vector is a(t) = 2j. This means that the acceleration is entirely in the y-direction, and there is no acceleration in the x- or z-directions.

The tangential component of the acceleration vector, a_T, is the component of the acceleration vector that is parallel to the velocity vector v(t). Since the velocity vector is i + 2tj + 3k and the acceleration vector is 2j, the tangential component is:

a_T = (a(t) · v(t) / ||v(t)||[tex]^{2}[/tex]) * v(t) = (0 + 4t + 0) / [tex](1 + 4t^{2} + 9)^{1/2}[/tex] * (i + 2tj + 3k)

Simplifying this expression, we get:

a_T = (4t / [tex](1 + 4t^{2} + 9 ^{1/2} )[/tex]i + (8t^2 / (1 + 4t^2 + 9)^(1/2))j + (12t / (1 + 4t^2 + 9)[tex]^{1/2}[/tex])k

The normal component of the acceleration vector, a_N, is the component of the acceleration vector that is perpendicular to the velocity vector. Since the acceleration vector is entirely in the y-direction, the normal component is:

a_N = a(t) - a_T = -4t / (1 + 4t² + 9)[tex]^{1/2}[/tex]* i + (2 - 8t² / (1 + 4t²+ 9)[tex]^{1/2}[/tex])j - 12t / (1 + 4t² + 9)[tex]^{1/2}[/tex]* k

Therefore, the tangential component of the acceleration vector is (4t / (1 + 4t² + 9)[tex]^{1/2}[/tex])i + (8t²/ (1 + 4t² + 9)[tex]^{1/2}[/tex])j + (12t / (1 + 4t² + 9)[tex]^{1/2}[/tex])k, and the normal component of the acceleration vector is -4t / (1 + 4t² + 9)[tex]^{1/2}[/tex] * i + (2 - 8t² / (1 + 4t² + 9)[tex]^{1/2}[/tex])j - 12t / (1 + 4t² + 9)[tex]^{1/2}[/tex] * k.

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The minimum pumping length of {} is any positive integer, of {a} is 1, {a, aaaa, aa}: 1, σ∗: 1 and of {ϵ} is not regular

To find the minimum pumping length for a given language, we need to consider the smallest possible strings in the language and find the smallest length at which we can apply the pumping lemma.

{} (the empty language): There are no strings in the language, so the pumping lemma vacuously holds for any pumping length. The minimum pumping length is any positive integer.

{a}: The smallest string in the language is "a". We can choose the pumping length to be 1, since any substring of "a" of length 1 is still "a". Thus, the minimum pumping length is 1.

{a, aaaa, aa}: The smallest string in the language is "a". We can choose the pumping length to be 1, since any substring of "a" of length 1 is still "a". Thus, the minimum pumping length is 1.

σ∗ (the Kleene closure of σ): Any string over {a} is in the language, including the empty string. We can choose the pumping length to be 1, since any substring of any string in the language of length 1 is still in the language. Thus, the minimum pumping length is 1.

{ϵ} (the language containing only the empty string): The smallest string in the language is the empty string, which has length 0. However, the pumping lemma requires that the pumping length be greater than 0. Since there are no other strings in the language, we cannot satisfy the pumping lemma for any pumping length. Thus, the language {ϵ} is not regular.

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1. Identify the linear equation. y = mx + b
2. Take (b) and plot it on the y axis. Since b is a positive 3, that means you plot a positive 3 on the y axis. This will be the number that your line crosses the y axis on.
3. Take (mx) and plot it in correlation to (b). mx = -x also known as -1. So, from +3 on the y axis, move once to the left and once down. Your coordinate should land on (2, 1).

From here on out, keep moving -1 on the y axis and +1 on the x axis. The ongoing coordinates should look something like (1, 2)(0, 3)(-1, 4) and so on.

Answer:

2.5 hours

Step-by-step explanation:

Let's call the time Jim spent on his bike "t", in hours.

We know that the total time of the trip was 4 hours, so the time he spent walking was 4 - t.

We can use the formula:

distance = rate x time

to set up two equations based on the distances traveled while biking and walking:

Distance biked = rate biking x time biking = 41t

Distance walked = rate walking x time walking = 5(4 - t) = 20 - 5t

The total distance of the trip is 110 miles, so:

Distance biked + distance walked = 110

Substituting the equations for distance biked and walked:

41t + 20 - 5t = 110

36t = 90

t = 2.5

So Jim spent 2.5 hours on his bike.

Hope this helps!

The cost of the film for the whole area of the figure is $73.6.

Given that,

Chris is covering a window with a decorative adhesive film to filter light.

The figure is a window in the shape of a parallelogram.

We have to find the area of the figure.

Area of parallelogram = Base × Height

Area = 8 × 4 = 32 feet²

Cost for the film per square foot = $2.3

Cost of the film for 32 square foot = 32 × $2.3 = $73.6

Hence the cost of the film is $73.6.

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Answer:

B' (-6, 7)

C' ( -8, 1)

Step-by-step explanation:

The rule is

(x,y) → (x -2, y - 1)

A( -2,4) → A' ( -4,3)

To get from A to A', the x value changed by -2 (-2-2 = -4).  The y changed by -1 ( 3-1 = 3)

Helping in the name of Jesus.

Answer:

26.95

Step-by-step explanation:

pass =  656.26 = (36 s + 2t) so 17.27 per person assuming teacher & student same price.

lunch = 348.48/36 =9.68/student

pass and lunch = 9.68 + 17.27 =26.95

Check the picture below.

[tex]\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h~~=height\\ a,b=\stackrel{parallel~sides}{bases~\hfill }\\[-0.5em] \hrulefill\\ a=8\\ b=10\\ h=30 \end{cases}\implies A=\cfrac{30(8+10)}{2}\implies A=270[/tex]

Answer:

5/5 es una fracción adecuada ya que el numerador es igual al denominador.

The probability of x ≤ 8 successes in 10 trials of a binomial experiment with probability of success p = 0.6 is option (c) 0.954.

We can use the cumulative distribution function (CDF) of the binomial distribution to calculate the probability of getting x ≤ 8 successes in 10 trials with a probability of success p = 0.6.

The CDF gives the probability of getting at most x successes in n trials, and is given by the formula

F(x) = Σi=0 to x (n choose i) p^i (1-p)^(n-i)

where (n choose i) represents the binomial coefficient, and is given by

(n choose i) = n! / (i! (n-i)!)

Plugging in the values, we get

F(8) = Σi=0 to 8 (10 choose i) 0.6^i (1-0.6)^(10-i)

Using a calculator or a software program, we can calculate this as

F(8) = 0.9544

So the probability of getting x ≤ 8 successes is 0.9544.

Therefore, the answer is (c) 0.954.

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The x-coordinate of the vertex of the parabola whose equation is given would be -3/2. Option C.

To find the x-coordinate of the vertex of the parabola, we need to use the formula:

In this case, a = 3 and b = 9, so:

x = -9/(2*3) = -3/2

Therefore, the x-coordinate of the vertex of the parabola is -3/2.

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The x-coordinate of the vertex of the parabola whose equation is given would be -3/2. Option C.

To find the x-coordinate of the vertex of the parabola, we need to use the formula:

In this case, a = 3 and b = 9, so:

x = -9/(2*3) = -3/2

Therefore, the x-coordinate of the vertex of the parabola is -3/2.

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Note that this is a vector calculus problem and the tabularized answers are attached accordingly. See the explanation below.

This is a vector calculus problem, which is a branch of mathematics that deals with vectors and functions of vectors. It involves the study of vector fields, which are functions that assign a vector to each point in a given region of space, and the operations that can be performed on them, such as gradient, divergence, and curl. It is often studied in the context of calculus, physics, and engineering.

To fill in the table, we need to calculate the curl and divergence of the given vector fields and determine if they are conservative. Here are the calculations:

F1 = (x - 2z)i + (x + 7y + z)j + (z - 2y)k

Curl F1 = (∂Q/∂y - ∂P/∂z)i + (∂R/∂z - ∂P/∂x)j + (∂P/∂y - ∂Q/∂x)k

= (1 - 0)i + (-2 - 0)j + (7 - 1)k

= i - 2j + 6k

Div F1 = ∂P/∂x + ∂Q/∂y + ∂R/∂z

= 1 + 7 - 2

= 6

Since the curl of F1 is not equal to zero, F1 is not a conservative vector field.

Therefore, the table for F1 would be:

F1 Curl F1 DivF1 is conservative (Y/N)?

(x-2z)i + (x+7y + z)j + (z-2y)k <i - 2j + 6k> 6 N

F2 = yzi + xzj + zyk

Curl F2 = (∂Q/∂y - ∂P/∂z)i + (∂R/∂z - ∂P/∂x)j + (∂P/∂y - ∂Q/∂x)k

= z i + 0j + x k

Div F2 = ∂P/∂x + ∂Q/∂y + ∂R/∂z

= z + z + 1

= 2z + 1

Since the curl of F2 is not equal to zero, F2 is not a conservative vector field.

Therefore, the table for F2 would be:

F2 Curl F2 DivF2 is conservative (Y/N)?

yzi + xzj + zyk <zi + 0k> 2z + 1 N

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Note that this is a vector calculus problem and the tabularized answers are attached accordingly. See the explanation below.

This is a vector calculus problem, which is a branch of mathematics that deals with vectors and functions of vectors. It involves the study of vector fields, which are functions that assign a vector to each point in a given region of space, and the operations that can be performed on them, such as gradient, divergence, and curl. It is often studied in the context of calculus, physics, and engineering.

To fill in the table, we need to calculate the curl and divergence of the given vector fields and determine if they are conservative. Here are the calculations:

F1 = (x - 2z)i + (x + 7y + z)j + (z - 2y)k

Curl F1 = (∂Q/∂y - ∂P/∂z)i + (∂R/∂z - ∂P/∂x)j + (∂P/∂y - ∂Q/∂x)k

= (1 - 0)i + (-2 - 0)j + (7 - 1)k

= i - 2j + 6k

Div F1 = ∂P/∂x + ∂Q/∂y + ∂R/∂z

= 1 + 7 - 2

= 6

Since the curl of F1 is not equal to zero, F1 is not a conservative vector field.

Therefore, the table for F1 would be:

F1 Curl F1 DivF1 is conservative (Y/N)?

(x-2z)i + (x+7y + z)j + (z-2y)k <i - 2j + 6k> 6 N

F2 = yzi + xzj + zyk

Curl F2 = (∂Q/∂y - ∂P/∂z)i + (∂R/∂z - ∂P/∂x)j + (∂P/∂y - ∂Q/∂x)k

= z i + 0j + x k

Div F2 = ∂P/∂x + ∂Q/∂y + ∂R/∂z

= z + z + 1

= 2z + 1

Since the curl of F2 is not equal to zero, F2 is not a conservative vector field.

Therefore, the table for F2 would be:

F2 Curl F2 DivF2 is conservative (Y/N)?

yzi + xzj + zyk <zi + 0k> 2z + 1 N

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The absolute minimum value of given trigonometric-function is 331.9 and absolute maximum value of the same function is 4403.

What is absolute value?

The non-negative value of x or its distance from zero on the number line, regardless of its sign, is the absolute value, modulus, or magnitude denoted by | x | for any real number x. When a function reaches its absolute minimum value, it has reached its lowest conceivable value, and when it reaches its absolute maximum value, it has reached its highest possible value.

Given that the trigonometric function is f(t) = 7t + [tex]7 cot\frac{t}{2}[/tex]

Also given the point at which the function has critical values= [[tex]\frac{\pi }{4} , \frac{7\pi }{2}[/tex] ]

Value of function at [tex]\frac{\pi }{4}[/tex] :

f( [tex]\frac{\pi }{4}[/tex] ) = 7( [tex]\frac{\pi }{4}[/tex] ) + 7 cot([tex]\frac{\pi }{4}.\frac{1}{2}[/tex])

       =[tex]\frac{7\pi }{4}[/tex]       + 7 cot ([tex]\frac{\pi }{8}[/tex])

       =315 + 7 cot 22.5

       =315  + 7(2.414)

       = 315 + 16.898

       =331.898

f( [tex]\frac{\pi }{4}[/tex] ) ≈ 331.9

Value of function at [tex]\frac{7\pi }{2}[/tex] :

f( [tex]\frac{7\pi }{2}[/tex] ) = 7( [tex]\frac{7\pi }{2}[/tex] ) + 7 cot([tex]\frac{7\pi }{2}.\frac{1}{2}[/tex])

        =[tex]\frac{49\pi }{2}[/tex]      + 7 cot ([tex]\frac{7\pi }{4}[/tex])

        =4410 + 7 cot 315

        =4410 + 7(-1)

        =4410-7

        =4403

f( [tex]\frac{7\pi }{2}[/tex] ) =4403

The minimum value=331.9 & maximum value is 4403

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If the "swimming-pool" for children is built with rectangular-prism and 2 halves of cylinder, then the total volume of pool is 312.64 m³.

From the figure, we observe that the swimming pool is made up of a rectangular prism, and 2 halves of cylinder,

the diameter of the half of cylinder is = 16m ,

So, radius of the half of cylinder is = 16/2 = 8m,

The volume of 2 halves of cylinder is = πr²h,

Substituting the values,

We get,

Volume of 2 halves of cylinder is = π × 8 × 8 × 0.6 ≈ 120.64 m³,

Now, the volume of the rectangular prism is = 20 × 16 × 0.6 = 192 m³,

So, the Volume of the swimming pool is = 192 + 120.64 = 312.64 m³.

Therefore, the total volume of swimming pool is 312.64 m³.

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The type of r.v.R is a continuous random variable, since its possible values form a continuous interval [0,1].

The sample space of R is the interval [0,1], since the distance from the origin to any point inside the unit circle is between 0 and 1.
To find P(R < r), we need to find the probability that the randomly selected point falls inside a circle of radius r centered at the origin. The area of this circle is πr^2, and the area of the entire unit circle is π, so the probability is P(R < r) = πr^2/π = r^2.
The cdf of R is the function F(r) = P(R ≤ r) = ∫0r 2πx dx / π = r^2, where the integral is taken over the interval [0,r]. This is because the probability that R is less than or equal to r is the same as the probability that the randomly selected point falls inside the circle of radius r centered at the origin, which has area πr^2. The cdf of R is a continuous and increasing function on the interval [0,1].

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The measure of angle A and B is 30° and 60° respectively.

Trigonometric Ratios are defined as the values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin(tetha) = opp/hyp

cos(tetha) = adj/hyp

tan(tetha) = opp/adj

16 is hypotenuse and 8 is opposite

therefore, sin(tetha) = 8/16

sin(tetha) = 0.5

tetha = sin^-1 ( 0.5)

= 30°

The sum of angle in a triangle is 180°. Therefore ,

angle B = 180-(90+30)

= 180-120 = 60°

therefore the measure of angle A and B is 30° and 60° respectively.

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The number which a value, 20 percent less than 120 is one-third more than is 72

A percentage is an expression of the ratio between quantities, expressed as a fraction with a denominator of 100.

The quantity 20 percent less than 120 can be expressed as follows;

20 percent less than 120 = ((100 - 20)/100) × 120 = 96

One-third more than a number = The number + (The number)/3

Let x represent the number, we get;

One-third more than the number = x + x/3

x + x/3 = 96

x·(1 + 1/3) = 96

4·x/3 = 96

x = 96 × 3/4 = 72

Therefore, 20 percent less than 120 is one-third more than 72

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Answer:

tanA = [tex]\frac{55}{48}[/tex]

Step-by-step explanation:

tanA = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{BC}{AC}[/tex] = [tex]\frac{55}{48}[/tex]

The overall reliability of the system through time t is approximately 0.370.

You have a system with four components and their reliabilities through time t are given as follows:

1. Component 1: 0.80
2. Component 2: 0.66
3. Component 3: 0.78
4. Component 4: 0.89

To find the overall reliability of the system, you'll need to multiply the reliabilities of each individual component:

Overall Reliability = Component 1 Reliability × Component 2 Reliability × Component 3 Reliability × Component 4 Reliability

Step-by-step calculation:

Overall Reliability = 0.80 × 0.66 × 0.78 × 0.89

Now, multiply the given reliabilities:

Overall Reliability ≈ 0.370

So, the overall reliability of the system through time t is approximately 0.370.

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The Laplace transform of a+bt+c is (a/s) + (b/s^2) + (c/s). The Laplace transform of A cos(t+Bsin(t)) is (s/(s^2+B^2)) (A cos(φ) + (B/sin(φ)) A sin(φ)), where φ = arctan(B/s).

For a function f(t), the Laplace transform F(s) is defined as ∫[0, ∞) e^(-st) f(t) dt, where s is a complex number.

To find the Laplace transform of a+bt+c, we use linearity and the Laplace transform of elementary functions:

L{a+bt+c} = L{a} + L{bt} + L{c} = a/s + bL{t} + c/s = a/s + b/s^2 + c/s

Therefore, the Laplace transform of a+bt+c is (a/s) + (b/s^2) + (c/s).

B. To find the Laplace transform of A cos(t+Bsin(t)), we use the following identity:

cos(t + Bsin(t)) = cos(t)cos(Bsin(t)) - sin(t)sin(Bsin(t))

Then, we apply the Laplace transform to both sides and use linearity and the Laplace transform of elementary functions:

L{cos(t + Bsin(t))} = L{cos(t)cos(Bsin(t))} - L{sin(t)sin(Bsin(t))}

Using the formula L{cos(at)} = s/(s^2 + a^2), we get:

L{cos(t + Bsin(t))} = (s/(s^2+B^2)) L{cos(t)} - (s/(s^2+B^2)) L{sin(t)}

Using the formula L{sin(at)} = a/(s^2 + a^2), we get:

L{cos(t + Bsin(t))} = (s/(s^2+B^2)) (1/s) - (B/(s^2+B^2)) (1/s)

Simplifying, we get:

L{cos(t + Bsin(t))} = (s/(s^2+B^2)) (A cos(φ) + (B/sin(φ)) A sin(φ)), where φ = arctan(B/s)

Therefore, the Laplace transform of A cos(t+Bsin(t)) is (s/(s^2+B^2)) (A cos(φ) + (B/sin(φ)) A sin(φ)), where φ = arctan(B/s).

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Answer:

c. 50(i-0.1)² + 210(i - 0.1) + 320.5.

Step-by-step explanation:

To find the equivalent expression that contains the future value for a monthly interest rate of i = 0.1, we simply substitute i = 0.1 into the equation V = 50(1 + i)² + 100(1 + i) + 150 and simplify.

V = 50(1 + 0.1)² + 100(1 + 0.1) + 150

V = 50(1.1)² + 100(1.1) + 150

V = 50(1.21) + 110 + 150

V = 60.5 + 110 + 150

V = 320.5

Therefore, the expression that contains the future value for a monthly interest rate of i = 0.1 is c. 50(i-0.1)² + 210(i - 0.1) + 320.5.

The problem involves finding the compositions of two functions f and g, their inverse functions, and the composition of the inverse functions. The solution demonstrates how to apply these concepts.

To find the compositions of functions and their inverse functions.

Using the given definitions of f and g.

We find their compositions and their inverse functions. Then we apply these results to find the compositions of inverse functions.

(a) fog: [tex]fog(x) = f(g(x)) = f(5x+7) = 2(5x+7) + 3 = 10x + 17[/tex]

(b) gof: [tex]gof(x) = g(f(x)) = g(2x+3) = 5(2x+3) + 7 = 10x + 22[/tex]

(c) [tex](fog)^-1:[/tex]

We first find fog(x) and then solve for x: [tex]fog(x) = 10x + 17[/tex]

[tex]x = (fog(x) - 17)/10[/tex]

[tex](fog)^-1(x) = (x - 17)/10[/tex]

[tex](d) f^-1 o g^-1:[/tex]

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

[tex]g^-1(x) = (x - 7)/5[/tex]

[tex](f^-1 o g^-1)(x) = f^-1(g^-1(x)) = f^-1((x-7)/5)[/tex] = [tex][(x-7)/5 - 3]/2 = (x-23)/10[/tex]

(e)[tex]g^1 o f^-1:[/tex] [tex]g^1(x) = (x-7)/5[/tex]

[tex](g^1 o f^-1)(x) = g^1(f^-1(x))[/tex]

=[tex]g^1((x-3)/2) = 5((x-3)/2) + 7[/tex]

=[tex](5x+2)/2[/tex]

Overall, the problem requires a solid understanding of function composition, inverse functions, and basic algebraic manipulation.

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The function with an amplitude of 3 and a phase shift of π/2 is h(x) = 3 cos (2x - π/2) + 3.

The amplitude of a function is the distance between the maximum and minimum values of the function, divided by 2. The phase shift of a function is the horizontal shift of the function from the standard position,

(y = cos(x) or y = sin(x)).
To find the function with an amplitude of 3 and a phase shift of π/2, we need to look for a function that has a coefficient of 3 on the cosine term and a horizontal shift of π/2.
Looking at the given options, we can eliminate option a) because it has a coefficient of -3 on the cosine term, which means that its amplitude is 3 but it is inverted.

Option b) has a coefficient of 3 on the cosine term but it has a phase shift of -π/2, which means it is shifted to the left instead of to the right. Option d) has a phase shift of π/2, but it has a coefficient of -2 on the cosine term, which means its amplitude is 2 and not 3.
A*cos(B( x - C)) + D

Where A is the amplitude, B affects the period, C is the phase shift, and D is the vertical shift.

f(x) = -3 cos(2x - π) + 4

Amplitude: |-3| = 3

Phase shift: π (not π/2) b) g(x) = 3cos(2x + π) -1

Amplitude: |3| = 3

Phase shift: -π (not π/2) c) h(x) = 3 cos (2x - π/2) + 3

Amplitude: |3| = 3

Phase shift: π/2 d) j(x) = -2cos(2x + π/2) + 3200

Amplitude: |-2| = 2 (not 3)

Phase shift: -π/2
Therefore, the only option left is c) h(x) = 3 cos (2x - π/2) + 3. This function has a coefficient of 3 on the cosine term and a horizontal shift of π/2, which means it has an amplitude of 3 and a phase shift of π/2.
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From the Venn diagram, the values of a, b, c and d are11,19,35,35 respectively

A Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things. Circles that overlap have a commonality while Circles that do not overlap do not share those traits.

The universal set is ∈ = 100

The languages are

Spanish = 30

Russian = 54

(S∪ R)¹ = 35 = d

a = Spanish only = a-b

30-b = a

Russia only = c-b

54 - b

Therefore, The universal set ∈ is

100 = (a-b) + (b)+ (c-b) +(d)

100 =  30-b + b + 54 - b + 35

100 = 119 - b = 119-100

b= 19

Therefore,

a = 30 -19 =11

b = 19

c = 59 - 19 35

d = 35

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The shortest distance from the point (3, 0, −2) to the plane x + y + z = 2 is √(3) or approximately 1.732 units.

To find the shortest distance, d, from the point (3, 0, −2) to the plane x + y + z = 2, we need to use the formula for the distance between a point and a plane.

First, we need to find the normal vector of the plane. The coefficients of x, y, and z in the plane equation (1, 1, 1) form the normal vector (since the plane is perpendicular to this vector).

Next, we can use the point-to-plane distance formula:

d = |(ax + by + cz - d) / √(a² + b² + c²)|

where (a, b, c) is the normal vector of the plane, (x, y, z) is the coordinates of the point, and d is the constant term in the plane equation.

Plugging in the values, we get:

d = |(1(3) + 1(0) + 1(-2) - 2) / √(1² + 1² + 1²)|

d = |(1 + 0 - 4) / √(3)|

d = |-3 / √(3)|

d = |-√(3)|

Therefore, the shortest distance from the point (3, 0, −2) to the plane x + y + z = 2 is √(3) or approximately 1.732 units.

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