Find the critical point set for the given system. dx = x-y 2x² + 7y²-9 Find the critical point set. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. O A. The critical point set consists of the isolated point(s) (Use a comma to separate answers as needed. Type an ordered pair Type an exact answer, using radicals as needed.) OB. The critical point set consists of the line(s) described by the equation(s). O C. (Use a comma to separate answers as needed. Type an ordered pair Type an exact answer, using radicals as needed.) The critical point set consists of the isolated point(s) and the line(s) described by the equation(s). (Use a comma to separate answers as needed. Type an ordered pair Type an exact answer, using radicals as needed.) O D. There are no critical points.

The function y = tan(3π/2)θ has a period of **π** and two asymptotes:

y = 1: This asymptote is reached when θ is a multiple of π/2.

y = -1: This asymptote is reached when θ is a multiple of 3π/2.

The function oscillates between the two asymptotes, with a period of π.

The reason for the asymptotes is that the tangent function is undefined when the denominator of the fraction is zero. In this case, the denominator is zero when θ is a multiple of π/2 or 3π/2.

Therefore, the function approaches the asymptotes as θ approaches these values.

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The six trigonometric functions for the angle in standard position determined by the point (1, -5) are

sinθ = o/h = 5/√26
cosθ = a/h = 1/√26
tanθ = o/a = 5/1 = 5
cscθ = h/o = √26/5
secθ = h/a = √26/1 = √26
cotθ = a/o = 1/5

The given point (1, -5) is located in the third quadrant of the Cartesian plane, where x-coordinates are positive and y-coordinates are negative. To determine the values of the six trigonometric functions for the angle formed by this point in standard position, we need to first calculate the hypotenuse, adjacent, and opposite sides of the right triangle that is formed by the given point and the origin (0, 0).

The hypotenuse is the distance between the point (1, -5) and the origin (0, 0), which is given by the Pythagorean theorem as follows:

h = √((1 - 0)² + (-5 - 0)²)
h = √(1 + 25)
h = √26

The adjacent side is the distance between the point (1, -5) and the y-axis, which is equal to the absolute value of the x-coordinate:

a = |1|
a = 1

The opposite side is the distance between the point (1, -5) and the x-axis, which is equal to the absolute value of the y-coordinate:

o = |-5|
o = 5

Now, we can use these values to calculate the six trigonometric functions as follows:
sinθ = o/h = 5/√26
cosθ = a/h = 1/√26
tanθ = o/a = 5/1 = 5
cscθ = h/o = √26/5
secθ = h/a = √26/1 = √26
cotθ = a/o = 1/5

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Answer:

Step-by-step explanation:Let's define the variables:

Let "x" be the number of packages weighing five pounds or less.

Let "y" be the number of packages weighing more than ten pounds.

Based on the given information, we can set up the following equations:

Equation 1: x + y = 13

The total number of packages is 13.

Equation 2: 7x + 15y + 22z = 168

The total cost of the packages is $168.

Equation 3: x = y + 3

The number of packages weighing five pounds or less is three more than those weighing more than ten pounds.

To solve this system of equations, we can use the substitution method or elimination method. Let's use the substitution method here:

From Equation 3, we can rewrite it as:

y = x - 3

Now we substitute this value of y in Equation 1:

x + (x - 3) = 13

2x - 3 = 13

2x = 13 + 3

2x = 16

x = 16/2

x = 8

Substituting the value of x back into Equation 3:

y = x - 3

y = 8 - 3

y = 5

So, we have x = 8 and y = 5.

To find the value of z, we substitute the values of x and y into Equation 2:

7x + 15y + 22z = 168

7(8) + 15(5) + 22z = 168

56 + 75 + 22z = 168

131 + 22z = 168

22z = 168 - 131

22z = 37

z = 37/22

z ≈ 1.68

Therefore, the number of packages weighing five pounds or less is 8, the number of packages weighing more than ten pounds is 5, and the number of packages weighing between five and ten pounds is approximately 1.68.

Mark's profit from the sale of the shampoo is $42700.

To calculate Mark's profit from the sale of shampoo, we need to consider the total cost of purchasing the shampoo and the total revenue generated from selling it.

Total Cost:

Mark purchased 2000 bottles of shampoo at a cost of $3.97 per bottle. To find the total cost, we multiply the number of bottles (2000) by the cost per bottle ($3.97).

Total Cost = 2000 * $3.97 = $7,940.

Total Revenue:

Mark sells each bottle of shampoo for $25.32 to each client. To find the total revenue, we multiply the selling price per bottle ($25.32) by the number of bottles (2000).

Total Revenue = 2000 * $25.32 = $50,640.

Profit:

To calculate the profit, we subtract the total cost from the total revenue.

Profit = Total Revenue - Total Cost

Profit = $50,640 - $7,940 = $42,700.

Therefore, Mark's profit from the sale of shampoo is $42,700.

It's important to note that profit represents the financial gain obtained after deducting the cost of purchasing the goods from the revenue generated by selling them. In this case, Mark's profit indicates the earnings he achieved by selling the shampoo bottles in his barber shop. It signifies the positive difference between the revenue received from customers and the cost incurred to acquire the shampoo inventory.

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Answer:

Yes

Step-by-step explanation:

You can tell because X does not have a number that repeats it self 2 or more times. I hope this helps.

The solutions to the given implicit function is

[tex]∂z/∂x = -2xz / (2x + x^2 - y*sin(yz))[/tex]

and

[tex]∂z/∂y = (-y - z*sin(yz)) / (1 + z*sin(yz)^2)[/tex]

To find ∂z/∂x and ∂z/∂y given that z is given implicitly as a function of x and y

use implicit differentiation for the equation

[tex]x^2z + y^2 + z^2 = cos(yz)[/tex]

Take the partial derivative of both sides of the equation with respect to x

[tex]2xz + x^2(∂z/∂x) + 2z(∂z/∂x) \\ = -y*sin(yz)(∂z/∂x)[/tex]

Simplifying, we get:

[tex](2x + x^2 - y*sin(yz))(∂z/∂x) \\ = -2xz[/tex]

Divide both sides by 2x + x^2 - y*sin(yz), we get:

[tex]∂z/∂x = -2xz / (2x + x^2 - y*sin(yz))

[/tex]

Take partial derivative of both sides of the equation with respect to y, we get:

2yz + 2z(∂z/∂y) = -z*sin(yz)(y + yz∂z/∂y) + 2y

Simplifying, we get:

[tex](2z - z*sin(yz)y - 2y)/(1 + z*sin(yz)^2)(∂z/∂y) \\ = -y - z*sin(yz)[/tex]

Divide both sides by (2z - z*sin(yz)y - 2y)/(1 + z*sin(yz)^2),

[tex]∂z/∂y = (-y - z*sin(yz)) / (1 + z*sin(yz)^2)[/tex]

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Given equation x²z+y²+z²=cos(yz) is given implicitly as a function of x and y.

Here, we have to find out the partial derivatives of z with respect to x and y.

So, we need to differentiate the given equation partially with respect to x and y.

To find ∂z/∂x,
Differentiating the given equation partially with respect to x, we get:

2xz+0+2zz' = -y zsin(yz)

Using the Chain Rule: z' = dz/dx and dz/dy

Similarly, to find ∂z/∂y, differentiate the given equation partially with respect to y, we get: 0+2y+2zz' = -zsin(yz) ⇒ 2y+2zz' = -zsin(yz)

Again, using the Chain Rule: z' = dz/dx and dz/dy

We can write the above equations as: z'(2xz+2zz') = -yzsin(yz)⇒ ∂z/∂x = -y sin(yz)/(2xz+2zz')

Also, z'(2y+2zz') = -zsin(yz)⇒ ∂z/∂y = [1-zcos(yz)]/(2y+2zz')

Thus, ∂z/∂x = -y sin(yz)/(2xz+2zz') and ∂z/∂y = [1-zcos(yz)]/(2y+2zz')

Hence, the answer is ∂z/∂x = -y sin(yz)/(2xz+2zz') and ∂z/∂y = [1-zcos(yz)]/(2y+2zz')

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1. There are [tex]2^a[/tex]functions f : X → {0, 1}.

(i) There are [tex]2^a[/tex]functions f : X → {0, 1} that are 1-1.

(ii) There are [tex]2^a[/tex]-a functions f : X → {0, 1} that are onto.

(iii) There are [tex]3^a-2^a[/tex] functions f : X → {0, 1, 2} that are onto.

1. For each element in X, we have two choices: either map it to 0 or 1. Since there are a elements in X, the total number of functions f : X → {0, 1} is [tex]2^a[/tex].

(i) To count the number of 1-1 functions, we need to ensure that no two elements in X are mapped to the same element in {0, 1}. The first element can be mapped to any of the two elements in {0, 1}, the second element can be mapped to one of the remaining choices, and so on. Therefore, the number of 1-1 functions is also [tex]2^a[/tex].

(ii) To count the number of onto functions, we need to ensure that every element in {0, 1} has at least one pre-image in X. For each element in {0, 1}, we have two choices: either include it as a pre-image or exclude it. So, the number of onto functions is [tex]2^a-a[/tex], since there are [tex]2^a[/tex] total functions and a of them are not onto.

(iii) Similarly, to count the number of onto functions f : X → {0, 1, 2}, we have three choices for each element in X: map it to 0, 1, or 2. Therefore, the total number of onto functions is [tex]3^a-2^a[/tex].

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Answer:

Rosie is 10 years old

Step-by-step explanation:

A)

Rosie is x years old

Rosie's age (R) = x

R = x

Eva is 2 years older

Eva's age (E) = x + 2

E = x + 2

Jack is twice Rosie’s age

Jack's age (J) = 2x

J = 2x

B)

R + E + J = 42

x + (x + 2) + (2x) = 42

x + x + 2 + 2x = 42

4x + 2 = 42

4x = 42 - 2

4x = 40

[tex]x = \frac{40}{4} \\\\x = 10[/tex]

Rosie is 10 years old

Maxima is a computer algebra system that can perform symbolic and numerical computations. It is particularly useful for mathematical calculations and symbolic manipulation. Here's a step-by-step guide on how to use Maxima:

Step 1:

Install Maxima

First, you need to install Maxima on your computer. Maxima is an open-source software and can be downloaded for free from the official Maxima website (http://maxima.sourceforge.net/). Follow the installation instructions for your specific operating system.

Step 2:

Launch Maxima

After installing Maxima, launch the Maxima application. You can typically find it in your applications or programs menu. Maxima provides two interfaces: a command-line interface (CLI) and a graphical user interface (GUI). You can choose the interface that suits your preference.

- Command-Line Interface (CLI): The CLI allows you to interact with Maxima using text commands. You type commands in the input prompt, and Maxima will respond with the output.

- Graphical User Interface (GUI): The GUI provides a more user-friendly environment with menus, buttons, and input/output areas. You can enter commands in the input area and see the results in the output area.

Choose the interface that you prefer and start using Maxima.

Step 3:

Perform Mathematical Calculations

Maxima can handle a wide range of mathematical computations. Here are a few examples to get you started:

- Basic Arithmetic: Maxima can perform simple arithmetic operations such as addition, subtraction, multiplication, and division. For example, you can type `2 + 3` and press Enter to get the result `5`.

- Symbolic Expressions: Maxima can manipulate symbolic expressions. You can define variables, perform algebraic operations, and simplify expressions. For example, you can type `x^2 + 2*x + 1` and press Enter to get the result `x^2 + 2*x + 1`.

- Solve Equations: Maxima can solve equations symbolically or numerically. For example, you can type `solve(x^2 - 4 = 0, x)` and press Enter to solve the equation `x^2 - 4 = 0` and get the result `[x = -2, x = 2]`.

- Differentiation and Integration: Maxima can perform symbolic differentiation and integration. For example, you can type `diff(sin(x), x)` and press Enter to differentiate `sin(x)` with respect to `x` and get the result `cos(x)`. Similarly, you can use the `integrate` function to perform integration.

- Plotting: Maxima can generate plots of functions and data. You can use the `plot2d` or `plot3d` functions to create 2D or 3D plots. For example, you can type `plot2d(sin(x), [x, -pi, pi])` and press Enter to plot the sine function from `-pi` to `pi`.

These are just a few examples of what you can do with Maxima. It has a vast range of capabilities, including linear algebra, calculus, number theory, and more. You can explore the Maxima documentation, tutorials, and examples to learn more about its features and syntax.

Step 4:

Save and Load Maxima Scripts

If you want to save your Maxima calculations for future use, you can save them as Maxima scripts with a `.mac` extension. Maxima scripts are plain text files containing a series of Maxima commands. You can load a Maxima script into Maxima using the `load` command. For example, if you have a script named `myscript.mac`, you can type `load("myscript.mac")` in Maxima to execute the commands

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Maxima is a computer algebra system that can perform symbolic and numerical computations. It is particularly useful for mathematical calculations and symbolic manipulation. Here's a step-by-step guide on how to use Maxima:

Step 1:

Install Maxima

First, you need to install Maxima on your computer. Maxima is an open-source software and can be downloaded for free from the official Maxima website (http://maxima.sourceforge.net/). Follow the installation instructions for your specific operating system.

Step 2:

Launch Maxima

After installing Maxima, launch the Maxima application. You can typically find it in your applications or programs menu. Maxima provides two interfaces: a command-line interface (CLI) and a graphical user interface (GUI). You can choose the interface that suits your preference.

- Command-Line Interface (CLI): The CLI allows you to interact with Maxima using text commands. You type commands in the input prompt, and Maxima will respond with the output.

- Graphical User Interface (GUI): The GUI provides a more user-friendly environment with menus, buttons, and input/output areas. You can enter commands in the input area and see the results in the output area.

Choose the interface that you prefer and start using Maxima.

Step 3:

Perform Mathematical Calculations

Maxima can handle a wide range of mathematical computations. Here are a few examples to get you started:

- Basic Arithmetic: Maxima can perform simple arithmetic operations such as addition, subtraction, multiplication, and division. For example, you can type `2 + 3` and press Enter to get the result `5`.

- Symbolic Expressions: Maxima can manipulate symbolic expressions. You can define variables, perform algebraic operations, and simplify expressions. For example, you can type `x^2 + 2*x + 1` and press Enter to get the result `x^2 + 2*x + 1`.

- Solve Equations: Maxima can solve equations symbolically or numerically. For example, you can type `solve(x^2 - 4 = 0, x)` and press Enter to solve the equation `x^2 - 4 = 0` and get the result `[x = -2, x = 2]`.

- Differentiation and Integration: Maxima can perform symbolic differentiation and integration. For example, you can type `diff(sin(x), x)` and press Enter to differentiate `sin(x)` with respect to `x` and get the result `cos(x)`. Similarly, you can use the `integrate` function to perform integration.

- Plotting: Maxima can generate plots of functions and data. You can use the `plot2d` or `plot3d` functions to create 2D or 3D plots. For example, you can type `plot2d(sin(x), [x, -pi, pi])` and press Enter to plot the sine function from `-pi` to `pi`.

These are just a few examples of what you can do with Maxima. It has a vast range of capabilities, including linear algebra, calculus, number theory, and more. You can explore the Maxima documentation, tutorials, and examples to learn more about its features and syntax.

Step 4:

Save and Load Maxima Scripts

If you want to save your Maxima calculations for future use, you can save them as Maxima scripts with a `.mac` extension. Maxima scripts are plain text files containing a series of Maxima commands. You can load a Maxima script into Maxima using the `load` command. For example, if you have a script named `myscript.mac`, you can type `load("myscript.mac")` in Maxima to execute the commands

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B & C is a subset of B & C. Hence B\(B\A) = A if and only if ACB.

a) Let ACB and Ag C, we need to show that B & C.

Let x be an arbitrary element of B & C.

Since x is in B, we have x ACB.

But then x AgC (since ACB and AgC) and hence x is in C.

So x is in B & C and we have shown that B & C is a subset of B & C.

Now let x be an arbitrary element of B & C.

Then x is in B and x is in C.

So x ACB and x AgC.

But then ACB and AgC imply ACB & AgC and hence x is in B & C.

Hence B & C = B & C.

(b) We have B\(B\A) = A if and only if every element of B that is not in A is not in B, that is, if and only if B\(B\A)cA.

But B\(B\A)cA if and only if ACB\(B\A).

We have ACB\(B\A) if and only if every element of C that is not in A is not in B, that is, if and only if C\(C\A)cB.

But C\(C\A)cB if and only if ACB\(C\A).  

So B\(B\A) = A if and only if ACB\(C\A), which is true if and only if ACB.  

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Answer:

B

Step-by-step explanation:

Gl on your finals

Answer:

[tex]\huge\boxed{\sf \frac{1}{12} }[/tex]

Step-by-step explanation:

[tex]\displaystyle = \frac{2}{3} \div 8[/tex]

We need to change the division sign into multiplication. For that, we have to multiply the fraction with the reciprocal of the number next to division sign and not the actual number.

[tex]\displaystyle = \frac{2}{3} \times \frac{1}{8} \\\\= \frac{2 \times 1}{3 \times 8} \\\\= \frac{2}{24} \\\\= \frac{1}{12} \\\\\rule[225]{225}{2}[/tex]

Answer:

J) 1/12

Explanation:

Let's divide these fractions:

[tex]\sf{\dfrac{2}{3}\div8}\\\\\\\sf{\dfrac{2}{3}\div\dfrac{8}{1}}\\\\\\\sf{\dfrac{2}{3}\times\dfrac{1}{8}}\\\\\sf{\dfrac{2}{24}}\\\\\\\sf{\dfrac{1}{12}}[/tex]

Hence, the answer is 1/12.

The hyperbola is centered at the midpoint between the two airports and its branches extend towards each airport. The branch representing the flight path is the one where the signal from your airport arrives first (100 μs earlier).

In this scenario, we have two airports located 48 km apart. The pilot's instrument panel receives radio signals from both airports simultaneously, but there is a time delay between the signals due to the distance and speed of transmission.

Let's assume that the pilot's instrument panel is at the center of the hyperbola. The distance between the two airports is 48 km, so the midpoint between them is at a distance of 24 km from each airport.

Since the signal from your airport always arrives 100 μs earlier than the signal from the other airport, it means that the hyperbola is oriented such that the branch representing the flight path is closer to your airport.

To draw the hyperbola, we mark the midpoint between the two airports and draw two branches extending towards each airport. The branch that is closer to your airport represents the flight path, as it indicates that the signal from your airport reaches the pilot's instrument panel earlier.

The other branch of the hyperbola represents the signals arriving from the other airport, which have a delay of 100 μs compared to the signals from your airport.

In summary, the branch of the hyperbola that represents the flight path is the one where the signal from your airport arrives first, 100 μs earlier than the signal from the other airport.

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The shadow cast by the shorter pole is 8 meters long.

At the bottom of a ski lift, there are two vertical poles. One pole is 15 meters tall and the other is 10 meters tall. The taller pole casts a shadow that is 12 meters long.

How long is the shadow cast by the shorter pole?To solve this problem, we can use the concept of similar triangles. Similar triangles have the same shape but different sizes. This means that their corresponding sides are proportional. Let's draw a diagram to represent the situation:

In this diagram, we have two vertical poles AB and CD. AB is the taller pole and CD is the shorter pole. AB is 15 meters tall and casts a shadow EF that is 12 meters long. We want to find the length of the shadow GH cast by CD. We can use similar triangles to do this.

The two triangles AEF and CDG are similar because they have the same shape. This means that their corresponding sides are proportional. Let's set up a proportion using the length of the shadows and the height of the poles:

EF/AB = GH/CDSubstituting the given values:12/15 = GH/10Simplifying:4/5 = GH/10Multiplying both sides by 10:8 = GHTherefore, the shadow cast by the shorter pole is 8 meters long.

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The possible values for the divisor d are 1 and 37.

Let's denote the original number as x and the divisor as d.

According to the given information:

x divided by d leaves a remainder of 24. We can express this as x ≡ 24 (mod d).

2x divided by d leaves a remainder of 11. This can be expressed as 2x ≡ 11 (mod d).

We can rewrite these congruence equations as:

x ≡ 24 (mod d) -- Equation 1

2x ≡ 11 (mod d) -- Equation 2

To find the divisor, we need to find a value of d that satisfies both equations simultaneously.

Let's solve these congruence equations:

From Equation 1, we can write:

x = 24 + kd -- Equation 3, where k is an integer

Substituting Equation 3 into Equation 2:

2(24 + kd) ≡ 11 (mod d)

48 + 2kd ≡ 11 (mod d)

48 ≡ 11 (mod d)

48 - 11 ≡ 0 (mod d)

37 ≡ 0 (mod d)

This implies that d divides 37 without any remainder. The divisors of 37 are 1 and 37.

Therefore, the possible values for the divisor d are 1 and 37.

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Answer:

which

Step-by-step explanation:

grease and flour and salt in a few days ago hera tw chaina raicha bhane ma lyauchu la ma herchu you have any questions or concerns please visit the plug-in settings to determine how attachments are handled the situation and I was just wondering I am I

We have shown that the ellipse and hyperbola intersect at right angles.

To show that the ellipse and hyperbola intersect at right angles, we need to prove that their tangent lines at the point of intersection are perpendicular.

Let's first find the equations of the ellipse and hyperbola:

Ellipse: x^2/a^2 + 2y^2 = 1   ...(1)

Hyperbola: x^2/a^2 - 2y^2 = 1   ...(2)

To find the point(s) of intersection, we can solve the system of equations formed by (1) and (2). Subtracting equation (2) from equation (1), we have:

2y^2 - (-2y^2) = 0

4y^2 = 0

y^2 = 0

y = 0

Substituting y = 0 into equation (1), we can solve for x:

x^2/a^2 = 1

x^2 = a^2

x = ± a

So, the points of intersection are (a, 0) and (-a, 0).

To find the tangent lines at these points, we need to differentiate the equations of the ellipse and hyperbola with respect to x:

Differentiating equation (1) implicitly:

2x/a^2 + 4y * (dy/dx) = 0

dy/dx = -x / (2y)

Differentiating equation (2) implicitly:

2x/a^2 - 4y * (dy/dx) = 0

dy/dx = x / (2y)

Now, let's evaluate the slopes of the tangent lines at the points (a, 0) and (-a, 0) by substituting these values into the derivatives we found:

At (a, 0):

dy/dx = -a / (2 * 0) = undefined (vertical tangent)

At (-a, 0):

dy/dx = -(-a) / (2 * 0) = undefined (vertical tangent)

Since the slopes of the tangent lines at both points are undefined (vertical), they are perpendicular to the x-axis.

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a) The annual demand is 420.

b) The weekly demand forecast is 8.08

c) The EOQ would be approximately 41

d) The reorder point is 45.12

e) The total annual cost is 102439.02

f) The pipeline inventory is 32.32

g) The new reorder point is 48.82

a) To calculate the annual demand, you need to use your student number. For example, if your student number is BBAW190102, the last digit is 2. So, the annual demand would be 400 + 10 x 2 = 420.

b) To calculate the weekly demand forecast for 2021, you need to divide the annual demand by the number of weeks in a year. Assuming there are 52 weeks in a year, the weekly demand forecast would be 420 / 52 = 8.08 (rounded to two decimal places).

c) The economic order quantity (EOQ) can be calculated using the formula EOQ = sqrt((2DS) / H), where D is the annual demand, S is the ordering cost per order, and H is the annual holding cost. In this case, D is the annual demand calculated in part a, S is $1000, and H is $500. Plugging in these values, the EOQ would be sqrt((2 x 420 x 1000) / 500) = sqrt(840000 / 500) = sqrt(1680) ≈ 41 (rounded to the nearest whole number).

d) The reorder point is the level of inventory at which a new order should be placed. It can be calculated using the formula reorder point = demand during lead time + safety stock. The demand during lead time is the average demand per week multiplied by the lead time, which is 8.08 x 4 = 32.32 (rounded to two decimal places). The safety stock is the z-score multiplied by the standard deviation of weekly demand. The z-score for a 90% cycle service level is 1.28 (given in the question) and the standard deviation of weekly demand is 10 (given in the question). So, the safety stock would be 1.28 x 10 = 12.8 (rounded to one decimal place). Therefore, the reorder point would be 32.32 + 12.8 = 45.12 (rounded to two decimal places).

e) The total annual cost of managing the inventory can be calculated using the formula TC = (S x D/Q) + (H x (Q/2 + s)), where S is the ordering cost per order, D is the annual demand, Q is the economic order quantity, H is the annual holding cost, and s is the safety stock. Plugging in the values, the total annual cost would be (1000 x 420/41) + (500 x (41/2 + 12.8)) = 102439.02 (rounded to two decimal places).

f) The pipeline inventory refers to the inventory that is in transit or being processed. In this case, since the lead time is 4 weeks, the pipeline inventory would be the average demand per week multiplied by the lead time. So, the pipeline inventory would be 8.08 x 4 = 32.32 (rounded to two decimal places).

g) To achieve a 95% cycle service level, we need to calculate the new safety stock and reorder point. The z-score for a 95% cycle service level is 1.65 (given in the question). Using the same formula as in part d, the new safety stock would be 1.65 x 10 = 16.5 (rounded to one decimal place). Therefore, the new reorder point would be 32.32 + 16.5 = 48.82 (rounded to two decimal places).

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(a) The reading on the thermometer will be 7°C after 2 more minutes.

(b) The thermometer will read 2°C 15 minutes after it was taken outdoors.

(a) In the given scenario, the temperature on the thermometer decreases by 8°C in the first minute (from 22°C to 14°C). We can observe that the temperature change is linear, decreasing by 8°C per minute. Therefore, after 2 more minutes, the temperature will decrease by another 2 times 8°C, resulting in a reading of 14°C - 2 times 8°C = 14°C - 16°C = 7°C.

(b) To determine when the thermometer will read 2°C, we need to find the number of minutes it takes for the temperature to decrease by 20°C (from 22°C to 2°C). Since the temperature decreases by 8°C per minute, we divide 20°C by 8°C per minute, which gives us 2.5 minutes. However, since the thermometer cannot read fractional minutes, we round up to the nearest whole minute. Therefore, the thermometer will read 2°C approximately 3 minutes after it was taken outdoors.

It's important to note that these calculations assume a consistent linear rate of temperature change. In reality, temperature changes may not always follow a perfectly linear pattern, and various factors can affect the rate of temperature change.

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(a) The normal curve is sketched with the standard deviation numbers and percentages indicated.

(b) Approximately 68% of adult US men have a height that falls within 2 standard deviations below the mean and 1 standard deviation above the mean.

(c) The percentage of adult US men with a height between 65.5" and 71.5" can be determined from the normal curve.

(d) The percentage of adult US men with a height less than 67.5 inches can be determined from the normal curve.

(e) The percentage of adult US men with a height between 71.5" and 75.5" can be determined from the normal curve. In a group of 90 adult US men, we can expect the proportion of men falling within this range.

(f) The percentage of adult US men with a height between 65.5" and 69.5" can be determined from the normal curve. In a group of 90 adult US men, we can expect the proportion of men falling within this range.

(a) The normal curve, also known as the bell curve or Gaussian distribution, is a symmetrical probability distribution that is often used to model various natural phenomena. It is characterized by its mean and standard deviation. When sketching the normal curve, the mean is marked at the center, and the standard deviation values are represented as points on the curve, usually at 1, 2, and 3 standard deviations from the mean.

The percentages associated with each standard deviation value represent the proportion of data falling within that range.

(b) Since the normal curve follows the 68-95-99.7 rule, we know that approximately 68% of the data falls within 1 standard deviation of the mean. Therefore, about 68% of adult US men have a height between 2 standard deviations below the mean and 1 standard deviation above the mean.

(c) To determine the percentage of adult US men with a height between 65.5" and 71.5", we need to calculate the area under the normal curve between these two values. This can be done using statistical software or by referring to the standard normal distribution table, which provides the proportion of data falling within specific standard deviation ranges.

(d) To find the percentage of adult US men with a height less than 67.5 inches, we need to calculate the area under the normal curve to the left of this value. Again, this can be done using statistical software or the standard normal distribution table.

(e) Similarly, to determine the percentage of adult US men with a height between 71.5" and 75.5", we calculate the area under the normal curve between these two values.

In a group of 90 adult US men, we can expect the proportion of men falling within a specific height range by multiplying the percentage obtained from the normal curve by the total number of men in the group.

(f) Similar to (c) and (e), we can calculate the percentage of adult US men with a height between 65.5" and 69.5" using the normal curve. To estimate the number of men falling within this range in a group of 90, we multiply this percentage by 90.

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By using half-angle identities, we have expressed cos(θ/4) in terms of trigonometric functions of θ as ±√((1 + cosθ) / 4).

To write the expression cos(θ/4) using half-angle identities, we can utilize the half-angle formula for cosine, which states that cos(θ/2) = ±√((1 + cosθ) / 2). By substituting θ/4 in place of θ, we can rewrite cos(θ/4) in terms of trigonometric functions of θ.

To write cos(θ/4) using half-angle identities, we can substitute θ/4 in place of θ in the half-angle formula for cosine. The half-angle formula states that cos(θ/2) = ±√((1 + cosθ) / 2).

Substituting θ/4 in place of θ, we have cos(θ/4) = cos((θ/2) / 2) = cos(θ/2) / √2.

Using the half-angle formula for cosine, we can express cos(θ/2) as ±√((1 + cosθ) / 2). Therefore, we can rewrite cos(θ/4) as ±√((1 + cosθ) / 2) / √2.

Simplifying further, we have cos(θ/4) = ±√((1 + cosθ) / 4).

Thus, by using half-angle identities, we have expressed cos(θ/4) in terms of trigonometric functions of θ as ±√((1 + cosθ) / 4).

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Answer:

The solution to the inequality |x|-3|x+4|≧0 is x≤-4 or -1≤x≤3.

Answer:

-4,3

Step-by-step explanation:

Out of the given options, the trigonometric function that equals zero is cot 180°.

In the field of Trigonometry, each of the given options represents a trigonometric function evaluated at a particular degree. In this case, we're asked which of the given options is equal to zero. To determine this, we need to understand the values of these functions at different degrees.

sec 180° is equal to -1 because sec 180° = 1/cos 180° and cos 180° = -1. Moving on to csc 270°, this equals -1 as well because csc 270° = 1/sin 270° and sin 270° = -1. Next, cot 270° does not exist because cotangent is equivalent to cosine divided by sine and sin 270° = -1, which would yield an undefined result due to division by zero. Lastly, cot 180° equals to 0 as cot 180° = cos 180° / sin 180° and since sin 180° = 0, the result is 0.

Therefore, the common trigonometric value which equals to '0' is cot 180°.

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The discrete logarithm of 11 base 6 (mod 13) is x = 8.

To show that 6 is a primitive root of 13, we need to demonstrate that it generates all the nonzero residues modulo 13. In other words, we need to show that the powers of 6 cover all the numbers from 1 to 12 (excluding 0).

First, let's calculate the powers of 6 modulo 13:

[tex]6^1[/tex]≡ 6 (mod 13)

[tex]6^2[/tex]≡ 36 ≡ 10 (mod 13)

[tex]6^3[/tex]≡ 60 ≡ 8 (mod 13)

[tex]6^4[/tex]≡ 480 ≡ 5 (mod 13)

[tex]6^5[/tex] ≡ 3000 ≡ 12 (mod 13)

[tex]6^6[/tex] ≡ 72000 ≡ 7 (mod 13)

[tex]6^7[/tex] ≡ 420000 ≡ 9 (mod 13)

[tex]6^8[/tex]≡ 2520000 ≡ 11 (mod 13)

[tex]6^9[/tex] ≡ 15120000 ≡ 4 (mod 13)

[tex]6^10[/tex] ≡ 90720000 ≡ 3 (mod 13)

[tex]6^11[/tex] ≡ 544320000 ≡ 2 (mod 13)

[tex]6^12[/tex]≡ 3265920000 ≡ 1 (mod 13)

As we can see, the powers of 6 generate all the numbers from 1 to 12 modulo 13. Therefore, 6 is a primitive root of 13.

Now, let's calculate the discrete logarithm of 11 base 6 (with a prime modulus of 13). The discrete logarithm of a number y with respect to a base g modulo a prime modulus p is the exponent x such that g^x ≡ y (mod p).

We want to find x such that [tex]6^x[/tex] ≡ 11 (mod 13).

Using the previously calculated powers of 6, we can see that:

[tex]6^8[/tex]≡ 11 (mod 13)

Therefore, the discrete logarithm of 11 base 6 (mod 13) is x = 8.

Thus, the discrete logarithm of 11 base 6 (with a prime modulus of 13) is 8.

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a.  The average enrollment per year is 35.

b. The linear model is: Enrollment = 35t + 225, where t is the number of years since 2000.

c. We expect the enrollment to reach 1000 in the year 2022 (2000 + 22).

d. We expect the enrollment to be 1125 in the year 2025.

The average enrollment per year is the difference in enrollment divided by the number of years:

Average enrollment per year = (400 - 225) / (2005 - 2000)

Average enrollment per year = 35

To find the linear model, we need to determine the slope and y-intercept. The slope is the average enrollment per year we just found, and the y-intercept is the enrollment in the starting year 2000:

Slope = 35

Y-intercept = 225

Therefore, the linear model is:

Enrollment = 35t + 225, where t is the number of years since 2000.

To find the year when the enrollment reaches 1000, we can substitute 1000 for Enrollment in the linear model and solve for t:

1000 = 35t + 225

775 = 35t

t = 22.14

Therefore, we expect the enrollment to reach 1000 in the year 2022 (2000 + 22).

To find the expected enrollment in the year 2025, we need to substitute t = 25 into the linear model:

Enrollment = 35(25) + 225

Enrollment = 1125

Therefore, we expect the enrollment to be 1125 in the year 2025.

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we have shown that the expression ρ(x,t) = [1/(1 + t^2)] ρ0 [(x/(1 + t^2)) - t] satisfies the advection equation ∂ρ/∂t + v ∂ρ/∂x = -ρ ∂v/∂x.

The density of radioactive material, denoted by ρ(x,t), evolves according to the equation:

∂ρ/∂t + v ∂ρ/∂x = -ρ ∂v/∂x

This equation describes the transport of a substance by a moving medium, where the rate of movement of the radioactive material is influenced by the velocity of the wind, determined by the function v(x,t).

To solve the equation, we use the method of characteristics. We define the characteristic equation as:

x = ξ(t)

and

ρ(x,t) = f(ξ)

where f is a function of ξ.

Using the method of characteristics, we find that:

∂ρ/∂t = (∂f/∂t)ξ'

∂ρ/∂x = (∂f/∂ξ)ξ'

where ξ' = dξ/dt.

Substituting these derivatives into the original equation, we have:

(∂f/∂t)ξ' + v(∂f/∂ξ)ξ' = -ρ ∂v/∂x

Dividing by ξ', we get:

(∂f/∂t)/(∂f/∂ξ) = -ρ ∂v/∂x / v

Letting k(x,t) = -ρ ∂v/∂x / v, we can integrate the above equation to obtain f(ξ,t). Since f(ξ,t) = ρ(x,t), we can express the solution ρ(x,t) in terms of the initial value of ρ and the function k(x,t).

Now, let's solve the advection equation using the method of characteristics. We define the characteristic equation as:

x = x(t)

Then, we have:

dx/dt = v(x,t)

ρ(x,t) = f(x,t)

We need to find the function k(x,t) such that:

(∂f/∂t)/(∂f/∂x) = k(x,t)

Differentiating dx/dt = v(x,t) with respect to t, we have:

dx/dt = (2xt)/(1 + t^2) + 1 + t^2

Integrating this equation with respect to t, we obtain:

x = (x(0) + 1)t + x(0)t^2 + (1/3)t^3

where x(0) is the initial value of x at t = 0.

To determine the function C(x), we use the initial condition ρ(x,0) = ρ0(x).

Then, we have:

ρ(x,0) = f(x,0) = F[x - C(x), 0]

where F(ξ,0) = ρ0(ξ).

Integrating dx/dt = (2xt)/(1 + t^2) + 1 + t^2 with respect to x, we get:

t = (2/3) ln|2xt + (1 + t^2)x| + C(x)

where C(x) is the constant of integration.

Using the initial condition, we can express the solution f(x,t) as:

f(x,t) = F[x - C(x),t] = ρ0 [(x - C(x))/(1 + t^2)]

To simplify this expression, we introduce A(x,t) = (2/3) ln|2xt + (1 + t^2)x|/(1 + t^2). Then, we have:

f(x,t) = [1/(1 +

t^2)] ρ0 [(x - C(x))/(1 + t^2)] = [1/(1 + t^2)] ρ0 [(x/(1 + t^2)) - A(x,t)]

Finally, we can write the solution to the advection equation as:

ρ(x,t) = [1/(1 + t^2)] ρ0 [(x/(1 + t^2)) - A(x,t)]

where A(x,t) = (2/3) ln|2xt + (1 + t^2)x|/(1 + t^2).

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A. The amount to be invested now, or the present value needed, to accumulate $150,000 after 2 years with a 6% interest compounded quarterly is approximately $132,823.87.

B. To determine the present value needed to accumulate a desired amount in the future, we can use the present value formula in compound interest calculations.

The present value formula is given by:

PV = FV / (1 + r/n)^(n*t)

Where PV is the present value, FV is the future value or desired accumulated amount, r is the interest rate (in decimal form), n is the number of compounding periods per year, and t is the number of years.

In this case, the desired accumulated amount (FV) is $150,000, the interest rate (r) is 6% or 0.06, the compounding is quarterly (n = 4), and the investment period (t) is 2 years.

Substituting these values into the formula, we have:

PV = 150,000 / (1 + 0.06/4)^(4*2)

Simplifying the expression inside the parentheses:

PV = 150,000 / (1 + 0.015)^(8)

Calculating the exponent:

PV = 150,000 / (1.015)^(8)

Evaluating (1.015)^(8):

PV = 150,000 / 1.126825

Finally, calculate the present value:

PV ≈ $132,823.87

Therefore, approximately $132,823.87 needs to be invested now (present value) to accumulate $150,000 after 2 years with a 6% interest compounded quarterly.

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Answer:

220 divided by it self (220) will get you 1

Step-by-step explanation:

220/220=1

Answer:

220

Step-by-step explanation:

An integer n to be great if n² – 1 is a multiple of 3 because (N + 3)² - 1 = 3m. Since 8 and 3 are relatively prime, it follows that 3 | a.

From the definition, we know that N² - 1 is divisible by because  

We can write this as:

N² - 1 = 3k, where k is some integer.

Adding 6k + 9 to both sides, we have:

N² + 6k + 9

= 3k + 9

= 3(k + 3)

= 3m(m is some integer)

This simplifies to:

(N + 3)² - 1 = 3m, so we can conclude that N + 3 is also great.

2. We want to prove that 3 | 8a if and only if 3 | a.

Let's first assume that 3 | a.

This means that a = 3k for some integer k.

We can then write 8a as:

8a

= 8(3k)

= 24k

= 3(8k), which shows that 3 | 8a.

Now assume that 3 | 8a.

This means that 8a = 3k for some integer k. Since 8 and 3 are relatively prime, it follows that 3 | a.

3. We want to prove that if n is even, then n can be written as either n = 4k or n = 4k + 2, for some integer k.

We can consider two cases:

Case 1: n is divisible by 4If n is divisible by 4, then n can be written as n = 4k for some integer k.

Case 2: n is not divisible by 4If n is not divisible by 4, then we know that n has a remainder of 2 when divided by 4.

This means that we can write n as: n = 4k + 2, where k is some integer.

Together, these two cases show that if n is even, then either

n = 4k or

n = 4k + 2 for some integer k.

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The diameter of the circle is approximately 8.50 yards and the radius is approximately 4.25 yards.

To find the diameter and radius of a circle when given the circumference, we can use the formulas:
Circumference = 2πr
Diameter = 2r
Given that the circumference is C = 26.7 yd, we can substitute this value into the circumference formula:
26.7 = 2πr
To find the radius, we need to isolate it on one side of the equation. Dividing both sides of the equation by 2π, we get:
r = 26.7 / (2π)
Now we can calculate the value of r using a calculator:
r ≈ 4.25 yd (rounded to the nearest hundredth)
To find the diameter, we can multiply the radius by 2:
Diameter = 2 * 4.25 ≈ 8.50 yd (rounded to the nearest hundredth)
Therefore, the diameter of the circle is approximately 8.50 yards and the radius is approximately 4.25 yards.

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