The United States and Canada have the production possibilities curves shown above. It is determined that the United States has the comparative advantage in peanuts. Will both nations gain from trade if the terms of trade that are offered are 1 Peanut= 3 Corn? Why or why not? Show your work.


This is economics

The inequality that represents the cost, c, that Mr. Daniels can spend on the class trip is; c < 300

We are told that;

Mr. Daniels is organizing a class trip.

Mr. Daniels wants to spend less than $300

Thus, if the cost that Mr. Daniels can spend on the class trip is given as c, then it means that c has to be less than the amount which he wants to spend which is $300.

Thus, since c has to be less than $300, then it means that the inequality to be used would be a less than sign which can be expressed as;

c < 300

Thus, we conclude that it is the required inequality.

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Answer:

the point where the two lines intersect is (-8, 4)

Answer:

isosceles triangle

Step-by-step explanation:

I'm pretty sure it's the last option, but don't count me on that.

The number of children are there in the given scenario are 400.

Given that, the ratio of children to adults on a school trip is initially 10:1.

Here, the given ratio can be written as 10x:1x

5 more children and 5 more adults join the trip so that the ratio is now 9:1

10x+5:1x+5

The new ratio is (10x+5)/(x+5) = 9/1

10x+5=9(x+5)

10x+5=9x+45

10x-9x=45-5

x=40

So, the number of children =10x=400

Therefore, the number of children are there in the given scenario are 400.

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You have to solve the given expression for x, first equal the expression to zero:

The car will cost $ 800 after a depreciation time of approximately 6 years.

Herein we are informed about the case of a car bought in 2011 at a cost of $ 36,000 and that depreciates linearly every year. Then, the depreciation function is described below:

c(t) = c' + m · t

Where:

If we know that c(0) = 36,000, c(4) = 12,000 and c(t) = 800, then the depreciation rate is:

m = (12,000 - 36,000) / (4 - 0)

m = - 24,000 / 4

m = - 6,000

800 = 36,000 - 6,000 · t

6,000 · t = 35,200

t = 35,200 / 6,000

t = 5.867

The expected depreciation time is approximately 6 years.

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The answer would be 0.5. You're welcome!

Answer:

Its either A or D

Step-by-step explanation:

Answer:

option a

explanation:

its the correct answer because if you know what a scientific notation it would help you a lot with problems like this one.

Answer:

B

Step-by-step explanation:

Answer:

He can only make 8 cheese pizzas

Step-by-step explanation:

4➗1/2=8

1/2+1/2+1/2+1.2+1.2+1.2+1.2+1.2=8

The required values of x, y and z are 58°, 29° and 60°

An angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle.

Given are two triangles with unknown angles, we are asked to find, value of x, y and z.

Since, the lines are given parallel,

Therefore,

2y = 58° [alternate interior angles]

y = 29°

Now,

2y = x [alternate interior angles]

x = 58°

Since, we know that, the sum of interior angles of a triangle is 180°

Therefore,

x + z + 62° = 180°

58° + 62° + z = 180°

z = 180° - 120°

z = 60°

Hence, the required values of x, y and z are 58°, 29° and 60°

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The answer is yes, (x + 6) is a possible length of a rectangle with area x² + x − 30.

To determine whether (x + 6) is a possible length of a rectangle with area x^² + x − 30, we can use an area model to visualize the situation.

First, we need to find the width of the rectangle.

The area of a rectangle is given by the formula A = lw,

where A is the area, l is the length, and w is the width.

We are given that the area is x² + x − 30, so we can set this equal to lw and solve for w:

x² + x − 30 = (x + 6)w

w = (x² + x − 30)/(x + 6)

The length of a rectangle must be greater than or equal to its width, so we need to check whether (x + 6) is greater than or equal to (x² + x − 30)/(x + 6). This simplifies to:

(x + 6) ≥ x² + x − 30

Expanding the left side and simplifying, we get:

x² + 12x + 36 ≥ x² + x − 30

11x ≥ -66

x ≥ -6

Since x must be a positive number (since we are dealing with lengths), we can conclude that (x + 6) is the possible length of the rectangle. Therefore, the answer is yes, (x + 6) is a possible length of a rectangle with an area x² + x − 30.

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Step-by-step explanation:

3x²+4=31

3x²=31-4

3x²=27

x²=27÷3

x²=9

x=±3

Answer:

I hope you got it. I tried my best to give answer

Answer:

11/2

Step-by-step explanation:

11 is bigger than 2.

Answer:

Area of the rectangular doormat = 2/5 m²

Step-by-step explanation:

Length of the rectangular doormat = 4/5m

Width of the rectangular doormat = 1/2m

What is the area of the doormat?

Area of a rectangle = length × width

= 4/5m × 1/2m

= (4×1) / (5×2)

= 4 / 10

= 2/5 m²

Area of the rectangular doormat = 2/5 m²

Answer:

1/4

Step-by-step explanation:

Sorry you had to wait sso long to get an answer, but here it is:

If you count out any of the cubes in a side of the original polygon, and then count the cubes in the same side of the smaller copy, you will see that there is a ratio. That ratio is 4 : 1, meaning that the scale factor is 1/4.

Hope this helps :)

Answer:

4:1

Step-by-step explanation:

first figure has 4 among the colomuns

second one has 1   so the differance is    first one is  4 times as big as the second one

Answer:

area= 83

Step-by-step explanation:

the formula for area of a tringale is HxB/2 so the area for the tringles are 28 and 15 and area for a rectangle is LxW so it is 40 and add it all togeteher and you get 83

Answer:

3.5 +2.2t +9.8 and 9.8 + 3.5 + 2.2t and 2.2t + 9.8 + 3.5

Answer:

Option 1, Option 4, and Option 5.

Step-by-step explanation:

So it's 3.5 + 2.2t + 9.8 , 9.8 + 3.5 + 2.2t , 2.2t + 9.8 + 3.5. :)

The value of the union of sets A and B is, P(A ∪ B) is 0.64.

The union of two sets means the total elements in both the sets combined.

Given: sets P(A) = 0.38, P(B) = 0.83, and  intersection P(A ∩ B) = 0.57

We need to find the union of P(A ∪ B).

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Let us substitute the given values in the formula.

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

=0.38+0.83-0.57

=1.21-0.57

=0.64

Therefore, the value of union of sets P(A ∪ B) is 0.64.

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Except than x = 1, all real numbers fall within the function's domain.

Since there are no limitations on what we can substitute for x, the domain of a function, f(x), is all real numbers because any real numbers would make f(x) a defined function. As a result, when this is not the case, the domain of a function, f(x), is not all real numbers.

The rational function r(x) = 2x/(x-1) is defined as follows.

So, we set the denominator to zero and solve for x in order to determine the domain of r(x):

x - 1 = 0

x = 1

Hence, x = 1 is the only value of x that causes the denominator to equal 0. R(x) therefore has a domain of all real numbers other than x = 1.

We can express the domain as follows in interval notation:

(-∞, 1) U (1, ∞)

Except than x = 1, all real numbers fall within the function's domain.

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Question:

Which of the following correctly describes the domain of the function shown below?

r(x) = 2x x-1

A. {x:x0}

B. {x: x = 1}

c. x all .real .numbers}

D. xx1}

Answer:

160

I hope it is helpful to you

Answer:

Step-by-step explanation:

A parallel line will have the same slope as the line on the graph. A perpendicular line will have a slope that is the opposite reciprocal of the slope of the graphed line.

The line on the graph rises 1 grid square for each run of 2 grid squares to the right. Its slope is ...

  m = rise/run = 1/2

The opposite reciprocal of this slope is ...

  -1/(1/2) = -2

A perpendicular line will have a slope of -2.

The slope-intercept form of the equation for a line is ...

  y = mx +b

where the slope is m and b is the y-intercept. For the purpose here, we don't care about the y-intercepts of any of the lines. We only care about the slope: the coefficient of x.

This means the equations we're looking for are of the form ...

  parallel line: y = 1/2x + b

  perpendicular line: y = 2x +b . . . . . for some constant b

Of the equations offered, the only one with an x-coefficient of 1/2 is ...

  y = 1/2x +3

Of the equations offered, the only one with an x-coefficient of -2 is ...

  y = -2x +1

Answer:

Step-by-step explanation:

you are welcome

The graph of the equation, x = 3, would be a vertical line that intersects the x axis at x = 3.

The slope of the graph would be undefined.

Vertical lines have an undefined slope because slopes are based on the change in the horizontal climb of a line. Seeing as vertical lines have no horizontal change, it means that the slope in undefined.

The line, x = 3 is a vertical line because the x - axis is the horizontal line. A line that would intersect it at x = 3 would therefore have to be a vertical line which is straight or it would not remain at x = 3 in perpetuity.

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Answer:

140

Step-by-step explanation:

one exterior = the sum of two interior

There are three left cosets of H in Z, given by:

H = {3k | k ∈ Z}

[1] + H = {3k + 1 | k ∈ Z}

[2] + H = {3k + 2 | k ∈ Z}

To find all the left cosets of H in Z, we need to find the equivalence classes of the relation x ~ y if and only if x - y ∈ H. That is, two integers are equivalent if their difference is a multiple of 3.

To find the left cosets, we choose a representative for each equivalence class. We can choose any integer in the class as the representative. Then, we add H to the representative to obtain the left coset.

For example, the equivalence class [0] consists of all integers that are multiples of 3. We can choose 0 as the representative. Then, the left coset is

[0] + H = {0 + 3k | k ∈ Z} = {3k | k ∈ Z}

Similarly, we can choose 1, 2 as the representatives for [1], [2], respectively, and obtain the left cosets:

[1] + H = {1 + 3k | k ∈ Z} = {3k + 1 | k ∈ Z}

[2] + H = {2 + 3k | k ∈ Z} = {3k + 2 | k ∈ Z}

We can continue this process to find all the left cosets. In general, there are three left cosets of H in Z, given by:

H = {3k | k ∈ Z}

[1] + H = {3k + 1 | k ∈ Z}

[2] + H = {3k + 2 | k ∈ Z}

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Answer:

x+15 or 15+x

Step-by-step explanation:

it doesn't really matter since the commutative property.

sum = addition

15 = one term

number x = second term

The sets of ordered pairs that could represent a linear function are A and D.

What is linear function ?

A linear function is a mathematical function that can be graphically represented as a straight line. It has a constant rate of change between any two points, which means that the slope (i.e., the steepness of the line) is constant.

To determine if a set of ordered pairs can represent a linear function, we need to check if there is a constant rate of change (i.e., a constant slope) between any two pairs of points. If there is a constant rate of change, then the set of ordered pairs could represent a linear function.

Let's check each set of ordered pairs:

A = {(-4, 2), (-2, 3), (2, 5), (4, 6)}

The rate of change between (-4, 2) and (-2, 3) is 1/2.

The rate of change between (-2, 3) and (2, 5) is 1/2.

The rate of change between (2, 5) and (4, 6) is 1/2.

Therefore, the set A could represent a linear function.

B = {(1, 2), (3, 4), (5, 6), (6, 8)}

The rate of change between (1, 2) and (3, 4) is 1.

The rate of change between (3, 4) and (5, 6) is 1.

The rate of change between (5, 6) and (6, 8) is 2.

Therefore, the set B could not represent a linear function.

C = {(-1, -1), (0, 2), (1, 4), (2, 6)}

The rate of change between (-1, -1) and (0, 2) is 3.

The rate of change between (0, 2) and (1, 4) is 2.

The rate of change between (1, 4) and (2, 6) is 2.

Therefore, the set C could not represent a linear function.

D = {(2, 9), (4, 6), (5, 3), (6, 0)}

The rate of change between (2, 9) and (4, 6) is -1.5.

The rate of change between (4, 6) and (5, 3) is -3.

The rate of change between (5, 3) and (6, 0) is -3.

Therefore, the set D could represent a linear function.

Therefore,  the sets of ordered pairs that could represent a linear function are A and D.

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The ____The correct answer is c. double.________ data type is used to store any number that might have a fractional part.

the double data type is used to store any number that might have a fractional part, including decimal numbers and scientific notation numbers. It has a higher precision than the float data type, which can lead to more accurate . In conclusion, if you need to store numbers with decimal points, the double data type is the best option.
The correct answer is c. double.

The double data type is used to store any number that might have a fractional part, such as decimals and real numbers. In contrast, a string is used to store text, an int is used to store whole numbers, and a boolean is used to store true or false values.

To store a number with a fractional part, you should use the double data type.

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Integrating the second term:

∫((y₂² * (y_c * exsec²x tanx)) / (Wronskian)) dx

= y_c * ∫((y₂² * (y₁ * exsec²x tanx)) / (Wronskian)) dx

A differential equation is an equation that relates a function to its derivatives. It involves one or more derivatives of an unknown function and specifies the relationship between the function and its derivatives.

To find the general solution of the differential equation y" - 2y' + y = exsec²x tanx using the variation of parameters method, we'll follow these steps:

Find the complementary solution (homogeneous solution) to the associated homogeneous equation y" - 2y' + y = 0.

The characteristic equation is r² - 2r + 1 = 0, which has a repeated root of r = 1.

Therefore, the complementary solution is y_c = c₁e^x + c₂xe^x.

Find the particular solution using the variation of parameters method.

We need to find the particular solutions y₁ and y₂, which are given by:

y₁ = -y_c ∫(y₂ * exsec²x tanx) / (Wronskian) dx

y₂ = y_c ∫(y₁ * exsec²x tanx) / (Wronskian) dx

Where the Wronskian is given by:

Wronskian = | y₁ y₂ |

| y₁' y₂' |

And the derivatives are:

y₁' = -(y_c * y₂ * exsec²x tanx) / (Wronskian)

y₂' = (y_c * y₁ * exsec²x tanx) / (Wronskian)

Calculate the Wronskian:

Wronskian = | c₁e^x + c₂xe^x y₂ |

| -c₁e^x - c₂xe^x y₂' |

Wronskian = (c₁e^x + c₂xe^x) * y₂' + (y₂ * (-c₁e^x - c₂xe^x))

= (c₁e^x + c₂xe^x) * ((y_c * y₁ * exsec²x tanx) / (Wronskian)) + (y₂ * (-(y_c * y₂ * exsec²x tanx) / (Wronskian)))

= (c₁e^x + c₂xe^x) * (y_c * y₁ * exsec²x tanx) / (Wronskian) - (y₂² * (y_c * exsec²x tanx)) / (Wronskian)

Integrate to find y₁:

y₁ = -y_c ∫((y₂ * exsec²x tanx) / (Wronskian)) dx

= -∫((c₁e^x + c₂xe^x) * (y_c * y₁ * exsec²x tanx) / (Wronskian)) dx + ∫((y₂² * (y_c * exsec²x tanx)) / (Wronskian)) dx

Integrating the first term:

-∫((c₁e^x + c₂xe^x) * (y_c * y₁ * exsec²x tanx) / (Wronskian)) dx

= -y_c * ∫((c₁e^x + c₂xe^x) * (y₁ * exsec²x tanx) / (Wronskian)) dx

Integrating the second term:

∫((y₂² * (y_c * exsec²x tanx)) / (Wronskian)) dx

= y_c * ∫((y₂² * (y₁ * exsec²x tanx)) / (Wronskian)) dx

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