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7.7: Variation


Learning Objectives

  • Solve applications involving direct variation.
  • Solve applications involving inverse variation.
  • Solve applications involving joint variation.

Direct Variation

Consider a freight train moving at a constant speed of 30 miles per hour. The equation that expresses the distance traveled at that speed in terms of time is given by

(D=30t)

After (1) hour the train has traveled (30) miles, after (2) hours the train has traveled (60) miles, and so on. We can construct a chart and graph this relation.

Time, (t), in hoursDistance (D=30t)
00
130
260
390
4120
Table (PageIndex{1})

In this example, we can see that the distance varies over time as the product of the constant rate, (30) miles per hour, and the variable, (t). This relationship is described as direct variation and (30) is called the variation constant. In addition, if we divide both sides of (D=30t) by (t) we have

(frac{D}{t}=30)

In this form, it is reasonable to say that (D) is proportional to (t), where (30) is the constant of proportionality. In general, we have

Key WordsTranslation
"(y) varies directly as (x)"[y=kx]
"(y) is directly proportional to (x)"
"(y) is proportional to (x)"
Table (PageIndex{2})

Here (k) is nonzero and is called the constant of variation or the constant of proportionality.

Example (PageIndex{1})

The circumference of a circle is directly proportional to its diameter, and the constant of proportionality is (π). If the circumference is measured to be (20) inches, then what is the radius of the circle?

Solution:

Let (C) represent the circumference of the circle.

Let (d) represent the diameter of a circle.

Use the fact that “the circumference is directly proportional to the diameter” to write an equation that relates the two variables.

(C=kd)

We are given that “the constant of proportionality is (π),” or (k=π). Therefore, we write

(C=πd)

Now use this formula to find (d) when the circumference is (20) inches.

(egin{aligned} 20&=pi d frac{20}{color{Cerulean}{pi}}&=frac{pi d}{color{Cerulean}{pi}} frac{20}{pi}&=d end{aligned})

The radius of the circle, (r), is one-half of its diameter.

(egin{aligned} r&=frac{d}{2} &=frac{color{OliveGreen}{frac{20}{pi}}}{2} &=frac{20}{pi}cdot frac{1}{2} &= frac{10}{pi} end{aligned})

Answer:

The radius is (frac{10}{π}) inches, or approximately (3.18) inches

Typically, we will not be given the constant of variation. Instead, we will be given information from which it can be determined.

Example (PageIndex{2})

An object’s weight on earth varies directly to its weight on the moon. If a man weighs (180) pounds on earth, then he will weigh (30) pounds on the moon. Set up an algebraic equation that expresses the weight on earth in terms of the weight on the moon and use it to determine the weight of a woman on the moon if she weighs (120) pounds on earth.

Solution:

Let (y) represent the weight on Earth.

Let (x) represent the weight on the Moon.

We are given that the “weight on earth varies directly to the weight on the moon.”

(y=kx)

To find the constant of variation (k), use the given information. A (180)-pound man on earth weighs (30) pounds on the moon, or (y=180) when (x=30).

(180 = k cdot 30)

Solve for (k).

(egin{aligned} frac{180}{30} &=k 6&=k end{aligned})

Next, set up a formula that models the given information.

(y=6x)

This implies that a person’s weight on earth is (6) times her weight on the moon. To answer the question, use the woman’s weight on earth, (y=120) pounds, and solve for (x).

(egin{aligned} 120&=6x frac{120}{6} &=x 20&=x end{aligned})

Answer:

The woman weighs (20) pounds on the mood.

Inverse Variation

Next, consider the relationship between time and rate,

(r=frac{D}{t})

If we wish to travel a fixed distance, then we can determine the average speed required to travel that distance in a given amount of time. For example, if we wish to drive 240 miles in 4 hours, we can determine the required average speed as follows:

(r=frac{240}{4}=6)

The average speed required to drive 240 miles in 4 hours is 60 miles per hour. If we wish to drive the 240 miles in 5 hours, then determine the required speed using a similar equation:

(r=frac{240}{5}=48)

In this case, we would only have to average 48 miles per hour. We can make a chart and view this relationship on a graph.

Time (t) in hoursSpeed (r=frac{240}{t})
2120
380
460
548
Table (PageIndex{3})

This is an example of an inverse relationship. We say that (r) is inversely proportional to the time (t), where (240) is the constant of proportionality. In general, we have

Key WordsTranslation
"(y) varies inversely as (x)"[y=frac{k}{x}]
"(y) is inversely proportional to (x)"
Table (PageIndex{4})

Again, (k) is nonzero and is called the constant of variation or the constant of proportionality.

Example (PageIndex{3})

If (y) varies inversely as (x) and (y=5) when (x=2), then find the constant of proportionality and an equation that relates the two variables.

Solution:

If we let (k) represent the constant of proportionality, then the statement “(y) varies inversely as (x)” can be written as follows:

(y=frac{k}{x})

Use the given information, (y=5) when (x=2), to find (k).

(5=frac{k}{2})

Solve for (k).

(egin{aligned} color{Cerulean}{2}color{black}{cdot 5}&=color{Cerulean}{2}color{black}{cdotfrac{k}{2}10&=k} end{aligned})

Therefore, the formula that models the problem is

(y=frac{10}{x})

Answer:

The constant of proportionality is (10), and the equation is (y=frac{10}{x}).

Example (PageIndex{4})

The weight of an object varies inversely as the square of its distance from the center of earth. If an object weighs (100) pounds on the surface of earth (approximately (4,000) miles from the center), then how much will it weigh at (1,000) miles above earth’s surface?

Solution:

Let (w) represent weight of the object.

Let (d) represent the object's distance from the center of Earth.

Since "(w) varies inversely as the square of (d)," we can write

(w=frac{k}{d^{2}})

Use the given information to find (k). An object weighs (100) pounds on the surface of earth, approximately (4,000) miles from the center. In other words, (w = 100) when (d = 4,000):

(100=frac{k}{(4000)^{2}})

Solve for (k):

(egin{aligned} color{Cerulean}{(4,000)^{2}}color{black}{cdot 100}&=color{Cerulean}{(4,000)^{2}}color{black}{cdotfrac{k}{(4,000)^{2}}} 1,600,000,000&=k 1.6 imes 10^{9} &=k end{aligned})

Therefore, we can model the problem with the following formula:

(w=frac{1.6 imes 10^{9}}{d^{2}})

To use the formula to find the weight, we need the distance from the center of earth. Since the object is (1,000) miles above the surface, find the distance from the center of earth by adding (4,000) miles:

(d=4,000+1,000=5,000) miles

To answer the question, use the formula with (d = 5,000).

(egin{aligned} y &= frac{1.6 imes 10^{9}}{(color{OliveGreen}{5,000}color{black}{)^{2}}} &=frac{1.6 imes 10^{9}}{25,000,000} &=frac{1.6 imes 10^{9}}{2.5 imes 10^{7}} &=0.64 imes 10^{2} &=64 end{aligned})

Answer:

The object will weigh (64) pounds at a distance (1,000) miles above the surface of earth.

Joint Variation

Lastly, we define relationships between multiple variables. In general, we have

VocabularyTranslation
"(y) varies jointly as (x) and (z)"[y=kyz]
"(y) is jointly proportional to (x) and (z)"
Table (PageIndex{5})

Here (k) is nonzero and is called the constant of variation or the constant of proportionality.

Example (PageIndex{5})

The area of an ellipse varies jointly as (a), half of the ellipse’s major axis, and (b), half of the ellipse’s minor axis. If the area of an ellipse is (300π ext{cm}^{2}), where (a=10) cm and (b=30) cm, then what is the constant of proportionality? Give a formula for the area of an ellipse.

Solution:

If we let (A) represent the area of an ellipse, then we can use the statement “area varies jointly as (a) and (b)” to write

(A=kab)

To find the constant of variation, (k), use the fact that the area is (300π) when (a=10) and (b=30).

(egin{aligned} 300pi &=k(color{OliveGreen}{10}color{black}{)(}color{OliveGreen}{30}color{black}{)} 300pi &=300k pi&=k end{aligned})

Therefore, the formula for the area of an ellipse is

(A=pi ab)

Answer:

The constant of proportionality is (π), and the formula for the area is (A=abπ).

Exercise (PageIndex{1})

Given that (y) varies directly as the square of (x) and inversely to (z), where (y = 2) when (x = 3) and (z = 27), find (y) when (x = 2) and (z = 16).

Answer

(frac{3}{2})

Key Takeaways

  • The setup of variation problems usually requires multiple steps. First, identify the key words to set up an equation and then use the given information to find the constant of variation (k). After determining the constant of variation, write a formula that models the problem. Once a formula is found, use it to answer the question.

Exercise (PageIndex{2}) Variation Problems

Translate the following sentences into a mathematical formula.

  1. The distance, (D), an automobile can travel is directly proportional to the time, (t), that it travels at a constant speed.
  2. The extension of a hanging spring, (d), is directly proportional to the weight, (w), attached to it.
  3. An automobile’s breaking distance, (d), is directly proportional to the square of the automobile’s speed, (v).
  4. The volume, (V), of a sphere varies directly as the cube of its radius, (r).
  5. The volume, (V), of a given mass of gas is inversely proportional to the pressure, (p), exerted on it.
  6. The intensity, (I), of light from a light source is inversely proportional to the square of the distance, (d), from the source.
  7. Every particle of matter in the universe attracts every other particle with a force, (F), that is directly proportional to the product of the masses, (m_{1}) and (m_{2}), of the particles and inversely proportional to the square of the distance, (d), between them.
  8. Simple interest, (I), is jointly proportional to the annual interest rate, (r), and the time, (t), in years a fixed amount of money is invested.
  9. The period, (T), of a pendulum is directly proportional to the square root of its length, (L).
  10. The time, (t), it takes an object to fall is directly proportional to the square root of the distance, (d), it falls.
Answer

1. (D=kt)

3. (d=kv^{2})

5. (V=frac{k}{p})

7. (F=frac{km_{1}⋅m_{2}}{d^{2}})

9. (T=ksqrt{L})

Exercise (PageIndex{3}) Variation Problems

Construct a mathematical model given the following.

  1. (y) varies directly as (x), and (y = 30) when (x = 6).
  2. (y) varies directly as (x), and (y = 52) when (x = 4).
  3. (y) is directly proportional to (x), and (y = 12) when (x = 3).
  4. (y) is directly proportional to (x), and (y = 120) when (x = 20).
  5. (y) varies directly as (x), and (y = 14) when (x = 10).
  6. (y) varies directly as (x), and (y = 2) when (x = 8).
  7. (y) varies inversely as (x), and (y = 5) when (x = 7).
  8. (y) varies inversely as (x), and (y = 12) when (x = 2).
  9. (y) is inversely proportional to (x), and (y = 3) when (x = 9).
  10. (y) is inversely proportional to (x), and (y = 21) when (x = 3).
  11. (y) varies inversely as (x), and (y = 2) when (x = frac{1}{8}).
  12. (y) varies inversely as (x), and (y = frac{3}{2}) when (x = frac{1}{9}).
  13. (y) varies jointly as (x) and (z), where (y = 8) when (x = 4) and (z = frac{1}{2}).
  14. (y) varies jointly as (x) and (z), where (y = 24) when (x =frac{1}{3}) and (z = 9).
  15. (y) is jointly proportional to (x) and (z), where (y = 2) when (x = 1) and (z = 3).
  16. (y) is jointly proportional to (x) and (z), where (y = 15) when (x = 3) and (z = 7).
  17. (y) varies jointly as (x) and (z), where (y = frac{2}{3}) when (x = frac{1}{2}) and (z = 12).
  18. (y) varies jointly as (x) and (z), where (y = 5) when (x = frac{3}{2}) and (z = frac{2}{9}).
  19. (y) varies directly as the square of (x), where (y = 45) when (x = 3).
  20. (y) varies directly as the square of (x), where (y = 3) when (x = frac{1}{2}).
  21. (y) is inversely proportional to the square of (x), where (y = 27) when (x = frac{1}{3}).
  22. (y) is inversely proportional to the square of (x), where (y = 9) when (x = frac{2}{3}).
  23. (y) varies jointly as (x) and the square of (z), where (y = 54) when (x = 2) and (z = 3).
  24. (y) varies jointly as (x) and the square of (z), where (y = 6) when (x = frac{1}{4}) and (z = frac{2}{3}).
  25. (y) varies jointly as (x) and (z) and inversely as the square of (w), where (y = 30) when (x = 8, z = 3), and (w = 2)
  26. (y) varies jointly as (x) and (z) and inversely as the square of (w), where (y = 5) when (x= 1, z = 3), and (w = frac{1}{2}).
  27. (y) varies directly as the square root of (x) and inversely as (z), where (y = 12) when (x= 9) and (z = 5).
  28. (y) varies directly as the square root of (x) and inversely as the square of (z), where (y = 15) when (x = 25) and (z = 2).
  29. (y) varies directly as the square of (x) and inversely as (z) and the square of (w), where (y = 14) when (x = 4, w = 2), and (z = 2).
  30. (y) varies directly as the square root of (x) and inversely as (z) and the square of (w), where (y = 27) when (x = 9, w = frac{1}{2}), and (z = 4).
Answer

1. (y=5x)

3. (y=4x)

5. (y=frac{7}{5}x)

7. (y=frac{35}{x})

9. (y=frac{27}{x})

11. (y=frac{1}{4x})

13. (y=4xz)

15. (y=frac{2}{3}xz)

17. (y=frac{1}{9}xz)

19. (y=5x^{2})

21. (y=3x^{2})

23. (y=3xz^{2})

25. (y=frac{5xz}{w^{2}})

27. (y=frac{20x}{sqrt{z}})

29. (y=frac{7x^{2}}{w^{2}z})

Exercise (PageIndex{4}) Variation Problems

Applications involving variation.

  1. Revenue in dollars is directly proportional to the number of branded sweat shirts sold. If the revenue earned from selling (25) sweat shirts is $(318.75), then determine the revenue if (30) sweat shirts are sold.
  2. The sales tax on the purchase of a new car varies directly as the price of the car. If an $(18,000) new car is purchased, then the sales tax is $(1,350). How much sales tax is charged if the new car is priced at $(22,000)?
  3. The price of a share of common stock in a company is directly proportional to the earnings per share (EPS) of the previous 12 months. If the price of a share of common stock in a company is $(22.55) and the EPS is published to be $(1.10), then determine the value of the stock if the EPS increases by $(0.20).
  4. The distance traveled on a road trip varies directly with the time spent on the road. If a (126)-mile trip can be made in (3) hours, then what distance can be traveled in (4) hours?
  5. The circumference of a circle is directly proportional to its radius. If the circumference of a circle with radius (7) centimeters is measured as (14π) centimeters, then find the constant of proportionality.
  6. The area of circle varies directly as the square of its radius. If the area of a circle with radius (7) centimeters is determined to be (49π) square centimeters, then find the constant of proportionality.
  7. The surface area of a sphere varies directly as the square of its radius. When the radius of a sphere measures (2) meters, the surface area measures (16π) square meters. Find the surface area of a sphere with radius (3) meters.
  8. The volume of a sphere varies directly as the cube of its radius. When the radius of a sphere measures (3) meters, the volume is (36π) cubic meters. Find the volume of a sphere with radius (1) meter.
  9. With a fixed height, the volume of a cone is directly proportional to the square of the radius at the base. When the radius at the base measures (10) centimeters, the volume is (200) cubic centimeters. Determine the volume of the cone if the radius of the base is halved.
  10. The distance, (d), an object in free fall drops varies directly with the square of the time, (t), that it has been falling. If an object in free fall drops (36) feet in (1.5) seconds, then how far will it have fallen in (3) seconds?
Answer

1. $(382.50 )

3. $(26.65 )

5. (2π)

7. (36π) square meters

9. (50) cubic centimeters

Exercise (PageIndex{5}) Variation Problems

Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.

  1. If a hanging spring is stretched (5) inches when a (20)-pound weight is attached to it, then determine its spring constant.
  2. If a hanging spring is stretched (3) centimeters when a (2)-kilogram weight is attached to it, then determine the spring constant.
  3. If a hanging spring is stretched (3) inches when a (2)-pound weight is attached, then how far will it stretch with a (5)-pound weight attached?
  4. If a hanging spring is stretched (6) centimeters when a (4)-kilogram weight is attached to it, then how far will it stretch with a (2)-kilogram weight attached?
Answer

1. (frac{1}{4})

3. (7.5) inches

Exercise (PageIndex{6}) Variation Problems

The breaking distance of an automobile is directly proportional to the square of its speed.

  1. If it takes (36) feet to stop a particular automobile moving at a speed of (30) miles per hour, then how much breaking distance is required if the speed is (35) miles per hour?
  2. After an accident, it was determined that it took a driver (80) feet to stop his car. In an experiment under similar conditions, it takes (45) feet to stop the car moving at a speed of (30) miles per hour. Estimate how fast the driver was moving before the accident.
Answer

1. (49) feet

Exercise (PageIndex{7}) Variation Problems

Boyle’s law states that if the temperature remains constant, the volume, (V), of a given mass of gas is inversely proportional to the pressure, (p), exerted on it.

  1. A balloon is filled to a volume of (216) cubic inches on a diving boat under (1) atmosphere of pressure. If the balloon is taken underwater approximately (33) feet, where the pressure measures (2) atmospheres, then what is the volume of the balloon?
  2. If a balloon is filled to (216) cubic inches under a pressure of (3) atmospheres at a depth of (66) feet, then what would the volume be at the surface, where the pressure is (1) atmosphere?
  3. To balance a seesaw, the distance from the fulcrum that a person must sit is inversely proportional to his weight. If a (72)-pound boy is sitting (3) feet from the fulcrum, then how far from the fulcrum must a (54)- pound boy sit to balance the seesaw?
  4. The current, (I), in an electrical conductor is inversely proportional to its resistance, (R). If the current is (frac{1}{4}) ampere when the resistance is (100) ohms, then what is the current when the resistance is (150) ohms?
  5. The number of men, represented by (y), needed to lay a cobblestone driveway is directly proportional to the area, (A), of the driveway and inversely proportional to the amount of time, (t), allowed to complete the job. Typically, (3) men can lay (1,200) square feet of cobblestone in (4) hours. How many men will be required to lay (2,400) square feet of cobblestone given (6) hours?
  6. The volume of a right circular cylinder varies jointly as the square of its radius and its height. A right circular cylinder with a (3)-centimeter radius and a height of (4) centimeters has a volume of (36π) cubic centimeters. Find a formula for the volume of a right circular cylinder in terms of its radius and height.
  7. The period, (T), of a pendulum is directly proportional to the square root of its length, (L). If the length of a pendulum is (1) meter, then the period is approximately (2) seconds. Approximate the period of a pendulum that is (0.5) meter in length.
  8. The time, (t), it takes an object to fall is directly proportional to the square root of the distance, (d), it falls. An object dropped from (4) feet will take (frac{1}{2}) second to hit the ground. How long will it take an object dropped from (16) feet to hit the ground?
Answer

1. (108) cubic inches

3. (4) feet

5. (4) men

7. (1.4) seconds

Exercise (PageIndex{8}) Variation Problems

Newton’s universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force, (F), that is directly proportional to the product of the masses, (m_{1}) and (m_{2}), of the particles and inversely proportional to the square of the distance, (d), between them. The constant of proportionality is called the gravitational constant.

  1. If two objects with masses (50) kilograms and (100) kilograms are (frac{1}{2}) meter apart, then they produce approximately (1.34×10^{−6}) newtons (N) of force. Calculate the gravitational constant.
  2. Use the gravitational constant from the previous exercise to write a formula that approximates the force, (F), in newtons between two masses (m_{1}) and (m_{2}), expressed in kilograms, given the distance (d) between them in meters.
  3. Calculate the force in newtons between earth and the moon, given that the mass of the moon is approximately (7.3×10^{22}) kilograms, the mass of earth is approximately (6.0×10^{24}) kilograms, and the distance between them is on average (1.5×10^{11}) meters.
  4. Calculate the force in newtons between earth and the sun, given that the mass of the sun is approximately (2.0×10^{30}) kilograms, the mass of earth is approximately (6.0×10^{24}) kilograms, and the distance between them is on average (3.85×10^{8}) meters.
  5. If (y) varies directly as the square of (x), then how does (y) change if (x) is doubled?
  6. If (y) varies inversely as square of (t), then how does (y) change if (t) is doubled?
  7. If (y) varies directly as the square of (x) and inversely as the square of (t), then how does (y) change if both (x) and (t) are doubled?
Answer

1. (6.7×10^{−11} frac{N m^{2}}{kg^{2}})

3. (1.98×10^{20}) N

5. (y) changes by a factor of (4)

7. (y) remains unchanged


Recognize and represent proportional relationships between quantities.

The constant of variation is the number that relates two variables that are directly proportional or inversely proportional to one another. Watch this tutorial to see how to find the constant of variation for a direct variation equation. Take a look!

How Do You Write an Equation for Direct Variation Given a Point?

Looking for some practice with direct variation? Watch this tutorial, and get that practice! This tutorial shows you how to take given information and turn it into a direct variation equation. Then, see how to use that equation to find the value of one of the variables.

How Do You Write an Equation for Direct Variation from a Table?

Looking for some practice with direct variation? Watch this tutorial, and get that practice! This tutorial shows you how to take a table of values and describe the relation using a direct variation equation.

What's the Constant of Variation?

The constant of variation is the number that relates two variables that are directly proportional or inversely proportional to one another. But why is it called the constant of variation? This tutorial answers that question, so take a look!

What Does Direct Variation Look Like on a Graph?

Want to know what a direct variation looks like graphically? Basically, it's a straight line that goes through the origin. To get a better picture, check out this tutorial!

How Do You Solve a Word Problem Using the Direct Variation Formula?

Word problems allow you to see math in action! Take a look at this word problem involving an object's weight on Earth compared to its weight on the Moon. See how the formula for direct variation plays an important role in finding the solution. Then use that formula to see how much you would weigh on the Moon!

What's the Direct Variation or Direct Proportionality Formula?

Ever heard of two things being directly proportional? Well, a good example is speed and distance. The bigger your speed, the farther you'll go over a given time period. So as one variable goes up, the other goes up too, and that's the idea of direct proportionality. But you can express direct proportionality using equations, and that's an important thing to do in algebra. See how to do that in the tutorial!

How Do You Use the Formula for Direct Variation?

If two things are directly proportional, you can bet that you'll need to use the formula for direct variation to solve! In this tutorial, you'll see how to use the formula for direct variation to find the constant of variation and then solve for your answer.

What's a Proportion?

The idea of proportions is that a ratio can be written in many ways and still be equal to the same value. That's why proportions are actually equations with equal ratios. This is a bit of a tricky definition, so make sure to watch the tutorial!

How Do You Find Equivalent Ratios by Making a Table?

To master equivalent ratios, you need to practice. Follow along with this tutorial to practice filling in a table with equivalent ratios.

How Do You Find Equivalent Ratios?

Ratios are used to compare numbers. When you're working with ratios, it's sometimes easier to work with an equivalent ratio. Equivalent ratios have different numbers but represent the same relationship. In this tutorial, you'll see how to find equivalent ratios by first writing the given ratio as a fraction. Take a look!

How Do You Know If Two Ratios are Proportional?

Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional. To see this process in action, check out this tutorial!

How Do You Determine if Two Ratios are Proportional Using Cross Products?

Trying to figure out if two ratios are proportional? If they're in fraction form, set them equal to each other to test if they are proportional. Cross multiply and simplify. If you get a true statement, then the ratios are proportional! This tutorial gives you a great example!

How Do You Determine Whether Values in a Table are Proportional?

To see if multiple ratios are proportional, you could write them as fractions, reduce them, and compare them. If the reduced fractions are all the same, then you have proportional ratios. To see this process step-by-step, check out this tutorial!

What are Equivalent Ratios?

Equivalent ratios are just like equivalent fractions. If two ratios have the same value, then they are equivalent, even though they may look very different! In this tutorial, take a look at equivalent ratios and learn how to tell if you have equivalent ratios.


Parsvottanasana Meaning

The name of parsvottanasana is derived from Sanskrit. It includes four terms that better clarify its meaning and significance.

  • ‘Parsva’ means ‘side’
  • ‘Ut’ refers to ‘intense’
  • ‘Tan’ means ‘to stretch’
  • ‘Asana’ is ‘pose’.

This is the crux of what the practitioner has to do for performing this pose, i.e. bending the body to one side and stretching immensely.

Parsvottanasana is better understood as a stage between Parivrtta Trikonasana and Utthita Trikonasana. Bending the body forward and balancing in this inversion gives the body a pyramid shape, hence the name.


Pour the whiskey into a highball glass filled with ice.

Garnish with a lemon wedge. Serve and enjoy.

What Is Seagram's 7?

Seagram's 7 is an 80-proof whiskey that you can find everywhere. It is not a bourbon, since it is not made in Bourbon County, Kentucky according to specific criteria. There are very few bars and liquor stores in the U.S. that do not have a bottle of affordable Seagram's 7 with its signature red label. Seagram's may not be the best whiskey out there, but it is surprisingly smooth for its affordable price, and it has many dedicated fans who savor every sip. It also does the job when you are looking for a good inexpensive drink like the seven and seven.

Is Seagram's 7 a Canadian Whiskey?

There is some debate about what style of whiskey Seagram's 7 is. The confusion stems from the fact that Seagram's is an old Canadian brand that has long produced whiskeys, vodkas, and gins at affordable prices. Pair that with the knowledge that Canada is known for producing blended whiskeys, and many drinkers are left with the impression that Seagram's 7 is a Canadian blended whiskey. But look at the label on a bottle of Seagram's 7 and it clearly states "American whiskey—a blend of distinctive character" that it is bottled and blended by The 7 Crown Distilling Company of Norwalk, Connecticut.

How Strong Is the Seven and Seven?

Everything about the seven and seven is laid out except how much 7-Up goes into the drink. You can pour as much or as little soda as you like to customize the drink's taste and strength. On average, it's relatively mild with an alcohol content in the 10 percent ABV (20 proof) range.


7.7: Variation

Standard deviations can be obtained from standard errors, confidence intervals, t values or P values that relate to the differences between means in two groups. The difference in means itself (MD) is required in the calculations from the t value or the P value. An assumption that the standard deviations of outcome measurements are the same in both groups is required in all cases, and the standard deviation would then be used for both intervention groups. We describe first how a t value can be obtained from a P value, then how a standard error can be obtained from a t value or a confidence interval, and finally how a standard deviation is obtained from the standard error. Review authors may select the appropriate steps in this process according to what results are available to them. Related methods can be used to derive standard deviations from certain F statistics, since taking the square root of an F value may produce the same t value. Care is often required to ensure that an appropriate F value is used, and advice of a knowledgeable statistician is recommended.

From P value to t value

Where actual P values obtained from t-tests are quoted, the corresponding t value may be obtained from a table of the t distribution. The degrees of freedom are given by N E + N C – 2, where N E and N C are the sample sizes in the experimental and control groups. We will illustrate with an example. Consider a trial of an experimental intervention (N E = 25) versus a control intervention (N C = 22), where the difference in means was MD = 3.8. It is noted that the P value for the comparison was P = 0.008, obtained using a two-sample t-test.

The t value that corresponds with a P value of 0.008 and 25+22 – 2=45 degrees of freedom is t = 2.78. This can be obtained from a table of the t distribution with 45 degrees of freedom or a computer (for example, by entering =tinv(0.008, 45) into any cell in a Microsoft Excel spreadsheet).

Difficulties are encountered when levels of significance are reported (such as P<0.05 or even P=NS which usually implies P>0.05) rather than exact P values. A conservative approach would be to take the P value at the upper limit (e.g. for P<0.05 take P=0.05, for P<0.01 take P=0.01 and for P<0.001 take P=0.001). However, this is not a solution for results which are reported as P=NS: see Section 7.7.3.7 .

From t value to standard error

The t value is the ratio of the difference in means to the standard error of the difference in means. The standard error of the difference in means can therefore be obtained by dividing the difference in means (MD) by the t value:

In the example, the standard error of the difference in means is obtained by dividing 3.8 by 2.78, which gives 1.37.

From confidence interval to standard error

If a 95% confidence interval is available for the difference in means, then the same standard error can be calculated as:

as long as the trial is large. For 90% confidence intervals 3.92 should be replaced by 3.29, and for 99% confidence intervals it should be replaced by 5.15. If the sample size is small then confidence intervals should have been calculated using a t distribution. The numbers 3.92, 3.29 and 5.15 need to be replaced with larger numbers specific to both the t distribution and the sample size, and can be obtained from tables of the t distribution with degrees of freedom equal to N E + N C – 2, where N E and N C are the sample sizes in the two groups. Relevant details of the t distribution are available as appendices of many statistical textbooks, or using standard computer spreadsheet packages.  For example, the t value for a 95% confidence interval from a comparison of a sample size of 25 with a sample size of 22 can be obtained by typing =tinv(1-0.95,25+22-2) in a cell in a Microsoft Excel spreadsheet.

From standard error to standard deviation

The within-group standard deviation can be obtained from the standard error of the difference in means using the following formula:

Note that this standard deviation is the average of the standard deviations of the experimental and control arms, and should be entered into RevMan twice (once for each intervention group).


7.7: Variation

In this section we outline a number of details concerning the implementation of the penalty method. These concern the derivatives of the functional with respect to the expansion coefficients for the support functions , the imposition of the normalisation constraint and the general outline of the scheme.

The support functions are expanded in spherical-wave basis functions:

For a functional of the support functions , the derivative with respect to the expansion coefficients is

For example, for the derivative of the total energy,

in which denotes the matrix element of the Kohn-Sham Hamiltonian with respect to the spherical-wave basis functions.

7.7.2 Normalisation constraint

We choose to minimise the total functional whilst constraining the normalisation of the density-matrix to the correct value. This is achieved firstly by projecting all gradients to be perpendicular to the gradient of the electron number, thus maintaining the electron number to first order, and secondly by re-converging the electron number to its correct value before each evaluation of the total functional.

In section 7.3 the electron number gradient with respect to the density-kernel, which is here denoted , is given as

The density-kernel along this search direction is parameterised by as

where denotes the initial density-matrix. The electron number, given by thus behaves linearly:

and it is a trivial matter to calculate the required value of to return the electron number to its correct value.

In general during the minimisation, the search direction is , and again this search can be parameterised by a single parameter :

We wish to project out from that component which is parallel to . The modified search direction can be written as

The variation of the electron number along this modified direction is

and we wish the coefficient of the linear term in to vanish, which defines the required value of to be

Since the electron number depends linearly upon the density-kernel, after this projection, the electron number is constant along the modified search direction, and the electron number need not be corrected after a trial step is taken.

The electron number gradient with respect to the support functions is given in section 7.3 and denoted by :

The support function variation is again parameterised by the parameter :

and this results in the following quadratic variation of the overlap matrix

which defines the matrices and . The variation of the electron number is therefore also quadratic

and the roots of this expression can be found to correct the electron number.

We consider a general search direction for the localised functions denoted and modify this search direction to obtain the direction which maintains the electron number to first order:

Now varying the localised functions according to

results in the following variation of the electron number:

where . Thus to maintain the electron number to first order, we choose

In this case, the electron number is not constant along the search direction, but will still vary quadratically, so that it is necessary to correct the density-matrix before evaluating the total functional.

The initialisation of the density-matrix is described in chapter 8. The functional minimisation consists of two nested loops. The inner loop consists of the minimisation with respect to the density-kernel, while keeping the support functions constant. As shown in section 7.4, this corresponds to making the density-matrix commute with the Hamiltonian in the representation of the current support functions. The outer loop consists of the support function minimisations, which in section 7.4 were shown to correspond to solving the Kohn-Sham equations. In general, we find that two or three cycles of the inner loop for each cycle of the outer loop suffice, and this is demonstrated in figure 7.1 in which three cycles of the inner loop appears to give the best performance to computational cost ratio.

Figure 7.1: Rate of convergence for different numbers of inner cycles. In the legend, corresponds to cycles of the inner loop for each iteration of the outer loop (horizontal axis).

The conjugate gradients scheme is used to determine the search directions from the gradients, and these gradients are then projected perpendicular to the electron number gradient as described in section 7.7.2. We approximate the total functional by a parabola along the search direction, using the initial value of the functional, the first derivative of the functional at the initial position (which is simply the scalar product of the steepest descent and search directions) and the value of the functional at some trial position. The value of the functional at the predicted minimum is then evaluated. If this value deviates significantly from the value predicted by the quadratic fit, a cubic fit is constructed using this new value of the functional. In general, this is only necessary for the first few steps, and the parabolic fit is very good. In the case of the support function variation, the support functions are altered to give the correct number of electrons before each evaluation of the functional. The conjugate gradients procedure is reset after a finite number of steps, and this is illustrated in figure 7.2 for the inner and outer loops. In both cases, we see that there is little advantage in conjugating more than eight gradients before resetting.

Figure 7.2: Performance of the conjugate gradients algorithm in the density-kernel variation (left) and the support function variation (right).

Once the functional has been minimised, the correction to the total energy is calculated. The whole calculation is generally repeated for a few different values of to ensure that the corrected energy has indeed converged.


Calculation of single locations

So let’s see how much the magnetic variation has changed for one specific place on the earth.

Let’s take this aerodrome declared variation on the eAIP from FMMI (Antanarivo International Airport), its magnetic variation published for 2015 says 15°W so let’s do the calculation for Epoch 2020.0 (1 January 2020)

retrieved from eAIP ASECNA December 16th,2019

We will be using the single point calculator from NOAA and first lets do a check on the value for 2015, the calculator is easy to use but since WMM is valid for 5 year periods only I will use one of the other available model IGRF to do the check getting the 15°W value.

To do the calculation let’s use the ARP of the airport and its height AMSL. The model changes with height but it wont make a huge difference.

Now let’s change the value to WMM and 1 January 2020

So you can see that in five yearst the change isn’t that much as the change per year is minimal, it usually takes a decade for shifting one degree in most parts of the world and since we just show a one degree resolution it doesn’t really show easily. Also as these are models you can check that the IGRF predicted a change of 1 minute 53 seconds per year while the new WMM model is predicting a change per year of four minutes and 20 seconds.

Basically in about two more years the change would be enough to cross the 30 minutes mark to change from 15°W to a 16°W magnetic declination

The following map shows the declination isolines where the change is in minutes per year


Japanese Rifle IdentificationMain Page

Modern Japanese rifles were produced in various configurations and calibers at several Arsenals located thoughout Japan, China, and Korea from about 1897 through 1945. Below are the markings on rifles in 6.5 Japanese Caliber manufactured from 1897 until the mid 1940's. These rifles include: The Type 30 Long Rifle and Carbine, the Type 35 Rifle, the Type 38 Long Rifle, Short Rifle, and Carbine, the Type 44 Carbine, the Type 97 Sniper Rifle, and the Italian Type I Long Rifle.

6.5 Caliber Japanese Rifle Receiver Markings


Type 30 - Model of 1897
Found in Long Rifle and Carbine configurations

Type 35 - Model of 1902
Found in Long Rifle Configuration only
adopted for Naval use

Type 38 - Model of 1905
Found in Long, Short Rifle,
and Carbine Configurations

Type 38 - Model of 1905
Double Circle Rifle or
Concentric Circle Rifle

Type 38 - Model of 1905
'Love of Country' marks
either side of Chrysanthemum

Mexican Type 38 - 1913
Rifle/Carbine configuration
No Type/Arsenal markings

Type 44 - Model of 1915
Found in Carbine Configuration only
3 Variations Identified by Bayonet Housings



Type 97 - Model of 1937
Kokura Arsenal
Sniper Configuration only

Type 97 - Model of 1937
Nagoya Arsenal
Sniper Configuration only

In the late 1930's the Japanese developed a rifle to compete in 'Modern Warfare'. It was a redesign of the Type 38 in a larger caliber, 7.7 Japanese. These rifles include: The Type 99 Long Rifle, the Type 99 Short Rifle, the Type 99 Carbine, the Type 99 Naval Special, the Type 100 Paratroop Rifle, and the Type 2 Paratroop Rifle. Receiver Markings of the 7.7 Caliber guns are below.

7.7 Caliber Japanese Rifle Receiver Markings


Early Type 99 - Model of 1939
Found in Long/Short Rifle
and Carbine configurations

Late Type 99 - Model of 1939
Double Circle Rifle or
Concentric Circle Rifle

Late Type 99 - Model of 1939
Last Ditch Receiver markings
Found on Short Rifles & Carbines

Naval Type 99 - Model of 1939
Naval Receiver markings
Found on Short Rifles & Carbines

Type 2 - Model of 1942
Found in Short Rifle configurations

The Model/Type markings are generally found on the top of the receiver, forward (towards to muzzle) of the chamber and generally indicate original caliber unless modified by another country at a later date. The serial number is found on the left side of the receiver on most standard rifles. The Arsenal mark on Japanese rifles is generally found to the right of the serial number on the left side of the receiver. Frequently there is a Series designator in a circle preceeding the serial number on the left side of the action.


Common Uses

The mean absolute deviation has a few applications. The first application is that this statistic may be used to teach some of the ideas behind the standard deviation. The mean absolute deviation about the mean is much easier to calculate than the standard deviation. It does not require us to square the deviations, and we do not need to find a square root at the end of our calculation. Furthermore, the mean absolute deviation is more intuitively connected to the spread of the data set than what the standard deviation is. This is why the mean absolute deviation is sometimes taught first, before introducing the standard deviation.

Some have gone so far as to argue that the standard deviation should be replaced by the mean absolute deviation. Although the standard deviation is important for scientific and mathematical applications, it is not as intuitive as the mean absolute deviation. For day-to-day applications, the mean absolute deviation is a more tangible way to measure how spread out data are.


3 Answers 3

In case anyone stumbles across this in the future: WooCommerce provides triggers throughout the add-to-cart-variation.js which allows you to hook into change events for the website. You can find all of them available in that file, but one which will likely help most in this case can be used as such

Where the trigger you want to hook into is the first argument inside .on() . Here's a few below to get you started:

woocommerce_variation_select_change Fires when the select is changed.

show_variation is fired when it finds a variation, and will actually pass you the variation object that is found so you can get direct access to the price without having to filter through the selects manually.


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