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## Equipartition Theorem
Using the equipartition theorem determine the specific heat capacity at constant volume for argon and nitrogen. Compare the resulting values with table values.
• #### Hint – Equipartition Theorem
The equipartition theorem is a statement from statistical physics and states that in equilibrium for every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadraticaly there is a mean energy of
$\frac{1}{2}kT,$
where k is the Boltzmann constant and T thermodynamic temperature of the system.
How is this fact related to a specific heat capacity (or molar heat capacity) and degrees of freedom of gas?
• #### Hint 2
Argon is a one-atom gas and nitrogen is a two-atom gas. What is their number of degrees of freedom?
• #### Given Values
cVi = ? specific heat capacity of gases at constant volume
Table values:
MmN = 28 g mol−1 = 0.028 kg mol−1 molar mass of nitrogen N2 (two-atom molecule) MmAr = 36 g mol−1 = 0.039 kg mol−1 molar mass of argon Ar (one-atom molecule) R = 8.31 Jmol−1K−1 molar gas constant
• #### Analysis
The equipartition theorem is a statement from statistical physics and states that in equilibrium, every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadratically corresponds to mean energy of 1/2kT, where k is the Boltzmann constant and T is thermodynamic temperature of the system.
This corresponds to the contribution of 1/2R to molar heat capacity CV at constant pressure.
This contribution belongs to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e. R. This is due to the fact that the energy of vibrational motion is composed of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.
The task is to determine the degrees of freedom for one-atom molecules of argon and two-atom molecules of nitrogen.
• #### Solution
The equipartition theorem is a statement from statistical physics according to which to each generalized momentum and each generalized coordinate acting in the formula for energy of the system quadratically corresponds to mean energy of $$\frac{1}{2}kT,$$ where k is the Boltzmann constant and T is thermodynamic temperature.
In other words each generalized momentum and generalized coordination acting in the formula for the energy of the system quadratically corresponds to a contribution of $$\frac{1}{2}R$$ to molar heat capacity CV.
This means that the state contribution corresponds to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e. R. This is due to the fact that the energy of vibrational motion constitutes of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.
Now we can easily determine heat capacities of given gases.
Argon is a one-atom gas which means that its molecules have only three translational degrees of freedom. Molar heat capacity CV at constant volume is therefore
$C_{VAr}=3\cdot\frac{1}{2}R=\frac{3}{2}R.$
Related specific heat capacity cV is determined from molar mass MmAr as follows:
$c_{VAr}=\frac{C_{VAr}}{M_{mAr}}=\frac{3R}{2M_{mAr}}.$
Nitrogen is a two-atom gas. It has six degrees of freedom in total. Three of them are translational, two are rotational and one is vibrational. According to the equipartition theorem the molar heat capacity CVN at constant volume is given by
$C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R+R=\frac{7}{2}R.$
For a corresponding specific heat capacity cVN we then obtain:
$c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{7R}{2M_{mN}},$
where MmN is molar mass of nitrogen.
• #### Numerical Solution
$c_{VAr}=\frac{3R}{2M_{mAr}}=\frac{3\cdot{8.31}}{2\cdot{0.039}}\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}$ $c_{VAr}=319.6\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}=0.32\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}$
$c_{VN}=\frac{7R}{2M_{mN}}=\frac{7\cdot{8.31}}{2\cdot{0.028}}\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}$ $c_{VN}=1038.8\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\dot{=}1.04\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}$
• #### Answer and Result Discussion
Calculated value Table value for t = 20 °C cVAr 0.32 kJ kg−1 K−1 0.32 kJ kg−1 K−1 cVN 1.04 kJ kg−1 K−1 0.74 kJ kg−1 K−1
By comparing calculated results with table values we can see that while the values for argon are the same, there is a big difference in the values for nitrogen. This is due to the fact that the equipartition theorem is derived within classical physics and thus does not respect the specifics of quantum mechanics that are manifested quite profoundly when dedscribing the vibrational motion at normal temperatures. In reality, the vibrations practically do not contribute to heat capacity under normal conditions. Therefore, to determine molar heat capacity CVN of nitrogen, we should use a more precise formula
$C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R=\frac{5}{2}R$
that involves only the translation and rotation.
If we calculate corresponding specific heat capacity cVN
$c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{5R}{2M_{mN}},$
we obtain a numerical value of
$c_{VN}\dot{=}742\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\dot{=}0.74\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}.$
This result is in a good agreement with the table value.
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## Equipartition Theorem. Using the equipartition theorem determine the specific heat capacity at constant volume for argon and nitrogen. Compare the resulting values with table values.. • #### Hint – Equipartition Theorem. The equipartition theorem is a statement from statistical physics and states that in equilibrium for every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadraticaly there is a mean energy of. $\frac{1}{2}kT,$. where k is the Boltzmann constant and T thermodynamic temperature of the system.. How is this fact related to a specific heat capacity (or molar heat capacity) and degrees of freedom of gas?. • #### Hint 2. Argon is a one-atom gas and nitrogen is a two-atom gas. What is their number of degrees of freedom?. • #### Given Values. cVi = ? specific heat capacity of gases at constant volume. Table values:. MmN = 28 g mol−1 = 0.028 kg mol−1 molar mass of nitrogen N2 (two-atom molecule) MmAr = 36 g mol−1 = 0.039 kg mol−1 molar mass of argon Ar (one-atom molecule) R = 8.31 Jmol−1K−1 molar gas constant. • #### Analysis. The equipartition theorem is a statement from statistical physics and states that in equilibrium, every generalized momentum or generalized coordinate acting in the formula for the energy of the system quadratically corresponds to mean energy of 1/2kT, where k is the Boltzmann constant and T is thermodynamic temperature of the system.. This corresponds to the contribution of 1/2R to molar heat capacity CV at constant pressure.. This contribution belongs to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e. R. This is due to the fact that the energy of vibrational motion is composed of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.. The task is to determine the degrees of freedom for one-atom molecules of argon and two-atom molecules of nitrogen.. • #### Solution. The equipartition theorem is a statement from statistical physics according to which to each generalized momentum and each generalized coordinate acting in the formula for energy of the system quadratically corresponds to mean energy of $$\frac{1}{2}kT,$$ where k is the Boltzmann constant and T is thermodynamic temperature.. In other words each generalized momentum and generalized coordination acting in the formula for the energy of the system quadratically corresponds to a contribution of $$\frac{1}{2}R$$ to molar heat capacity CV.. This means that the state contribution corresponds to each translational and rotational degree of freedom. It is different with vibrational degrees of freedom. The contribution is twice as big, i.e. R. This is due to the fact that the energy of vibrational motion constitutes of kinetic and potential energy, and not only kinetic energy as it is with translational and rotational motion.
|
Now we can easily determine heat capacities of given gases.. Argon is a one-atom gas which means that its molecules have only three translational degrees of freedom. Molar heat capacity CV at constant volume is therefore. $C_{VAr}=3\cdot\frac{1}{2}R=\frac{3}{2}R.$. Related specific heat capacity cV is determined from molar mass MmAr as follows:. $c_{VAr}=\frac{C_{VAr}}{M_{mAr}}=\frac{3R}{2M_{mAr}}.$. Nitrogen is a two-atom gas. It has six degrees of freedom in total. Three of them are translational, two are rotational and one is vibrational. According to the equipartition theorem the molar heat capacity CVN at constant volume is given by. $C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R+R=\frac{7}{2}R.$. For a corresponding specific heat capacity cVN we then obtain:. $c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{7R}{2M_{mN}},$. where MmN is molar mass of nitrogen.. • #### Numerical Solution. $c_{VAr}=\frac{3R}{2M_{mAr}}=\frac{3\cdot{8.31}}{2\cdot{0.039}}\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}$ $c_{VAr}=319.6\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}=0.32\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}$. $c_{VN}=\frac{7R}{2M_{mN}}=\frac{7\cdot{8.31}}{2\cdot{0.028}}\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}$ $c_{VN}=1038.8\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\dot{=}1.04\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}$. • #### Answer and Result Discussion. Calculated value Table value for t = 20 °C cVAr 0.32 kJ kg−1 K−1 0.32 kJ kg−1 K−1 cVN 1.04 kJ kg−1 K−1 0.74 kJ kg−1 K−1. By comparing calculated results with table values we can see that while the values for argon are the same, there is a big difference in the values for nitrogen. This is due to the fact that the equipartition theorem is derived within classical physics and thus does not respect the specifics of quantum mechanics that are manifested quite profoundly when dedscribing the vibrational motion at normal temperatures. In reality, the vibrations practically do not contribute to heat capacity under normal conditions. Therefore, to determine molar heat capacity CVN of nitrogen, we should use a more precise formula. $C_{VN}=3\cdot\frac{1}{2}R+2\cdot\frac{1}{2}R=\frac{5}{2}R$. that involves only the translation and rotation.. If we calculate corresponding specific heat capacity cVN. $c_{VN}=\frac{C_{VN}}{M_{mN}}=\frac{5R}{2M_{mN}},$. we obtain a numerical value of. $c_{VN}\dot{=}742\,\mathrm{Jkg}^{-1}\mathrm{K}^{-1}\dot{=}0.74\,\mathrm{kJkg}^{-1}\mathrm{K}^{-1}.$. This result is in a good agreement with the table value.
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# Permutations and Combinations | Counting | Infinity Learn | Summary and Q&A
3.5M views
July 13, 2015
by
Infinity Learn NEET
Permutations and Combinations | Counting | Infinity Learn
## TL;DR
Learn the key rules of counting to understand permutations and combinations, making problem-solving easier.
## Key Insights
• 🔄 The understanding of counting principles is crucial for mastering permutations and combinations.
• 📏 The 'OR' rule represents addition, while the 'AND' rule represents multiplication in counting techniques.
• 👻 Counting techniques can be applied to various scenarios, allowing for the calculation of different combinations and permutations.
• 🔄 Visual representations, such as the example of counting circles in a grid, can simplify counting and understanding permutations and combinations.
• 🧑🎓 Many students find permutations and combinations challenging, but with a solid understanding of counting principles, it becomes a manageable topic.
• 🔄 Formulae for permutations and combinations are useful but not as important as understanding counting principles.
• 👻 Developing counting skills allows for efficient problem-solving in permutations and combinations.
## Transcript
The formula for N C R is 'N factorial, divided by R' factorial times 'N minus R' factorial. What is this? This is the formula for combinations, but it is not the best way to understand it ! And what is NPR. It's N factorial over 'N minus R' factorial. And this again, is not the best way to understand permutations! In many books, and videos you woul... Read More
### Q: What is the formula for calculating permutations and combinations?
The formula for permutations is N! / (N-R)! while combinations use N! / (R!(N-R)!). However, understanding counting principles is more important than relying solely on these formulas.
### Q: How do counting techniques relate to permutations and combinations?
Counting techniques, such as understanding 'OR' and 'AND' principles, are essential to solving permutations and combinations problems. 'OR' represents addition while 'AND' represents multiplication.
### Q: How can counting principles be applied to simple scenarios?
Counting principles can be applied to various scenarios. For example, if you have 3 pens and 2 markers, you can count the total number of ways to select any one item as 3 (pens) + 2 (markers) = 5.
### Q: How can counting principles be used to determine the ways to select one pen and one marker?
To calculate the ways to select one pen and one marker, you multiply the number of ways to select a pen (3) with the number of ways to select a marker (2), resulting in 3 (pens) x 2 (markers) = 6.
## Summary & Key Takeaways
• Understanding permutations and combinations requires a solid grasp of counting techniques.
• Permutations can be calculated using the formula N! / (N-R)! while combinations use N! / (R!(N-R)!).
• The key to mastering permutations and combinations is to understand the rules of counting, which involve addition (OR) and multiplication (AND) principles.
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# Permutations and Combinations | Counting | Infinity Learn | Summary and Q&A. 3.5M views. July 13, 2015. by. Infinity Learn NEET. Permutations and Combinations | Counting | Infinity Learn. ## TL;DR. Learn the key rules of counting to understand permutations and combinations, making problem-solving easier.. ## Key Insights. • 🔄 The understanding of counting principles is crucial for mastering permutations and combinations.. • 📏 The 'OR' rule represents addition, while the 'AND' rule represents multiplication in counting techniques.. • 👻 Counting techniques can be applied to various scenarios, allowing for the calculation of different combinations and permutations.. • 🔄 Visual representations, such as the example of counting circles in a grid, can simplify counting and understanding permutations and combinations.. • 🧑🎓 Many students find permutations and combinations challenging, but with a solid understanding of counting principles, it becomes a manageable topic.. • 🔄 Formulae for permutations and combinations are useful but not as important as understanding counting principles.. • 👻 Developing counting skills allows for efficient problem-solving in permutations and combinations.. ## Transcript. The formula for N C R is 'N factorial, divided by R' factorial times 'N minus R' factorial.
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What is this? This is the formula for combinations, but it is not the best way to understand it ! And what is NPR. It's N factorial over 'N minus R' factorial. And this again, is not the best way to understand permutations! In many books, and videos you woul... Read More. ### Q: What is the formula for calculating permutations and combinations?. The formula for permutations is N! / (N-R)! while combinations use N! / (R!(N-R)!). However, understanding counting principles is more important than relying solely on these formulas.. ### Q: How do counting techniques relate to permutations and combinations?. Counting techniques, such as understanding 'OR' and 'AND' principles, are essential to solving permutations and combinations problems. 'OR' represents addition while 'AND' represents multiplication.. ### Q: How can counting principles be applied to simple scenarios?. Counting principles can be applied to various scenarios. For example, if you have 3 pens and 2 markers, you can count the total number of ways to select any one item as 3 (pens) + 2 (markers) = 5.. ### Q: How can counting principles be used to determine the ways to select one pen and one marker?. To calculate the ways to select one pen and one marker, you multiply the number of ways to select a pen (3) with the number of ways to select a marker (2), resulting in 3 (pens) x 2 (markers) = 6.. ## Summary & Key Takeaways. • Understanding permutations and combinations requires a solid grasp of counting techniques.. • Permutations can be calculated using the formula N! / (N-R)! while combinations use N! / (R!(N-R)!).. • The key to mastering permutations and combinations is to understand the rules of counting, which involve addition (OR) and multiplication (AND) principles.
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https://www.vedantu.com/question-answer/write-one-million-in-scientific-notatio-class-5-maths-cbse-5ff721257c2bba6bf9e5c9ef
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# How do you write one million in scientific notation?
Last updated date: 12th Aug 2024
Total views: 395.1k
Views today: 10.95k
Verified
395.1k+ views
Hint:In the given question, we have been given the scientific notation of a number written in a phrase. To solve this, we must know what a scientific notation is – it is a way of writing extremely large numbers including lots of zeroes, in the form of a single digit followed by a decimal point, followed by the rest of non-zero numbers, followed by multiplication with ten raised to the power of condensed numbers, including the zeroes.
First, one million is one followed by six zeroes, or, $1,000,000$.
$1 \times {10^6}$
Hence, one million can be written as $1 \times {10^6}$.
The way of writing in the scientific notation is especially helpful to physicists, scientists, researchers who are working with very large numbers. For example, writing $1$ followed by twenty-nine zeros is very space and time consuming, so it is better to write it as $1 \times {10^{29}}$. This small notation saved so many zeroes’ space and time.
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Courses. Courses for Kids. Free study material. Offline Centres. More. Store. # How do you write one million in scientific notation?. Last updated date: 12th Aug 2024. Total views: 395.1k. Views today: 10.95k.
|
Verified. 395.1k+ views. Hint:In the given question, we have been given the scientific notation of a number written in a phrase. To solve this, we must know what a scientific notation is – it is a way of writing extremely large numbers including lots of zeroes, in the form of a single digit followed by a decimal point, followed by the rest of non-zero numbers, followed by multiplication with ten raised to the power of condensed numbers, including the zeroes.. First, one million is one followed by six zeroes, or, $1,000,000$.. $1 \times {10^6}$. Hence, one million can be written as $1 \times {10^6}$.. The way of writing in the scientific notation is especially helpful to physicists, scientists, researchers who are working with very large numbers. For example, writing $1$ followed by twenty-nine zeros is very space and time consuming, so it is better to write it as $1 \times {10^{29}}$. This small notation saved so many zeroes’ space and time.
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https://www.thestudentroom.co.uk/showthread.php?t=1622984
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# Core 2 help - AQAWatch
#1
Hey could anybody help me differentiate this:
y = x + 3 + 8/x^4 (x^4 meaning x to the power 4)
I got an answer, which is -24x^-5 +1 however I am not confident with it and I am fairly certain it's wrong
Thanks for any help!
0
7 years ago
#2
Yes, you are slightly wrong, its 1 -32x^-5
because -4 times 8 is -32 not -24 :P.
Multiply by the power, and then reduce the power by 1.
1
7 years ago
#3
(Original post by raveen789)
Hey could anybody help me differentiate this:
y = x + 3 + 8/x^4 (x^4 meaning x to the power 4)
I got an answer, which is -24x^-5 +1 however I am not confident with it and I am fairly certain it's wrong
Thanks for any help!
put more brackets in or write it in latex (google it) so i know what ur asking
0
7 years ago
#4
(Original post by In One Ear)
Yes, you are slightly wrong, its 1 -32x^-5
because -4 times 8 is -32 not -24 :P.
Multiply by the power, and then reduce the power by 1.
ThIs.
0
#5
(Original post by In One Ear)
Yes, you are slightly wrong, its 1 -32x^-5
because -4 times 8 is -32 not -24 :P.
Multiply by the power, and then reduce the power by 1.
Ahh yeah thanks lol I always make stupid mistakes, thanks for the help!
0
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# Core 2 help - AQAWatch. #1. Hey could anybody help me differentiate this:. y = x + 3 + 8/x^4 (x^4 meaning x to the power 4). I got an answer, which is -24x^-5 +1 however I am not confident with it and I am fairly certain it's wrong. Thanks for any help!. 0. 7 years ago. #2. Yes, you are slightly wrong, its 1 -32x^-5. because -4 times 8 is -32 not -24 :P.. Multiply by the power, and then reduce the power by 1.. 1. 7 years ago. #3. (Original post by raveen789). Hey could anybody help me differentiate this:. y = x + 3 + 8/x^4 (x^4 meaning x to the power 4). I got an answer, which is -24x^-5 +1 however I am not confident with it and I am fairly certain it's wrong. Thanks for any help!. put more brackets in or write it in latex (google it) so i know what ur asking. 0. 7 years ago. #4. (Original post by In One Ear). Yes, you are slightly wrong, its 1 -32x^-5. because -4 times 8 is -32 not -24 :P.. Multiply by the power, and then reduce the power by 1.. ThIs.. 0. #5. (Original post by In One Ear). Yes, you are slightly wrong, its 1 -32x^-5. because -4 times 8 is -32 not -24 :P.
|
Multiply by the power, and then reduce the power by 1.. Ahh yeah thanks lol I always make stupid mistakes, thanks for the help!. 0. X. new posts. Latest. My Feed. ### Oops, nobody has postedin the last few hours.. Why not re-start the conversation?. see more. ### See more of what you like onThe Student Room. You can personalise what you see on TSR. Tell us a little about yourself to get started.. ### University open days. • Cranfield University. Cranfield Forensic MSc Programme Open Day Postgraduate. Thu, 25 Apr '19. • University of the Arts London. Open day: MA Footwear and MA Fashion Artefact Postgraduate. Thu, 25 Apr '19. • Cardiff Metropolitan University. Sat, 27 Apr '19. ### Poll. Join the discussion. #### Have you registered to vote?. Yes! (100). 40.16%. No - but I will (12). 4.82%. No - I don't want to (16). 6.43%. No - I can't vote (<18, not in UK, etc) (121). 48.59%.
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# Graphing Linear Functions in Slope-Intercept Form
There are different ways to graph a linear function. Sometimes, the way the rule of the function is written, can dictate the simplest way to graph it. Below, the graphs of linear functions given in slope-intercept form will be explored.
## Slope-Intercept Form
One way to express linear function rules is called slope-intercept form.
y=mx+b
Sometimes, f(x)=mx+b is used. In either case, m and b describe the general characteristics of the line. m indicates the slope, and b indicates the y-intercept. The linear function graphed below can be expressed as
f(x)=2x+1,
because it has a slope of 2 and a y-intercept at (0,1).
## Graphing a Linear Function in Slope-Intercept Form
To graph a line in slope-intercept form, the slope, m, and the y-intercept, b, are both needed. Consider the linear function y=2x3. Since the rule is written as y=mx+b, it can be seen that
To graph the line, plot the y-intercept, then use the slope to find another point on the line. Specifically, from (0,-3), move up 2 units and right 1 unit.
Next, draw a line through both points to create the graph of the linear function.
fullscreen
Exercise
In Clear Lake, Iowa, during a particular evening, there is a 3-inch layer of snow on the ground. At midnight, it begins to snow. Each hour, one inch of snow falls. Graph a function that shows the amount of snow on the ground from midnight to 6 AM.
Show Solution
Solution
To begin, we can define the quantities that x and y represent.
• Let x represent the number of hours it's been snowing.
• Let y represent the number of inches of snow on the ground.
It's been given that there is a 3-inch layer of snow on the ground before it begins to snow. Thus, when x=0,y=3. In other words, the y-intercept is b=3. It is also given that the amount of snow on the ground increases by 1 inch every 1 hour. Thus, the slope of the line is We can write the rule for this function as
To graph the function, we can plot a point at (0,3), then move up 1 unit and right 1 unit to find another point. The line that connects these points is the graph of the function.
The graph above shows the function f(x)=x+3. It can be seen that, at 6AM, there is a total of 9 inches of snow on the ground.
## Describing Key Features of Linear Functions
Key features are certain characteristics of the graphs of functions that are noteworthy. For linear functions, a graph's intercepts, increasing/decreasing intervals, and end behavior are important.
fullscreen
Exercise
The graph shows the function Describe the intercepts, increasing/decreasing intervals, and end behavior. Then, show the features on the graph.
Show Solution
Solution
To begin, we'll describe each of the features. A graph's x- and y-intercepts are the points where the graph intersects the x- and y-axes, respectively. It can be seen that f intersects the x-axis at (6,0) and the y-axis at (0,4). Thus,
Since f is a linear function, it has a constant rate of change. This means the line will always be increasing or always decreasing. Looking from left to right on the graph, it can be seen that, as x increases, f(x) decreases infinitely. Thus,
Since f always decreases, we can say f is negative. Notice that this corresponds with f's negative slope.
Looking at the graph, we can see that the left end extends upward and the right end extends downward. Thus, the end behavior of f can be written as follows.
Lastly, we'll label the features on the graph.
## Solving Linear Equations Graphically
The function y=f(x) is comprised of all the (x,y) points that satisfy its function rule. For example, some of the points that lie on the line y=x+3 are
All of the x-y pairs that satisfy an equation are the points on the corresponding graph. Thus, the solution(s) to an equation can be found using a graph.
fullscreen
Exercise
Use the graph of y=2.5x1.5 to solve the equation 6=2.5x1.5.
Show Solution
Solution
The given graph shows all of the x-y pairs that satisfy y=2.5x1.5. Let's begin by comparing the function rule with the equation.
Since the right-hand side of both the function rule and the equation are the same, it follows then that
y=6.
Thus, solving the equation 6=2.5x1.5 means finding the x-coordinate of the point on the graph of y=2.5x1.5 whose y-coordinate is 6. We'll find this point by first locating the point on the graph where y=6.
We can move down from this point on the graph to find the x-coordinate.
Since the desired point on the graph is (3,6), the solution to 6=2.5x1.5 is x=3. We can verify this value by substituting it into the equation to see if a true statement is made.
6=2.5x1.5
6=6
Since a true statement is made, we can confirm that the solution to the equation is x=3.
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mathleaks.com mathleaks.com Start chapters home Start History history History expand_more Community. Community expand_more. {{ filterOption.label }}. {{ item.displayTitle }}. {{ item.subject.displayTitle }}. arrow_forward. {{ searchError }}. search. {{ courseTrack.displayTitle }}. {{ printedBook.courseTrack.name }} {{ printedBook.name }}. # Graphing Linear Functions in Slope-Intercept Form. There are different ways to graph a linear function. Sometimes, the way the rule of the function is written, can dictate the simplest way to graph it. Below, the graphs of linear functions given in slope-intercept form will be explored.. ## Slope-Intercept Form. One way to express linear function rules is called slope-intercept form.. y=mx+b. Sometimes, f(x)=mx+b is used. In either case, m and b describe the general characteristics of the line. m indicates the slope, and b indicates the y-intercept. The linear function graphed below can be expressed as. f(x)=2x+1,. because it has a slope of 2 and a y-intercept at (0,1).. ## Graphing a Linear Function in Slope-Intercept Form. To graph a line in slope-intercept form, the slope, m, and the y-intercept, b, are both needed. Consider the linear function y=2x3. Since the rule is written as y=mx+b, it can be seen that. To graph the line, plot the y-intercept, then use the slope to find another point on the line. Specifically, from (0,-3), move up 2 units and right 1 unit.. Next, draw a line through both points to create the graph of the linear function.. fullscreen. Exercise. In Clear Lake, Iowa, during a particular evening, there is a 3-inch layer of snow on the ground. At midnight, it begins to snow. Each hour, one inch of snow falls. Graph a function that shows the amount of snow on the ground from midnight to 6 AM.. Show Solution. Solution. To begin, we can define the quantities that x and y represent.. • Let x represent the number of hours it's been snowing.. • Let y represent the number of inches of snow on the ground.. It's been given that there is a 3-inch layer of snow on the ground before it begins to snow. Thus, when x=0,y=3. In other words, the y-intercept is b=3. It is also given that the amount of snow on the ground increases by 1 inch every 1 hour. Thus, the slope of the line is We can write the rule for this function as. To graph the function, we can plot a point at (0,3), then move up 1 unit and right 1 unit to find another point.
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The line that connects these points is the graph of the function.. The graph above shows the function f(x)=x+3. It can be seen that, at 6AM, there is a total of 9 inches of snow on the ground.. ## Describing Key Features of Linear Functions. Key features are certain characteristics of the graphs of functions that are noteworthy. For linear functions, a graph's intercepts, increasing/decreasing intervals, and end behavior are important.. fullscreen. Exercise. The graph shows the function Describe the intercepts, increasing/decreasing intervals, and end behavior. Then, show the features on the graph.. Show Solution. Solution. To begin, we'll describe each of the features. A graph's x- and y-intercepts are the points where the graph intersects the x- and y-axes, respectively. It can be seen that f intersects the x-axis at (6,0) and the y-axis at (0,4). Thus,. Since f is a linear function, it has a constant rate of change. This means the line will always be increasing or always decreasing. Looking from left to right on the graph, it can be seen that, as x increases, f(x) decreases infinitely. Thus,. Since f always decreases, we can say f is negative. Notice that this corresponds with f's negative slope.. Looking at the graph, we can see that the left end extends upward and the right end extends downward. Thus, the end behavior of f can be written as follows.. Lastly, we'll label the features on the graph.. ## Solving Linear Equations Graphically. The function y=f(x) is comprised of all the (x,y) points that satisfy its function rule. For example, some of the points that lie on the line y=x+3 are. All of the x-y pairs that satisfy an equation are the points on the corresponding graph. Thus, the solution(s) to an equation can be found using a graph.. fullscreen. Exercise. Use the graph of y=2.5x1.5 to solve the equation 6=2.5x1.5.. Show Solution. Solution. The given graph shows all of the x-y pairs that satisfy y=2.5x1.5. Let's begin by comparing the function rule with the equation.. Since the right-hand side of both the function rule and the equation are the same, it follows then that. y=6.. Thus, solving the equation 6=2.5x1.5 means finding the x-coordinate of the point on the graph of y=2.5x1.5 whose y-coordinate is 6. We'll find this point by first locating the point on the graph where y=6.. We can move down from this point on the graph to find the x-coordinate.. Since the desired point on the graph is (3,6), the solution to 6=2.5x1.5 is x=3. We can verify this value by substituting it into the equation to see if a true statement is made.. 6=2.5x1.5. 6=6. Since a true statement is made, we can confirm that the solution to the equation is x=3.
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https://music.stackexchange.com/questions/58531/can-we-use-any-note-without-caring-much-on-rule-of-particular-meter-as-long-as-n/58533
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Can we use any note without caring much on rule of particular meter as long as note value fits in the meter?
As we know there are two main common meter :- simple and compound. We all know that simple meter divides a beat into two and compound meter into three. We also know that a measure holding two beat is duple, three beat triple and four beat quadruple so on. Now my question rise here. I saw some where in internet a music sheet using 2/4 meter which seems "simple duple". 2 beat in each measure "duple" and it would have been normal if a quarter note or eighth note use those beat space because they can be divided into two which make sense to be a simple. But a triplet? I found they using two triplet of eight note which we know 1-triplet = 1 beat so, 2-triplet = 2 beat. It fitted fine with the 2/4 time signature in the view of "duple" I think we can say two dotted quarter note made a triplet (if i'm wrong at this point tell me guys). But in view of "simple" a triplet seems to be divided in three part of a note then how is it fitted on "simple" ? If triplet is a note divided in three then isn't it suitable to be in compound? Is 2/4 meter really a compound duple or simple duple? Please help me to clear my misunderstanding.
• Could I ask you to work on the formatting of this question. It is very hard to read as is and that may very easily affect the quality of the answers. – Neil Meyer Jun 19 '17 at 9:58
• I'm not that great on english never mind if you can edit my question keeping in care what I'm trying to say that will be great @NeilMeyer – SOuřaan Gřg Jun 19 '17 at 10:45
• @SOuřaanGřg I think your English is off to a good start; I understood your question no problem! – Richard Jun 19 '17 at 13:20
If ALL the notes relate to triplets, then you will have six quaver triplets per bar, and it would be better written in 6/8 (compound duple). If only some bars have triplets, and others contain the normal 4 quavers, 2 crotchets, or a combination, the time sig will probably stay as simple duple, 2/4. So, it depends on the writer and the tune itself. One will always be a better fit than the other.
• I have less reputation for now but submitted a up vote for your kind help! When we have a dot with any note value is the dot left uncount ? I mean doesn't the dot take effect on the number of beat per measure of a meter? A two "quarter dotted" note is on 2/4 meter which seem only "quarter" note been counted for beat per measure but "dot" doesn't seem to be counted? – SOuřaan Gřg Jun 19 '17 at 2:17
• A dot after a note increases its value or length by an extra half of its original. BUT - triplets are actually three identical notes that are played in the time of two of those same notes. A kind of opposite to what the dot does! – Tim Jun 19 '17 at 6:16
2/4 is a simple duple meter. It doesn't matter how many triplets, quintuplets, septuplets, or any other tuplets are in the music--it's still a simple duple meter.
Now, if you notate all eighth-note triplets and no other tuplets (including bog-standard duplets) in the entire 2/4 piece, people may argue that your music should be in 6/8 (a compound duple meter) instead.
(I'm answering based on my best understanding of your question. I hope I'm interpreting your words right...)
• I have less reputation for now but submitted a up vote for your kind help! It mean as long as a note value fit in the meter we can add it without worrying about "simple or compound" ? Just keeping in mind with duple,triple,quadruple etc.. ? – SOuřaan Gřg Jun 19 '17 at 2:00
• You can put in any notes you want as long as they fit in the meter (such as a 16th-note quintuplet and an 8th-note triplet in the same 2/4 bar). Consistently put triplets instead of any other type of tuplet in a piece, though, and people will wonder why the piece is written in a simple meter instead of a compound one. – Dekkadeci Jun 19 '17 at 7:57
I think you have a very cursory understanding of what a Time Signature is. The beats of time signatures can have any number of notes in them. Compound vs Simple is much more an issue of what beats consist of.
Compound time has dotted note values for beats where Simple time simply has notes without dots for beats, this has very little to do with what particular notes a beat can or should consist of.
A compound time signature can have two notes in a bar trough the use of duplets, just in the same way as a simple time signature can have three notes in a beat trough the use of triplets
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Can we use any note without caring much on rule of particular meter as long as note value fits in the meter?. As we know there are two main common meter :- simple and compound. We all know that simple meter divides a beat into two and compound meter into three. We also know that a measure holding two beat is duple, three beat triple and four beat quadruple so on. Now my question rise here. I saw some where in internet a music sheet using 2/4 meter which seems "simple duple". 2 beat in each measure "duple" and it would have been normal if a quarter note or eighth note use those beat space because they can be divided into two which make sense to be a simple. But a triplet? I found they using two triplet of eight note which we know 1-triplet = 1 beat so, 2-triplet = 2 beat. It fitted fine with the 2/4 time signature in the view of "duple" I think we can say two dotted quarter note made a triplet (if i'm wrong at this point tell me guys). But in view of "simple" a triplet seems to be divided in three part of a note then how is it fitted on "simple" ? If triplet is a note divided in three then isn't it suitable to be in compound? Is 2/4 meter really a compound duple or simple duple? Please help me to clear my misunderstanding.. • Could I ask you to work on the formatting of this question. It is very hard to read as is and that may very easily affect the quality of the answers. – Neil Meyer Jun 19 '17 at 9:58. • I'm not that great on english never mind if you can edit my question keeping in care what I'm trying to say that will be great @NeilMeyer – SOuřaan Gřg Jun 19 '17 at 10:45. • @SOuřaanGřg I think your English is off to a good start; I understood your question no problem! – Richard Jun 19 '17 at 13:20. If ALL the notes relate to triplets, then you will have six quaver triplets per bar, and it would be better written in 6/8 (compound duple). If only some bars have triplets, and others contain the normal 4 quavers, 2 crotchets, or a combination, the time sig will probably stay as simple duple, 2/4. So, it depends on the writer and the tune itself. One will always be a better fit than the other.. • I have less reputation for now but submitted a up vote for your kind help! When we have a dot with any note value is the dot left uncount ? I mean doesn't the dot take effect on the number of beat per measure of a meter? A two "quarter dotted" note is on 2/4 meter which seem only "quarter" note been counted for beat per measure but "dot" doesn't seem to be counted? – SOuřaan Gřg Jun 19 '17 at 2:17.
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• A dot after a note increases its value or length by an extra half of its original. BUT - triplets are actually three identical notes that are played in the time of two of those same notes. A kind of opposite to what the dot does! – Tim Jun 19 '17 at 6:16. 2/4 is a simple duple meter. It doesn't matter how many triplets, quintuplets, septuplets, or any other tuplets are in the music--it's still a simple duple meter.. Now, if you notate all eighth-note triplets and no other tuplets (including bog-standard duplets) in the entire 2/4 piece, people may argue that your music should be in 6/8 (a compound duple meter) instead.. (I'm answering based on my best understanding of your question. I hope I'm interpreting your words right...). • I have less reputation for now but submitted a up vote for your kind help! It mean as long as a note value fit in the meter we can add it without worrying about "simple or compound" ? Just keeping in mind with duple,triple,quadruple etc.. ? – SOuřaan Gřg Jun 19 '17 at 2:00. • You can put in any notes you want as long as they fit in the meter (such as a 16th-note quintuplet and an 8th-note triplet in the same 2/4 bar). Consistently put triplets instead of any other type of tuplet in a piece, though, and people will wonder why the piece is written in a simple meter instead of a compound one. – Dekkadeci Jun 19 '17 at 7:57. I think you have a very cursory understanding of what a Time Signature is. The beats of time signatures can have any number of notes in them. Compound vs Simple is much more an issue of what beats consist of.. Compound time has dotted note values for beats where Simple time simply has notes without dots for beats, this has very little to do with what particular notes a beat can or should consist of.. A compound time signature can have two notes in a bar trough the use of duplets, just in the same way as a simple time signature can have three notes in a beat trough the use of triplets.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2006-04-11 07:11:33
Chemist
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### Line Integral
I just can't figure out what's the difference b/w the following two methods for solving line integrals.
In the book, they say for evaluating a line integral for a function f(x,y,z) = x - 3y2 + z over the line segment C joining the origin and the point (1,1,1) , we must 1st determine the parametric equations in terms of a parameter t.
r = t i + t j + t k 0 ≤ t ≤ 1 ( smooth parameter , 1st derivative is cont and never zero )
they give the line integral form : ∫ f(x,y,z)ds = ∫ f(t,t,t) |v(t)| dt 0 ≤ t ≤ 1
In class, however, this method is never used.
We start from ∫ F.dr ( both vectors ) where F = vector field = M i + N j + O k
∫ F.Tds ( T is the tangent unit vector, F.T----> dot product ) = ∫ F.dr = ∫ F.drdt /dt = ∫ ( M dx/dt + N dy/dt + O dz/dt ) dt
We've used this form to determine the line integral in all the problems we have encountered so far. If I try the 1st way, the answer is not always the same for the 2nd, actually, rarely is it the same. I tried to see the common point between the two:
∫ F.Tds = ∫ F.dr = ∫ F.dr dt / dr = ∫ F.V(t) dt , here obviously V is a velocity vector.
so, ∫ F.V dt = ∫ F |V(t)| cosθ dt ( θ b/w F and V )
In the book , they're using ∫ f(t,t,t) |V(t)| dt , so perhaps they assume that V is parallel to F ? Why would they do that, and what's the significance of either method?
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #2 2006-04-11 08:35:58
Ricky
Moderator
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### Re: Line Integral
The main point of below is that you must do different things when your function goes from R³⇒R and R³⇒ R³
For a Path integral, you have the integral over a path c = (x(t), y(t), z(t)) of a function f(x, y, z), where f goes from R^3 -> R.
To integrate this, you integrate from a to b (the start and end points of your path) of f(c(t)) ||c'(t)|| dt, where ||c'(t)|| is the length of the derivative of c(t).
But a Line integral is completely different (sort of...). Here, F goes from R^3 -> R^3, not to just R. Then you need to do:
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
## #3 2006-04-11 08:38:05
Ricky
Moderator
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### Re: Line Integral
If you tried to use both formulas on the same problem, then you either:
Tried to integrate a vector
or
Tried to take the dot product of a scalar.
Of course, neither makes sense.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
## #4 2006-04-11 09:21:14
Chemist
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### Re: Line Integral
Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?
Tried to integrate a vector
or
Tried to take the dot product of a scalar.
Of course, neither makes sense.
Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #5 2006-04-11 11:07:04
Chemist
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### Re: Line Integral
If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering ...
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #6 2006-04-11 11:12:25
Ricky
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### Re: Line Integral
Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?
Well, take the example: f(x, y, z) = x + y + z
That's from R³⇒R. You could define it from R³⇒R³ by:
f(x, y, z) = (x + y + z, 0, 0)
But these are two completely different things with completely different properties. They look the same, but they aren't.
Is this what you are talking about?
Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?
They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.
If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
## #7 2006-04-11 15:02:10
George,Y
Super Member
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### Re: Line Integral
The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.
The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity. i cannot figure out what the product means.
So you'd better use the second defaultly.
So far, i haven't encountered the situation to use your first integral.
X'(y-Xβ)=0
## #8 2006-04-11 19:34:48
Chemist
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### Re: Line Integral
Well, take the example: f(x, y, z) = x + y + z
That's from R³⇒R. You could define it from R³⇒R³ by:
f(x, y, z) = (x + y + z, 0, 0)
I'm not sure what this is, what's the 0,0 for?
But these are two completely different things with completely different properties. They look the same, but they aren't.
Is this what you are talking about?
No , I was simply saying that if we have a scalar function f(x,y,z) = x +y +z , we can change it to a vector field or to its total differential form. This is what I did in calculating the path integral of a function f, I changed it to its vector form and calculated the scalar product.
They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.
If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.
Thanks for the explanation.
If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering ...
I think I may be right here. I was just solving some problems, it turns out that whenever the total diff form of a vector field is exact ( rotor = 0 ), the line integral is independent of the path, and I can simply compute it F(B) - F(A) if we assume the line integral is from A to B across an arc. This really makes the computation easy for a line integral.
But if the total diff form is exact, continuous and its 1st derivative is continuous too, the line integral in a closed domain would be zero. Why's that?
And can anyone prove Green-Rieman's formula? I have an idea but I think it's incorrect.
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #9 2006-04-11 19:36:37
Chemist
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### Re: Line Integral
#### George,Y wrote:
The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.
The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity. i cannot figure out what the product means.
So you'd better use the second defaultly.
So far, i haven't encountered the situation to use your first integral.
Thanks for the info , I think I got the difference between the two. And yes |V(t)| is the velocity magnitude or speed.
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #10 2006-04-12 07:54:48
Chemist
Member
Offline
### Re: Line Integral
We can obtain Green's theorem area formula by going backward :
A = ∫ ∫ dydx = ∫ ∫ ( 1/2 + 1/2 ) dydx = 1/2 ∫ ( xdy - ydx )
, ofcourse the domain is closed.
But it seems sometimes we can compute the area directly from A = ∫ xdy
Example:
Determine the area b/w y=x2 and y = x across C+
A = 1/2 ( ∫ xdy - ydx )
I divided the line integral into arc OA and a straight line OA where A (1,1), the point of intersection of the parabola and the straight line, and O is the origin ofcourse.
I chose the parametirc equations,
For [OA] : y = x , x=x 0 ≤ x ≤ 1
For arc OA : x=y2 , y=y 1 ≤ y ≤ 0
I1 = 1/2 ∫ (x-x)dx = 0 ----> for [OA]
I2 = 1/2 ∫ (y2 - 2y2)dy = 1/6 ---> for arc OA
A = I1 + I2 = 1/6
Now notice that the area could have been calculated directly from:
A = ∫ xdy ( line int across C+ ) = ∫ xdy ( across [OA] ) + ∫ xdy ( across arc OA )
Using the same variables, A = 1/6
The second way is quicker, but I can't seem to figure out when to use it.
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
## #11 2006-04-12 12:45:06
George,Y
Super Member
Offline
### Re: Line Integral
to compute the area enclosed by an ellipse.
X'(y-Xβ)=0
## #12 2006-04-12 23:40:24
Chemist
Member
Offline
### Re: Line Integral
You mean the area A = 1/2 ∫ ( xdy - ydx ) can be simplified to A = ∫ xdy in the case of an ellipse? It was correct for the closed region between x=y2 and y=x. There was no ellipse there ...
"Fundamentally one will never be able to renounce abstraction."
Werner Heisenberg
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °. You are not logged in.. ## #1 2006-04-11 07:11:33. Chemist. Member. Offline. ### Line Integral. I just can't figure out what's the difference b/w the following two methods for solving line integrals.. In the book, they say for evaluating a line integral for a function f(x,y,z) = x - 3y2 + z over the line segment C joining the origin and the point (1,1,1) , we must 1st determine the parametric equations in terms of a parameter t.. r = t i + t j + t k 0 ≤ t ≤ 1 ( smooth parameter , 1st derivative is cont and never zero ). they give the line integral form : ∫ f(x,y,z)ds = ∫ f(t,t,t) |v(t)| dt 0 ≤ t ≤ 1. In class, however, this method is never used.. We start from ∫ F.dr ( both vectors ) where F = vector field = M i + N j + O k. ∫ F.Tds ( T is the tangent unit vector, F.T----> dot product ) = ∫ F.dr = ∫ F.drdt /dt = ∫ ( M dx/dt + N dy/dt + O dz/dt ) dt. We've used this form to determine the line integral in all the problems we have encountered so far. If I try the 1st way, the answer is not always the same for the 2nd, actually, rarely is it the same. I tried to see the common point between the two:. ∫ F.Tds = ∫ F.dr = ∫ F.dr dt / dr = ∫ F.V(t) dt , here obviously V is a velocity vector.. so, ∫ F.V dt = ∫ F |V(t)| cosθ dt ( θ b/w F and V ). In the book , they're using ∫ f(t,t,t) |V(t)| dt , so perhaps they assume that V is parallel to F ? Why would they do that, and what's the significance of either method?. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #2 2006-04-11 08:35:58. Ricky. Moderator. Offline. ### Re: Line Integral. The main point of below is that you must do different things when your function goes from R³⇒R and R³⇒ R³. For a Path integral, you have the integral over a path c = (x(t), y(t), z(t)) of a function f(x, y, z), where f goes from R^3 -> R.. To integrate this, you integrate from a to b (the start and end points of your path) of f(c(t)) ||c'(t)|| dt, where ||c'(t)|| is the length of the derivative of c(t).. But a Line integral is completely different (sort of...). Here, F goes from R^3 -> R^3, not to just R. Then you need to do:. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now...". ## #3 2006-04-11 08:38:05. Ricky. Moderator. Offline. ### Re: Line Integral. If you tried to use both formulas on the same problem, then you either:. Tried to integrate a vector. or. Tried to take the dot product of a scalar.. Of course, neither makes sense.. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now...". ## #4 2006-04-11 09:21:14. Chemist. Member. Offline. ### Re: Line Integral. Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?. Tried to integrate a vector. or. Tried to take the dot product of a scalar.. Of course, neither makes sense.. Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #5 2006-04-11 11:07:04. Chemist. Member. Offline. ### Re: Line Integral. If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering .... "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #6 2006-04-11 11:12:25. Ricky. Moderator. Offline. ### Re: Line Integral. Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?. Well, take the example: f(x, y, z) = x + y + z. That's from R³⇒R. You could define it from R³⇒R³ by:. f(x, y, z) = (x + y + z, 0, 0). But these are two completely different things with completely different properties. They look the same, but they aren't.. Is this what you are talking about?. Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?. They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.. If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now...". ## #7 2006-04-11 15:02:10. George,Y. Super Member. Offline. ### Re: Line Integral. The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.. The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity.
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i cannot figure out what the product means.. So you'd better use the second defaultly.. So far, i haven't encountered the situation to use your first integral.. X'(y-Xβ)=0. ## #8 2006-04-11 19:34:48. Chemist. Member. Offline. ### Re: Line Integral. Well, take the example: f(x, y, z) = x + y + z. That's from R³⇒R. You could define it from R³⇒R³ by:. f(x, y, z) = (x + y + z, 0, 0). I'm not sure what this is, what's the 0,0 for?. But these are two completely different things with completely different properties. They look the same, but they aren't.. Is this what you are talking about?. No , I was simply saying that if we have a scalar function f(x,y,z) = x +y +z , we can change it to a vector field or to its total differential form. This is what I did in calculating the path integral of a function f, I changed it to its vector form and calculated the scalar product.. They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.. If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.. Thanks for the explanation.. If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering .... I think I may be right here. I was just solving some problems, it turns out that whenever the total diff form of a vector field is exact ( rotor = 0 ), the line integral is independent of the path, and I can simply compute it F(B) - F(A) if we assume the line integral is from A to B across an arc. This really makes the computation easy for a line integral.. But if the total diff form is exact, continuous and its 1st derivative is continuous too, the line integral in a closed domain would be zero. Why's that?. And can anyone prove Green-Rieman's formula? I have an idea but I think it's incorrect.. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #9 2006-04-11 19:36:37. Chemist. Member. Offline. ### Re: Line Integral. #### George,Y wrote:. The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.. The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity. i cannot figure out what the product means.. So you'd better use the second defaultly.. So far, i haven't encountered the situation to use your first integral.. Thanks for the info , I think I got the difference between the two. And yes |V(t)| is the velocity magnitude or speed.. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #10 2006-04-12 07:54:48. Chemist. Member. Offline. ### Re: Line Integral. We can obtain Green's theorem area formula by going backward :. A = ∫ ∫ dydx = ∫ ∫ ( 1/2 + 1/2 ) dydx = 1/2 ∫ ( xdy - ydx ). , ofcourse the domain is closed.. But it seems sometimes we can compute the area directly from A = ∫ xdy. Example:. Determine the area b/w y=x2 and y = x across C+. A = 1/2 ( ∫ xdy - ydx ). I divided the line integral into arc OA and a straight line OA where A (1,1), the point of intersection of the parabola and the straight line, and O is the origin ofcourse.. I chose the parametirc equations,. For [OA] : y = x , x=x 0 ≤ x ≤ 1. For arc OA : x=y2 , y=y 1 ≤ y ≤ 0. I1 = 1/2 ∫ (x-x)dx = 0 ----> for [OA]. I2 = 1/2 ∫ (y2 - 2y2)dy = 1/6 ---> for arc OA. A = I1 + I2 = 1/6. Now notice that the area could have been calculated directly from:. A = ∫ xdy ( line int across C+ ) = ∫ xdy ( across [OA] ) + ∫ xdy ( across arc OA ). Using the same variables, A = 1/6. The second way is quicker, but I can't seem to figure out when to use it.. "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg. ## #11 2006-04-12 12:45:06. George,Y. Super Member. Offline. ### Re: Line Integral. to compute the area enclosed by an ellipse.. X'(y-Xβ)=0. ## #12 2006-04-12 23:40:24. Chemist. Member. Offline. ### Re: Line Integral. You mean the area A = 1/2 ∫ ( xdy - ydx ) can be simplified to A = ∫ xdy in the case of an ellipse? It was correct for the closed region between x=y2 and y=x. There was no ellipse there .... "Fundamentally one will never be able to renounce abstraction.". Werner Heisenberg.
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Pre-Calc Homework Solutions 92
# Pre-Calc Homework Solutions 92 - 92 Section 3.6 d sin 2x dx...
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28. Continuous: Note that g (0) 5 lim x 0 1 g ( x ) 5 lim x 0 1 cos x 5 cos (0) 5 1, and lim x 0 2 g ( x ) 5 lim x 0 2 ( x 1 b ) 5 b . We require lim x 0 2 g ( x ) 5 g (0), so b 5 1. The function is continuous if b 5 1. Differentiable: For b 5 1, the left-hand derivative is 1 and the right-hand derivative is 2 sin (0) 5 0, so the function is not differentiable. For other values of b , the function is discontinuous at x 5 0 and there is no left-hand derivative. So, there is no value of b that will make the function differentiable at x 5 0. 29. Observe the pattern: } d d x } cos x 5 2 sin x } d d x 5 5 } cos x 5 2 sin x } d d x 2 2 } cos x 5 2 cos x } d d x 6 6 } cos x 5 2 cos x } d d x 3 3 } cos x 5 sin x } d d x 7 7 } cos x 5 sin x } d d x 4 4 } cos x 5 cos x } d d x 8 8 } cos x 5 cos x Continuing the pattern, we see that } d d x n n } cos x 5 sin x when n 5 4 k 1 3 for any whole number k . Since 999 5 4(249) 1 3, } d d x 9 9 9 9 9 9 } cos x 5 sin x . 30. Observe the pattern: } d d x } sin x 5 cos x } d d x 5 5 } sin x 5 cos x } d d x 2 2 } sin x 5 2 sin x } d d x 6 6 } sin x 5 2 sin x } d
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{[ promptMessage ]}. Bookmark it. {[ promptMessage ]}. Pre-Calc Homework Solutions 92. # Pre-Calc Homework Solutions 92 - 92 Section 3.6 d sin 2x dx.... This preview shows page 1. Sign up to view the full content.. 28. Continuous: Note that g (0) 5 lim x 0 1 g ( x ) 5 lim x 0 1 cos x 5 cos (0) 5 1, and lim x 0 2 g ( x ) 5 lim x 0 2 ( x 1 b ) 5 b . We require lim x 0 2 g ( x ) 5 g (0), so b 5 1. The function is continuous if b 5 1. Differentiable: For b 5 1, the left-hand derivative is 1 and the right-hand derivative is 2 sin (0) 5 0, so the function is not differentiable.
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For other values of b , the function is discontinuous at x 5 0 and there is no left-hand derivative. So, there is no value of b that will make the function differentiable at x 5 0. 29. Observe the pattern: } d d x } cos x 5 2 sin x } d d x 5 5 } cos x 5 2 sin x } d d x 2 2 } cos x 5 2 cos x } d d x 6 6 } cos x 5 2 cos x } d d x 3 3 } cos x 5 sin x } d d x 7 7 } cos x 5 sin x } d d x 4 4 } cos x 5 cos x } d d x 8 8 } cos x 5 cos x Continuing the pattern, we see that } d d x n n } cos x 5 sin x when n 5 4 k 1 3 for any whole number k . Since 999 5 4(249) 1 3, } d d x 9 9 9 9 9 9 } cos x 5 sin x . 30. Observe the pattern: } d d x } sin x 5 cos x } d d x 5 5 } sin x 5 cos x } d d x 2 2 } sin x 5 2 sin x } d d x 6 6 } sin x 5 2 sin x } d. This is the end of the preview. Sign up to access the rest of the document.. {[ snackBarMessage ]}. Ask a homework question - tutors are online.
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# Multi-step Problems
## Solve multi-step problems in contexts, deciding which operations and methods to use and why.
##### Menu Level 1Level 2Level 3More Arithmetic
This is level 2; Multiplication and division multi-step problems. You can earn a trophy if you get at least 7 questions correct.
1. I have 6 crayons which are each 17cm long. I also have a ruler which is 31cm long. How far would they reach if I arranged them on the floor in a long line? Working: cm 2. The angles inside an irregular hexagon add up to 720°. If three of the angles are 119° and two of them are 132° what is the size of the other angle? Working: ° 3. Frank has two boxes of screws. The first box contains 58 screws and the second box contains 170 screws. If he needs four screws for every model he makes, how many models can he make with the screws he has? Working: 4. Using the most efficient method you can think of multiply together the numbers from 3 to 8 then divide the product by seven? Working: 5. Farmer Brown has two fields of sheep. The first field contains 17 sheep and the second field contains 64 sheep. He can only transport the sheep to the shearing station nine at a time. What is the minimum number of journeys he needs to make? Working: 6. A 50p coin weighs 8 grams. What is the value in pounds of two kilograms of 50p coins? Working: £ 7. If four soldiers can stand on one square metre. How many soldiers could stand on a ractanguler parade ground measuting 50 metres long and 20 metres wide? Working: 8. Six people share the cost of a meal in a restaurant. The total cost of the starters was £32, The total cost of the main courses was £34 and the total cost of the deserts was £78. How much does each person pay? Working: £ 9. Ayden makes Christmas crackers. Each cracker contains a toy, 2 sweets and a party hat. If each toy costs 46p, each sweet costs 8p and each hat costs 25p, how much change will he have from £50 if he makes 12 crackers? Working: £ 10. Six security guards are deciding on the times they should be on duty so that one of them is guarding the bank at every moment of every day of the week. They multiply the number of days in a week by the number hours in a day then divide the result by six. How many hours will each guard have to work so that between them the whole week is covered? Working: hours
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# Multi-step Problems. ## Solve multi-step problems in contexts, deciding which operations and methods to use and why.. ##### Menu Level 1Level 2Level 3More Arithmetic. This is level 2; Multiplication and division multi-step problems. You can earn a trophy if you get at least 7 questions correct.. 1. I have 6 crayons which are each 17cm long. I also have a ruler which is 31cm long. How far would they reach if I arranged them on the floor in a long line? Working: cm 2. The angles inside an irregular hexagon add up to 720°. If three of the angles are 119° and two of them are 132° what is the size of the other angle? Working: ° 3. Frank has two boxes of screws. The first box contains 58 screws and the second box contains 170 screws. If he needs four screws for every model he makes, how many models can he make with the screws he has? Working: 4. Using the most efficient method you can think of multiply together the numbers from 3 to 8 then divide the product by seven? Working: 5. Farmer Brown has two fields of sheep. The first field contains 17 sheep and the second field contains 64 sheep. He can only transport the sheep to the shearing station nine at a time. What is the minimum number of journeys he needs to make? Working: 6. A 50p coin weighs 8 grams. What is the value in pounds of two kilograms of 50p coins? Working: £ 7. If four soldiers can stand on one square metre. How many soldiers could stand on a ractanguler parade ground measuting 50 metres long and 20 metres wide? Working: 8. Six people share the cost of a meal in a restaurant. The total cost of the starters was £32, The total cost of the main courses was £34 and the total cost of the deserts was £78. How much does each person pay? Working: £ 9. Ayden makes Christmas crackers. Each cracker contains a toy, 2 sweets and a party hat. If each toy costs 46p, each sweet costs 8p and each hat costs 25p, how much change will he have from £50 if he makes 12 crackers? Working: £ 10. Six security guards are deciding on the times they should be on duty so that one of them is guarding the bank at every moment of every day of the week. They multiply the number of days in a week by the number hours in a day then divide the result by six. How many hours will each guard have to work so that between them the whole week is covered? Working: hours. Check. This is Multi-step Problems level 2. You can also try:. Level 1 Level 3. ## Instructions. Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.. When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file.. ## More Activities:. Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician?. Comment recorded on the 14 October 'Starter of the Day' page by Inger Kisby, Herts and Essex High School:. "Just a quick note to say that we use a lot of your starters. It is lovely to have so many different ideas to start a lesson with.
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Thank you very much and keep up the good work.". Comment recorded on the 28 September 'Starter of the Day' page by Malcolm P, Dorset:. "A set of real life savers!!. Keep it up and thank you!". Each month a newsletter is published containing details of the new additions to the Transum website and a new puzzle of the month.. The newsletter is then duplicated as a podcast which is available on the major delivery networks. You can listen to the podcast while you are commuting, exercising or relaxing.. Transum breaking news is available on Twitter @Transum and if that's not enough there is also a Transum Facebook page.. #### Beat The Clock. It is a race against the clock to answer 30 mental arithmetic questions. There are nine levels to choose from to suit pupils of different abilities.. There are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer.. A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves.. Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members.. If you would like to enjoy ad-free access to the thousands of Transum resources, receive our monthly newsletter, unlock the printable worksheets and see our Maths Lesson Finishers then sign up for a subscription now:. Subscribe. ## Go Maths. Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths page is an alphabetical list of free activities designed for students in Secondary/High school.. ## Maths Map. Are you looking for something specific? An exercise to supplement the topic you are studying at school at the moment perhaps. Navigate using our Maths Map to find exercises, puzzles and Maths lesson starters grouped by topic.. ## Teachers. If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:. Alternatively, if you use Google Classroom, all you have to do is click on the green icon below in order to add this activity to one of your classes.. It may be worth remembering that if Transum.org should go offline for whatever reason, there is a mirror site at Transum.info that contains most of the resources that are available here on Transum.org.. When planning to use technology in your lesson always have a plan B!. Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.. For Students:. For All:. ## Description of Levels. Close. Level 1 - Addition and subtraction multi-step problems. Level 2 - Multiplication and division multi-step problems. Level 3 - Mixed multi-step problems. More Arithmetic including lesson Starters, visual aids, investigations and self-marking exercises.. Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.. ## Curriculum Reference. See the National Curriculum page for links to related online activities and resources.
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PROBLEM 09 – 0359: A uniform circular disk of radius 25 cm is pivoted at a point
20 cm from its center and allowed to oscillate in a vertical
plane under the action of gravity. What is the period of its
small oscillations? Take g as π2 m.s−2.
Solution: The moment of inertia of a uniform circular disk of radius R and
mass M about an axis through its center perpendicular to its plane is
1/2 MR2. By the parallel-axes theorem, the moment of inertia about a
parallel axis a distance h from the first is
I = 1/2 MR2 + Mh2.
The disk is acting as a physical pendulum, and hence its period for
small oscillations is given by
T = 2π √(1/Mgh)
where I is the moment of inertia of the physical pendulum about its axis of
suspension. Hence,
T = 2π √[(1/2 MR2 + Mh2)/Mgh]
= 2π √[(1/2 × 0.252 m2 + 0.022 m2)/(π2 m∙s−2 × 0.2 m)]
= 2 √(0.35625 s)
= 1.193 s.
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PROBLEM 09 – 0359: A uniform circular disk of radius 25 cm is pivoted at a point. 20 cm from its center and allowed to oscillate in a vertical. plane under the action of gravity. What is the period of its. small oscillations? Take g as π2 m.s−2.. Solution: The moment of inertia of a uniform circular disk of radius R and. mass M about an axis through its center perpendicular to its plane is. 1/2 MR2. By the parallel-axes theorem, the moment of inertia about a. parallel axis a distance h from the first is. I = 1/2 MR2 + Mh2.
|
The disk is acting as a physical pendulum, and hence its period for. small oscillations is given by. T = 2π √(1/Mgh). where I is the moment of inertia of the physical pendulum about its axis of. suspension. Hence,. T = 2π √[(1/2 MR2 + Mh2)/Mgh]. = 2π √[(1/2 × 0.252 m2 + 0.022 m2)/(π2 m∙s−2 × 0.2 m)]. = 2 √(0.35625 s). = 1.193 s.
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https://astronomy.stackexchange.com/questions/53152/how-much-of-the-variance-of-matter-is-due-to-scales
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# How much of the variance of matter is due to scales
Imagine a Universe in which the matter power spectrum behaves as $$\mathcal{P}\sim k^{0.8}$$
How much of the variance (not variance squared!) of matter is due to scales around $$k=k_1=10^{2.9}\mathrm{Mpc}^{-1}$$, versus scales around $$k=k_2=10^{-3.0}\mathrm{Mpc}^{-1}$$
To be explicit, how much is: $$\frac{\sigma(k\sim k_1)}{\sigma(k\sim k_2)}?$$
• I'd recommend formatting this with latex, currently, with images cutting through the text, it looks very confusing.. Mar 14, 2023 at 17:07
• Please explain some of the magic numbers here. Why 0.8? Why 2.9? Why -3.0? Mar 14, 2023 at 18:08
• They are random values assigned, nothing theoretical just to obtain certain value.
– Alba
Mar 14, 2023 at 18:38
I'll assume $$\mathcal{P}(k)$$ is the "dimensionless" power spectrum (also known as $$\Delta^2(k)$$), i.e. $$\mathcal{P}(k)\equiv k^3/(2\pi^2) P(k)$$. The density variance is $$\sigma^2=\int_0^\infty \frac{\mathrm{d}k}{k}\mathcal{P}(k).\tag{1}\label{1}$$
"How much of $$\sigma^2$$ is contributed by scales around $$k$$" is ambiguous, but the most natural interpretation (to me) is "how much of $$\sigma^2$$ is contributed by a logarithmic interval around $$k$$", in which case the answer is just (proportional to) the integrand, $$\mathcal{P}(k)$$. To see this explicitly, we can define $$\sigma^2(k)=\int_0^k\frac{\mathrm{d}k^\prime}{k^\prime}\mathcal{P}(k^\prime)\tag{2}\label{2}$$ as the variance contributed by wavenumbers smaller than $$k$$ (i.e. scales larger than $$k$$). Then the differential variance contributed by scales near $$k$$, per logarithmic interval in $$k$$, is $$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma^2(k)=\mathcal{P}(k).\tag{3}\label{3}$$
Now, you are not asking about $$\sigma^2$$ but about $$\sigma$$. In this case there is further ambiguity to the question, because $$\sigma$$ is no longer simply a sum over the power contributed at different scales and is instead a nonlinear function thereof. In general, the answer to "how much rms variance is contributed by scales near $$k$$" depends on the order in which you want to add up the power at different scales. If we add up power in order from the largest scales (smallest $$k$$) to the smallest scales (largest $$k$$), as in equation \eqref{2}, then the answer is $$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma(k) =\frac{\mathrm{d}}{\mathrm{d}\log k}[\sigma^2(k)]^{1/2} =\frac{1}{2}[\sigma^2(k)]^{-1/2}\frac{\mathrm{d}\sigma^2(k)}{\mathrm{d}\log k} =\frac{\mathcal{P}(k)}{2\sigma(k)}.\tag{4}\label{4}$$ To add up the scales in a different order, just reorder the integral in equation \eqref{2}; equation \eqref{4} remains accurate with the redefined $$\sigma(k)$$.
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# How much of the variance of matter is due to scales. Imagine a Universe in which the matter power spectrum behaves as $$\mathcal{P}\sim k^{0.8}$$. How much of the variance (not variance squared!) of matter is due to scales around $$k=k_1=10^{2.9}\mathrm{Mpc}^{-1}$$, versus scales around $$k=k_2=10^{-3.0}\mathrm{Mpc}^{-1}$$. To be explicit, how much is: $$\frac{\sigma(k\sim k_1)}{\sigma(k\sim k_2)}?$$. • I'd recommend formatting this with latex, currently, with images cutting through the text, it looks very confusing.. Mar 14, 2023 at 17:07. • Please explain some of the magic numbers here. Why 0.8? Why 2.9? Why -3.0? Mar 14, 2023 at 18:08. • They are random values assigned, nothing theoretical just to obtain certain value.. – Alba.
|
Mar 14, 2023 at 18:38. I'll assume $$\mathcal{P}(k)$$ is the "dimensionless" power spectrum (also known as $$\Delta^2(k)$$), i.e. $$\mathcal{P}(k)\equiv k^3/(2\pi^2) P(k)$$. The density variance is $$\sigma^2=\int_0^\infty \frac{\mathrm{d}k}{k}\mathcal{P}(k).\tag{1}\label{1}$$. "How much of $$\sigma^2$$ is contributed by scales around $$k$$" is ambiguous, but the most natural interpretation (to me) is "how much of $$\sigma^2$$ is contributed by a logarithmic interval around $$k$$", in which case the answer is just (proportional to) the integrand, $$\mathcal{P}(k)$$. To see this explicitly, we can define $$\sigma^2(k)=\int_0^k\frac{\mathrm{d}k^\prime}{k^\prime}\mathcal{P}(k^\prime)\tag{2}\label{2}$$ as the variance contributed by wavenumbers smaller than $$k$$ (i.e. scales larger than $$k$$). Then the differential variance contributed by scales near $$k$$, per logarithmic interval in $$k$$, is $$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma^2(k)=\mathcal{P}(k).\tag{3}\label{3}$$. Now, you are not asking about $$\sigma^2$$ but about $$\sigma$$. In this case there is further ambiguity to the question, because $$\sigma$$ is no longer simply a sum over the power contributed at different scales and is instead a nonlinear function thereof. In general, the answer to "how much rms variance is contributed by scales near $$k$$" depends on the order in which you want to add up the power at different scales. If we add up power in order from the largest scales (smallest $$k$$) to the smallest scales (largest $$k$$), as in equation \eqref{2}, then the answer is $$\frac{\mathrm{d}}{\mathrm{d}\log k}\sigma(k) =\frac{\mathrm{d}}{\mathrm{d}\log k}[\sigma^2(k)]^{1/2} =\frac{1}{2}[\sigma^2(k)]^{-1/2}\frac{\mathrm{d}\sigma^2(k)}{\mathrm{d}\log k} =\frac{\mathcal{P}(k)}{2\sigma(k)}.\tag{4}\label{4}$$ To add up the scales in a different order, just reorder the integral in equation \eqref{2}; equation \eqref{4} remains accurate with the redefined $$\sigma(k)$$.
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https://calculus.subwiki.org/w/index.php?title=Sinc_function&oldid=832
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# Sinc function
This article is about a particular function from a subset of the real numbers to the real numbers. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article.
View a complete list of particular functions on this wiki
For functions involving angles (trigonometric functions, inverse trigonometric functions, etc.) we follow the convention that all angles are measured in radians. Thus, for instance, the angle of $90\,^\circ$ is measured as $\pi/2$.
## Definition
This function, denoted $\operatorname{sinc}$, is defined as follows:
$\operatorname{sinc} x := \left\lbrace \begin{array}{rl} 1, & x = 0 \\ \frac{\sin x}{x} & x \ne 0\end{array}\right.$
## Key data
Item Value
default domain all real numbers, i.e., all of $\R$.
range the closed interval $[\alpha,1]$ where $\alpha$ is approximately $-2/(3\pi)$.
period none; the function is not periodic
horizontal asymptotes $y = 0$, i.e., the $x$-axis. This is because $\lim_{x \to \pm \infty} \frac{\sin x}{x} = 0$, which in turn can be deduced from the fact that the numerator is bounded while the magnitude of the denominator approaches $\infty$.
local maximum values and points of attainment The local maxima occur at points $x$ satisfying $\tan x = x$ and $x \in [2n\pi,(2n + 1)\pi]$ or $x \in [-(2n + 1)\pi,-2n\pi]$ for $n$ a positive integer.
There is an anomalous local maximum at $x = 0$ with value 1. Apart from that, the other local maxima occur at points of the form $\pm(2n\pi + \alpha_n)$ where $\alpha_n$ is fairly close to $\pi/2$ for all $n > 0$. The local maximum value at this point is slightly more than $1/(2n\pi + \pi/2)$.
local minimum values and points of attainment The local minima occur at points $x$ satisfying $\tan x = -x$ and $x \in [(2n - 1)\pi,2n\pi]$ or $x \in [-2n\pi,-(2n - 1)\pi]$ for $n$ a positive integer.
The local minima occur at points of the form $\pm(2n\pi - \alpha_n)$ where $\alpha_n$ is fairly close to $\pi/2$ for all $n > 0$. The local minimum value at this point is slightly less than $-1/(2n\pi - \pi/2)$.
points of inflection Fill this in later
important symmetries The function is an even function, i.e., the graph has symmetry about the $y$-axis.
derivative $\operatorname{sinc}'x = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2}, & x \ne 0\\\end{array}\right.$
antiderivative the sine integral (this is defined as the antiderivative of the sinc function that takes the value 0 at 0)
power series and Taylor series The power series about 0 (which is also the Taylor series) is
$\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k + 1)!} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \dots$
The power series converges globally to the function.
## Graph
Below is a graph of the function for the domain restricted to $[-3\pi,3\pi]$:
The picture is a little unclear, so we consider an alternative depiction of the graph where the $x$-axis and $y$-axis are scaled differently to make it clearer:
## Differentation
### First derivative
Computation at $x = 0$:
WARNING ON ERRONEOUS DIFFERENTIATION APPROACH: Suppose a function is given a separate definition at an isolated point from the definition at points surrounding it (on its immediate left or immediate right). To compute the derivative, it is not correct to simply differentiate the definition at the point and arrive at the derivative. Rather, we need to compute the derivative from first principles as a limit of a difference quotient, where the function value at the point is taken as the specified value and the function value at nearby points is given by the expression used to define the function around the point.
Here, we need to compute the derivative using first principles, as a limit of a difference quotient:
$\operatorname{sinc}' 0 := \lim_{x \to 0} \frac{\operatorname{sinc} x - \operatorname{sinc} 0}{x - 0} = \lim_{x \to 0} \frac{(\sin x)/x - 1}{x} = \lim_{x \to 0} \frac{\sin x - x}{x^2}$
This limit can be computed in many ways. For instance, it can be computed using the L'Hopital rule:
$\lim_{x \to 0} \frac{\sin x - x}{x^2} \stackrel{*}{=} \lim_{x \to 0} \frac{\cos x - 1}{2x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\sin x}{2} = \frac{-\sin 0}{2} = 0$
DERIVATIVE OF EVEN FUNCTION AT ZERO: For an even function, if the function is differentiable at zero, then the derivative at zero is zero. However, it is possible for an even function to not be differentiable at zero, so we cannot directly conclude merely from the function being even that its derivative at zero is zero.
Computation at $x \ne 0$: At any such point, we know that the function looks like $(\sin x)/x$ in an open interval about the point, so we can use the quotient rule for differentiation, which states that:
$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$
In this case, $f(x) = \sin x$ and $g(x) = x$, so we get:
$\frac{d}{dx}(\operatorname{sinc}\ x) = \frac{x \cos x - \sin x}{x^2}$
Combining the two computations, we get:
$\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2}, & x \ne 0 \\\end{array}\right.$
### Second derivative
Computation at $x = 0$:
WARNING ON ERRONEOUS DIFFERENTIATION APPROACH: Suppose a function is given a separate definition at an isolated point from the definition at points surrounding it (on its immediate left or immediate right). To compute the derivative, it is not correct to simply differentiate the definition at the point and arrive at the derivative. Rather, we need to compute the derivative from first principles as a limit of a difference quotient, where the function value at the point is taken as the specified value and the function value at nearby points is given by the expression used to define the function around the point.
Here, we need to compute the derivative using first principles, as a limit of a difference quotient:
$\operatorname{sinc}'' 0 := \lim_{x \to 0} \frac{\operatorname{sinc}' x - \operatorname{sinc}' 0}{x - 0} = \lim_{x \to 0} \frac{(x \cos x - \sin x)/x^2 - 0}{x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{x^3}$
The latter limit can be computed in many ways. One way is to use the L'Hopital rule:
$\lim_{x \to 0} \frac{x \cos x - \sin x}{x^3} \stackrel{*}{=} \lim_{x \to 0} \frac{-x \sin x + \cos x - \cos x}{3x^2} = \lim_{x \to 0} \frac{-\sin x}{3x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\cos x}{3} = \frac{-1}{3}$
Thus, $\operatorname{sinc}'' 0 = -1/3$.
The computation at other points can be done using the usual method for computing derivatives, i.e., with the quotient rule for differentiation combined with other rules. Overall, we get:
$\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} \frac{-1}{3}, & x = 0 \\ \frac{(2 - x^2)\sin x - 2x\cos x}{x^3}, & x \ne 0 \\\end{array}\right.$
## Taylor series and power series
### Computation of power series
We use the power series for the sine function (see sine function#Computation of power series):
$\! \sin x := x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k + 1)!}$
Dividing both sides by $x$ (valid when $x \ne 0$), we get:
$\! \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k+1)!}$
We note that the power series also works at $x = 0$ (because $\operatorname{sinc} \ 0 = 1$), hence it works globally, and is the power series for the sinc function.
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# Sinc function. This article is about a particular function from a subset of the real numbers to the real numbers. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article.. View a complete list of particular functions on this wiki. For functions involving angles (trigonometric functions, inverse trigonometric functions, etc.) we follow the convention that all angles are measured in radians. Thus, for instance, the angle of $90\,^\circ$ is measured as $\pi/2$.. ## Definition. This function, denoted $\operatorname{sinc}$, is defined as follows:. $\operatorname{sinc} x := \left\lbrace \begin{array}{rl} 1, & x = 0 \\ \frac{\sin x}{x} & x \ne 0\end{array}\right.$. ## Key data. Item Value. default domain all real numbers, i.e., all of $\R$.. range the closed interval $[\alpha,1]$ where $\alpha$ is approximately $-2/(3\pi)$.. period none; the function is not periodic. horizontal asymptotes $y = 0$, i.e., the $x$-axis. This is because $\lim_{x \to \pm \infty} \frac{\sin x}{x} = 0$, which in turn can be deduced from the fact that the numerator is bounded while the magnitude of the denominator approaches $\infty$.. local maximum values and points of attainment The local maxima occur at points $x$ satisfying $\tan x = x$ and $x \in [2n\pi,(2n + 1)\pi]$ or $x \in [-(2n + 1)\pi,-2n\pi]$ for $n$ a positive integer.. There is an anomalous local maximum at $x = 0$ with value 1. Apart from that, the other local maxima occur at points of the form $\pm(2n\pi + \alpha_n)$ where $\alpha_n$ is fairly close to $\pi/2$ for all $n > 0$. The local maximum value at this point is slightly more than $1/(2n\pi + \pi/2)$.. local minimum values and points of attainment The local minima occur at points $x$ satisfying $\tan x = -x$ and $x \in [(2n - 1)\pi,2n\pi]$ or $x \in [-2n\pi,-(2n - 1)\pi]$ for $n$ a positive integer.. The local minima occur at points of the form $\pm(2n\pi - \alpha_n)$ where $\alpha_n$ is fairly close to $\pi/2$ for all $n > 0$. The local minimum value at this point is slightly less than $-1/(2n\pi - \pi/2)$.. points of inflection Fill this in later. important symmetries The function is an even function, i.e., the graph has symmetry about the $y$-axis.. derivative $\operatorname{sinc}'x = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2}, & x \ne 0\\\end{array}\right.$. antiderivative the sine integral (this is defined as the antiderivative of the sinc function that takes the value 0 at 0). power series and Taylor series The power series about 0 (which is also the Taylor series) is. $\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k + 1)!} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \dots$. The power series converges globally to the function.. ## Graph. Below is a graph of the function for the domain restricted to $[-3\pi,3\pi]$:. The picture is a little unclear, so we consider an alternative depiction of the graph where the $x$-axis and $y$-axis are scaled differently to make it clearer:. ## Differentation. ### First derivative. Computation at $x = 0$:. WARNING ON ERRONEOUS DIFFERENTIATION APPROACH: Suppose a function is given a separate definition at an isolated point from the definition at points surrounding it (on its immediate left or immediate right). To compute the derivative, it is not correct to simply differentiate the definition at the point and arrive at the derivative.
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Rather, we need to compute the derivative from first principles as a limit of a difference quotient, where the function value at the point is taken as the specified value and the function value at nearby points is given by the expression used to define the function around the point.. Here, we need to compute the derivative using first principles, as a limit of a difference quotient:. $\operatorname{sinc}' 0 := \lim_{x \to 0} \frac{\operatorname{sinc} x - \operatorname{sinc} 0}{x - 0} = \lim_{x \to 0} \frac{(\sin x)/x - 1}{x} = \lim_{x \to 0} \frac{\sin x - x}{x^2}$. This limit can be computed in many ways. For instance, it can be computed using the L'Hopital rule:. $\lim_{x \to 0} \frac{\sin x - x}{x^2} \stackrel{*}{=} \lim_{x \to 0} \frac{\cos x - 1}{2x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\sin x}{2} = \frac{-\sin 0}{2} = 0$. DERIVATIVE OF EVEN FUNCTION AT ZERO: For an even function, if the function is differentiable at zero, then the derivative at zero is zero. However, it is possible for an even function to not be differentiable at zero, so we cannot directly conclude merely from the function being even that its derivative at zero is zero.. Computation at $x \ne 0$: At any such point, we know that the function looks like $(\sin x)/x$ in an open interval about the point, so we can use the quotient rule for differentiation, which states that:. $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$. In this case, $f(x) = \sin x$ and $g(x) = x$, so we get:. $\frac{d}{dx}(\operatorname{sinc}\ x) = \frac{x \cos x - \sin x}{x^2}$. Combining the two computations, we get:. $\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} 0, & x = 0 \\ \frac{x \cos x - \sin x}{x^2}, & x \ne 0 \\\end{array}\right.$. ### Second derivative. Computation at $x = 0$:. WARNING ON ERRONEOUS DIFFERENTIATION APPROACH: Suppose a function is given a separate definition at an isolated point from the definition at points surrounding it (on its immediate left or immediate right). To compute the derivative, it is not correct to simply differentiate the definition at the point and arrive at the derivative. Rather, we need to compute the derivative from first principles as a limit of a difference quotient, where the function value at the point is taken as the specified value and the function value at nearby points is given by the expression used to define the function around the point.. Here, we need to compute the derivative using first principles, as a limit of a difference quotient:. $\operatorname{sinc}'' 0 := \lim_{x \to 0} \frac{\operatorname{sinc}' x - \operatorname{sinc}' 0}{x - 0} = \lim_{x \to 0} \frac{(x \cos x - \sin x)/x^2 - 0}{x} = \lim_{x \to 0} \frac{x \cos x - \sin x}{x^3}$. The latter limit can be computed in many ways. One way is to use the L'Hopital rule:. $\lim_{x \to 0} \frac{x \cos x - \sin x}{x^3} \stackrel{*}{=} \lim_{x \to 0} \frac{-x \sin x + \cos x - \cos x}{3x^2} = \lim_{x \to 0} \frac{-\sin x}{3x} \stackrel{*}{=} \lim_{x \to 0} \frac{-\cos x}{3} = \frac{-1}{3}$. Thus, $\operatorname{sinc}'' 0 = -1/3$.. The computation at other points can be done using the usual method for computing derivatives, i.e., with the quotient rule for differentiation combined with other rules. Overall, we get:. $\operatorname{sinc}'(x) = \left\lbrace \begin{array}{rl} \frac{-1}{3}, & x = 0 \\ \frac{(2 - x^2)\sin x - 2x\cos x}{x^3}, & x \ne 0 \\\end{array}\right.$. ## Taylor series and power series. ### Computation of power series. We use the power series for the sine function (see sine function#Computation of power series):. $\! \sin x := x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k + 1)!}$. Dividing both sides by $x$ (valid when $x \ne 0$), we get:. $\! \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots = \sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{(2k+1)!}$. We note that the power series also works at $x = 0$ (because $\operatorname{sinc} \ 0 = 1$), hence it works globally, and is the power series for the sinc function.
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https://mathematica.stackexchange.com/questions/185596/finding-the-best-representation-of-a-numerically-inverted-function-via-interpola?noredirect=1
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# Finding the best representation of a numerically-inverted function via InterpolatingPolynomial and/or variations
Below is the routine I am using to sort of represent the numerically-inverted function TP. Basically I am finding a necessary interpolating polynomial TPint that fits with the data points given by TPtab, so that I can represent TP as TPint and use in my other calculations. I just wonder, is this by far the best and proper way to find a polynomial that fits my data points to properly represent my numerically-inverted function? The output looks like a gross. Is there other routine wherein the output is compact and yet the accuracy of the function will not be compromised?
Thank you.
b = 1;
q = -1;
rat = 10^-30;
rho[r_,b_,q_]:=(2b/(1-q))(1-(b/r)^(1-q))^(1/2)Hypergeometric2F1[1/2,1-1/(q-1),3/2,1-(b/r)^(1-q)]
TP=Rationalize[InverseFunction[Function[{r,b,q},rho[r,b,q]],1,3]];
TPtab=Table[{ρ,TP[ρ,b,q]},{ρ,0,1,.1}]
TPint=Rationalize[InterpolatingPolynomial[TPtab,ρ],rat]//Simplify//Expand
Why not use NDSolveValue to compute an interpolating function of the inverse? Basically, given $$y = f(x)$$ we want to find $$x = f^{-1}(y)$$. To do this differentiate $$y = f(x(y))$$ with respect to $$y$$ to obtain $$1 = f'(x(y)) x'(y)$$
Hence, the ODE to be solved is: $$x'(y) = 1/f'(x(y))$$
and the following NDSolveValue call should produce your desired result:
if = NDSolveValue[
{x'[y] == 1/Derivative[1,0,0][rho][x[y], b, q], x[1] == Sqrt[2]},
x,
{y, 0, 1}
];
(where I used rho[Sqrt[2], b, q] == 1)
Visualization:
GraphicsRow[{Plot[if[ρ], {ρ, 0, 1}], Plot[TPint, {ρ, 0, 1}]}]
• That's fine also. But my problem is that further in my calculations, i am to use inverse function which is an interpolating function. So how would I able to execute algebraic calculations involving an interpolating function? That's basically my problem, that's why i need to find for an interpolating polynomial to best represent my inverse function. Commented Nov 9, 2018 at 0:26
• Notice that the output that TPint gives is a polynomial of certain degree. Is there other way to obtain a reasonable fit with corresponding polynomial to the inverse function? Commented Nov 9, 2018 at 0:36
• @user583893 What calculations are you doing that need a polynomial approximation to the inverse? Perhaps there's a way to modify those calculations to work with an interpolation function instead. Also, note that approximations using InterpolatingPolynomial can be very bad, see the Possible issues section in the documentation for InterpolatingPolynomial. Commented Nov 9, 2018 at 0:40
Interpolation over a regular grid has known problems. Over the Chebyshev extreme grid, however, there no such problems in interpolating smooth functions, or even continuous functions although convergence can be slow. A Chebyshev interpolation will be near minimax (best possible with respect to the infinity norm) for a smooth function. With machine-precision input, a near machine-precision approximation is possible. With higher precision input, higher precision approximations are possible.
I was introduced to this approach here and one can use adaptiveChebSeries to automatically determine the degree of the Chebyshev interpolation needed to approximate a function within a given error. For more, see Trefethen, Approximation Theory and Approximation Practice and Boyd, Solving Transcendental Equations. There are two methods of implementation, barycentric interpolation (see Berrut & Trefethen (2004) for its advantages) and Chebyshev series. Both methods are generally numerically better than an interpolation based on a power basis expansion. In this case, the difference is negligible.
ClearAll[rho, TP];
b = 1;
q = -1;
rat = 10^-30;
rho[r_, b_,
q_] := (2 b/(1 - q)) (1 - (b/r)^(1 - q))^(1/2) Hypergeometric2F1[1/2,
1 - 1/(q - 1), 3/2, 1 - (b/r)^(1 - q)];
TP[ρ_?NumericQ, b_?NumericQ, q_?NumericQ] :=
Block[{r}, With[{ρ0 = SetPrecision[ρ, 32]},
r /. FindRoot[rho[r, b, q] == ρ0, {r, 1001/1000},
WorkingPrecision -> 32] // Re]];
(* Chebyshev nodes and points *)
deg = 24;
chebnodes = N[Rescale[Sin[Pi/2 Range[-deg, deg, 2]/deg]], 32];
TPtab = Table[{ρ, TP[ρ, b, q]}, {ρ, chebnodes}];
(* Barycentric interpolation *)
rif = StatisticsLibraryBarycentricInterpolation[N@TPtab[[All, 1]],
N@TPtab[[All, 2]],
"Weights" ->
ReplacePart[
Table[(-1)^k, {k, 0, Length@TPtab - 1}], {1 -> 1/2, -1 -> 1/2}]];
Plot[rho[rif[ρ], b, q] - ρ // RealExponent, {ρ, 0, 1}]
(* Power basis interpolation *)
(* To diminish round-off error, increase precision *)
TPint = SetPrecision[InterpolatingPolynomial[TPtab, ρ], 50] // Expand // N
Plot[rho[N@TPint, b, q] - ρ // RealExponent, {ρ, 0, 1}]
(* Chebyshev series coefficients via FFT/DCT *)
cc = Sqrt[2/(Length@TPtab - 1)] FourierDCT[Reverse@TPtab[[All, 2]], 1];
cc[[{1, -1}]] /= 2;
rCS[ρ_] := cc.Cos[Range[0, Length@cc - 1] ArcCos[2 ρ - 1]];
Plot[rho[rCS[ρ], b, q] - ρ // RealExponent, {ρ, 0, 1}]
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# Finding the best representation of a numerically-inverted function via InterpolatingPolynomial and/or variations. Below is the routine I am using to sort of represent the numerically-inverted function TP. Basically I am finding a necessary interpolating polynomial TPint that fits with the data points given by TPtab, so that I can represent TP as TPint and use in my other calculations. I just wonder, is this by far the best and proper way to find a polynomial that fits my data points to properly represent my numerically-inverted function? The output looks like a gross. Is there other routine wherein the output is compact and yet the accuracy of the function will not be compromised?. Thank you.. b = 1;. q = -1;. rat = 10^-30;. rho[r_,b_,q_]:=(2b/(1-q))(1-(b/r)^(1-q))^(1/2)Hypergeometric2F1[1/2,1-1/(q-1),3/2,1-(b/r)^(1-q)]. TP=Rationalize[InverseFunction[Function[{r,b,q},rho[r,b,q]],1,3]];. TPtab=Table[{ρ,TP[ρ,b,q]},{ρ,0,1,.1}]. TPint=Rationalize[InterpolatingPolynomial[TPtab,ρ],rat]//Simplify//Expand. Why not use NDSolveValue to compute an interpolating function of the inverse? Basically, given $$y = f(x)$$ we want to find $$x = f^{-1}(y)$$. To do this differentiate $$y = f(x(y))$$ with respect to $$y$$ to obtain $$1 = f'(x(y)) x'(y)$$. Hence, the ODE to be solved is: $$x'(y) = 1/f'(x(y))$$. and the following NDSolveValue call should produce your desired result:. if = NDSolveValue[. {x'[y] == 1/Derivative[1,0,0][rho][x[y], b, q], x[1] == Sqrt[2]},. x,. {y, 0, 1}. ];. (where I used rho[Sqrt[2], b, q] == 1). Visualization:. GraphicsRow[{Plot[if[ρ], {ρ, 0, 1}], Plot[TPint, {ρ, 0, 1}]}]. • That's fine also. But my problem is that further in my calculations, i am to use inverse function which is an interpolating function. So how would I able to execute algebraic calculations involving an interpolating function? That's basically my problem, that's why i need to find for an interpolating polynomial to best represent my inverse function. Commented Nov 9, 2018 at 0:26. • Notice that the output that TPint gives is a polynomial of certain degree. Is there other way to obtain a reasonable fit with corresponding polynomial to the inverse function? Commented Nov 9, 2018 at 0:36. • @user583893 What calculations are you doing that need a polynomial approximation to the inverse? Perhaps there's a way to modify those calculations to work with an interpolation function instead. Also, note that approximations using InterpolatingPolynomial can be very bad, see the Possible issues section in the documentation for InterpolatingPolynomial. Commented Nov 9, 2018 at 0:40. Interpolation over a regular grid has known problems. Over the Chebyshev extreme grid, however, there no such problems in interpolating smooth functions, or even continuous functions although convergence can be slow. A Chebyshev interpolation will be near minimax (best possible with respect to the infinity norm) for a smooth function. With machine-precision input, a near machine-precision approximation is possible. With higher precision input, higher precision approximations are possible.
|
I was introduced to this approach here and one can use adaptiveChebSeries to automatically determine the degree of the Chebyshev interpolation needed to approximate a function within a given error. For more, see Trefethen, Approximation Theory and Approximation Practice and Boyd, Solving Transcendental Equations. There are two methods of implementation, barycentric interpolation (see Berrut & Trefethen (2004) for its advantages) and Chebyshev series. Both methods are generally numerically better than an interpolation based on a power basis expansion. In this case, the difference is negligible.. ClearAll[rho, TP];. b = 1;. q = -1;. rat = 10^-30;. rho[r_, b_,. q_] := (2 b/(1 - q)) (1 - (b/r)^(1 - q))^(1/2) Hypergeometric2F1[1/2,. 1 - 1/(q - 1), 3/2, 1 - (b/r)^(1 - q)];. TP[ρ_?NumericQ, b_?NumericQ, q_?NumericQ] :=. Block[{r}, With[{ρ0 = SetPrecision[ρ, 32]},. r /. FindRoot[rho[r, b, q] == ρ0, {r, 1001/1000},. WorkingPrecision -> 32] // Re]];. (* Chebyshev nodes and points *). deg = 24;. chebnodes = N[Rescale[Sin[Pi/2 Range[-deg, deg, 2]/deg]], 32];. TPtab = Table[{ρ, TP[ρ, b, q]}, {ρ, chebnodes}];. (* Barycentric interpolation *). rif = StatisticsLibraryBarycentricInterpolation[N@TPtab[[All, 1]],. N@TPtab[[All, 2]],. "Weights" ->. ReplacePart[. Table[(-1)^k, {k, 0, Length@TPtab - 1}], {1 -> 1/2, -1 -> 1/2}]];. Plot[rho[rif[ρ], b, q] - ρ // RealExponent, {ρ, 0, 1}]. (* Power basis interpolation *). (* To diminish round-off error, increase precision *). TPint = SetPrecision[InterpolatingPolynomial[TPtab, ρ], 50] // Expand // N. Plot[rho[N@TPint, b, q] - ρ // RealExponent, {ρ, 0, 1}]. (* Chebyshev series coefficients via FFT/DCT *). cc = Sqrt[2/(Length@TPtab - 1)] FourierDCT[Reverse@TPtab[[All, 2]], 1];. cc[[{1, -1}]] /= 2;. rCS[ρ_] := cc.Cos[Range[0, Length@cc - 1] ArcCos[2 ρ - 1]];. Plot[rho[rCS[ρ], b, q] - ρ // RealExponent, {ρ, 0, 1}].
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https://doubtnut.com/question-answer/let-a1-a2-a3-an-1an-be-an-ap-statement-1-a1-a2-a3-ann-2a1-an-statement-2-ak-an-k-1a1-an-for-k123-n-53796475
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or
# Let a_(1)+a_(2)+a_(3), . . . ,a_(n-1),a_(n) be an A.P. <br> Statement -1: a_(1)+a_(2)+a_(3)+ . . . +a_(n)=(n)/(2)(a_(1)+a_(n)) <br> Statement -2 a_(k)+a_(n-k+1)=a_(1)+a_(n)" for "k=1,2,3, . . . , n
Question from Class 11 Chapter Sequences And Series
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Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.Statement -1 is true, Statement -2 is False.Statement -1 is False, Statement -2 is True.
A
Solution :
Let `S_(n)=a_(1)+a_(2)+a_(3)+ . . . +a_(n-1)+a_(n)`. Then, <br> `S_(n)=a_(n)+a_(n-1)+a_(n-2)+ . . . +a_(2)+a_(1)` <br> `rArr" "2S_(n)=n(a_(1)+a_(n))" [Using statement-2]"` <br> `rArr" "S_(n)=(n)/(2)(a_(1)+a_(2))` <br> So, statement -2 correct explanation for statement -1.
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or. # Let a_(1)+a_(2)+a_(3), . . . ,a_(n-1),a_(n) be an A.P. <br> Statement -1: a_(1)+a_(2)+a_(3)+ . . . +a_(n)=(n)/(2)(a_(1)+a_(n)) <br> Statement -2 a_(k)+a_(n-k+1)=a_(1)+a_(n)" for "k=1,2,3, . . . , n. Question from Class 11 Chapter Sequences And Series. Apne doubts clear karein ab Whatsapp par bhi. Try it now.. Watch 1000+ concepts & tricky questions explained!. 300+ views | 23.6 K+ people like this. Share. Share. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.Statement -1 is true, Statement -2 is False.Statement -1 is False, Statement -2 is True.. A. Solution :. Let `S_(n)=a_(1)+a_(2)+a_(3)+ . . . +a_(n-1)+a_(n)`. Then, <br> `S_(n)=a_(n)+a_(n-1)+a_(n-2)+ . . . +a_(2)+a_(1)` <br> `rArr" "2S_(n)=n(a_(1)+a_(n))" [Using statement-2]"` <br> `rArr" "S_(n)=(n)/(2)(a_(1)+a_(2))` <br> So, statement -2 correct explanation for statement -1.. Related Video.
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# How to Solve Trig Ratios of General Angles? (+FREE Worksheet!)
Learn trigonometric ratios of general angles and how to solve math problems related to trig ratios by the following step-by-step guide.
## Step by step guide to solve Trig Ratios of General Angles
• Learn common trigonometric functions:
### Trig Ratios of General Angles – Example 1:
Find the trigonometric function: $$cos$$ $$120^\circ$$
Solution:
$$cos$$ $$120^{\circ}$$
Use the following property: $$cos$$$$(x)=$$ $$sin$$$$(90^{\circ}-x)$$
$$cos$$ $$120^{\circ} =$$ $$sin$$ $$( 90^{\circ} -120^{\circ})=$$ $$sin ( -30^{\circ})$$
Now use the following property: $$sin (-x)$$$$=- sin (x)$$
Then: $$sin ( -30^{\circ})=-sin (30^{\circ}$$)$$=-\frac{1}{2 }$$
### Trig Ratios of General Angles – Example 2:
Find the trigonometric function: $$sin$$ $$135^\circ$$
Solution:
Use the following property: $$sin$$$$(x)=$$ $$cos$$$$(90^\circ-x)$$
$$sin$$ $$135^\circ=$$ $$cos$$$$(90^\circ-135^\circ)=$$ $$cos$$$$(-45^\circ)$$
Now use the following property: $$cos$$$$(-x)=cos x$$
Then: $$cos$$$$(-45^\circ)=$$ $$cos$$$$(45^\circ)=\frac{\sqrt{2}}{2 }$$
### Trig Ratios of General Angles – Example 3:
Find the trigonometric function: $$sin$$ $$-120^\circ$$
Solution:
Use the following property: $$sin$$$$(-x)=-$$ $$sin$$$$(x)$$
$$sin$$$$-120^\circ=-$$ $$sin$$ $$120^\circ$$ , $$sin$$$$120^\circ=\frac{\sqrt{3}}{2}$$
Then: $$sin$$$$-120^\circ=-\frac{\sqrt{3}}{2}$$
### Trig Ratios of General Angles – Example 4:
Find the trigonometric function: $$cos$$ $$150^\circ$$
Solution:
$$cos$$ $$150^{\circ}$$
Use the following property: $$cos$$$$(x)=$$ $$sin$$$$(90^{\circ}-x)$$
$$cos$$ $$150^{\circ} =$$ $$sin$$ $$( 90^{\circ} -150^{\circ})=$$ $$sin ( -60^{\circ})$$
Now use the following property: $$sin (-x)$$$$=- sin (x)$$
Then: $$sin ( -60^{\circ})=-sin (60^{\circ}$$)$$= -\frac{\sqrt{3}}{2}$$
## Exercises
### Use a calculator to find each. Round your answers to the nearest ten–thousandth.
• $$\color{blue}{sin \ – 120^\circ}$$
• $$\color{blue}{sin \ 150^\circ}$$
• $$\color{blue}{cos \ 315^\circ}$$
• $$\color{blue}{cos \ 180^\circ}$$
• $$\color{blue}{sin \ 120^\circ}$$
• $$\color{blue}{sin \ – 330^\circ }$$
### Download Trig Ratios of General Angles Worksheet
• $$\color{blue}{-\frac{\sqrt{3}}{2}}$$
• $$\color{blue}{\frac{1}{2}}$$
• $$\color{blue}{\frac{\sqrt{2}}{2}}$$
• $$\color{blue}{-1}$$
• $$\color{blue}{\frac{\sqrt{3}}{2}}$$
• $$\color{blue}{\frac{1}{2}}$$
.
### What people say about "How to Solve Trig Ratios of General Angles? (+FREE Worksheet!)"?
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# How to Solve Trig Ratios of General Angles? (+FREE Worksheet!). Learn trigonometric ratios of general angles and how to solve math problems related to trig ratios by the following step-by-step guide.. ## Step by step guide to solve Trig Ratios of General Angles. • Learn common trigonometric functions:. ### Trig Ratios of General Angles – Example 1:. Find the trigonometric function: $$cos$$ $$120^\circ$$. Solution:. $$cos$$ $$120^{\circ}$$. Use the following property: $$cos$$$$(x)=$$ $$sin$$$$(90^{\circ}-x)$$. $$cos$$ $$120^{\circ} =$$ $$sin$$ $$( 90^{\circ} -120^{\circ})=$$ $$sin ( -30^{\circ})$$. Now use the following property: $$sin (-x)$$$$=- sin (x)$$. Then: $$sin ( -30^{\circ})=-sin (30^{\circ}$$)$$=-\frac{1}{2 }$$. ### Trig Ratios of General Angles – Example 2:. Find the trigonometric function: $$sin$$ $$135^\circ$$. Solution:. Use the following property: $$sin$$$$(x)=$$ $$cos$$$$(90^\circ-x)$$. $$sin$$ $$135^\circ=$$ $$cos$$$$(90^\circ-135^\circ)=$$ $$cos$$$$(-45^\circ)$$. Now use the following property: $$cos$$$$(-x)=cos x$$. Then: $$cos$$$$(-45^\circ)=$$ $$cos$$$$(45^\circ)=\frac{\sqrt{2}}{2 }$$. ### Trig Ratios of General Angles – Example 3:. Find the trigonometric function: $$sin$$ $$-120^\circ$$. Solution:. Use the following property: $$sin$$$$(-x)=-$$ $$sin$$$$(x)$$. $$sin$$$$-120^\circ=-$$ $$sin$$ $$120^\circ$$ , $$sin$$$$120^\circ=\frac{\sqrt{3}}{2}$$. Then: $$sin$$$$-120^\circ=-\frac{\sqrt{3}}{2}$$. ### Trig Ratios of General Angles – Example 4:. Find the trigonometric function: $$cos$$ $$150^\circ$$. Solution:. $$cos$$ $$150^{\circ}$$.
|
Use the following property: $$cos$$$$(x)=$$ $$sin$$$$(90^{\circ}-x)$$. $$cos$$ $$150^{\circ} =$$ $$sin$$ $$( 90^{\circ} -150^{\circ})=$$ $$sin ( -60^{\circ})$$. Now use the following property: $$sin (-x)$$$$=- sin (x)$$. Then: $$sin ( -60^{\circ})=-sin (60^{\circ}$$)$$= -\frac{\sqrt{3}}{2}$$. ## Exercises. ### Use a calculator to find each. Round your answers to the nearest ten–thousandth.. • $$\color{blue}{sin \ – 120^\circ}$$. • $$\color{blue}{sin \ 150^\circ}$$. • $$\color{blue}{cos \ 315^\circ}$$. • $$\color{blue}{cos \ 180^\circ}$$. • $$\color{blue}{sin \ 120^\circ}$$. • $$\color{blue}{sin \ – 330^\circ }$$. ### Download Trig Ratios of General Angles Worksheet. • $$\color{blue}{-\frac{\sqrt{3}}{2}}$$. • $$\color{blue}{\frac{1}{2}}$$. • $$\color{blue}{\frac{\sqrt{2}}{2}}$$. • $$\color{blue}{-1}$$. • $$\color{blue}{\frac{\sqrt{3}}{2}}$$. • $$\color{blue}{\frac{1}{2}}$$. .. ### What people say about "How to Solve Trig Ratios of General Angles? (+FREE Worksheet!)"?. No one replied yet.. X. 30% OFF. Limited time only!. Save Over 30%. SAVE $5 It was$16.99 now it is \$11.99.
|
http://www.numbersaplenty.com/50358
| 1,597,276,783,000,000,000
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text/html
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crawl-data/CC-MAIN-2020-34/segments/1596439738950.31/warc/CC-MAIN-20200812225607-20200813015607-00400.warc.gz
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Search a number
50358 = 23711109
BaseRepresentation
bin1100010010110110
32120002010
430102312
53102413
61025050
7266550
oct142266
976063
1050358
1134920
1225186
1319bc9
14144d0
15edc3
hexc4b6
50358 has 32 divisors (see below), whose sum is σ = 126720. Its totient is φ = 12960.
The previous prime is 50341. The next prime is 50359. The reversal of 50358 is 85305.
It can be divided in two parts, 503 and 58, that added together give a triangular number (561 = T33).
50358 is digitally balanced in base 2 and base 4, because in such bases it contains all the possibile digits an equal number of times.
It is a Harshad number since it is a multiple of its sum of digits (21).
It is a nialpdrome in base 15.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (50359) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 408 + ... + 516.
It is an arithmetic number, because the mean of its divisors is an integer number (3960).
250358 is an apocalyptic number.
It is a practical number, because each smaller number is the sum of distinct divisors of 50358, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (63360).
50358 is an abundant number, since it is smaller than the sum of its proper divisors (76362).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
50358 is a wasteful number, since it uses less digits than its factorization.
50358 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 132.
The product of its (nonzero) digits is 600, while the sum is 21.
The square root of 50358 is about 224.4058822758. The cubic root of 50358 is about 36.9280315200.
The spelling of 50358 in words is "fifty thousand, three hundred fifty-eight".
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Search a number. 50358 = 23711109. BaseRepresentation. bin1100010010110110. 32120002010. 430102312. 53102413. 61025050. 7266550. oct142266. 976063. 1050358. 1134920. 1225186. 1319bc9. 14144d0. 15edc3. hexc4b6. 50358 has 32 divisors (see below), whose sum is σ = 126720. Its totient is φ = 12960.. The previous prime is 50341. The next prime is 50359.
|
The reversal of 50358 is 85305.. It can be divided in two parts, 503 and 58, that added together give a triangular number (561 = T33).. 50358 is digitally balanced in base 2 and base 4, because in such bases it contains all the possibile digits an equal number of times.. It is a Harshad number since it is a multiple of its sum of digits (21).. It is a nialpdrome in base 15.. It is a congruent number.. It is not an unprimeable number, because it can be changed into a prime (50359) by changing a digit.. It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 408 + ... + 516.. It is an arithmetic number, because the mean of its divisors is an integer number (3960).. 250358 is an apocalyptic number.. It is a practical number, because each smaller number is the sum of distinct divisors of 50358, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (63360).. 50358 is an abundant number, since it is smaller than the sum of its proper divisors (76362).. It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.. 50358 is a wasteful number, since it uses less digits than its factorization.. 50358 is an evil number, because the sum of its binary digits is even.. The sum of its prime factors is 132.. The product of its (nonzero) digits is 600, while the sum is 21.. The square root of 50358 is about 224.4058822758. The cubic root of 50358 is about 36.9280315200.. The spelling of 50358 in words is "fifty thousand, three hundred fifty-eight".
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https://math.answers.com/Q/What_is_42_out_of_50_in_percent
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0
# What is 42 out of 50 in percent?
Updated: 9/23/2023
Wiki User
8y ago
84%
Wiki User
8y ago
Earn +20 pts
Q: What is 42 out of 50 in percent?
Submit
Still have questions?
Related questions
### What is 50 percent of 84?
50 percent of 84 = .50 x 84 = 42
### What number is 42 percent of 50?
What is 42% of 50:= 42% * 50= 0.42 * 50= 21
### What is 42 percent percent written as a fraction?
42 percent written as a fraction is 42/100, which simplifies to 21/50.
### What whole number is 50 42 percent of?
50 is 42% of= 50 / 42= 50 / 0.42= 119
### What is 42 out of 50 as a percent?
42 out of 50 can be expressed as the fraction 42/50. 42 divided by 50 = 0.84To convert 0.84 to percent multiply by 100: 0.84 × 100 = 84 %
50% of 84 is 42
### What is the percentage of 50 percent of 42?
If you mean 50% of 42 then it is 21
### What is the fraction for 42 percent?
42% := 42/100= 21/50
42% = 21/50
42% = 21/50
It is 42%
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0. # What is 42 out of 50 in percent?. Updated: 9/23/2023. Wiki User. 8y ago. 84%. Wiki User. 8y ago. Earn +20 pts. Q: What is 42 out of 50 in percent?. Submit. Still have questions?. Related questions. ### What is 50 percent of 84?. 50 percent of 84 = .50 x 84 = 42. ### What number is 42 percent of 50?. What is 42% of 50:= 42% * 50= 0.42 * 50= 21.
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### What is 42 percent percent written as a fraction?. 42 percent written as a fraction is 42/100, which simplifies to 21/50.. ### What whole number is 50 42 percent of?. 50 is 42% of= 50 / 42= 50 / 0.42= 119. ### What is 42 out of 50 as a percent?. 42 out of 50 can be expressed as the fraction 42/50. 42 divided by 50 = 0.84To convert 0.84 to percent multiply by 100: 0.84 × 100 = 84 %. 50% of 84 is 42. ### What is the percentage of 50 percent of 42?. If you mean 50% of 42 then it is 21. ### What is the fraction for 42 percent?. 42% := 42/100= 21/50. 42% = 21/50. 42% = 21/50. It is 42%.
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https://math.answers.com/questions/If_f_of_x_equals_2x_squared_what_is_the_positive_value_of_x_for_which_f_of_x_equals_136_With_the_explanation_please_Thanks
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0
If f of x equals 2x squared what is the positive value of x for which f of x equals 136 With the explanation please Thanks?
Wiki User
2015-04-25 12:04:58
f(x) = 2x²
f(x) = 136
→ 2x² = 136
→ x² = 68
→ x = ±√68
As the positive value is required, x = √68 ≈ 8.246
Wiki User
2015-04-25 12:04:58
Study guides
20 cards
A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
3.74
1190 Reviews
Wiki User
2015-01-29 11:34:48
f(x) = 2*x^2 = 136that is x^2 = 136/2 = 68
and so x = + or - sqrt(68).
But x is said to be the positive value so x = +sqrt(68).
Earn +20 pts
Q: If f of x equals 2x squared what is the positive value of x for which f of x equals 136 With the explanation please Thanks?
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0. If f of x equals 2x squared what is the positive value of x for which f of x equals 136 With the explanation please Thanks?. Wiki User. 2015-04-25 12:04:58. f(x) = 2x². f(x) = 136. → 2x² = 136. → x² = 68. → x = ±√68. As the positive value is required, x = √68 ≈ 8.246. Wiki User. 2015-04-25 12:04:58. Study guides. 20 cards.
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A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials. ➡️. See all cards. 3.74. 1190 Reviews. Wiki User. 2015-01-29 11:34:48. f(x) = 2*x^2 = 136that is x^2 = 136/2 = 68. and so x = + or - sqrt(68).. But x is said to be the positive value so x = +sqrt(68).. Earn +20 pts. Q: If f of x equals 2x squared what is the positive value of x for which f of x equals 136 With the explanation please Thanks?.
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Associated Topics || Dr. Math Home || Search Dr. Math
### An Ordinary Differential Equation in Everyday Life?
```Date: 03/26/2011 at 05:33:53
From: Hassan
Subject: ODE's
I just want to ask about ordinary differential equations. Where do these
find practical applications?
```
```
Date: 03/26/2011 at 20:45:22
From: Doctor Jordan
Subject: Re: ODE's
Hi Hassan,
Let's say I have a small cube of metal of mass m sitting on a table. Say
the cube of metal is attached to a spring in such a way that the cube will
move back and forth on the table.
Suppose we want to know the position of this cube of metal at time t. How
can we obtain this?
This is the kind of problem that call for ordinary differential equations.
Methods for solving ODE let us write down an equation using data
that we have, and then the solution to the ODE will give us the
information that we want -- in this case, the position of the cube at
time t.
To get the differential equation for this mechanical system, we need to
know the forces acting on the object at any time -- namely, the restoring
force of the spring, which is approximately -kx, where x is the distance
from the rest position of the cube and k is an experimentally determined
number that depends on the spring's material composition and
manufacturing; and the damping force, which is approximately cx', where c
is called the viscous damping coefficient. Here, x' is the derivative of x
with respect to t.
Then by Newton's second law here, the force cx' - kx is equal to mx'',
where x'' is the second derivative of x with respect to t -- in other
words, the acceleration of the metal cube. So here the differential
equation is
mx'' = cx' - kx
It is called a differential equation because it involves derivatives. It
says how the system is moving at each point in time.
We have mathematical techniques for solving ordinary differential
equations. We apply these techniques to this equation and we get a
solution, x(t). But what is x(t)? x(t) is the position of the metal cube
at time t. Therefore, solving this differential equation gives us a
function which tells us where the metal cube is at any point in time.
We started knowing only the forces acting on the cube at each point in
time and put this together into a differential equation. By applying
mathematical techniques, we ended up with a function that told us the
exact arrangement of our system at each point in time.
- Doctor Jordan, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 03/27/2011 at 00:27:38
From: Hassan
Subject: Thank you (ODE's)
Nice job; very grateful.
May Allah bless you more.
Jazak Allah.
```
Associated Topics:
High School Calculus
Search the Dr. Math Library:
Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words
Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search
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Associated Topics || Dr. Math Home || Search Dr. Math. ### An Ordinary Differential Equation in Everyday Life?. ```Date: 03/26/2011 at 05:33:53. From: Hassan. Subject: ODE's. I just want to ask about ordinary differential equations. Where do these. find practical applications?. ```. ```. Date: 03/26/2011 at 20:45:22. From: Doctor Jordan. Subject: Re: ODE's. Hi Hassan,. Let's say I have a small cube of metal of mass m sitting on a table. Say. the cube of metal is attached to a spring in such a way that the cube will. move back and forth on the table.. Suppose we want to know the position of this cube of metal at time t. How. can we obtain this?. This is the kind of problem that call for ordinary differential equations.. Methods for solving ODE let us write down an equation using data. that we have, and then the solution to the ODE will give us the. information that we want -- in this case, the position of the cube at. time t.. To get the differential equation for this mechanical system, we need to. know the forces acting on the object at any time -- namely, the restoring. force of the spring, which is approximately -kx, where x is the distance. from the rest position of the cube and k is an experimentally determined. number that depends on the spring's material composition and. manufacturing; and the damping force, which is approximately cx', where c. is called the viscous damping coefficient. Here, x' is the derivative of x. with respect to t.. Then by Newton's second law here, the force cx' - kx is equal to mx'',. where x'' is the second derivative of x with respect to t -- in other.
|
words, the acceleration of the metal cube. So here the differential. equation is. mx'' = cx' - kx. It is called a differential equation because it involves derivatives. It. says how the system is moving at each point in time.. We have mathematical techniques for solving ordinary differential. equations. We apply these techniques to this equation and we get a. solution, x(t). But what is x(t)? x(t) is the position of the metal cube. at time t. Therefore, solving this differential equation gives us a. function which tells us where the metal cube is at any point in time.. We started knowing only the forces acting on the cube at each point in. time and put this together into a differential equation. By applying. mathematical techniques, we ended up with a function that told us the. exact arrangement of our system at each point in time.. - Doctor Jordan, The Math Forum. http://mathforum.org/dr.math/. ```. ```. Date: 03/27/2011 at 00:27:38. From: Hassan. Subject: Thank you (ODE's). Nice job; very grateful.. May Allah bless you more.. Jazak Allah.. ```. Associated Topics:. High School Calculus. Search the Dr. Math Library:. Find items containing (put spaces between keywords): Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words. Submit your own question to Dr. Math. Math Forum Home || Math Library || Quick Reference || Math Forum Search.
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It is currently Tue Oct 16, 2018 5:58 pm
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Post subject: GMAT Geometry (Data Sufficiency)Posted: Fri Aug 20, 2010 4:09 am
Joined: Sun May 30, 2010 3:15 am
Posts: 424
In the figure above, three segments are drawn to connect opposite vertices of a hexagon, forming six triangles. All three of these segments intersect at point A. What is the area of the hexagon?
(1) One of the triangles has an area of 12.
(2) All sides of the hexagon are of equal length.
A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.
(E) Statement (1) is not sufficient because we do not know whether the hexagon is a regular hexagon (meaning that all 6 sides
have the same length). We could only calculate the area if the hexagon could be divided into six congruent triangles.
Statement (2) is also insufficient because we are not given any numbers. So, it is impossible to calculate the area. Combined, the two statements are still insufficient. A hexagon could have six equal sides, yet not be a regular hexagon. Imagine pushing opposite sides of a regular hexagon closer to each other: the
hexagon would flatten, while all the sides would remain the same length. One of the triangles within this flattened hexagon could have an area of 12, but not all of the triangles would have the same area. Therefore, we can't calculate the area, even by combining the statements, and the correct answer is choice (E).
-------------
In my opinion the the answer should be (A).
I will attempt to give a detailed explanation below:
Since it is a Hexagon, sum of internal angles is 720 using the formula: (n – 2) × 180, where 'n' is the number of sides of the polygon.
Therefore each internal angle is 120 degree. From this I say each triangle formed inside is an equilateral triangle. From (1), the area is 12 for each triangle. This is SUFFICIENT to conclude the hexagon having six triangles having an area of 12 each, has an area of
72.
Please confirm
.
Attachments: hexagon1.gif [3.85 KiB] Not downloaded yet
Top
Post subject: Re: math (test 1, question 25): geometry, data sufficiencyPosted: Fri Aug 20, 2010 4:48 am
Joined: Sun May 30, 2010 2:23 am
Posts: 498
When you solve any data sufficiency geometry question you should avoid graphics distorting your reasoning.
In this case the hexagon seems to be regular (all sides and angles are equal) and it affects the reasoning. Therefore you may conclude that all angles are equal (120⁰) but we have no facts that this conclusion can be based on.
The question statement gives us just the fact that we have a hexagon. It is just some arbitrary hexagon and can look in many different ways:
or
and many other different shapes.
That gives us a clear idea why statement (1) is insufficient.
Statement (2) also doesn't imply that hexagon is regular (regular hexagon must have all equal sides AND angles). If we know that all sides are equal then hexagon can still have diverse shapes:
or
and many other variants.
So be careful with geometry graphics and always check if what you see is based on facts.
Attachments: hexagon5.gif [4.27 KiB] Not downloaded yet hexagon4.gif [3.52 KiB] Not downloaded yet hexagon3.gif [4.3 KiB] Not downloaded yet hexagon2.gif [3.82 KiB] Not downloaded yet
Top
Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 7:59 am
Joined: Sun May 30, 2010 3:15 am
Posts: 424
Hi,
As per my understanding, every polygon with equal sides is always cyclic in nature.
Could you, please, correct if I am wrong.
Best regards.
Top
Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 7:59 am
Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
As per my understanding, every polygon with equal sides is always cyclic in nature.
A polygon with equal sides is cyclic if all its angles are equal. In other words such polygon is a regular polygon (all sides are equal and all angles are equal).
Here is an example of a polygon with equal sides but not with equal angles. It is not cyclic.
Top
Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:00 am
Joined: Sun May 30, 2010 3:15 am
Posts: 424
On the gmat, will it specifically state whether the hexagon is a regular
hexagon or not? I assumed that when you mentioned hexagon it was a regular
hexagon.
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:01 am
Joined: Sun May 30, 2010 2:23 am
Posts: 498
With the problem at hand you don't need to know if the hexagon is regular or not. All of the necessary information is provided. However, on the GMAT when you actually encounter a problem that involves a geometric figure that is regular, and, it is required that you know the latter in order to solve the problem, then yes, the GMAT question will provide you with that information.
Take care.
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:01 am
Joined: Sun May 30, 2010 3:15 am
Posts: 424
I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:02 am
Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!
No. You can see that if you draw various cases of 3 arbitrary segments that intersect in one point, then you just need to connect the ends of the segments to create various hexagons.
Attachments: arbitrary_hexagon.gif [4.44 KiB] Not downloaded yet
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Thu Jun 20, 2013 4:03 pm
Joined: Sun May 30, 2010 3:15 am
Posts: 424
If all sides are equal, its a regular hexagon. An area of a regular hexagon can be considered 6 × area of the equilateral triangles. Thus using both the statements we could arrive at the solution . Hence option C is correct.
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Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Thu Jun 20, 2013 4:08 pm
Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
If all sides are equal, its a regular hexagon.
A hexagon is regular if all sides are equal AND all interior angles are equal.
Think of 6 equal sticks connected in form of a hexagon and imagine moving those sticks. Here is an examples of a non-regular hexagon, which sides are all equal:
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It is currently Tue Oct 16, 2018 5:58 pm. All times are UTC - 5 hours [ DST ]. Page 1 of 1 [ 10 posts ]. Print view Previous topic | Next topic. Author Message. Post subject: GMAT Geometry (Data Sufficiency)Posted: Fri Aug 20, 2010 4:09 am. Joined: Sun May 30, 2010 3:15 am. Posts: 424. In the figure above, three segments are drawn to connect opposite vertices of a hexagon, forming six triangles. All three of these segments intersect at point A. What is the area of the hexagon?. (1) One of the triangles has an area of 12.. (2) All sides of the hexagon are of equal length.. A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.. B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.. C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.. D. Either statement BY ITSELF is sufficient to answer the question.. E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.. (E) Statement (1) is not sufficient because we do not know whether the hexagon is a regular hexagon (meaning that all 6 sides. have the same length). We could only calculate the area if the hexagon could be divided into six congruent triangles.. Statement (2) is also insufficient because we are not given any numbers. So, it is impossible to calculate the area. Combined, the two statements are still insufficient. A hexagon could have six equal sides, yet not be a regular hexagon. Imagine pushing opposite sides of a regular hexagon closer to each other: the. hexagon would flatten, while all the sides would remain the same length. One of the triangles within this flattened hexagon could have an area of 12, but not all of the triangles would have the same area. Therefore, we can't calculate the area, even by combining the statements, and the correct answer is choice (E).. -------------. In my opinion the the answer should be (A).. I will attempt to give a detailed explanation below:. Since it is a Hexagon, sum of internal angles is 720 using the formula: (n – 2) × 180, where 'n' is the number of sides of the polygon.. Therefore each internal angle is 120 degree. From this I say each triangle formed inside is an equilateral triangle. From (1), the area is 12 for each triangle. This is SUFFICIENT to conclude the hexagon having six triangles having an area of 12 each, has an area of. 72.. Please confirm. .. Attachments: hexagon1.gif [3.85 KiB] Not downloaded yet. Top. Post subject: Re: math (test 1, question 25): geometry, data sufficiencyPosted: Fri Aug 20, 2010 4:48 am. Joined: Sun May 30, 2010 2:23 am. Posts: 498. When you solve any data sufficiency geometry question you should avoid graphics distorting your reasoning.. In this case the hexagon seems to be regular (all sides and angles are equal) and it affects the reasoning. Therefore you may conclude that all angles are equal (120⁰) but we have no facts that this conclusion can be based on.. The question statement gives us just the fact that we have a hexagon. It is just some arbitrary hexagon and can look in many different ways:. or. and many other different shapes.. That gives us a clear idea why statement (1) is insufficient.. Statement (2) also doesn't imply that hexagon is regular (regular hexagon must have all equal sides AND angles). If we know that all sides are equal then hexagon can still have diverse shapes:. or. and many other variants.. So be careful with geometry graphics and always check if what you see is based on facts.. Attachments: hexagon5.gif [4.27 KiB] Not downloaded yet hexagon4.gif [3.52 KiB] Not downloaded yet hexagon3.gif [4.3 KiB] Not downloaded yet hexagon2.gif [3.82 KiB] Not downloaded yet. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 7:59 am. Joined: Sun May 30, 2010 3:15 am. Posts: 424. Hi,. As per my understanding, every polygon with equal sides is always cyclic in nature.
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Could you, please, correct if I am wrong.. Best regards.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 7:59 am. Joined: Sun May 30, 2010 2:23 am. Posts: 498. Quote:. As per my understanding, every polygon with equal sides is always cyclic in nature.. A polygon with equal sides is cyclic if all its angles are equal. In other words such polygon is a regular polygon (all sides are equal and all angles are equal).. Here is an example of a polygon with equal sides but not with equal angles. It is not cyclic.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:00 am. Joined: Sun May 30, 2010 3:15 am. Posts: 424. On the gmat, will it specifically state whether the hexagon is a regular. hexagon or not? I assumed that when you mentioned hexagon it was a regular. hexagon.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:01 am. Joined: Sun May 30, 2010 2:23 am. Posts: 498. With the problem at hand you don't need to know if the hexagon is regular or not. All of the necessary information is provided. However, on the GMAT when you actually encounter a problem that involves a geometric figure that is regular, and, it is required that you know the latter in order to solve the problem, then yes, the GMAT question will provide you with that information.. Take care.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:01 am. Joined: Sun May 30, 2010 3:15 am. Posts: 424. I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Wed Apr 17, 2013 8:02 am. Joined: Sun May 30, 2010 2:23 am. Posts: 498. questioner wrote:. I understand the question, but I think "three segments connecting opposite vertices of a hexagon intersect at a point" means that the hexagon is a regular hexagon!. No. You can see that if you draw various cases of 3 arbitrary segments that intersect in one point, then you just need to connect the ends of the segments to create various hexagons.. Attachments: arbitrary_hexagon.gif [4.44 KiB] Not downloaded yet. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Thu Jun 20, 2013 4:03 pm. Joined: Sun May 30, 2010 3:15 am. Posts: 424. If all sides are equal, its a regular hexagon. An area of a regular hexagon can be considered 6 × area of the equilateral triangles. Thus using both the statements we could arrive at the solution . Hence option C is correct.. Top. Post subject: Re: GMAT Geometry (Data Sufficiency)Posted: Thu Jun 20, 2013 4:08 pm. Joined: Sun May 30, 2010 2:23 am. Posts: 498. questioner wrote:. If all sides are equal, its a regular hexagon.. A hexagon is regular if all sides are equal AND all interior angles are equal.. Think of 6 equal sticks connected in form of a hexagon and imagine moving those sticks. Here is an examples of a non-regular hexagon, which sides are all equal:. Top. Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending. Page 1 of 1 [ 10 posts ]. All times are UTC - 5 hours [ DST ]. #### Who is online. Users browsing this forum: No registered users and 4 guests. You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum. Search for:. Jump to: Select a forum ------------------ GMAT GMAT: Quantitative Section (Math) GMAT: Verbal Section GMAT: Integrated Reasoning GMAT: General Questions GRE GRE: Quantitative Reasoning (Math) GRE: Verbal Reasoning GRE: General Questions General questions Other questions. Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group.
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https://opencourser.com/course/h1dxgd/khanacademy-introduction-to-polynomials
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# Polynomials
This course is a part of Algebra I, a 17-course Topic series from Khan Academy.
Learn how to add, subtract, and multiply polynomial expressions. For example, write (2x+3)(x-1) as 2x²+x-3.
This course contains 9 segments:
Intro to polynomials
Learn about polynomial expressions: What are they? How are they constructed? What can we do with them?
Learn how to add and subtract polynomial expressions with one variable.
Adding and subtracting polynomials: two variables
Learn how to add and subtract polynomial that involve two variables. For example, x^3 + xy + 3y - (x^3 + 6xy + 2y^2).
Multiplying monomials
As an intro to more elaborate polynomial multiplication, learn how to multiply monomials (which are polynomials with a single term). For example, multiply 2x³ and 5x⁷.
Multiplying monomials by polynomials
Learn how to multiply a polynomial expression by a monomial expression. Monomials are just polynomials with a single term!
Multiplying binomials
Learn how to multiply two binomials together. For example, (3x - 7) * (10x + 2).
Multiplying binomials by polynomials
Learn how to multiply a polynomial expression by a binomial expression.
Special products of binomials
Learn about the special types of products of binomials: perfect squares and the difference of two squares. These will be very helpful once you tackle more advanced expressions in Algebra.
Polynomials word problems
See a few examples of how we can represent real-world situations with polynomials.
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# Polynomials. This course is a part of Algebra I, a 17-course Topic series from Khan Academy.. Learn how to add, subtract, and multiply polynomial expressions. For example, write (2x+3)(x-1) as 2x²+x-3.. This course contains 9 segments:. Intro to polynomials. Learn about polynomial expressions: What are they? How are they constructed? What can we do with them?. Learn how to add and subtract polynomial expressions with one variable.. Adding and subtracting polynomials: two variables. Learn how to add and subtract polynomial that involve two variables. For example, x^3 + xy + 3y - (x^3 + 6xy + 2y^2).. Multiplying monomials. As an intro to more elaborate polynomial multiplication, learn how to multiply monomials (which are polynomials with a single term). For example, multiply 2x³ and 5x⁷.. Multiplying monomials by polynomials. Learn how to multiply a polynomial expression by a monomial expression. Monomials are just polynomials with a single term!. Multiplying binomials. Learn how to multiply two binomials together. For example, (3x - 7) * (10x + 2).. Multiplying binomials by polynomials. Learn how to multiply a polynomial expression by a binomial expression.. Special products of binomials.
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Learn about the special types of products of binomials: perfect squares and the difference of two squares. These will be very helpful once you tackle more advanced expressions in Algebra.. Polynomials word problems. See a few examples of how we can represent real-world situations with polynomials.. Rating Not enough ratings 9 segments On Demand (Start anytime) Free Khan Academy On all desktop and mobile devices English Mathematics Math Algebra I. ## Careers. An overview of related careers and their average salaries in the US. Bars indicate income percentile.. Brand Expressions Leader, US Brand Team \$129k. Brand Expressions Leader, US Brand Team \$129k. ## Write a review. Your opinion matters. Tell us what you think.. ## Similar Courses. Sorted by relevance. ## Like this course?. Here's what to do next:. • Save this course for later. • Get more details from the course provider. • Enroll in this course.
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# Power Lab - In the power lab, are group thought that Eric would do the most work because he has the most mass and thought that Ashley would do the least work because she had the lowest mass.
Extracts from this document...
Introduction
Ben Fitzgerald 9/21/09 Mr. Thorndike Power Lab Hypothesis: In the power lab, are group thought that Eric would do the most work because he has the most mass and thought that Ashley would do the least work because she had the lowest mass. We guessed that Ashley would generate the most power because she had to work harder than Eric who we thought would produce the least amount of energy. As we added the books to the current mass of our bodies, it required more work. Machines would be more efficient than humans when it comes to using energy. We guessed that we would have to eat a little amount of food to be able to climb the stairs with books, we guessed that Ashley would need one piece of cereal, I would need a piece of bread and Eric would need a granola bar. These estimations were guess to how well we would do in the power lab exercise. Procedure: 1. Obtain four books from Mr. Thorndike 2. Go to the staircase under Mr. Thorndike's room between the 2nd floor and the 1st floor mezzanine 3. ...read more.
Middle
Work without pack = 63.6 kg x 3.34 sec. = 212.424 Newton's x 3.75 meters = 796.59 joules 2. Work with pack = 74.5 kg x 3.44 sec. = 256.28 Newton's x 3.75 meters = 961.05 joules 3. Power generated without pack = 63.6 kg x 9.81 m/g x 3.75 m = 2339.7 joules / 3.34 seconds = 651.7 watts 4. Power generated with pack = 74.5 kg x 9.81 m/g x 3.75 m = 2740.66875 joules / 3.44 seconds = 796.7 watts Group Results: Ashley's Data: 5. Work without pack = 49 kg x 4.04 sec. = 197.96 Newton's x 3.75 meters = 742.35 joules 6. Work with pack = 63.6 x 3.34 sec. = 212.424 Newton's x 3.75 meters = 796.59 joules 7. Power generated without pack = 49 kg x 9.81 m/g x 3.75 m = 1802.6 joules / 4.04 seconds = 446.2 8. Power generated with pack = 63.6 x 9.81 m/g x 3.75 m = 2203.6 joules / 3.34 seconds = 508.9 watts Eric's Data: 9. ...read more.
Conclusion
This makes me think that if you do more work, then you also produce more power. We thought that as we added more books, that it would require us to work harder and use more energy, this is true to the actual because our work and our energy increased when we had books compared to when we did not have books. We said that machines are more efficient at using energy than humans, and that is completely true because our body's can only convert so much food or power into energy, compared to machines. We said that for us to walk up the stairs, I would have to eat a piece of bread, Eric would have to eat a granola bar and Ashley would have to eat a piece of cereal, we were way off, all three of us would need a piece of cereal because me and Ashley used less than a calorie on each trial and Eric used less than 1.5 calories with his trials. Some possible errors with this lab were things like were we all going at the same pace, did we skip steps, or touch every one, these things could in the end effect how much energy and power we produced. ...read more.
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• Join over 1.2 million students every month. • Accelerate your learning by 29%. • Unlimited access from just £6.99 per month. Page. 1. 1. 1. 2. 2. 2. 3. 3. 3. # Power Lab - In the power lab, are group thought that Eric would do the most work because he has the most mass and thought that Ashley would do the least work because she had the lowest mass.. Extracts from this document.... Introduction. Ben Fitzgerald 9/21/09 Mr. Thorndike Power Lab Hypothesis: In the power lab, are group thought that Eric would do the most work because he has the most mass and thought that Ashley would do the least work because she had the lowest mass. We guessed that Ashley would generate the most power because she had to work harder than Eric who we thought would produce the least amount of energy. As we added the books to the current mass of our bodies, it required more work. Machines would be more efficient than humans when it comes to using energy. We guessed that we would have to eat a little amount of food to be able to climb the stairs with books, we guessed that Ashley would need one piece of cereal, I would need a piece of bread and Eric would need a granola bar. These estimations were guess to how well we would do in the power lab exercise. Procedure: 1. Obtain four books from Mr. Thorndike 2. Go to the staircase under Mr. Thorndike's room between the 2nd floor and the 1st floor mezzanine 3. ...read more.. Middle. Work without pack = 63.6 kg x 3.34 sec. = 212.424 Newton's x 3.75 meters = 796.59 joules 2. Work with pack = 74.5 kg x 3.44 sec. = 256.28 Newton's x 3.75 meters = 961.05 joules 3. Power generated without pack = 63.6 kg x 9.81 m/g x 3.75 m = 2339.7 joules / 3.34 seconds = 651.7 watts 4. Power generated with pack = 74.5 kg x 9.81 m/g x 3.75 m = 2740.66875 joules / 3.44 seconds = 796.7 watts Group Results: Ashley's Data: 5. Work without pack = 49 kg x 4.04 sec. = 197.96 Newton's x 3.75 meters = 742.35 joules 6. Work with pack = 63.6 x 3.34 sec. = 212.424 Newton's x 3.75 meters = 796.59 joules 7. Power generated without pack = 49 kg x 9.81 m/g x 3.75 m = 1802.6 joules / 4.04 seconds = 446.2 8. Power generated with pack = 63.6 x 9.81 m/g x 3.75 m = 2203.6 joules / 3.34 seconds = 508.9 watts Eric's Data: 9. ...read more.. Conclusion. This makes me think that if you do more work, then you also produce more power. We thought that as we added more books, that it would require us to work harder and use more energy, this is true to the actual because our work and our energy increased when we had books compared to when we did not have books. We said that machines are more efficient at using energy than humans, and that is completely true because our body's can only convert so much food or power into energy, compared to machines. We said that for us to walk up the stairs, I would have to eat a piece of bread, Eric would have to eat a granola bar and Ashley would have to eat a piece of cereal, we were way off, all three of us would need a piece of cereal because me and Ashley used less than a calorie on each trial and Eric used less than 1.5 calories with his trials. Some possible errors with this lab were things like were we all going at the same pace, did we skip steps, or touch every one, these things could in the end effect how much energy and power we produced. ...read more.. The above preview is unformatted text. This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.
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## Found what you're looking for?. • Start learning 29% faster today. • 150,000+ documents available. • Just £6.99 a month. Not the one? Search for your essay title.... • Join over 1.2 million students every month. • Accelerate your learning by 29%. • Unlimited access from just £6.99 per month. # Related International Baccalaureate Physics essays. 1. ## The Affect of Mass on the Time It Takes an Object To Fall. Therefore, the equation that includes x to the power of -0.5 is within our uncertainty. Furthermore, when the inverse of the square root of mass was then taken, the graph became very close to linear. The positive slope of this linear line, 3.3568, shows that as the inverse of the. 2. ## Oscillating Mass. I also predict that the elongation of the spring l will not have an effect on the period of the oscillating mass. 1.3 Variables Independent: Mass m of the hanging object, measured by weighing it on the analytical balance, and the length l of elongation of the spring, measured using the meter stick.. 1. ## The Affect of Mass on the Period. relationship between the mass and period of a pendulum because the slope of the graph is so close to zero. Therefore, the period constantly stays exceptionally close 1.1s despite the change in mass. The uncertainty in mass is not represented in the graph because it is so miniscule that it cannot be seen without misrepresenting data.. 2. ## Physics Lab. and g is the acceleration due to gravity that every body on the earth feels. g is taken to be around 10ms-2. For example: 1Kg = 1000g � 100g � 1000g = 0.1Kg 1Kg = 10N � 0.1Kg � 10N = 1N.. 1. ## Physics Lab. 1000 = 5550 = 144.58 seconds around the planet C = C =5550 C = 34871.678 = 241.19 = p = 2938734.519 M = 1.74 x 10^-4 1.74 x 10^-4 x 10^25 = 1.74 x 10^21 Miranda X = 8.3 Y = 12.3 Radius = = 10.3 10.3 x 1000. 2. ## HL Physics Revision Notes. body A human body is a conducting medium, so when it is moving in an alternating magnetic field at extra-low-frequency, then electric field is induced, hence inducing current in human body. Discuss some of the possible risks involved in living and working near high-voltage power lines.. 1. ## Physics Lab: Images formed by a plane mirror. In this case, the image will be inverted. The image dimensions are equal to the object dimensions. Thus, the absolute value of magnification is exactly 1. The image is again real and can be projected onto the screen at the image location. Case 3: The pin object is located between 2f and f: When the object is located. 2. ## Gravity lab using mass and force meters. ±0.1 N 1.0 2.0 3.0 4.0 5.0 6.0 7.0 7.9 8.9 9.9 5.8 *In the mass row, we have 1% uncertainties because the producer company who has built those masses, says that there may be a 1% percent difference in the masses of those objects.. • Over 160,000 pieces. of student written work. • Annotated by. experienced teachers. • Ideas and feedback to.
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