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If you have come to this post looking for an easy, no-hard-work method to scale furniture, I’m afraid you’re going to be disappointed. There is no magic bullet. No simple number that can be used as a multiplier to arrive at the final sizes needed to enlarge or scale down a plan. However, there are a couple ways that work if you put in the time and effort.
The easiest, most simplistic method to scale furniture is to take a scaled drawing to your local copy shop and have them enlarge the piece to the size you are after. The drawings on the blanket chest in our August 2009 issue are to scale, so you would need to take the drawings to your copy shop and have them enlarge the plans by a certain percentage. To adjust the plan from a box size of 32″ to the larger box size of 42″, you should ask that the plan be enlarged by 31.25 percent (32″ x 1.3125 percent = 42″). In turn, the other measurements would also adjust accordingly. The overall height adjusts from 20″ to 26-1/4″ and the feet go to 5″ from 3-3/4″.
While this option allows you to work from a drawing, I find it much more helpful to learn how to scale from photos , photos that are shown with full, mostly front-on views and not necessarily shown from angles. To do this, I use ratios of measurements taken off the photo.
As an example, the blanket chest photo I’m using measures 6-3/4″ wide and in the description of the chest the actual width is 53″. To find other measurements along the width of the chest such as the width of one drawer, I’ll set up a ratio of 6.75/53. I can use this ratio to find the width of any other part of the chest so long as I take all my measurements from the same photo.
If the width of one of the long drawers in my photo equals 2-1/4″, then I would set up the following where X is the actual width of the drawer front:
6.75/53 , 2.25/X
(Read as 6.75 is to 53 as 2.25 is to X)
Solve for x with cross-multiplication
2.25 x 53 = 6.75X
119.25 = 6.75X
119.25/6.75 = X
17.666 = X
So the actual measurement of the drawer front, based on the photo, is 17-5/8″.
I would find another ratio for any measurements of height. Using my example, the photo measurement of the height of the chest is 3-7/16″ while the height of the actual chest is listed at 29″. My ratio is 3.4375/29. Measuring the drawer height in the photo at 9/16″ and solving for X in the ratio finds the actual drawer height.
3.4375/29 , .5625/x
(Read as 3.4375 is to 29 as .5625 is to X)
.5625 x 29 = 3.4375X
16.3125 = 3.4375X
16.3125/3.4375 = X
4.745 = X
The drawer height is 4.75 or 4-3/4″.
This process, along with a general knowledge about furniture construction, should give you a way to scale furniture from photos. I would use this ratio information to establish the sizes of the box of the blanket chest, and any interior-piece sizes would be determined off the actual box , as it should be. Begin with the main structure, be it a box for a blanket chest or the case for a chest of drawers, then fit any parts to that structure.
Additionally, there are ways to manipulate the drawings in SketchUp that allow the sizes to be scaled up and/or down , use the scale tool. But in doing so, the thicknesses of the parts also change and there might be other issues of which I’m not aware. If you scale in SketchUp, I would suggest you work only with the elevations of the drawings. And work on a copy of the project. That way you can compare your changes to that of the original.
– Glen D. Huey
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Recent Posts
• I’d love to know how you can perform this work if the object in the photograph is angled away from the camera!
• Robert Whitlock
I was actually hoping you would explain how to scale up or down to make different sizes of the same piece. I make Adirondack chairs, but I want to be able to make the same chair for kids, or for the "big and tall" market.
• Mike Siemsen
Glen,
I take a photograph of the picture in the book and then project it on a piece of glass with paper attached to the back. By placing the projector at the same position as the photo was shot it will project the image back to being square (see parallax, keystoning). move the projector closer to make the image smaller and back it off to make the image larger. Now go behind the glass with the room darkened and there is your piece of furniture, just trace, you won’t even be working in your own shadow! If you want more info talk to me in St. Charles! There is no math involved.
Mike
• Hey Glen:
Make your life easier. Switch to metric for these types of measurements. The math is simplified and you could switch back to inches at the end of your calculations. Look at the rule you used in your pictures. Just flip it over for centimeters.
Randy
• Gene
I think you have just answered the age-old question of middle-school students everywhere: When am I ever going to use math in real life?!
• Rob Millard
I have used the same method to scale all but a few of the pieces I have made. I usually use a dial caliper reading to .001” for the highest accuracy. Scaling this way has worked very well, but I’ve had a few unpleasant surprises, where the finished piece didn’t exactly match the original.
Rob Millard
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0
# What are the factors and prime factors of 321?
Updated: 4/28/2022
Wiki User
14y ago
Best Answer
321 is a composite number because it has factors other than 1 and itself. It is not a Prime number.
The 4 factors of 321 are 1, 3, 107, and 321.
The factor pairs of 321 are 1 x 321 and 3 x 107.
The proper factors of 321 are 1, 3, and 107 or,
if the definition you are using excludes 1, they are 3 and 107.
The prime factors of 321 are 3 and 107.
The distinct prime factors (listing each prime factor only once) of 321 are also 3 and 107.
The prime factorization of 321 is 3 x 107.
NOTE: There cannot be common factors, a greatest common factor, or a least common multiple because "common" refers to factors or multiples that two or more numbers have in common.
Wiki User
14y ago
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# Maths GCSE Question Watch
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1. I was doing a past paper question and found this I've gotten part of the way through looked at the mark scheme and tried but failed to work it out. Hopefully someone can help me understand how to do this. The question is to solve:
3 - 4
------ -------
X-1 X+2
I got to
3x+6-4x+4
--------------
X^2+X-2
Help is greatly appreciated many thanks .
2. i got 2-x all over (x-1) (x+2) but i have a feeling i am wrong
3. (Original post by Snowystar)
I was doing a past paper question and found this I've gotten part of the way through looked at the mark scheme and tried but failed to work it out. Hopefully someone can help me understand how to do this. The question is to solve:
3 - 4
------ -------
X-1 X+2
I got to
3x+6-4x+4
--------------
X^2+X-2
Help is greatly appreciated many thanks .
You can't solve it unless it is equal to something. In terms of the numerator, you can simplify that by collecting the like terms...
The denominator is fine, although you should leave it in its factorised form.
Posted from TSR Mobile
4. (Original post by Chittesh14)
You can't solve it unless it is equal to something. In terms of the numerator, you can simplify that by collecting the like terms...
The denominator is fine, although you should leave it in its factorised form.
Posted from TSR Mobile
Oh whoops it's equal to 2.
5. (Original post by Snowystar)
Oh whoops it's equal to 2.
So, if the numerator is equal to -X + 10. The denominator is (X-1)(X+2).
-X + 10
-------------- = 2
(X-1)(X+2)
So, to get rid of the denominator and equate two equations in X, what do you have to do next?
Posted from TSR Mobile
6. (Original post by Chittesh14)
So, if the numerator is equal to -X + 10. The denominator is (X-1)(X+2).
-X + 10
-------------- = 2
(X-1)(X+2)
So, to get rid of the denominator and equate two equations in X, what do you have to do next?
Posted from TSR Mobile
No clue
7. (Original post by Snowystar)
No clue
Multiply the denominator by 2 to get rid off it. Now, it's on the right hand side.
-X + 10 = 2[(X-1)(X+2)]
Posted from TSR Mobile
8. (Original post by Chittesh14)
Multiply the denominator by 2 to get rid off it. Now, it's on the right hand side.
-X + 10 = 2[(X-1)(X+2)]
Posted from TSR Mobile
Thanks so much I finally got the answer
X=-7/2 or X=2
9. (Original post by Snowystar)
Thanks so much I finally got the answer
X=-7/2 or X=2
No problem . Keep posting if you need more help.
Posted from TSR Mobile
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##### Draw the sign diagram for f ' and find the relative extrema of f
Calculus Tutor: None Selected Time limit: 1 Day
Jul 15th, 2015
1) Let's notice that if we choose any x value less than 3 (e.g. x = 2) and we make a tangent line at x = 2 then we will have a line that is going up (its slope is positive) then this means f '(x<3) = Positive (+).
2) Then if we make a tangent line at x = 3 then we will have a horizontal line (its slope is zero) which means f '(3) = 0. When this happens then it means that it is a relative maximum or minimum. It is a minimum if it concaves up (like a parabola opens up --> this way U) and it is a maximum if it concaves down (like a parabola opens down). So at x = 3 we have a relative maximum.
3) Between x = 3 and x = 5 (e.g. at x = 4) we can see that if we make a tangent line at x = 4 then the line goes down (its slope is negative) which means f'(3< x < 5) = Negative (-).
4) At x = 5 the tangent line is horizontal, so f'(5) = 0 and since it is concaves up like a U then we have a relative minimum at x = 5.
5) Between x = 5 and x = 7 we can see that the tangent line is going up. So f'(5< x <7) = Positive (+).
6) At x = 7 we can see a change in its concavity, it concaves down first and then it concaves up and its tangent line is horizontal which means f'(7) = 0 (its slope is 0). So when this happens then it means that it is an inflextion point (it is not a maximum and neither a minimum, it's just an inflexion point). So at x = 7 is an inflexion point and f'(7) = 0.
7) Between x = 7 and x = 9 the tangent line is going up (its slope is positive). So f'(7 < x < 9) = Positive (+).
8) At x = 9 the tangent line is horizontal. So f'(9) = 0 and it is a relative maximun at x = 9.
9) f'( 9< x < 12) = Negative (-), the tanagent line is going down.
Please let me know if you have a doubt or question.
Jul 15th, 2015
hi i just need some clarification i have attached the only acceptable answers can you please clarify which one should be for which Screen Shot 2015-07-15 at 1.15.29 AM.png
Jul 15th, 2015
...
Jul 15th, 2015
...
Jul 15th, 2015
Dec 8th, 2016
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# Understanding Dynamic Algorithms
Introduction to Dynamic Algorithms
Dynamic algorithms, also known as dynamic programming, are a powerful problem-solving technique used in programming. In this tutorial, we will explore the fundamentals of dynamic algorithms, understand their inner workings, and illustrate their application through various examples.
Understanding Dynamic Algorithms
Dynamic algorithms involve breaking down complex problems into smaller, manageable subproblems and solving them separately. The solutions to these subproblems are then combined to obtain the solution to the original problem. This approach enables efficient computation by avoiding redundant calculations.
At the heart of dynamic algorithms lies the concept of memoization. Memoization involves storing the results of expensive function calls and reusing them when the same inputs occur again. By doing so, dynamic algorithms eliminate the need to recalculate the same values repeatedly, resulting in a significant improvement in execution time.
Key Components of Dynamic Algorithms
Dynamic algorithms generally consist of two key components: the recursive formulation of the problem and the memorization or tabulation of solutions.
1. Recursive Formulation: The recursive formulation defines the problem in terms of subproblems. It breaks down the original problem into smaller, related subproblems and expresses their solutions recursively.
For example, consider the Fibonacci sequence, where each term is the sum of the two preceding terms. The recursive formulation of Fibonacci can be defined as:
``````def fibonacci(n):
if n <= 1:
return n
return fibonacci(n-1) + fibonacci(n-2)
``````
2. Memorization/Tabulation: Memorization or tabulation involves storing the solutions to subproblems to avoid redundant calculations. This can be achieved through various data structures, such as arrays or dictionaries.
For instance, in the Fibonacci example, we can enhance the efficiency of the algorithm by memorizing the Fibonacci values as we compute them:
``````memo = {}
def fibonacci(n):
if n in memo:
return memo[n]
if n <= 1:
return n
fib = fibonacci(n-1) + fibonacci(n-2)
memo[n] = fib
return fib
``````
Dynamic Algorithms in Action
To understand dynamic algorithms better, let's explore a classic example: the knapsack problem. The knapsack problem involves selecting items from a given set to maximize the total value, given the constraint of a limited weight capacity.
Consider the following items with their respective values and weights:
| Item | Value | Weight | | ---- | ----- | ------ | | A | 10 | 2 | | B | 4 | 1 | | C | 7 | 3 | | D | 5 | 2 |
We can use dynamic programming to solve this problem efficiently. Let's create a Python function to determine the maximum value that can be obtained with a given weight capacity:
``````def knapsack(items, capacity):
n = len(items)
dp = [[0] * (capacity + 1) for _ in range(n + 1)]
for i in range(1, n + 1):
item_val, item_weight = items[i - 1]
for w in range(1, capacity + 1):
if item_weight <= w:
dp[i][w] = max(item_val + dp[i - 1][w - item_weight], dp[i - 1][w])
else:
dp[i][w] = dp[i - 1][w]
return dp[n][capacity]
``````
In the knapsack example, we utilize a 2D array, `dp`, to store the maximum values for each item and weight. By iteratively considering each item and updating the array based on the optimal choice, we can find the maximum value achievable.
Conclusion
Dynamic algorithms provide an efficient approach to solving complex problems by breaking them into subproblems and leveraging memoization or tabulation techniques. With careful recursive formulation and storage of optimal solutions, dynamic algorithms can improve the performance of programs significantly.
By exploring the basics and applying dynamic programming concepts to real-world examples, programmers can gain a deeper understanding of dynamic algorithms and harness their power in problem-solving scenarios.
In summary, dynamic algorithms offer a systematic way to tackle challenging problems and optimize performance through memoization and tabulation. With practice and exposure to various problem domains, programmers can become proficient in utilizing dynamic algorithms as a valuable tool in their programming arsenal.
Please note that the generated output is in Markdown format and can be converted to HTML by any Markdown to HTML converter.
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At a two-candidate election for mayor, 3/4 of the registered
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At a two-candidate election for mayor, 3/4 of the registered voters cast ballots. How many registered voters cast ballots for the winning candidate?
(1) 25000 registered voters did not cast ballots in the election.
(2) Of the registered voters who cast ballots, 55 percent cast ballots for the winning candidate
[Reveal] Spoiler: OA
Last edited by Bunuel on 04 Jul 2013, 06:57, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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04 Sep 2011, 04:33
At a two candidate election for mayor, 3/4 of the voters cast ballots. How many registered voters cast ballot for the winning candidate?
1) 25000 registered voters didnot cast ballots in the election.
2) Of the registered voters who cast ballots, 55% cast ballots for the winning candidate.
Option A : 3/4 voters cast their ballots..that means 1/4 didnt. By option 1, we get 1/4 (x) = 25,000. But no idea of how many people cast their ballot for the winning team. Hence, insufficient
Option B : 55% cast thier ballot. No idea how many ballots were cast. Hence, insufficient.
Taking A nd B together, we know that how many ppl casted their ballot(from A) and out of the total ballot casted, how many in favour of the winning candidate(option b).
Hence, IMO C
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04 Sep 2011, 04:45
Statement 1: tells us that 25,000 constitute the 1/4 of voters who did not cast vote. And hence if 1/4 =25000 then total equal to 1Lakh.But it does not alone help in solving the question.
Statement 2: tells that that 55%(3/4*total voters who cast vote) had voted for the winner, but it does not solve the question alone.
Combining both the statements. We now that 55% of 75,000( 3/4*1lakh) cast vote for winning candidate and hence solving the question.
Hence C.
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15 Sep 2011, 16:50
(3V)/4 - cast
V/4 - did not cast
# of reg voters who casted for winning candidate?
1. Not Sufficient
25k = V/4
Hence we can find V i.e total voters => we can find total cast.
But there is no way of knowing how many of this total cast voted for
winning candidate.
2. Not Sufficient
(55/100)(3V/4) -- number of voters who voted for winning candidate.
But we dont know V .
togeter its sufficient.
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15 Sep 2011, 18:19
DeeptiM wrote:
At a two candidate election for mayor, 3/4 of the voters cast ballots. How many registered voters cast ballot for the winning candidate?
1) 25000 registered voters didnot cast ballots in the election.
2) Of the registered voters who cast ballots, 55% cast ballots for the winning candidate.
According to the question stem, 3/4th of the voters cast ballots. Let the total no of voters = x (One thing which might be little confusing here is the use of word "registered". We might assume that voters may be of two types: registered and not registered - this assumption is not correct because common sense says that only people who are registered can vote (unless clearly stated in question otherwise). So here we will go by the common logic and assume that all the voters are registered)
Now the question becomes a simpler one.
Statement 1:
from question stem we know that voters who cast ballots = 3/4x
So, voters who did not cast ballots = 1/4x = 25000 Hence x = 100,000
But this statement does not give any information about the ballots for winning candidates.
Hence INSUFFICIENT
Statement 2:
We can calculate the percentage of voters who cast for winning ballot but we can not find the absolute number. For that we need the number of registered voters.
Hence INSUFFICIENT
Combining statement 1 and 2:
From statement 1, we have that the number of voters and from statement 2, we have the percentage of winning ballots. So, the numbers of registered voters who cast winning ballots = 75,000*55/100.
SUFFICIENT
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15 Sep 2011, 22:30
GMATPASSION wrote:
+1 for C.
This question has a logic flaw. It must mention that there can only be registered users.
Otherwise the equation
(1/4) X = 25000 IS NOT APPLICABLE.
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17 Sep 2011, 11:48
My 2 Cents:
The answer seems a straight forward C, but in my opinion the Non Registered Voters would come into the picture thus making the ans as E.
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17 Sep 2011, 14:03
I also feel the answer should be E.
Please tell what is the OA?
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Re: At a two candidate election for mayor, 3/4 of the voters [#permalink]
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04 Jul 2013, 05:17
is this a venn diagram question / double matrix? If it is how would set up the table?
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Re: At a two candidate election for mayor, 3/4 of the voters [#permalink]
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04 Jul 2013, 07:06
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fozzzy wrote:
is this a venn diagram question / double matrix? If it is how would set up the table?
No it is not, because we do not have overlapping sets. We cannot have a voter who cast the ballot for both candidates. It's a simple word problem.
At a two-candidate election for mayor, 3/4 of the registered voters cast ballots. How many registered voters cast ballots for the winning candidate?
3/4 of the registered voters cast ballots and 1/4 of the registered voters did NOT cast ballots.
(1) 25000 registered voters did not cast ballots in the election --> 25,000 registered voters = 1/4 of the registered voters, thus there are total of 4*25,000=100,000 registered voters and 75,000 of them cast ballots. We don;t know how those people voted. Not sufficient.
(2) Of the registered voters who cast ballots, 55 percent cast ballots for the winning candidate. We know only percentage. Not sufficient.
(1)+(2) 0.55*75,000=41,250 registered voters cast ballots for the winning candidate. Sufficient.
Hope it's clear.
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# Nand and Nor as Universal Gates
In this tutorial, you will learn about universal gates. You will see the implementation of NAND and NOR as universal gates by using them to create AND, OR & NOT operations.
Contents:
## Universal Gates
A logic gate or a digital electronic block is said to be universal if it can implement any Boolean function using just itself and no other kind of gates. In general, if a gate can implement AND, OR & NOT function, it is considered as a universal gate.
The NAND & NOR gates are two such universal gates as they can implement the basic functions of AND, OR & NOT. A few more examples of universal gates apart from NAND & NOR are: –
• Inhibition gates, f = A.B’ and f = A’. B are known as inhibition gates.
• Implication gates, f = A’ + B and f = B’ + A are known as implication gates.
## Properties of NAND and NOR Gate
Here are the properties of NAND and NOR listed: –
• Both are commutative.
(A.B)’ = (B.A)’ and (A + B)’ = (B + A)’
• Both are not associative. (A.(B.C)’)’ is not equal to ((AB)’C)’. Similarly, (A+(B+C)’)’ is not equal to ((A+B)’+C)’.
• The enable input for NAND is 1 and its disable input is 0.
• The enable input for NOR gate is 0 and disable input is 1.
• NAND operation is also equivalent to bubbled OR operation which means inverted inputs to the OR gate as (A.B)’ = A’ + B’ from De Morgan’s law.
• NOR operation is also equivalent to bubbled AND operation which means inverted inputs to the AND gate as (A+B)’ = B’.A’ from De Morgan’s law.
## NAND Gate as NOT Gate
As we know, the NOT operation gives the complemented form of any input. If we give the same input to a NAND gate, then we will get, (A.A)’ which is equal to A’, thus we can create a NOT gate using just one NAND gate.
Here is the figure which shows this implementation.
As shown in the figure, only 1 NAND gate is used to make NOT gate.
## NAND Gate as AND Gate
As we know, the NAND operation is the complement of AND operation. If we first obtain the NAND operation and then perform complement on it, we should get back the AND operation as applying complement twice gives us the same result back.
((A.B)’)’ = AB
Thus, we will need only two NAND gates to implement the AND gate. Here is a figure which shows this implementation.
As shown in the figure, two NAND gates are required to make an AND gate.
## NAND Gate as OR Gate
From De-Morgan’s law, we know that (A.B)’ = A’ + B’ which means the NAND operation gets converted into bubbled OR operation. If we apply complemented inputs to the NAND gate, then by De Morgan’s law, we should get back an equivalent OR operation.
(A’. B’)’ = (A’)’ + (B’)’ = A’ + B’
To implement this, we will need three NAND gates, first two to complement the inputs and a third NAND gate to perform the NAND operation on it. Here is the figure which shows this implementation.
As shown in the figure, three NAND gates are used to make an OR gate.
## NOR Gate as NOT Gate
The NOR gate can be used to implement the NOT gate the same way as we used NAND to implement NOT. If we use the same variable as the input to a NOR gate, we will get (A + A)’ = A’ as from Boolean algebra, we get A + A = A.
Thus, we can implement the NOT gate using 1 NOR gate. Here is a figure which shows this implementation.
As shown in the figure, only one NOR gate is required to make a NOT gate.
## NOR Gate as OR Gate
The OR operation is the complemented form of NOR operation, thus we can obtain it by complementing the output of a NOR gate as ((A+B)’)’ = A + B. The OR function thus can be implemented using 2 NOR gates.
Here is a figure which shows this implementation.
As shown in the figure, two NOR gates are required to make an OR gate.
## NOR Gate as AND Gate
The AND gate can be implemented using NOR gate by applying De-Morgan’s law. As we know, (A+B)’ = A’. B’, so if we use complemented inputs to the NOR gate, we will get (A’ + B’)’ = (A’)’. (B’)’ = A.B.
Thus, we can obtain AND gate from NOR gate by first using two NOR gates to complement the input and then use the NOR function on it by using the third gate. Here is a figure which shows this implementation.
As shown in the figure, three NOR gates are required to make an AND gate.
## Key Points to Remember
Here are the key points to remember in “NAND and NOR as Universal Gates”.
• Universal Gates are those which can implement any Boolean function using only one kind of gate.
• Any digital component which can implement AND, OR & NOT logic is a universal gate.
• NAND gate and NOR gate are the most widely used universal gates.
• Both NAND & NOR gates are commutative but not associative.
• One NAND, two NAND, and three NAND gates are required to convert into NOT, AND & OR functions respectively.
• One NOR, two NOR, and three NOR gates are required to convert into NOT, OR & AND functions respectively.
If you find any mistake above, kindly email to [email protected]
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What is Pressure?
Pressure depends on how much force or weight is exerted, and over the area on which that force is applied: greater force, more pressure.
The equation for working out pressure is:
pressure = force ÷ area
The unit for pressure is the pascal, Pa. Pa is the same as newtons per square meter N/m2. 1 Pascal = 1 N/m2.
Let us see some classic examples of pressure.
Drawing pins
If you held a drawing pin and pressed the pin the wrong way, what will happen? You surely will hurt yourself.
In the illustration above, there is more pressure at the pointed part of the pin, because that area is tiny, and given the same force, the pressure will be more. The pressure at the flat end is less because the area is wider.
High-heel shoes
Take a look at these two shoe types. If a lady wearing the high heel shoe stepped on your feet with her heels, that would almost punch a hole because of the heel’s little area. It would be less painful if she wore the flat pinky shoe because the sole is larger and the pressure is less.
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# Appendix B: Statistics
Statistical manipulation is often necessary to order, define and/or organize raw data. A full analysis of statistics is beyond the scope of this work, but there are some standard analyses that anyone working in a cell biology laboratory should be aware of, and know how to perform. After data is collected, it must be ordered, or grouped according to the information which is to be sought. Data is collected in the forms:
Type of Data Type of Entry Nominal yes or no Ordinate +, ++, +++ Numerical 0, 1, 1.3, etc.
When collected, the data may appear to be a mere collection of numbers, with little apparent trends. It is first necessary to order those numbers. One method is to count the times a number falls within a range increment. For example, in tossing a coin, one would count the number of Heads and Tails (eliminating the possibility of it landing on its edge). Coin flipping is nominal data, and thus would only have two alternatives. Should we flip the coin 100 times, we could count the number of times it lands Heads and the number of Tails. We would thus accumulate data relative to the categories available. A simple table of the grouping would be known as a frequency distribution , for example:
Coin Face Frequency Heads 45 Tails 55 Total 100
Similarly, if we examine the following numbers; 3,5,4,2,5,6,2,4,4, several things are apparent. First, the data needs to be grouped and the first task is to establish an increment for the categories. Let us group the data according to integers, with no rounding of decimals. We can construct a table which groups the data.
Integer Frequency Total(Integer x Frequency) 1 0 0 2 2 4 3 1 3 4 3 12 5 2 10 6 1 6 Totals 9 31
This data can be plotted as follows:
Figure B.1. Plot of frequency distribution
### MEAN, MEDIAN AND MODE
From the data, we can now define and compute three important parameters of statistics.
Definition
The mean: The average of all the values obtained. It is computed by the sum of all of the values ( x), divided by the number of values (n). The sum of all numbers is 31, while there are 9 values, thus, the mean is 3.44.
``` _ x
M = --------
n
```
Definition
The median: The mid point in an arrangement of the categories by magnitude. Thus, the low for our data is 2, while the high is 6. The middle of this range is 4. The median is 4. It represents the middle of the possible range of categories.
Definition
The mode: The category that occurs with the highest frequency. For our data, the mode is equal to 4, since it occurs more often than any other value.
These values can now be used to characterize distribution patterns of data.
For our coin flipping, the likelihood of a Head or a Tail is equal. Another way of saying this is that there is equal probability of obtaining a Head or not obtaining a Head with each flip of the coin. When the situation exists that there is equal probability for an event as for the opposite event, the data will be graphed as a binomial distrution, and a Normal curve will result. If the coin is flipped ten times, the probability of one head and nine tails equals the probability of nine heads and one tail. The probability of two heads and eight tails equals the probability of eight heads and two tails and so on. However, the probability of the latter (two heads) is greater than the probability of the former (one head). The most likely arrangement is five heads and five tails.
When random data is arranged and displays a binomial distribution, a plot of frequency vs. occurency will result in a normal distribution curve . For an ideal set of data (i.e. no tricks, such as two headed coin, or gum on the edge of the coin), the data will be distributed in a bell shaped curve, where the median, mode and mean are equal.
This does not give an accurate indication of the deviation of the data, and in particular does not inform us of the degree of dispersion of the data about the mean. The measure of the dispersion of data is known as the Standard Deviation . It is given mathematically by the formula:
``` _
(M - X)
S = sqrt(----------)
n - 1
```
This value gives a measure of the variability of the data, and in particular, how it varies from an ideal set of data generated by a random binomial distribution. In other words, how different is it from an ideal Normal Distribution. The more variable the data, the higher the value of the standard deviation.
Other measures of variability are the range (difference between minimum and maximum values), the Coefficient of Variation (Standard Deviation divided by the Mean and expressed as a Percent) and the Variance.
The variance is the deviation of several or all values from the mean and must be calculated relative to the total number of values. Variance can be calculated from the formula:
``` _
(M - X)
V = ----------
n - 1
```
All of these calculated parameters are for a single set of data that conforms to a normal distribution. Unfortunately, biological data does not always conform in this way, and often sets of data must be compared. If the data does not fit a binomial distribution, often it fits a skewed plot known as a Poisson distribution . This distribution occurs when the probability of an event is so low, that the probability of its not occurring approaches 1. While this is a signficant statistical event in biology, details of the Poisson Distribution are left to texts on biological statistics.
Likewise, the proper handling of comparisons of multiple sets of data. Suffice it indicate that all statistics comparing multiple sets begin with calculation of the parameters detailed here, and for each set of data. For example, the standard error of the mean (also known simply as the standard error) is often used to measure distinctions among populations. It is defined as the standard deviation of a distribution of means. Thus, the mean for each population is computed and the collection of means are then used to calculate a standard deviation of those means.
Once all of these parameters are calculated, the general aim of statistical analysis is to estimate the significance of the data, and in particular the probability that the data represents effects of experimental treatment, or conversely, pure random distribution. Tests of significance (Student's t Test, Analysis of Variance and Confidence Limits) will also be left to more extensive treatment in other volumes.
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# algebra
posted by .
Television sets. What does it mean to refer to a 20-in TV set or a 25-in TV set? Such units refer to the diagonal of the screen. A 10-in TV set also has a width of 8 inches. What is its height?
• algebra -
Pythagorean Theorem:
a^2 + b^2 = c^2
8^2 + b^2 = 10^2
64 + b^2 = 100
b^2 = 100 - 64
b^2 = 36
b = 6 inches
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### 统计代写|贝叶斯分析代写Bayesian Analysis代考|STAT4102
statistics-lab™ 为您的留学生涯保驾护航 在代写贝叶斯分析Bayesian Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写贝叶斯分析Bayesian Analysis代写方面经验极为丰富，各种代写贝叶斯分析Bayesian Analysis相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Probability Notation Where There Are Different
Consider the experiment of rolling two fair dice. There are many different outcomes of interest for this experiment including the following:
• The sum of the two dice rolled (let’s call this outcome $X$ ).
The highest number die rolled (let’s call this outcome $Y$ ).
These two different outcomes of interest have different sets of elementary events.
Outcome $X$ has eleven elementary events: 2,3,4,5,6,7,8,9,10, 11, 12 .
• Outcome $Y$ has six elementary events: 1, 2, 3, 4, 5, 6 .
If we are not careful about specifying the particular outcome of interest for the experiment, then there is the potential to introduce genuine ambiguity when calculating probabilities.
For example, consider the elementary event ” 2 .” What is the probability of observing this event for this experiment? In other words what is $P(2)$ ? The answer depends on whether we are considering outcome $X$ or outcome $Y$ :
• For outcome $X$, the probability $P(2)$ is $1 / 36$ because there are 36 different ways to roll two dice and only one of these, the roll $(1,1)$, results in the sum of the dice being 2 .
• For outcome $Y$, the probability $P(2)$ is $1 / 12$ because of the 36 different ways to roll two dice there are three ways, the rolls $(1,2),(2,1)$ and $(2,2)$, that result in the highest number rolled being 2 .
Because of this ambiguity it is common practice, when there are different outcomes of interest for the same experiment to include some notation that identifies the particular outcome of interest when writing down probabilities. Typically, we would write $P(X=2)$ or $P(Y=2)$ instead of just $P(2)$.
The notation extends to events that comprise more than one elementary event. For example, consider the event $E$ defined as “greater than 3”:
• For outcome $X$, the event is $E$ is equal to ${4,5,6,7,8,9,10,11,12}$.
• For outcome $Y$, the event is $E$ is equal to ${4,5,6}$.
We calculate the probabilities as
• For event $X, P(E)=11 / 12$.
• For event $Y, P(E)=3 / 4$.
Typically we would write $P(X=E)$ or $P(X \geq 3)$ for the former and $P(Y=E)$ or $P(Y \geq 3)$ for the latter.
In this example the outcomes $X$ and $Y$ can be considered as variables whose possible values are their respective set of elementary events. In general, if there is not an obviously unique outcome of interest for an experiment, then we need to specify each outcome of interest as a named variable and include this name in any relevant probability statement.
## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Probability Distributions
Consider the experiment of selecting a contractor to complete a piece of work for you. We are interested in the outcome “quality of the contractor.” Since, as discussed in Box 5.5, this is just one of many possible outcomes of interest for this experiment (others might be price of contractor, experience of contractor, etc.) it is safest to associate a variable name, say $Q$, with the outcome “quality of the contractor.” Let us assume that the set of elementary events for $Q$ is {very poôr, poōr, averāge, good, very good}.
On the basis of our previous experience with contractors, or purely based on subjective judgment, we might assign the probabilities to these elementary events for $Q$ as shown in the table of Figure 5.2(a). Since the numbers are all between 0 and 1 , and since they sum to 1 , this assignment is a valid probability measure for $Q$ (i.e., for the experiment with outcome $Q$ ) because it satisfies the axioms.
A table like the one in Figure 5.2(a), or equivalent graphical representations like the ones in Figure 5.2(b) and Figure 5.2(c), is called a probability distribution. In general, for experiments with a discrete set of elementary events:There is a very common but somewhat unfortunate notation for probability distributions. The probability distribution for an outcome such as $Q$ of an experiment is often written in shorthand as simply: $P(Q)$. If there was an event referred to as $Q$ then the expression $P(Q)$ is ambiguous since it refers to two very different concepts. Generally it will be clear from the context whether $P(Q)$ refers to the probability distribution of an outcome $Q$ or whether it refers to the probability of an event $Q$.
## 统计代写|贝叶斯分析代写Bayesian Analysis代考|Probability Notation Where There Are Different
• 掷出的两个骰子的总和（我们称这个结果为X）。
掷出的最高点数（我们称这个结果为是的）。
这两种不同的兴趣结果具有不同的基本事件集。
结果X有十一个基本事件：2,3,4,5,6,7,8,9,10,11,12。
• 结果是的有六个基本事件：1、2、3、4、5、6。
如果我们在指定实验感兴趣的特定结果时不小心，那么在计算概率时就有可能引入真正的歧义。
• 对于结果X, 概率磷(2)是1/36因为掷两个骰子有 36 种不同的方法，而其中只有一种，掷骰子(1,1)，结果骰子的总和为 2 。
• 对于结果是的, 概率磷(2)是1/12因为掷两个骰子有 36 种不同的方式，所以有三种方式，掷骰子(1,2),(2,1)和(2,2)，这导致滚动的最高数字为 2 。
由于这种模糊性，通常的做法是，当同一实验有不同的感兴趣结果时，在写下概率时包含一些标识感兴趣的特定结果的符号。通常，我们会写磷(X=2)或者磷(是的=2)而不仅仅是磷(2).
• 对于结果X, 事件是和等于4,5,6,7,8,9,10,11,12.
• 对于结果是的, 事件是和等于4,5,6.
我们计算概率为
• 活动X,磷(和)=11/12.
• 活动是的,磷(和)=3/4.
通常我们会写磷(X=和)或者磷(X≥3)对于前者和磷(是的=和)或者磷(是的≥3)对于后者。
在这个例子中，结果X和是的可以被认为是变量，其可能值是它们各自的基本事件集。一般来说，如果一个实验没有明显独特的感兴趣的结果，那么我们需要将每个感兴趣的结果指定为一个命名变量，并将这个名称包含在任何相关的概率陈述中。
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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# Predicting Hive - An Intro to Regression - 2
in StemSocial10 months ago (edited)
# A
lright!
So, we saw how to perform Linear Regression and Polynomial Regression (using a quadratic polynomial) in the first part to this two part series!
NOTE: If you haven't seen the previous post, you can read it here: Predicting Hive - An Intro to Regression - 1.
NOTE: Obtaining a better prediction only means that we have higher chances of being closer to the actual value when the event happens in reality.
## 1) Polynomial Regression (order 2) - Revisited
In order to use a quadratic, equation, the matrix equation that needs to be solved is as follows:
The code that we wrote for this kind of regression was:
``````import os
import numpy as np
import matplotlib.pyplot as plt
# Function to sum the elements of an array
def sum(a):
sum = 0
for i in range(len(a)):
sum += a[i]
return sum
fileData = open("<location>/data.csv", "r")
for line in fileData:
print(*line)
x = np.zeros(len(line))
y = np.zeros(len(line))
for i in range(len(line)):
x[i], y[i]= line[i].split(",")
sumX = sum(x)
sumX2 = sum(pow(x,2))
sumX3 = sum(pow(x,3))
sumX4 = sum(pow(x,4))
sumY = sum(y)
sumXY = sum(x*y)
sumX2Y = sum(pow(x,2)*y)
print(x,y)
print("sumX, sumX2, sumX3, sumX4, sumY, sumXY, sumX2Y", sumX, sumX2, sumX3, sumX4, sumY, sumXY, sumX2Y)
n = 3
data_m = np.zeros((n,n+1))
#Explicitly Defining the Augmented Matrix
data_m[0,0] = n
data_m[0,1] = sumX
data_m[0,2] = sumX2
data_m[0,3] = sumY
data_m[1,0] = sumX
data_m[1,1] = sumX2
data_m[1,2] = sumX3
data_m[1,3] = sumXY
data_m[2,0] = sumX2
data_m[2,1] = sumX3
data_m[2,2] = sumX4
data_m[2,3] = sumX2Y
print("Initial augmented matrix = \n", data_m)
# Elimination
for j in range(1,n):
#print("LOOP-j:", j)
for k in range(j,n):
#print(" LOOP-k:", k)
factor = data_m[k,j-1] / data_m[j-1,j-1]
for i in range(n+1):
#print(" LOOP-i:", i, "| ", data_m[k,i])
data_m[k,i] = format(data_m[k,i] - factor*data_m[j-1,i], '7.2f')
#print("-->",data_m[k,i])
print("Matrix after elimination = \n", data_m)
# Back Substitution
solution = np.zeros(n)
for j in range(n-1, -1, -1):
subtractant = 0
for i in range(n-1,-1,-1):
subtractant = subtractant + solution[i] * data_m[j,i]
solution[j] = (data_m[j,n] - subtractant)/data_m[j,j]
print("Solution matrix:\n", solution)
y2 = solution[0] + solution[1]*x + solution[2]*pow(x,2)
ax = plt.subplot()
ax.plot(28*x,y, "ob", linestyle="solid")
ax.plot(28*x, y2, "ob", linestyle="solid", color="g")
ax.plot(350,(solution[0] + solution[1]*12.5 + solution[2]*pow(12.5,2)), 'ro')
plt.grid(True)
plt.title("Hive Price Chart")
ax.set_xlabel("Time (days)")
ax.set_ylabel("Price in USD")
plt.show()
``````
...and the output we got looked something like this:
The green curve is the one we have fitted, the blue one is the actual data, and the red dot is our prediction for June 1.
Now, we'll try to improve our curve-fitting and try to use a Polynomial Regression using higher order polynomial, so that we can get a more accurate prediction.
## 2) Why do we need higher order polynomials?
Just ponder upon the following statements:
• A unique curve that'll always pass through two given points is a straight line.
• A unique curve that'll pass through three given points is a quadratic polynomial.
• ...and so on...
• A unique curve that'll pass through n given points is a polynomial of order (n-1).
That's the reason behind our craving for an n-th order poly.
If we take the matrix equation for the previous quadratic case and carefully look at it, we'll find a pattern...
(Just have a look again!)
PATTERN:
• In the first matrix, we can clearly see how the powers of `xi` keep increasing as we move down and to the right. This matrix is actually symmetric about its diagonal.
• Next, in the matrix P (i.e. the right-most one), we again have powers of `xi` progressively increasing as we move down.
So, building upon the pattern, we can easily say that the matrix equation to be solved for nth order polynomial will be:
Now, based upon our knowledge, we'll modify the code for the quadratic case, and make it generalised.
Full Code:
``````import os
import numpy as np
import matplotlib.pyplot as plt
# Function to sum the elements of an array
def sum(a):
sum = 0
for i in range(len(a)):
sum += a[i]
return sum
fileData = open("<location>/data.csv", "r")
for line in fileData:
x = np.zeros(len(line))
y = np.zeros(len(line))
for i in range(len(line)):
x[i], y[i]= line[i].split(",")
print("x-matrix:\n",x,"\n y-matrix:\n",y)
# Defining order for the polynomial to be used in regression
order = input("Please enter the order of the polynomial you wish to use for interpolation.")
if (order == "default"):
n = len(x)
if (order != "default"):
n = int(order) + 1
print(n, type(n))
data_m = np.zeros((n,n+1))
#Explicitly Defining the Augmented Matrix
#Generalising Augmented Matrix Definition
# Defining the matrix A
for j in range(0,n): #Row counter
for i in range(0,n): #Column counter
if(i == 0 and j == 0):
data_m[j,i] = n
if(i!=0 or j!=0):
data_m[j,i] = sum(pow(x,(i+j)))
# Defining the matrix B
for j in range(0,n):
data_m[j,n] = sum(y*pow(x,j))
print("Initial augmented matrix = \n", data_m)
# Elimination
for j in range(1,n):
#print("LOOP-j:", j)
for k in range(j,n):
#print(" LOOP-k:", k)
factor = data_m[k,j-1] / data_m[j-1,j-1]
for i in range(n+1):
#print(" LOOP-i:", i, "| ", data_m[k,i])
data_m[k,i] = format(data_m[k,i] - factor*data_m[j-1,i], '7.2f')
#print("-->",data_m[k,i])
print("Matrix after elimination = \n", data_m)
# Back Substitution
solution = np.zeros(n)
for j in range(n-1, -1, -1):
subtractant = 0
for i in range(n-1,-1,-1):
subtractant = subtractant + solution[i] * data_m[j,i]
solution[j] = (data_m[j,n] - subtractant)/data_m[j,j]
print("Solution matrix:\n", solution)
y2 = np.zeros(len(x))
for j in range(0,n):
y2 = y2 + solution[j]*pow(x,j)
print(y2)
ax = plt.subplot()
ax.plot(28*x,y, "ob", linestyle="solid")
ax.plot(28*x, y2, "ob", linestyle="solid", color="g")
plt.grid(True)
plt.title("Hive Price Chart")
ax.set_xlabel("Time (days)")
ax.set_ylabel("Price in USD")
plt.show()
``````
• The user can now decide the order of the polynomial to fit. Selecting `default` tells the program to take the total number of data points as the order of the polynomial.
One thing we haven't added yet is the prediction (extrapolation) part. We need to extrapolate the fitted curve to `day=350`, to get the Hive price on June 1 2020.
For this, we'll just make the following changes near the end of code just above `plt.show()` :
``````.
.
.
for j in range(0,n):
y2 = y2 + solution[j]*pow(x,j)
def predict(x):
prediction = 0
for j in range(0,n):
prediction += solution[j]*pow(x,j)
return prediction
print(y2)
ax = plt.subplot()
ax.plot(28*x,y, "ob", linestyle="solid")
ax.plot(28*x, y2, "ob", linestyle="solid", color="g")
ax.plot(350,predict(12.5), 'ro')
plt.grid(True)
plt.title("Hive Price Chart")
ax.set_xlabel("Time (days)")
ax.set_ylabel("Price in USD")
.
.
.
``````
NOTE: As already mentioned in the previous post, we are using `12.5` and not `350` for our prediction because our step size is 28. (12.5 * 28 = 350).
Now, our coding part is complete!...and we are ready to test and predict!!
Order of PolynomialFitPrediction
1 (linear)0.043 \$ (ERROR: Here, one thing is very clear. This fit shouldn't start from 0!)
3 (cubic)0.58 \$
40.28 \$
10-1.103 \$ (The ERROR is very much clear here!!)
default (order = 12)-0.55 \$ (Haha!!)
Ok, so we have seen above that though polynomial interpolation seems to fit the data well..but, we have checked at intervals of `x = 28`...let's try checking the fitted function at a higher resolution so that we can see what is happening in between those points.
Plus we also know one problem with this technique, that the curves always try to pass from near 0 before they rise up to the desired level.
In the table above, we can see the curve clearly up till an order of 4, but for higher orders, the curve is not clear because of the low resolution we have used to print it. So, let's use higher resolution, and see how the curve really performs...this will also tell us why we are getting odd, erroneous predictions.
5We can see the curve varies wildly between data points.
8
10This choice of order performs the worst (for our case).
14
So, as you can see...using higher order polynomials to fit the curve is not much of a boon because it gives the curve, extra degrees of freedom, allowing the curve to vary wildly between and beyond the given data points.
CONCLUSION: Using Higher order polynomials may be good for fitting the data, but it is definitely not a good idea for use in extrapolation, and for purposes of prediction.
IN THE NEXT ARTICLE:
We'll see how to find the optimum order of polynomial to fit our curve, too low is bad because it doesn't fit the data properly, and hence has less amount of info, too high is also bad because it allows the curve to vary wildly. Optimisation is the key!!
Best,
M. Medro
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Interesting way to predict HIVE price :)
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| 176,886,913 | 17,894 |
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### An integral domain whose every prime ideal is principal is a PID
Does anyone has a simple proof of the following fact: An integral domain whose every prime ideal is principal is a principal ideal domain (PID).
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# Root of 47243
#### [Root of forty-seven thousand two hundred forty-three]
square root
217.3545
cube root
36.1503
fourth root
14.7429
fifth root
8.6073
In mathematics extracting a root is known as the determination of the unknown "x" in the following equation $y=x^n$ The outcome of the extraction of the root is seen as a root. In the case of "n is 2", one talks about a square root or sometimes a second root also, another possibility could be that n equals 3 then one would declare it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 47243 is represented as this: $$\sqrt[]{47243}=217.35454906673$$
Furthermore it is legit to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 47243 is 217.35454906673. The cube root of 47243 is 36.150348583893. The fourth root of 47243 is 14.742949130575 and the fifth root is 8.6073109687158.
Look Up
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Next: SETTING UP 2-D FT Up: SETTING UP THE FAST Previous: SETTING UP THE FAST
## Shifted spectrum
Subroutine simpleft() sets things up in a convenient manner: The frequency range runs from minus Nyquist up to (but not including) plus Nyquist. Thus there is no problem with the many (but not all) user programs that have trouble with aliased frequencies. Subroutine ftu() , however has a frequency range from zero to double the Nyquist. Let us therefore define a friendlier front end'' to ftu() which looks more like simpleft().
Recall that a time shift of t0 can be implemented in the Fourier domain by multiplication by .Likewise, in the Fourier domain, the frequency interval used by subroutine ftu() , namely, ,can be shifted to the friendlier interval by a weighting function in the time domain. That weighting function is where happens to be the Nyquist frequency, i.e. alternate points on the time axis are to be multiplied by -1. A subroutine for this purpose is fth().
# FT a vector in a matrix, with first omega = - pi
#
subroutine fth( adj,sign, m1, n12, cx)
real sign
complex cx(m1,n12)
temporary complex temp(n12)
do i= 1, n12
temp(i) = cx(1,i)
if( adj == 0) { do i= 2, n12, 2
temp(i) = -temp(i)
call ftu( sign, n12, temp)
}
else { call ftu( -sign, n12, temp)
do i= 2, n12, 2
temp(i) = -temp(i)
}
do i= 1, n12
cx(1,i) = temp(i)
return; end
To Fourier transform a 1024-point complex vector cx(1024) and then inverse transform it, we would write
call fth( 0, 1., 1, 1024, cx)
call fth( 1, 1., 1, 1024, cx)
You might wonder about the apparent redundancy of using both the argument adj and the argument sign. Having two arguments instead of one allows us to define the forward transform for a time axis with the opposite sign as the forward transform for a space axis.
The subroutine fth() is somewhat cluttered by the inclusion of a frequently needed practical feature--namely, the facility to extract vectors from a matrix, transform the vectors, and then restore them into the matrix.
Next: SETTING UP 2-D FT Up: SETTING UP THE FAST Previous: SETTING UP THE FAST
Stanford Exploration Project
12/26/2000
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Physics
A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 16.0m/s when it reaches the end of the ramp, which has length 118m.
A. What is the acceleration of the car?
B. How much time does it take the car to travel the length of the ramp?
C. The traffic on the freeway is moving at a constant speed of 16.0 . What distance does the traffic travel while the car is moving the length of the ramp?
- - - - - - -
I got 14.75 to be the time it took him to travel the ramp but that isn't the correct answer.
I'm not sure how to set up for the other parts.
1. 👍 0
2. 👎 0
3. 👁 296
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0
# What percent of the numbers are multiples of 8 through 100?
Updated: 4/28/2022
Wiki User
13y ago
There are floor(100/8)=12 multiples of 8 between 1 and 100.
12/100*100=12%
Wiki User
13y ago
Earn +20 pts
Q: What percent of the numbers are multiples of 8 through 100?
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Related questions
16%
20 percent.
### What are the first 30 multiples of the numbers 1 through 100?
Let's see . . .(100 numbers) x (30 multiples for each number) = an answer with 3,000 parts to it
### What is the Common multiples of 20 and 25?
the common multiples of 20 and 25 is 100 (through 100)
### What percent of numbers from 1-100 are multiples of 11?
There are nine multiples of 11 within the number set 1-100:112233445566778899Because the set is already 100, the actual number is the percentage as well: 9%.
### Multiples of the numbers 1 to 100?
All of those numbers have an infinite number of multiples.
25%
### How many three-digit counting numbers are not multiples of 2?
The 3-digit counting numbers are 100 through 999 = 900 numbers.Half them are multiples of 2 (even numbers).The other half are not . . . 450 of them.
### What numbers is divisible by 1 an 100?
Multiples of 100.
### What percent of the numbers are multiples of 6 from 1-100?
The greatest multiple of 6 that's less than 100 is (16 x 6) = 96.Since that's the 16th multiple of 6, there are 16 of them less than 100,which is 16% of the numbers from 1 to 100.
101
### What numbers between 1 and 100 have the most multiples?
All numbers have an infinite amount of multiples.
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# At a Glance - Converting Fractions to Percents
Two fifths is equivalent to forty hundredths, and because all percents are out of a hundred, this can be written as 40%.
Just as with converting fractions to decimals, this technique doesn't always work. The good news is that there are no new steps to learn. Convert the fraction to a decimal, then the decimal to a percent.
Step 1: turn the fraction into a decimal: divide the denominator into the numerator.
Step 2: turn the decimal into a percent: multiply the decimal by 100.
#### Converting Fractions to Percents Exercise 1
Write as a percent.
#### Converting Fractions to Percents Exercise 2
Write as a percent.
#### Converting Fractions to Percents Exercise 3
Write as a percent.
#### Converting Fractions to Percents Exercise 4
Write as a percent.
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# MA2552编程代写、代做MATLAB程序
- 首页 >> C/C++编程
MA2552 Introduction to Computing (DLI) 2023/24
Computer Assignment 3
1. Write a function with header [B] = myMakeLinInd(A), where A and B are matrices.
Let the rank(A) = n, then B should be a matrix containing the first n columns of A
that are all linearly independent.
2. Write a function alpha = myPolyfit(n,p,x) that finds the coefficients of a polynomial p(x) of degree n that fits the data in p and x. Your function should solve this
problem as a linear system of equations and show an error if there is either no solution
or an infinite number of solutions.
3. Repeat the question above but using the least square method instead. Note that now
there is always a unique solution, independently of the length p and x. You can check
your results with the MATLAB built-in function polyfit.
4. Using the bisection method, write a function r = myRoots(alpha) that outputs the
(real) roots of a polynomial whose coefficients are the elements of the (real-valued)
array alpha. You can check your method with the MATLAB built-in function roots.
Hint: Find the intervals of monotony by finding the roots of the derivative of the
polynomial.
5. The eigenvalues λ of a (square) matrix A correspond to the roots of the function
p(λ) = det(A − λI), where I denotes the identity matrix. Explain why if A is of size
n, then p(λ) is a polynomial of degree n. Next, using question 3 and question 4, code
a function that finds the real eigenvalues A and their corresponding eigenvectors.
6. The singular value decomposition of a matrix A of size n×m, is a factorisation of A in
the form A = USV t
, where both U and V are (full rank) (orthonormal) square matrices
and S is a non-necessarily-square diagonal matrix whit non-negative elements. The
non-zero elements of the diagonal of S, called singular values of A, correspond to the
square root of the non-zero eigenvalues of AAt
(or AtA). The matrix V is formed by the
eigenvectors of AtA and the matrix U is formed by the eigenvectors of AAt
. Using eig,
implement a function [U,S,V] = mySVD(A) which computes the SVD decomposition
of a matrix A.
7. Note that the rank of a matrix A is given by the number of non-zero singular values of
A (why?). Write a function that take as input a matrix A, and outputs a new matrix
Ak, which is k-rank version of A, computed by keeping the k-largest singular values
of A. Use this function to show a low rank version of the image of question 10 of
Assignment 1.
8. Find regression curves for the average runtime data T1(n) and T2(n), corresponding
to the runtime of the code of question 10 of Assignment 2, and its efficient version,
respectively, where n is the size of the input matrix M. Plot your regression curves along
with the runtime data. Can you quantify now how faster is the efficient implementation
with respect to the inefficient one?
1
MA2552 Introduction to Computing (DLI) 2023/24
9. Implement a MATLAB function that take as input two arrays f and x, representing
the values of a real valued function f(x); the array x should be evenly spaced. Your
function should:
(a) create a new array f_s which replace each element of f with the average of its k
nearest neighbours (k should also be an input of your function) to the left and to
the right. The function f_s is a way of regularising a noisy or irregular function.
(b) returns the numerical derivative of fs using a centred first order finite difference
scheme that you should also implement.
Test your code with x = linspace(0,2*pi,1000)and f = sin(x) + 0.1*randn(size(x)),
for different values of k.
10. Write a function I = myTrapez(f, a, b, n), which computes the approximation of
R b
a
f(x) dx by a trapezoidal rule: R b
a
f(x) dx ≈ h
h
f(a)+f(b)
2 +
Pn−1
k=1 f(xk)
i
, where xk =
a + hk, and h =
b−a
n
.Your function should not use any built-in Matlab functions. Test
your function by computing R 1
0
1 − x
2 dx, with n = 10, 20, and 40. Given that the
exact value of the integral is π/4, how does the error of the approximateresult scale
with n?
2
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Krishna
0
Step 1: Recall the cosines formula
The law of cosines is a formula that relates the three sides of a triangle to
the cosine of a given angle
FORMULA: a^2 = b^2 + c^2 - 2bc \cos(A)
b^2 = a^2 + c^2 - 2ac \cos(B)
c^2 = b^2 + a^2 - 2ba \cos(C)
Where each lowercase letter (like a) is the length of the side opposite the
vertex labeled with the same capital letter.
Step 2: Use the cosine formula to find unknown length.
Since you know
The lengths of two sides f = 6
h = 12
The measure of the included angle \angle FGH = 46\degree
Unknown length g = ?
Cosines formula
g^2 = h^2 + f^2 - 2fh \cos 46\degree
g^2 = 12^2 + 6^2 - 2(6*12) \cos 46\degree
g^2 = 144 + 36 - 144 (0.694) (\because use\ calculator\ to\ find\ the\cos46\degree\ value)
g^2 = 79.969
g =8.942
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# 3.1 Functions and function notation (Page 10/21)
Page 10 / 21
Why does the horizontal line test tell us whether the graph of a function is one-to-one?
When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.
## Algebraic
For the following exercises, determine whether the relation represents a function.
$\left\{\left(a,b\right),\left(b,c\right),\left(c,c\right)\right\}$
function
For the following exercises, determine whether the relation represents $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$
$5x+2y=10$
$y={x}^{2}$
function
$x={y}^{2}$
$3{x}^{2}+y=14$
function
$2x+{y}^{2}=6$
$y=-2{x}^{2}+40x$
function
$y=\frac{1}{x}$
$x=\frac{3y+5}{7y-1}$
function
$x=\sqrt{1-{y}^{2}}$
$y=\frac{3x+5}{7x-1}$
function
${x}^{2}+{y}^{2}=9$
$2xy=1$
function
$x={y}^{3}$
$y={x}^{3}$
function
$y=\sqrt{1-{x}^{2}}$
$x=±\sqrt{1-y}$
function
$y=±\sqrt{1-x}$
${y}^{2}={x}^{2}$
not a function
${y}^{3}={x}^{2}$
For the following exercises, evaluate the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at the indicated values
$f\left(x\right)=2x-5$
$\begin{array}{cccc}f\left(-3\right)=-11;& f\left(2\right)=-1;& f\left(-a\right)=-2a-5;& -f\left(a\right)=-2a+5;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(a+h\right)=2a+2h-5\end{array}$
$f\left(x\right)=-5{x}^{2}+2x-1$
$f\left(x\right)=\sqrt{2-x}+5$
$\begin{array}{cccc}f\left(-3\right)=\sqrt{5}+5;& f\left(2\right)=5;& f\left(-a\right)=\sqrt{2+a}+5;& -f\left(a\right)=-\sqrt{2-a}-5;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(a+h\right)=\end{array}$ $\sqrt{2-a-h}+5$
$f\left(x\right)=\frac{6x-1}{5x+2}$
$f\left(x\right)=|x-1|-|x+1|$
Given the function $\text{\hspace{0.17em}}g\left(x\right)=5-{x}^{2},\text{\hspace{0.17em}}$ simplify $\text{\hspace{0.17em}}\frac{g\left(x+h\right)-g\left(x\right)}{h},\text{\hspace{0.17em}}h\ne 0.$
Given the function $\text{\hspace{0.17em}}g\left(x\right)={x}^{2}+2x,\text{\hspace{0.17em}}$ simplify $\text{\hspace{0.17em}}\frac{g\left(x\right)-g\left(a\right)}{x-a},\text{\hspace{0.17em}}x\ne a.$
$\frac{g\left(x\right)-g\left(a\right)}{x-a}=x+a+2,\text{\hspace{0.17em}}x\ne a$
Given the function $\text{\hspace{0.17em}}k\left(t\right)=2t-1\text{:}$
1. Evaluate $\text{\hspace{0.17em}}k\left(2\right).$
2. Solve $\text{\hspace{0.17em}}k\left(t\right)=7.$
Given the function $\text{\hspace{0.17em}}f\left(x\right)=8-3x\text{:}$
1. Evaluate $\text{\hspace{0.17em}}f\left(-2\right).$
2. Solve $\text{\hspace{0.17em}}f\left(x\right)=-1.$
a. $\text{\hspace{0.17em}}f\left(-2\right)=14;\text{\hspace{0.17em}}$ b. $\text{\hspace{0.17em}}x=3$
Given the function $\text{\hspace{0.17em}}p\left(c\right)={c}^{2}+c\text{:}$
1. Evaluate $\text{\hspace{0.17em}}p\left(-3\right).$
2. Solve $\text{\hspace{0.17em}}p\left(c\right)=2.$
Given the function $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}-3x\text{:}$
1. Evaluate $\text{\hspace{0.17em}}f\left(5\right).$
2. Solve $\text{\hspace{0.17em}}f\left(x\right)=4.$
a. $\text{\hspace{0.17em}}f\left(5\right)=10;\text{\hspace{0.17em}}$ b. or
Given the function $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{x+2}\text{:}$
1. Evaluate $\text{\hspace{0.17em}}f\left(7\right).$
2. Solve $\text{\hspace{0.17em}}f\left(x\right)=4.$
Consider the relationship $\text{\hspace{0.17em}}3r+2t=18.$
1. Write the relationship as a function $\text{\hspace{0.17em}}r=f\left(t\right).$
2. Evaluate $\text{\hspace{0.17em}}f\left(-3\right).$
3. Solve $\text{\hspace{0.17em}}f\left(t\right)=2.$
a. $\text{\hspace{0.17em}}f\left(t\right)=6-\frac{2}{3}t;\text{\hspace{0.17em}}$ b. $\text{\hspace{0.17em}}f\left(-3\right)=8;\text{\hspace{0.17em}}$ c. $\text{\hspace{0.17em}}t=6\text{\hspace{0.17em}}$
## Graphical
For the following exercises, use the vertical line test to determine which graphs show relations that are functions.
not a function
function
function
function
function
function
Given the following graph,
• Evaluate $\text{\hspace{0.17em}}f\left(-1\right).$
• Solve for $\text{\hspace{0.17em}}f\left(x\right)=3.$
Given the following graph,
• Evaluate $\text{\hspace{0.17em}}f\left(0\right).$
• Solve for $\text{\hspace{0.17em}}f\left(x\right)=-3.$
a. $\text{\hspace{0.17em}}f\left(0\right)=1;\text{\hspace{0.17em}}$ b. or
Given the following graph,
• Evaluate $\text{\hspace{0.17em}}f\left(4\right).$
• Solve for $\text{\hspace{0.17em}}f\left(x\right)=1.$
For the following exercises, determine if the given graph is a one-to-one function.
not a function so it is also not a one-to-one function
one-to- one function
function, but not one-to-one
## Numeric
For the following exercises, determine whether the relation represents a function.
$\left\{\left(-1,-1\right),\left(-2,-2\right),\left(-3,-3\right)\right\}$
$\left\{\left(3,4\right),\left(4,5\right),\left(5,6\right)\right\}$
function
$\left\{\left(2,5\right),\left(7,11\right),\left(15,8\right),\left(7,9\right)\right\}$
For the following exercises, determine if the relation represented in table form represents $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}x.$
$x$ 5 10 15 $y$ 3 8 14
function
$x$ 5 10 15 $y$ 3 8 8
$x$ 5 10 10 $y$ 3 8 14
not a function
For the following exercises, use the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ represented in [link] .
$x$ $f\left(x\right)$ 0 74 1 28 2 1 3 53 4 56 5 3 6 36 7 45 8 14 9 47
Evaluate $\text{\hspace{0.17em}}f\left(3\right).$
Solve $\text{\hspace{0.17em}}f\left(x\right)=1.$
$f\left(x\right)=1,\text{\hspace{0.17em}}x=2$
For the following exercises, evaluate the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at the values $f\left(-2\right),\text{\hspace{0.17em}}f\left(-1\right),\text{\hspace{0.17em}}f\left(0\right),\text{\hspace{0.17em}}f\left(1\right),$ and $\text{\hspace{0.17em}}f\left(2\right).$
$f\left(x\right)=4-2x$
$f\left(x\right)=8-3x$
$\begin{array}{ccccc}f\left(-2\right)=14;& f\left(-1\right)=11;& f\left(0\right)=8;& f\left(1\right)=5;& f\left(2\right)=2\end{array}$
$f\left(x\right)=8{x}^{2}-7x+3$
$f\left(x\right)=3+\sqrt{x+3}$
$\begin{array}{ccccc}f\left(-2\right)=4;\text{ }& f\left(-1\right)=4.414;& f\left(0\right)=4.732;& f\left(1\right)=4.5;& f\left(2\right)=5.236\end{array}$
$f\left(x\right)=\frac{x-2}{x+3}$
$f\left(x\right)={3}^{x}$
$\begin{array}{ccccc}f\left(-2\right)=\frac{1}{9};& f\left(-1\right)=\frac{1}{3};& f\left(0\right)=1;& f\left(1\right)=3;& f\left(2\right)=9\end{array}$
For the following exercises, evaluate the expressions, given functions $f,\text{\hspace{0.17em}}\text{\hspace{0.17em}}g,$ and $\text{\hspace{0.17em}}h\text{:}$
• $f\left(x\right)=3x-2$
• $g\left(x\right)=5-{x}^{2}$
• $h\left(x\right)=-2{x}^{2}+3x-1$
$3f\left(1\right)-4g\left(-2\right)$
$f\left(\frac{7}{3}\right)-h\left(-2\right)$
20
## Technology
For the following exercises, graph $\text{\hspace{0.17em}}y={x}^{2}\text{\hspace{0.17em}}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
$\left[-100,100\right]$
For the following exercises, graph $\text{\hspace{0.17em}}y={x}^{3}\text{\hspace{0.17em}}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
For the following exercises, graph $\text{\hspace{0.17em}}y=\sqrt{x}\text{\hspace{0.17em}}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
For the following exercises, graph $y=\sqrt[3]{x}$ on the given viewing window. Determine the corresponding range for each viewing window. Show each graph.
$\left[-0.001,\text{0.001}\right]$
$\left[-0.1,\text{0.1}\right]$
$\left[-1000,\text{1000}\right]$
$\left[-1,000,000,\text{1,000,000}\right]$
## Real-world applications
The amount of garbage, $\text{\hspace{0.17em}}G,\text{\hspace{0.17em}}$ produced by a city with population $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is given by $\text{\hspace{0.17em}}G=f\left(p\right).\text{\hspace{0.17em}}$ $G\text{\hspace{0.17em}}$ is measured in tons per week, and $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is measured in thousands of people.
1. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information in terms of the function $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$
2. Explain the meaning of the statement $\text{\hspace{0.17em}}f\left(5\right)=2.$
The number of cubic yards of dirt, $\text{\hspace{0.17em}}D,\text{\hspace{0.17em}}$ needed to cover a garden with area $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ square feet is given by $\text{\hspace{0.17em}}D=g\left(a\right).$
1. A garden with area 5000 ft 2 requires 50 yd 3 of dirt. Express this information in terms of the function $\text{\hspace{0.17em}}g.$
2. Explain the meaning of the statement $\text{\hspace{0.17em}}g\left(100\right)=1.$
a. $\text{\hspace{0.17em}}g\left(5000\right)=50;$ b. The number of cubic yards of dirt required for a garden of 100 square feet is 1.
Let $\text{\hspace{0.17em}}f\left(t\right)\text{\hspace{0.17em}}$ be the number of ducks in a lake $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ years after 1990. Explain the meaning of each statement:
1. $f\left(5\right)=30$
2. $f\left(10\right)=40$
Let $\text{\hspace{0.17em}}h\left(t\right)\text{\hspace{0.17em}}$ be the height above ground, in feet, of a rocket $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds after launching. Explain the meaning of each statement:
1. $h\left(1\right)=200$
2. $h\left(2\right)=350$
a. The height of a rocket above ground after 1 second is 200 ft. b. the height of a rocket above ground after 2 seconds is 350 ft.
Show that the function $\text{\hspace{0.17em}}f\left(x\right)=3{\left(x-5\right)}^{2}+7\text{\hspace{0.17em}}$ is not one-to-one.
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
yeah
Morosi
prime number?
Morosi
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### MEAP Preparation - Grade 7 Mathematics2.68 Multiplication of Monomials - I
Method: 1. Rearrange the factors using commutative and associative properties, so that all numerical factors are at one place and all the literal factors (variables) are at another place in their alphabetical order. 2. Multiply all the numerical factors together. 3. Multiply all the literal factors. Example: (2x)*4 = _____? Solution: (2x)*4 = (2*x)*4 = 2*(x*4) ........ (Associative Law) = 2*(4*x) ........ (Commutative Law) = (2*4)*x ........ (Associative Law) = 8*x 8x Directions: Solve the following multiplication problems and enter your answers in the space provided. For example, 3x * 4 = 12x. Also write at least 10 examples of your own.
Q 1: 15x * 9 = _____ ?Answer: Q 2: 15x * 5 = _____ ?Answer: Q 3: 50x * 4 = _____ ?Answer: Q 4: 7x * 8 = _____ ?Answer: Q 5: 12x * 6 = _____ ?Answer: Q 6: 25x * 25 = _____ ?Answer: Q 7: 7x * 15 = _____ ?Answer: Q 8: 19x * 17 = _____ ?Answer: Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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profit.calculator.zone
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# Profit Calculator
After selecting the calculation type in the profit calculator below, enter the required information and press the calculate button.
Profit Calculator
Information: Profit margin will also be calculated.
1. e.g 1650.75
2. e.g 1768.25
## What is profit?
The dictionary literally means money from shopping. In trade, it means the difference between the cost amount and the selling price.
## What is the profit rate and how is it calculated?
It is the ratio of the positive difference between the revenue (sale) amount and the cost (purchase) amount obtained from the sale of a product or service to the cost amount.
It is obtained by subtracting the amount of gross profit from the income amount and multiplying it by 100. For example, the gain rate obtained from a good with a purchase price of USD 500 and a selling price of USD 750 is (750 - 500) X 100/500 = 250 X 100/500 = 50 (ie 50%).
## What is the profit margin and how is it calculated?
It is the ratio of the positive difference between the revenue amount and the cost amount obtained from the sale of a product or service to the sales amount.
It is calculated by subtracting the cost amount from the income amount and dividing the gross profit amount by the income amount and multiplying it by 100. For example, the profit margin in the sale of a product with a purchase price of USD 500 and a selling price of USD 750 is (750 - 500) X 100/500 = 250 X 100/750 = 33.33 (ie 33.33%).
## How is profit calculated?
It is calculated with the positive difference obtained by subtracting the cost amount from the income amount. For example, the money obtained from a good with a purchase price of USD 100 and a selling price of USD 120 is calculated as 120 - 100 = USD 20.
## How is the sales price calculated?
The selling price is determined by increasing the purchase price (or cost amount) by the gain rate. For example, if you want to gain 20% from the sale of a good with a purchase price of USD 200, the selling price will be calculated as 200 x (100 + 20) / 100 = 200 x 1.2 = USD 240.
## How is the purchase price calculated?
The purchase price is calculated by multiplying the sales price by 100 multiplied by the sum of the sales ratio by 100. For example, if a 20% profit is obtained from the sale of a good with a sales price of USD 240, the purchase price (or cost) of that good will be calculated as 240 x 100 / (100 + 20) = 240 / 1.2 = USD 200.
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# Problem: The combustion of propane may be described by the chemical equation C3H8(g) + 5 O2(g) ⟶ 3 CO2(g) + 4 H2O(g) How many grams of O2(g) are needed to completely burn 86.6 g C3H8(g)?
###### FREE Expert Solution
We're being asked to calculate the grams of O2(g) needed to completely burn 86.6 g C3H8(g).
We're given the following balanced reaction:
C3H8(g) + 5 O2(g) ⟶ 3 CO2(g) + 4 H2O(g)
From the balanced reaction: 1 mole of C3H8(g) reacts with 5 moles of O2(g)
mass C3H8 (molar mass C3H8) → moles C3H8 (mole-to-mole comparison) → moles O2 (molar mass) → mass O2
83% (484 ratings)
###### Problem Details
The combustion of propane may be described by the chemical equation
C3H8(g) + 5 O2(g) ⟶ 3 CO2(g) + 4 H2O(g)
How many grams of O2(g) are needed to completely burn 86.6 g C3H8(g)?
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# TCS Aptitude Questions with solutions
1 . A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?
Sol) x= 2 kg Packs
y= 1 kg packs
x + y = 150 ………. Eqn 1
2x + y = 264 ………. Eqn 2
Solve the Simultaneous equation; x = 114
so, y = 36
ANS : Number of 2 kg Packs = 114.
2 . My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
a)6:00 am b) 6:40am c) 7:40 d) 7:00 e) 8:00
Sol) The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
Hence, the time at the destination place is 12 noon – 5:20 hours = 6: 40 AM
3 . The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent?
Sol) see it doesn’t matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms
4 . Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X?
Sol) Let no. of fish x catches=p
no. caught by y =r
r=5p.
r+p=42
then p=7,r=35
5 . Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings?
suppose total income is 100
so amount x is getting is 80
y is 70
z =60
total=210
but total money is 300
300-210=90
so they are getting 90 rs less
90 is 30% of 300 so they r getting 30% discount
6 . The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income?
Sol) incomes:3:4
expenditures:4:5
3x-4y=1/4(3x)
12x-16y=3x
9x=16y
y=9x/16
(3x-4(9x/16))/((4x-5(9x/16)))
ans:12/19
7 . If G(0) = -1 G(1)= 1 and G(N)=G(N-1) – G(N-2) then what is the value of G(6)?
ans: -1
bcoz g(2)=g(1)-g(0)=1+1=2
g(3)=1
g(4)=-1
g(5)=-2
g(6)=-1
8 . what’s the answer for that :
A, B and C are 8 bit no’s. They are as follows:
A -> 1 1 0 0 0 1 0 1
B -> 0 0 1 1 0 0 1 1
C -> 0 0 1 1 1 0 1 0 ( – =minus, u=union)
Find ((A – C) u B) =?
sol ) To find A-C, We will find 2’s compliment of C and them add it with A,
That will give us (A-C)
2’s compliment of C=1’s compliment of C+1
=11000101+1=11000110
A-C=11000101+11000110
=10001001
Now (A-C) U B is .OR. logic operation on (A-C) and B
10001001 .OR . 00110011
Whose decimal equivalent is 187.
9 . One circular array is given(means memory allocation tales place in circular fashion) dimension(9X7) and starting add. is 3000, What is the address of (2,3)……..
Sol) it’s a 9×7 int array so it reqiure a 126 bytes for storing.b’ze integer value need 2 byes of memory allocation. and starting add is 3000
so starting add of 2×3 will be 3012.
10 . A bus started from bustand at 8.00am at the speed of 18mph and after 30 min staying at destination, it returned back to the bustand. the destination is 27 miles from the bustand. the speed of the bus 50 percent fast IN SPEED. SO AT what time it returns to the bustand?
sol) a bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min. and it wait at stand =30 min.
after this speed of return increase by 50% so 50%of 18 mph=9mph
Total speed of returnig=18+9=27
Then in return it take 27/27=1 hour
then total time in joureny=1+1:30+00:30 =3 hour
so it will come at 8+3 hour=11 a.m.
So Ans==11 a.m
#### Sixface
I'm a always needed to know whats happening in technology world. I believe that "Technology" is the only word which can able to change the world and which should be more used by normal people to do their things easier.
## HCL Campus Interview Questions & Answers 2012
• deborah
June 29, 2011 at 3:31 PM
nice 6face….. can u explain 9th and 10th que….and one more thing…plz avoid spelling mistakes as in 9thquestion…..i like it….as u said….keep moving forward
• Sixface
June 30, 2011 at 6:11 PM
And in 9 th question its important that we have to see two things one is Starting address that is 3000 second one is what we need to find it is (2,3)
First we need to multiply it to get how many locations we need 2 * 3 = 6
And integer needs 2 bytes for each value for that 6 * 2 = 12 memory location so 3000 + 12 = 3012 ans
For 10 th question there is a value missing and it is bus speed is 18 MPH. i hope now you can understand that problem.
this two answers are brought you by Kirthika,
Thank you for your valuable comment deborah it makes us to work more to get new good articles. I will take extra care to avoid spelling mistakes.
And Keep looking yours comments. Keep moving Forward.
• Amrut Chavan
September 19, 2011 at 5:42 AM
what is the address of (3,2);(6,1);(1,6) in question 9??
• farook
September 30, 2011 at 1:52 PM
now get started ya……….
• deborah
July 7, 2011 at 2:36 PM
thanks kirthi and 6face….for the explanation….and thanks for considering my comments
• pramod
August 17, 2011 at 11:58 AM
can u plzzz send more TCS aptitude questions on my email id…
• pranjali balasaheb patil
August 23, 2011 at 4:18 PM
plzz can u send me more TCS aptitude questions on my email id plzzz.
• R.Nisha
August 25, 2011 at 5:37 AM
its full of intelligent question
• R.Nisha
August 25, 2011 at 5:40 AM
plzzz send more questions on my mail id………… bcozz aftr 3years i want to become a gud job in tcs company……..
• Sixface
August 26, 2011 at 5:40 PM
Thanks for comments……We will be posting about tcs interview questions part-2 in one week keep watching…….
• Arvind Patel
August 29, 2011 at 8:32 AM
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August 30, 2011 at 5:17 AM
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August 30, 2011 at 3:13 PM
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September 4, 2011 at 12:04 PM
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September 9, 2011 at 6:28 AM
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September 15, 2011 at 10:14 AM
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September 22, 2011 at 2:32 PM
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September 23, 2011 at 4:22 PM
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• shashank
September 26, 2011 at 11:06 AM
ans of the question does not seems to be right…..address of (2,3) cant be 3012 if you are considering array as 2 bytes….what i think is (2,3) means element number 18 in the array…how come it will have an address of 3012 in memory…
• ekta
September 28, 2011 at 12:04 PM
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i am B.TECH student
• Sixface
September 29, 2011 at 3:36 PM
@ekta we appreciate your courtesy . Be strong in base langs. keep yourself updated with new technology. All the best for your future.
Regards,
Tech Twinklers Team.
• AMUTHA A
October 5, 2011 at 5:09 AM
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• karthik
October 5, 2011 at 12:46 PM
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• Yuvan
December 4, 2011 at 10:57 AM
Nice!
January 3, 2012 at 3:08 PM
PLEASE SEND ME MORE APTITUDE QUESTIONS TO MY E-MAIL ID.
• Vaibhav Tiwari
January 4, 2012 at 3:58 PM
I m Sorry 2 say that ,I attended TCS campus placement few months ago.But none of the question was from given the question above.I have questions that were ask and I want 2 share them but dont know how???
• Sixface
January 18, 2012 at 7:18 PM
Thanks for your feedback dude. this is model questions which were asked in one campus interview. Using this models we need to work out similar problems. we trying to publish all latest questions. If you like to share pls send your questions to us. we will share with your name. [email protected]
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February 9, 2012 at 5:18 PM
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February 16, 2012 at 3:54 PM
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September 16, 2012 at 4:54 PM
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September 21, 2013 at 6:24 AM
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#### Welcome To TechTwinklers!
Tech Twinklers is a Tech Blog run by Students with an aspiring enthusiasm in Technology and Gaming. This Blog will bring News about the Modern Technology, Educational Advances, Campus Interviews, etc. So make sure you subscribe to our blog because we don't want you to miss a thing that is posted here..
June 4, 2017
November 3, 2012
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## CSC 241 Lab 9
Submission: Submit the python or text file that contains all your answers to the submission folder for this lab by the end of the lab. To simplify Drew's (our tutor) work, please include your name and the number of the lab at the top of the file (if you're submitting a Python file, use comments: start the line with a #).
1. (While Loop) Implement the factorial function factorial(n) which calculates 1*2* ... * n. We've done this before, here the difference is: don't use the built-in factorial, and don't use a for loop. Work with a while loop.
2. (While Loop - Statistics) We want to find out how many throws of a die it takes to get a 6. To do this, write two functions: die() and stats(n):
a) A call of the die() function repeatedly rolls a die until it comes out 6. The function counts how often you have to roll the die before you get the 6 and returns that count. Here are some test-runs:
So in the last call to die() the very first roll gave a 6. In the call before that, it took 9 times. Hint: use the randrange(a,b) function in the random module to simulate a die-roll (remember that a is inclusive, and b is exclusive), and a while loop to keep going until you have a 6.
b) Write a function stats(n) that calls the die() function n times, and records each outcome. At the end, it then reports the average value of all outcomes. We'd expect it to 6 rolls on average to get a six, so the larger n, the closer we should get to 6. This is borne out by the following test-runs (the last one took a couple of seconds to complete).
3. (While Loop) Here's a somewhat silly game: you alternate naming numbers that always have to be strictly larger than the previous one. Write a function bigger() that supports playing this game. The game starts with 0; if anybody enters a number lower than the previously highest number the game ends with the player losing. Entering nothing simply ends the game.
Marcus Schaefer
Last updated:May 22nd, 2019.
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Poll
Hmmm. Very interesting 2 votes (66.66%) Meh. I learned this in school No votes (0%) I thought this thread would be about Amazon Prime No votes (0%) Is nerdiness contagious? No votes (0%) Next, will Gordon start riding unicycles? 1 vote (33.33%) This is less interesting than sex No votes (0%)
3 members have voted
gordonm888
Joined: Feb 18, 2015
• Posts: 2825
November 11th, 2020 at 7:18:38 PM permalink
Quote: ThatDonGuy
Challenge accepted!
Doesn't your method make the assumption that every nonprime number can be expressed as the difference of two squares?
Try your method with X = 10. What happens?
I should have said that every ODD non prime number can be expressed as the difference of two squares. Even composite numbers are definitely not of the form (n^2 -m^2)
Since odd numbers are the only ones we would really be concerned with, I think that is sufficiently interesting.
Quote: ThatDonGuy
Also, you need to make sure that if you do find two squares, they aren't consecutive, as otherwise one of your factors will be 1.
Excellent point. Example: 8^2-7^2 =15 which has factors (8+7) and (8-7) which isn't very interesting.
So: X = n2 - m2; n>m+1
What is interesting is mathematicians have forever been unable to find a polynomial expression that defines prime numbers. But the above polynomial expression defines all odd numbers that aren't prime.
Quote: ThatDonGuy
Note that every positive odd number can be expressed as the difference of two (consecutive) primes.
You lost me with this comment. The difference between two consecutive primes (ignoring 2) must be an even number. What did you mean to say?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888
Joined: Feb 18, 2015
• Posts: 2825
November 11th, 2020 at 8:06:02 PM permalink
Okay, let's continue to assume that you have an integer X and you want to test to see whether there is a solution to
n2 - m2 = X
as a way to find factors of X.
In my last post I used modular arithmetic to develop selection rules so that we don't need to test all values of n and n2 to see if an m2 solution exists. Namely:
1. If X ends in 7 then n cannot have a 0, 2, 3, 5, 7 or 8 as its last digit.
2. If X ends in 1 then n cannot have a 2, 3, 7 or 8 as its last digit.
3. If X ends in 3 then n cannot have a 0, 1, 4, 5, 6 or 8 as its last digit.
4. If X ends in 9 then n cannot have a 1, 4, 6, or 9 as its last digit.
I have just started to look at the last two digits of X.
Basically n2 is congruent with (0, 1, 4, 9, 16, 21, 24, 25, 29, 36, 41, 49, 56, 61 64, 69, 76, 81, 84, 89, 96) mod100.
That is 22 of 100 numbers as residues of mod100.
When the last two digits of X are 07 then n is limited to the following two last digits: 4, 6, 14, 16, 24, 26, 34, 36, 44, 46, 54, 56, 64, 66, 74, 76, 84, 86, 94, 96. Basically, that means that n is limited to ending in 4 or 6! That reduces the possible values of n by 80%!
When the last two digits on X are 17, then n is limited to the following two last digits: 9, 19, 21, 29, 31, 41, 59, 69, 71, 79, 81 and 91. That only 12 numbers out of a possible 100 for the two last digits of n -an 88% reduction in numbers to sieve through.
That's all I've done so far. But it shows that we may be able to look at the last n digits of the number to be factorized and use selection rules to gain very significant efficiencies in terms of reducing the number of values of n that must be sieved through to solve n^2 - m^2 = X.
Update: I just looked at the rules for when the last two digits of X are 27 -and they are identical to the rules when the last two digits of X are 07.
Last edited by: gordonm888 on Nov 11, 2020
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4639
Thanks for this post from:
November 12th, 2020 at 6:27:20 AM permalink
Quote: gordonm888
You lost me with this comment. The difference between two consecutive primes (ignoring 2) must be an even number. What did you mean to say?
Every positive odd number can be expressed as the difference of the squares of two consecutive numbers.
Simple enough proof: every positive odd number is of the form 2N+1, where N is a nonnegative integer, and (N+1)2 - N2 = 2N+1.
gordonm888
Joined: Feb 18, 2015
• Posts: 2825
November 12th, 2020 at 3:58:54 PM permalink
Okay, let's continue to assume that you have an integer X and you want to test to see whether there is a solution to
n2 - m2 = X
as a way to find factors of X which are (n+m) and (n-m)
I have now worked out the rules for the last two digits of X; namely, dependent on the last two digits of X the possible values of n are limited to what is shown in the table below. These means that one can use the last two digits of the number you are trying to factor to speed up the sieve search for a suitable set of (m,n).
Last Two Digits of X
Last Two digits of n
01
x5, 01,49,51,99
03, 23, 43, 63, 83
x2, x8
07, 27, 47, 67, 87
x4, x6
09
x5, 03,47,53,97
11
x0, 06,44,56,94
13, 33, 53, 73, 93
x3, x7
17, 37, 57, 77, 97
x1, x9
19
x0, 12,38,62,88
21
x5, 11,39,61,89
29
x5, 23,27,73,77
31
x0, 16,34,66,84
39
x0, 8,42,58,92
41
x5, 21,29,71,79,
49
x5, 7,43,57,93
51
x0, 24,26,74,76
59
x0, 22,28,72,78
61
x5, 19,31,69,81
69
x5, 13,37,63,87
71
x0, 14,36,64,86
79
x0, 02,48,52,98
81
x5, 09,41,59,91
89
x5, 17,33,67,83
91
x0, 3,47,53,97
99
x0, 18,32,68,82
Lots of symmetry in that table, isn't there?
So for some X the search for a suitable pair of (m,n) will be sped up by 80%, while for other X the search will be faster by 86%.
I think the reason I am so intrigued is because the sieve search can be made more efficient by looking at the last n digits of X.
I mean, in an Erasthosthenes sieve, one knows that numbers ending in 0,2,4,5,6, and 8 are not prime and that numbers whose digits sum to a multiple of 3 are also not prime. So, that even when if you don't have a priori knowledge of which potential divisors are prime, you can readily eliminate about 73% of all numbers as possible divisors to be tested by just looking at the digits. But that's about the end of the efficiency for Erasthosthenes. This method is already eliminating more than 73% and will be even more efficient if we develop rules based on the last 3 digits of X (or 4 or more.)
Of course, my proposed method has some other inefficiencies not common with Erasthosthenes. But let's keep plugging away.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888
Joined: Feb 18, 2015
• Posts: 2825
November 13th, 2020 at 10:15:21 AM permalink
I have added an additional column to the table from my last post. I am finding that modular nomenclature is an efficient way to represent the constraints on n.
Last Two Digits of X
Last Two digits of n
Required form of n
01
x5, 01,49,51,99
5(mod10), ±1(mod 50)
03, 23, 43, 63, 83
x2, x8
2, 8(mod10)
07, 27, 47, 67, 87
x4, x6
4, 6(mod10)
09
x5, 03,47,53,97
5(mod10), ±3(mod 50)
11
x0, 06,44,56,94
0(mod10), ±6(mod 50)
13, 33, 53, 73, 93
x3, x7
3,7(mod10)
17, 37, 57, 77, 97
x1, x9
1, 9(mod10)
19
x0, 12,38,62,88
0(mod10), ±12(mod 50)
21
x5, 11,39,61,89
5(mod10), ±11(mod 50)
29
x5, 23,27,73,77
5(mod10), ±23(mod 50)
31
x0, 16,34,66,84
0(mod10), ±16(mod 50)
39
x0, 8,42,58,92
0(mod10), ±8(mod 50)
41
x5, 21,29,71,79,
5(mod10), ±21(mod 50)
49
x5, 7,43,57,93
5(mod10), ±7(mod 50)
51
x0, 24,26,74,76
0(mod10), ±24(mod 50)
59
x0, 22,28,72,78
0(mod10), ±22(mod 50)
61
x5, 19,31,69,81
5(mod10), ±19(mod 50)
69
x5, 13,37,63,87
5(mod10), ±13(mod 50)
71
x0, 14,36,64,86
0(mod10), ±14(mod 50)
79
x0, 02,48,52,98
0(mod10), ±2(mod 50)
81
x5, 09,41,59,91
5(mod10), ±9(mod 50)
89
x5, 17,33,67,83
5(mod10), ±17(mod 50)
91
x0, 3,47,53,97
0(mod10), ±3(mod 50)
99
x0, 18,32,68,82
0(mod10), ±18(mod 50)
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888
Joined: Feb 18, 2015
• Posts: 2825
November 13th, 2020 at 1:02:25 PM permalink
Well, I see that I have taken a wrong turn. I should be searching for values of m rather than values of n.
Again, the basic idea is, for a given integer X to search for values of (m,n) that satisfy X=n2 - m2; n-m >1 because that means that (n+m) and (n-m) are factors of X.
Now obviously n2 > X and n2 > m2 but is m2 < X? If m<sqrt(X) then searching for m between 0 and sqrt(X) will be sufficient to find m.
Ex: 10002 - 9892 = 21 879 and sqrt(21879) ~ 147.92. Here m=989 >147.92 so searching for m between 0 and sqrt(X) is not going to find this value of m.
However, 21 879 = 3*3*11*13*17 and thus will have many sets of (m,n) that satisfy X=n2 - m2; n-m >1. This is because we can combine the prime factors in different ways to make multiple values of (m+n,m-n). For example, 21 879 = 1602 - 612 and also 1522 - 352. And both of those (m,n) pairs have an m< sqrt(X).
What if X is a semiprime, i.e., a product of only two prime factors? Then there will only be one (m,n) pair that satisfies X=n2 - m2; n-m >1. And I think I have convinced myself that in case of a semiprime X that m< sqrt(X). And if that is the case then any X with many prime factors must also have one or more (m,n) pairs where m < sqrt(X).
Thus we can restrict ourselves to the range 0 to sqrt(x) when looking for at least one suitable m that solves our equation. And that should be an easier task than looking for n in the range of numbers greater than sqrt(X)!
The selection rules for searching for m are closely related to the rules for searching for n, but are not the same. They will likely involve the same efficiencies though. But I will need to recalculate new rules.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888
Joined: Feb 18, 2015
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+0
# help
0
242
3
Find the number of positive integers $$n$$ so that $$4 < \sqrt{n} < 10$$
Apr 14, 2019
#1
+2856
+3
Since $$\sqrt{n}$$ can be 5,6,7,8,9.
You sqaure the possible solutions to get 25,36,49,64,81.
These are the ONLY solutions because the problem said "positive integers"
So there are 5 possible solutions
Apr 14, 2019
#1
+2856
+3
Since $$\sqrt{n}$$ can be 5,6,7,8,9.
You sqaure the possible solutions to get 25,36,49,64,81.
These are the ONLY solutions because the problem said "positive integers"
So there are 5 possible solutions
CalculatorUser Apr 14, 2019
#2
+111433
+1
Very nice, CU !!!!!
CPhill Apr 14, 2019
#3
+219
0
Thanks CalculatorUser and CPhill!
Apr 14, 2019
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# RD Sharma Solutions For Class 12 Maths Exercise 11.3 Chapter 11 Differentiation
RD Sharma Solutions for Class 12 Maths Exercise 11.3 Chapter 11 Differentiation is provided here for students to study and excel in their board exams. Differentiation by using trigonometrical substitutions is the main topic covered in this exercise. The solutions for RD Sharma book, solved by BYJU’S experts, explain the steps most easily without missing out on essential aspects of solving a question.
Students can easily view and download the RD Sharma Solutions for Class 12 Exercise 11.3 of Chapter 11 Differentiation from the below-provided links. These RD Sharma Solutions primarily improve their speed in solving problems, which builds time management skills, which is key from the exam perspective.
## RD Sharma Solutions for Class 12 Chapter 11 – Differentiation Exercise 11.3:
### Access other exercises of RD Sharma Solutions for Class 12 Chapter 11 – Differentiation
Exercise 11.1 Solutions
Exercise 11.2 Solutions
Exercise 11.4 Solutions
Exercise 11.5 Solutions
Exercise 11.6 Solutions
Exercise 11.7 Solutions
Exercise 11.8 Solutions
### Access answers to Maths RD Sharma Solutions for Class 12 Chapter 11 – Differentiation Exercise 11.3
Exercise 11.3 Page No: 11.62
Differentiate the following functions with respect to x:
Solution:
Solution:
Solution:
Let,
Solution:
Let,
Solution:
Solution:
7. Sin-1 (2x2 – 1), 0 < x < 1
Solution:
8. Sin-1 (1 – 2x2), 0 < x < 1
Solution:
Solution:
Solution:
Solution:
Solution:
Solution:
Solution:
Solution:
Solution:
Solution:
Solution:
Solution:
Let,
Solution:
Solution:
Solution:
Solution:
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# Relative Velocity
1. Feb 3, 2010
### casemeister06
1. The problem statement, all variables and given/known data
Two piers, A and B are located on a river: B is 1500 m downstream from A. Two friends must make round trips from pier A to B and back. One rows a boat at a constant speed of 4 km/hr relative to the water; the other walks on the shore at a constant speed of 4 km/hr. The velocity of the river is 2.80 km/hr in the direction from A to B. How much time does it take each person to make the round trip.
2. Relevant equations
Vavg = $$\Delta x / \Delta t$$
VP/A = VP/B + VB/A
3. The attempt at a solution
I'm assuming we would have to find a single average velocity for both the boat and the person then take each velocity and let it be the dividend of the total distance to solve for the time. The person is relatively easy to solve for. For the boat though. I'm having trouble setting up the relative velocity equation. I get confused on what my point of reference should be and if the speeds are with respect to the water or the earth.
2. Feb 3, 2010
### casemeister06
Sorry about the double post. Computer lagged out.
3. Feb 3, 2010
### ByronT
First, make sure you convert km/hr to m/s.
4 km/hr = ~ 1.1 m/s
2.8 km/hr = ~ .78 m/s
We know that friend 1 is going to travel fast from A to B on the first part of the trip because he has a current behind him. His velocity of rowing will just be added with the velocity of the water. When he rows back, he'll be working against the current, so we will subtract the water's speed from his rowing speed, and he'll move more slowly.
It might be easier for you to think in terms of splitting the problem up. They go from A to B at a certain speed, and it takes them a certain time. Then, they go from B to A at a certain speed, and it takes the boater longer than his first part of the trip, since he's rowing against a current rather than with it. The other friend walks the same speed the whole time.
Use the equation time = distance/velocity. For each person, add their times for each part of the trip.
It doesn't make sense to find the average velocity for the boater because you don't know how long he's rowing downstream and upstream. It's obviously easy finding the average velocity for the walker, though.
The speeds are with respect to an observer on the earth. It wouldn't make sense for the problem to give you the speed of the water with respect to the moving water. Don't try to think too deeply about problems like these. The important thing is to break them up into multiple problems if you are having trouble understanding.
Last edited: Feb 3, 2010
4. Feb 3, 2010
### casemeister06
Thanks for the help. I guess I was thinking way to hard.
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# Fraction Activities Hands-On Fun
Hands-on fraction activities give kids a chance to visualize how fractions work! Understanding fractions is a very important concept that kids will need from here on out!
Check out this first fraction activity below...
Just grab a regular deck of cards, and you're ready to get the fraction fun started!
"Order Me Some Fractions"
Materials: A regular deck of playing cards; remove the jokers and face cards.
Number of Players: 2 or more.
Math Skills: Ordering fractions from least to greatest
How to Play
1) Pick one player to shuffle and deal the cards.
2) The dealer deals out four cards face-down to each player. After dealing each player their four cards, the dealer places the remaining cards face-down in the middle of the table.
3) The players look at their four cards and create any two proper fractions (smaller number in the numerator) from their four cards.
Object of Game:
4) Each player places the two fractions they created in order, from least to greatest, the smallest fraction on the left.
5) All of the players check the other player's two fractions to see if they are in the correct order: from least to greatest
• Each player earns 1 point if their two factions are in the correct order.
• Whichever players discover another player's fractions are nit in the correct order, they also receive one point.
After the round is over, players place their used cards in a discard pile, which will be shuffled and re-used after all the face-down cards in the middle of the table have been used up.
6) The first player to earn 10 points, wins the game!
## More Fun Fraction Activities
"Cup of Water" - fraction Activity
Measure Up:
Give your child measuring spoons and measuring cups. Have them pour 1 cup of water into a glass. Then have them guess:
• How many half-cups will it take to fill another glass the same size?
• How many quarter cups?
• How many one-third cups?
Let Pizza Night turn into a delicious, and fraction-filled time of learning with the kids. Have your child figure out how many pieces of pizza each person in the group should have if each one is to have the same number of pieces.
• For the pizza in the picture above, how many slices will each person get if there is a group of 4 people, and each person gets the same amount of pizza?
• What fraction of the pizza does each of the four people get?
Kids learn fractions better with visual aids and hands-on activities. In other words, mix it up a bit and have some fun.
## More Fraction Activities
• Equivalent Fractions
• Common Denominator
• Improper Fractions
• Mixed Numbers
• Subtracting Fractions
Fraction Concepts can be hard to grasp for some kids, but having a variety of fraction activities at your disposal will help make learning fractions fun.
Using different activities that are fun and engaging will help students build a solid foundation in their knowledge of fractions.
Check out these online fraction games as well.
It's pretty common for elementary students in the same grade to be at different levels when it comes to their fraction skills, so as best as you can, to try match the fraction activity with where the student is in their understanding of fractions.
Stay tuned, as I'll also be organizing the fraction activities / games into grade levels as you can see below:
Return from Fraction Activities to Fraction Games
Go to Learn With Math Games Home
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A301376 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that x^2-(3*y)^2 = 4^k for some k = 0,1,2,.... 25
1, 1, 2, 1, 1, 3, 1, 1, 4, 2, 2, 3, 3, 3, 3, 1, 5, 6, 2, 2, 10, 5, 4, 3, 2, 7, 7, 3, 5, 4, 3, 1, 12, 8, 2, 6, 4, 5, 10, 2, 7, 13, 8, 5, 10, 6, 6, 3, 8, 4, 7, 7, 8, 11, 4, 3, 17, 9, 5, 4, 8, 5, 9, 1, 8, 14, 8, 8, 13, 5, 8, 6, 11, 10, 7, 5, 13, 15, 7, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Conjecture: a(n) > 0 for all n > 0. Moreover, any positive square n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers and y even such that x^2 - (3*y)^2 = 4^k for some k = 0,1,2,.... We have verifed this for all n = 1..10^7. Compare this conjecture with the conjectures in A299537. As 3*A001353(n)^2 + 1 = A001075(n)^2, the conjecture in A300441 implies that any positive square can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x^2 - 3*y^2 = 4^k for some k = 0,1,2,.... See also A301391 for a similar conjecture. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018. EXAMPLE a(1) = 1 since 1^2 = 1^2 + 0^2 + 0^2 + 0^2 with 1^2 - (3*0)^2 = 4^0. a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4^2 - (3*0)^2 = 4^2. a(7) = 1 since 7^2 = 2^2 + 0^2 + 3^2 + 6^2 with 2^2 - (3*0)^2 = 4^1. a(31) = 3 since 31^2 = 10^2 + 2^2 + 4^2 + 29^2 with 10^2 - (3*2)^2 = 4^3, and 31^2 = 20^2 + 4^2 + 4^2 + 23^2 = 20^2 + 4^2 + 16^2 + 17^2 with 20^2 - (3*4)^2 = 4^4. MATHEMATICA f[n_]:=f[n]=FactorInteger[n]; g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n], i], 1]-3, 4]==0&&Mod[Part[Part[f[n], i], 2], 2]==1], {i, 1, Length[f[n]]}]==0; QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]); SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; tab={}; Do[r=0; Do[If[SQ[4^k+9y^2]&&QQ[n^2-4^k-10y^2], Do[If[SQ[n^2-(4^k+10y^2)-z^2], r=r+1], {z, 0, Sqrt[(n^2-4^k-10y^2)/2]}]], {k, 0, Log[2, n]}, {y, 0, Sqrt[(n^2-4^k)/10]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab] CROSSREFS Cf. A000118, A000290, A000302, A299537, A299794, A299924, A300219, A300396, A300441, A300510, A301391. Sequence in context: A224838 A030272 A157128 * A307828 A280698 A217667 Adjacent sequences: A301373 A301374 A301375 * A301377 A301378 A301379 KEYWORD nonn AUTHOR Zhi-Wei Sun, Mar 19 2018 STATUS approved
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# Newton’s Law of Universal Gravitation
### The Force of Gravity and Gravitational Potential
The law of universal gravitation was formulated by Isaac Newton $$\left(1643-1727\right)$$ and published in $$1687.$$
In accordance with this law, two point masses attract each other with a force that is directly proportional to the masses of these bodies $${m_1}$$ and $${m_2},$$ and inversely proportional to the square of the distance between them:
$F = G\frac{{{m_1}{m_2}}}{{{r^2}}}.$
Here, $$r$$ is the distance between the centers of mass of the bodies, $$G$$ is the gravitational constant, whose value found by experiment is $$G =$$ $$6,67 \times {10^{ – 11}}{\large\frac{{{\text{m}^3}}}{{\text{kg} \cdot {\text{s}^2}}}\normalsize}.$$
The force of gravitational attraction is a central force, that is directed along a line passing through the centers of the interacting bodies.
In a system of two bodies (Figure $$2$$), the attraction force $${\mathbf{F}_{12}}$$ of the second body acts on the first body of mass $${m_1}.$$
Similarly, the attraction force $${\mathbf{F}_{21}}$$ of the first body acts on the second body of mass $${m_2}.$$ Both the forces $${\mathbf{F}_{12}}$$ and $${\mathbf{F}_{21}}$$ are equal and directed along $$\mathbf{r},$$ where
$\mathbf{r} = {\mathbf{r}_2} – {\mathbf{r}_1}.$
Using the Newton’s second law we can write the following differential equations describing the motion of each body:
${{m_1}\frac{{{d^2}{\mathbf{r}_1}}}{{d{t^2}}} = G\frac{{{m_1}{m_2}}}{{{r^3}}}\mathbf{r},\;\;\;}\kern-0.3pt {{m_2}\frac{{{d^2}{\mathbf{r}_2}}}{{d{t^2}}} = – G\frac{{{m_1}{m_2}}}{{{r^3}}}\mathbf{r}}$
or
${\frac{{{d^2}{\mathbf{r}_1}}}{{d{t^2}}} = G\frac{{{m_2}}}{{{r^3}}}\mathbf{r},\;\;\;}\kern-0.3pt {\frac{{{d^2}{\mathbf{r}_2}}}{{d{t^2}}} = – G\frac{{{m_1}}}{{{r^3}}}\mathbf{r}.}$
It follows from the last two equations that
${\frac{{{d^2}{\mathbf{r}_1}}}{{d{t^2}}} – \frac{{{d^2}{\mathbf{r}_2}}}{{d{t^2}}} }={ G\frac{{{m_2}}}{{{r^3}}}\mathbf{r} + G\frac{{{m_1}}}{{{r^3}}}\mathbf{r},\;\;}\Rightarrow {\frac{{{d^2}\mathbf{r}}}{{d{t^2}}} = -G\frac{{{m_1} + {m_2}}}{{{r^3}}}\mathbf{r}.}$
This differential equation describes the change of the vector $$\mathbf{r}\left( t \right),$$ i.e. the relative motion of two bodies under the force of gravitational attraction.
With a large difference in mass of the bodies, we can neglect the smaller body mass in the right side of this equation. For example, the mass of the Sun is $$333,000$$ times greater than the mass of the Earth. In this case, the differential equation can be written in a simpler form:
$\frac{{{d^2}\mathbf{r}}}{{d{t^2}}} = – G\frac{{{M_\text{C}}}}{{{r^3}}}\mathbf{r},$
where $${M_\text{S}}$$ is the mass of the Sun.
The gravitational interaction of bodies takes place through a gravitational field, which can be described by a scalar potential $$\varphi.$$ The force acting on a body of mass $$m,$$ placed in a field with potential $$\varphi,$$ is equal to
${\mathbf{F} = m\mathbf{a} }={ – m\,\mathbf{\text{grad}}\,\varphi .}$
In the case of a point mass $$M,$$ the potential of the gravitational field is given by
$\varphi = – \frac{{GM}}{r}.$
The latter formula is also valid for distributed bodies with central symmetry, such as a planet or star.
### Kepler’s Laws
The basic laws of planetary motion were established by Johannes Kepler $$\left(1571-1630\right)$$ based on the analysis of astronomical observations of Tycho Brahe $$\left(1546-1601\right)$$. In $$1609,$$ Kepler formulated the first two laws. The third law was discovered in $$1619.$$ Later, in the late $$17$$th century, Isaac Newton proved mathematically that all three laws of Kepler are a consequence of the law of universal gravitation.
### Kepler’s First Law
The orbit of each planet in the solar system is an ellipse, one focus of which is the Sun (Figure $$3$$).
### Kepler’s Second Law
The radius vector connecting the Sun and the planet describes equal areas in equal intervals of time. Figure $$4$$ shows the two sectors of the ellipse corresponding to the same time intervals.
According to Kepler’s second law, the areas of these sectors are equal.
### Kepler’s Third Law
The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit:
${T^2} \propto {a^3}.$
The proportionality coefficient is the same for all planets in the solar system. Therefore, for any two planets, one can write the relationship
$\frac{{T_2^2}}{{T_1^2}} = \frac{{a_2^3}}{{a_1^3}}.$
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
A small cosmic body starts to fall to Earth from rest under the action of gravitational force. The initial distance to the centre of the Earth is equal to $$L.$$ Determine the velocity and time of the drop.
Page 1
Concept
Page 2
Problem 1
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Derivation of an estimate for CDF given ordered sample
Given a sample $x_1,\ldots,x_n$ one may order them $x_{(1)},\ldots,x_{(n)}$ and an estimate for the CDF $P(X\leq x_{(i)})$ is $\frac{i}{n}$ which is valid given that the CDF of a sample is uniform. However, the text I'm reading claims the following estimate to be a valid (and better) estimate for the CDF
$$P(X\leq x_{(i)}) \approx \frac{i-\frac{3}{8}}{n+\frac{1}{4}}$$
but it doesn't explain how to derive this estimate, so I was wondering how this estimate can be found?
The text later assumes $x_i$ to be normally distributed, if that is of relevance.
I found the answer after searching the web: This is due to Blom (1954), who used a general estimate for the empirical distribution function given by
$$P(X\leq x_{(i)}) = F(x_{(i)}) \approx \hat{F}_n(x_{(i)}) = \frac{i-\alpha}{n+1-2\alpha}$$
where $\alpha \in [0,0.5]$. He attempted to find $\alpha$ by solving $$\operatorname{E}(x_{(i)}) = \Phi^{-1}\left(\frac{i-\alpha}{n+1-2\alpha}\right)$$
which yields
$$\alpha_{i,n} = \frac{1-(n+1)\Phi(E(x_{(i)})}{1-2\Phi(E(x_{(i)})}$$
Blom conjectured this value to lie in the interval $(0.33,0.5)$ which he found was true for small $n$, so a good compromise was $0.375$, or $3/8$. Inserting this value for $\alpha$ in the original estimate yields
$$\hat{F}_n(x) = \frac{i-3/8}{n+1/4}$$
Sources:
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# Solving a problem
#### werehk
PHF Hall of Fame
When some objects collide with each other,
how should I define the system(i.e. what objects to be included) to determine whether a force is external or internal force?
When would earth be considered inside the system?
What is the meaning of transfer of momentum from one object to another?
#### topsquark
Forum Staff
When some objects collide with each other,
how should I define the system(i.e. what objects to be included) to determine whether a force is external or internal force?
When would earth be considered inside the system?
What is the meaning of transfer of momentum from one object to another?
What system to use can sometimes be confusing. The rule of thumb is to choose the system that makes your problem as simple as possible. For example, in most cases of projectile motion we ignore the Earth as being part of the system. The reason for this is that the motion of the Earth (toward the falling object) introduces an extra complication to the problem and is, frankly, negligible anyway.
In other cases we find that redefining the system (with care!) can be to our advantage.
Consider a string pulling with a force F a combination of two masses, M and m, which are in turn connected by a string. What is the acceleration of the two masses?
By choosing the system to be composed of both masses (and their connecting string) we can very easily find that
$$\displaystyle a = \frac{T}{m + M}$$
a result that we can verify by taking each mass and the connecting string as separate systems and solve a system of equations for taking much more time and effort.
Rule of thumb: If the Earth is not specifically mentioned in the problem, the odds are that you don't need to consider it.
A momentum transfer takes place during a collision. Typically the momentum of two (or more) objects in a collision is transfered in some way, such that the sum of the changes in the momentum of all the objects is 0 kg m/s. There is always some such transfer of momentum in a collision, no matter how small.
If this didn't answer your questions well enough, just say so.
-Dan
#### werehk
PHF Hall of Fame
For the transfer of momentum, can it be interpreted as change in speed during collision?
#### topsquark
Forum Staff
For the transfer of momentum, can it be interpreted as change in speed during collision?
Typicaslly, yes. However, not that a change in momoentum is defined as the change in the quantity mv, so if the mass of the objects change during the collision (if they were, say, putty) then this too would affect the change in momentum. (Most Intro Physics problems do not take this into account for simplicity.)
-Dan
werehk
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Circle
Triangle with circles (red)
The three excircles belong to the district and the inscribed circle to the special circles of a triangle that already in the ancient times were studied by Greek mathematicians.
The circles are defined as circles that are touched tangentially by one side of the triangle from the outside and by the extensions of the other two sides . Any triangle has three circles. The center points of the circle each lie on the bisector of an interior angle and on the bisector of the two exterior angles that do not belong to the interior angle.
The radius of the circle that touches the side ( ) inside results from ${\ displaystyle a}$${\ displaystyle [BC]}$
${\ displaystyle \ rho _ {a} = {\ frac {A} {sa}}}$,
it stands for the area and for half the circumference of the triangle . ${\ displaystyle A}$${\ displaystyle s}$${\ displaystyle s = {\ tfrac {1} {2}} (a + b + c)}$
The radii and the two other circles are calculated in the same way . ${\ displaystyle \ rho _ {b}}$${\ displaystyle \ rho _ {c}}$
If one expresses the area according to Heron's theorem by the lengths of the sides, one obtains
${\ displaystyle \ rho _ {a} = {\ sqrt {\ frac {s (sb) (sc)} {sa}}}}$.
The same applies to the other two arrivals
${\ displaystyle \ rho _ {b} = {\ sqrt {\ frac {s (sa) (sc)} {sb}}}}$and .${\ displaystyle \ rho _ {c} = {\ sqrt {\ frac {s (sa) (sb)} {sc}}}}$
Contact point distances
Triangle, contact point distances of the circles, distances of the same color have the same lengths
designation
• ${\ displaystyle c_ {a}}$is the distance from to the points of contact of the circle with the side and with the extension of the side${\ displaystyle C}$${\ displaystyle a}$${\ displaystyle b.}$
• ${\ displaystyle b_ {a}}$is the distance from to the points of contact of the circle with the side and with the extension of the side${\ displaystyle B}$${\ displaystyle a}$${\ displaystyle c.}$
The index means that the circle is considered that touches the side in the triangle and not in the extension. The designation for the other two circles is chosen in the same way. ${\ displaystyle a}$${\ displaystyle a}$
The following applies:
${\ displaystyle c_ {a} = a_ {c} = sb}$
${\ displaystyle c_ {b} = b_ {c} = sa,}$
${\ displaystyle a_ {b} = b_ {a} = sc.}$
Here is half the circumference of the triangle. ${\ displaystyle s}$
If you add a side length with a contact point distance of the circle on the side extension, this results ${\ displaystyle s.}$
example
${\ displaystyle c + b_ {a} = b + c_ {a} = s}$
Midpoints
The centers of the circles on their respective sides have the following barycentric coordinates, with the center of the circle representing side a: ${\ displaystyle \ displaystyle I_ {a}}$
• ${\ displaystyle \ displaystyle I_ {a} = (- a: b: c)}$
• ${\ displaystyle \ displaystyle I_ {b} = (a: -b: c)}$
• ${\ displaystyle \ displaystyle I_ {c} = (a: b: -c)}$
Construction of the circle centers
Triangle, construction of the circle centers
The following can be concluded from the introduction and the triangle with circles (red) above . The three center points of the circle can also be found solely by halving three outer angles which, as angle legs, each have one side and an extension of an adjacent side.
It starts with the extensions of the sides of the triangle beyond its corner points. Then follows z. As the bisector of the exterior angle at the apex of the angle leg side and extension of the side from the bisector of the external angle at the apex of the angle leg side and extension of the side from joins while delivering, as the intersection with the first Ankreismittelpunkt If all three Ankreismittelpunkte searched, the bisector of the outer angle at the apex with the angle legs side and extension of the side down is finally required. This results in the intersection points with the already existing bisector and also the two circle centers and${\ displaystyle ABC}$${\ displaystyle w_ {1}}$ ${\ displaystyle C}$${\ displaystyle a}$${\ displaystyle b}$${\ displaystyle C.}$${\ displaystyle w_ {2}}$${\ displaystyle B}$${\ displaystyle a}$${\ displaystyle c}$${\ displaystyle B}$${\ displaystyle w_ {1}}$${\ displaystyle I_ {a}.}$${\ displaystyle w_ {3}}$${\ displaystyle A}$${\ displaystyle c}$${\ displaystyle b}$${\ displaystyle A}$${\ displaystyle w_ {1}}$${\ displaystyle w_ {2},}$${\ displaystyle I_ {b}}$${\ displaystyle I_ {c}.}$
Other properties
Triangle inscribed center${\ displaystyle I_ {a} \; I_ {b} \; I_ {c},}$
• The circle centers and the triangle form a triangle whose height intersection is the inscribed center of the triangle .${\ displaystyle I_ {a}, I_ {b}}$${\ displaystyle I_ {c}}$${\ displaystyle ABC}$ ${\ displaystyle H}$${\ displaystyle ABC}$
• If you connect the corners of a triangle with the opposite contact points of the circles, the connecting straight lines intersect at one point, the nail point .
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# 85.31 kg to lbs - 85.31 kilograms to pounds
Do you need to know how much is 85.31 kg equal to lbs and how to convert 85.31 kg to lbs? You are in the right place. You will find in this article everything you need to make kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to highlight that whole this article is devoted to one number of kilograms - exactly one kilogram. So if you need to know more about 85.31 kg to pound conversion - read on.
Before we get to the more practical part - that is 85.31 kg how much lbs calculation - we will tell you some theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 85.31 kg to lbs? 85.31 kilograms it is equal 188.0763557122 pounds, so 85.31 kg is equal 188.0763557122 lbs.
## 85.31 kgs in pounds
We will start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, known also as International System of Units (in short form SI).
At times the kilogram could be written as kilogramme. The symbol of this unit is kg.
First definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. First definition was simply but totally impractical to use.
Then, in 1889 the kilogram was defined by the International Prototype of the Kilogram (in short form IPK). The IPK was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was switched by another definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It could be also divided to 100 decagrams and 1000 grams.
## 85.31 kilogram to pounds
You learned some information about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to point out that there are not only one kind of pound. What does it mean? For instance, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is used in the British and United States customary systems of measurements. To be honest, this unit is used also in other systems. The symbol of this unit is lb or “.
The international avoirdupois pound has no descriptive definition. It is just equal 0.45359237 kilograms. One avoirdupois pound is divided into 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of the pound was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 85.31 kg?
85.31 kilogram is equal to 188.0763557122 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 85.31 kg in lbs
The most theoretical part is already behind us. In next part we are going to tell you how much is 85.31 kg to lbs. Now you learned that 85.31 kg = x lbs. So it is time to get the answer. Just see:
85.31 kilogram = 188.0763557122 pounds.
This is an accurate result of how much 85.31 kg to pound. You can also round it off. After it your outcome will be exactly: 85.31 kg = 187.682 lbs.
You learned 85.31 kg is how many lbs, so see how many kg 85.31 lbs: 85.31 pound = 0.45359237 kilograms.
Obviously, this time it is possible to also round it off. After it your outcome is exactly: 85.31 lb = 0.45 kgs.
We also want to show you 85.31 kg to how many pounds and 85.31 pound how many kg outcomes in tables. See:
We are going to begin with a table for how much is 85.31 kg equal to pound.
### 85.31 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
85.31 188.0763557122 187.6820
Now see a table for how many kilograms 85.31 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
85.31 0.45359237 0.45
Now you know how many 85.31 kg to lbs and how many kilograms 85.31 pound, so it is time to go to the 85.31 kg to lbs formula.
### 85.31 kg to pounds
To convert 85.31 kg to us lbs you need a formula. We are going to show you a formula in two different versions. Let’s start with the first one:
Number of kilograms * 2.20462262 = the 188.0763557122 outcome in pounds
The first version of a formula will give you the most correct outcome. In some cases even the smallest difference could be considerable. So if you need a correct result - this formula will be the best for you/option to know how many pounds are equivalent to 85.31 kilogram.
So let’s move on to the another version of a formula, which also enables calculations to learn how much 85.31 kilogram in pounds.
The another version of a formula is as following, see:
Amount of kilograms * 2.2 = the outcome in pounds
As you can see, the second formula is simpler. It can be the best solution if you need to make a conversion of 85.31 kilogram to pounds in fast way, for instance, during shopping. You only have to remember that your outcome will be not so correct.
Now we want to learn you how to use these two formulas in practice. But before we will make a conversion of 85.31 kg to lbs we are going to show you easier way to know 85.31 kg to how many lbs totally effortless.
### 85.31 kg to lbs converter
Another way to check what is 85.31 kilogram equal to in pounds is to use 85.31 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. It is based on first version of a formula which we gave you above. Due to 85.31 kg pound calculator you can effortless convert 85.31 kg to lbs. Just enter amount of kilograms which you need to calculate and click ‘calculate’ button. You will get the result in a second.
So try to convert 85.31 kg into lbs using 85.31 kg vs pound converter. We entered 85.31 as an amount of kilograms. It is the result: 85.31 kilogram = 188.0763557122 pounds.
As you can see, this 85.31 kg vs lbs converter is intuitive.
Now we can go to our primary issue - how to convert 85.31 kilograms to pounds on your own.
#### 85.31 kg to lbs conversion
We are going to start 85.31 kilogram equals to how many pounds calculation with the first formula to get the most accurate result. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 188.0763557122 the outcome in pounds
So what have you do to check how many pounds equal to 85.31 kilogram? Just multiply number of kilograms, in this case 85.31, by 2.20462262. It is exactly 188.0763557122. So 85.31 kilogram is 188.0763557122.
You can also round off this result, for instance, to two decimal places. It is exactly 2.20. So 85.31 kilogram = 187.6820 pounds.
It is high time for an example from everyday life. Let’s convert 85.31 kg gold in pounds. So 85.31 kg equal to how many lbs? And again - multiply 85.31 by 2.20462262. It is exactly 188.0763557122. So equivalent of 85.31 kilograms to pounds, if it comes to gold, is 188.0763557122.
In this case it is also possible to round off the result. This is the outcome after rounding off, in this case to one decimal place - 85.31 kilogram 187.682 pounds.
Now we are going to examples converted with short formula.
#### How many 85.31 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Amount of kilograms * 2.2 = 187.682 the result in pounds
So 85.31 kg equal to how much lbs? And again, you have to multiply number of kilogram, in this case 85.31, by 2.2. Let’s see: 85.31 * 2.2 = 187.682. So 85.31 kilogram is exactly 2.2 pounds.
Let’s do another conversion using shorer version of a formula. Now calculate something from everyday life, for instance, 85.31 kg to lbs weight of strawberries.
So calculate - 85.31 kilogram of strawberries * 2.2 = 187.682 pounds of strawberries. So 85.31 kg to pound mass is exactly 187.682.
If you know how much is 85.31 kilogram weight in pounds and can convert it with use of two different formulas, we can move on. Now we are going to show you these outcomes in tables.
#### Convert 85.31 kilogram to pounds
We realize that outcomes presented in charts are so much clearer for most of you. We understand it, so we gathered all these outcomes in tables for your convenience. Thanks to this you can easily make a comparison 85.31 kg equivalent to lbs outcomes.
Let’s start with a 85.31 kg equals lbs table for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
85.31 188.0763557122 187.6820
And now let’s see 85.31 kg equal pound table for the second version of a formula:
Kilograms Pounds
85.31 187.682
As you see, after rounding off, if it comes to how much 85.31 kilogram equals pounds, the outcomes are not different. The bigger amount the more significant difference. Please note it when you need to make bigger number than 85.31 kilograms pounds conversion.
#### How many kilograms 85.31 pound
Now you learned how to calculate 85.31 kilograms how much pounds but we want to show you something more. Are you interested what it is? What about 85.31 kilogram to pounds and ounces calculation?
We want to show you how you can calculate it step by step. Begin. How much is 85.31 kg in lbs and oz?
First thing you need to do is multiply amount of kilograms, in this case 85.31, by 2.20462262. So 85.31 * 2.20462262 = 188.0763557122. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To convert how much 85.31 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your outcome is equal 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then your outcome is 2 pounds and 33 ounces.
As you can see, calculation 85.31 kilogram in pounds and ounces simply.
The last calculation which we want to show you is calculation of 85.31 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters it is needed another formula. Before we give you this formula, see:
• 85.31 kilograms meters = 7.23301385 foot pounds,
• 85.31 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 85.31 foot pounds to kilograms meters you have to multiply 85.31 by 0.13825495. It is equal 0.13825495. So 85.31 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for example, to two decimal places. Then 85.31 foot pounds will be exactly 0.14 kilogram meters.
We hope that this calculation was as easy as 85.31 kilogram into pounds calculations.
We showed you not only how to make a calculation 85.31 kilogram to metric pounds but also two other calculations - to check how many 85.31 kg in pounds and ounces and how many 85.31 foot pounds to kilograms meters.
We showed you also other solution to make 85.31 kilogram how many pounds conversions, it is using 85.31 kg en pound converter. This will be the best option for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.
We hope that now all of you are able to make 85.31 kilogram equal to how many pounds calculation - on your own or with use of our 85.31 kgs to pounds calculator.
Don’t wait! Let’s calculate 85.31 kilogram mass to pounds in the way you like.
Do you need to do other than 85.31 kilogram as pounds conversion? For example, for 10 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so easy as for 85.31 kilogram equal many pounds.
### How much is 85.31 kg in pounds
We want to sum up this topic, that is how much is 85.31 kg in pounds , we gathered answers to the most frequently asked questions. Here you can see all you need to know about how much is 85.31 kg equal to lbs and how to convert 85.31 kg to lbs . Have a look.
How does the kilogram to pound conversion look? To make the kg to lb conversion it is needed to multiply 2 numbers. Let’s see 85.31 kg to pound conversion formula . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
See the result of the conversion of 85.31 kilogram to pounds. The correct answer is 188.0763557122 pounds.
It is also possible to calculate how much 85.31 kilogram is equal to pounds with another, easier version of the equation. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So in this case, 85.31 kg equal to how much lbs ? The answer is 188.0763557122 pounds.
How to convert 85.31 kg to lbs quicker and easier? It is possible to use the 85.31 kg to lbs converter , which will make the rest for you and give you an accurate answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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# Toss One 21
## Introduction
I first saw Toss One 21 at the Atlantis Casino in Reno on January 20, 2017. It is a very simple blackjack variant requiring almost no skill.
## Rules
1. A single 52-card deck is used.
2. All cards are scored as in blackjack (ace = 1 or 11, 2-10=pip value, face cards = 10).
3. After making a wager the player and dealer are each dealt four cards.
4. The player must toss one card. His object is to have the other three cards total as many points as possible without going over 21.
5. The best possible hand is a blackjack, which is composed of an ace and any two 10-point cards.
6. A player blackjack is an automatic winner and pays 3 to 2.
7. If the player can't form a three-card hand of 21 points or less, then he loses.
8. After the player discards one card, the dealer shall examine his card and toss the card resulting in the greatest number of points in the other three cards, without going over 21.
9. The player and dealer hands are compared, the one with more points, without going over 21, wins.
10. If the player has more points, then the player shall win and his wager is paid even money.
11. If the dealer has more points, then the player shall lose.
12. In the event the player and dealer have the same number of points, then the two toss cards shall be considered to break the tie. Toss cards shall be ranked as in blackjack. I do not know whether aces are counted as high or low for toss card purposes.
13. If the toss cards are needed to break a tie and they each have the same blackjack value, then the dealer shall win.
14. In addition, there is a side bet called the "All 4," which pays based on the total points between the four player cards.
If my rules were not clear, maybe the rack card will be more clear. Click on the image for a larger version.
## Analysis
Following is my analysis of Toss One 21. The lower right cell shows a house edge of 1.03%.
### Toss One 21
Event Pays Permutations Probability Return
Player blackjack 1.5 2,053,270,425,600 0.067670 0.101505
Dealer bust 1 4,681,480,581,504 0.154289 0.154289
Player beats dealer 1 7,767,078,848,064 0.255982 0.255982
Player bust -1 6,123,273,822,720 0.201806 -0.201806
Dealer blackjack -1 1,556,351,907,840 0.051293 -0.051293
Dealer beats player -1 7,767,078,848,064 0.255982 -0.255982
Tie -1 393,803,774,208 0.012979 -0.012979
Total 30,342,338,208,000 1.000000 -0.010284
Following is my analysis of the All 4 wager. The lower right cell shows a house edge of 8.83%.
### All 4
Event Pays Permutations Probability Return
4 250 24 0.000004 0.000923
5 150 384 0.000059 0.008865
6 to 8 50 10,200 0.001570 0.078493
9 or 10 10 25,920 0.003989 0.039893
11 to 16 4 376,368 0.057926 0.231704
17 to 21 2 1,057,560 0.162767 0.325533
22 to 40 -1 5,026,944 0.773685 -0.773685
Total 6,497,400 1.000000 -0.088274
## Strategy
The strategy is quite obvious -- discard the card that results in the most points, without going over 21. With a house edge of 8.83%, I would avoid the All 4 bet.
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How to calculate compound growth rate formula
Geometric Mean Formula. As we said in the last section, the geometric mean is based on the CAGR or compound annual growth rate is method to calculate the growth rate of a particular amount annually, by default we do not have any inbuilt formula in
CAGR is defined as: C A G Actual or normalized values may be used for calculation as long as they retain the same mathematical 13 Jun 2019 Compound Annual Growth Rate. Formula and Calculation of CAGR. What CAGR Can Tell You. Example of How to Use CAGR. Additional The CAGR calculator is a useful tool when determining an annual growth rate on an investment whose value has fluctuated widely from one period to the next. 11 Jul 2019 Likewise, when you know the rate per compound period (r) and the number of compound periods per year (n), you can calculate the effective Note: in other words, to calculate the CAGR of an investment in Excel, divide the value of the investment at the end by the value of the investment at the start. Next, The Compound Annual Growth Rate formula requires only the ending value of the investment, the beginning value, and the number of compounding years to
8 Aug 2016 One of the options amongst your quick table calculations in to compute the ' compound growth rate' (CGR). The CGR is a measure of growth
23 Jul 2019 Using the formula above, we can calculate the return based on quarterly calculation. We can check our outcome by using the CAGR calculation. 2 Oct 2019 Calculate the Reverse Compound Annual Growth Rate in Excel. This calculation is used to determine the future value of your investment with Guide to Compounded Annual Growth Rate Formula. Here we discuss how to calculate CAGR Using Formula with example,Calculator and downloadable excel 7 Apr 2011 Calculating Compound Growth (CAGR). CAGR stands for compound average growth rate. The active word there is “compound.” It means that Exponential Growth and Decay on MathHelp.com · Exponential Growth and Decay. One very important exponential equation is the compound-interest formula: For instance, let the interest rate r be 3%, compounded monthly, and let the
The CAGR calculator is a useful tool when determining an annual growth rate on an investment whose value has fluctuated widely from one period to the next.
Geometric Mean Formula. As we said in the last section, the geometric mean is based on the CAGR or compound annual growth rate is method to calculate the growth rate of a particular amount annually, by default we do not have any inbuilt formula in Using the formula for compound annual growth rate can help you answer these and other Plugging these known quantities into our equation then yields:. Compound Annual Growth Rate (CAGR); The CAGR Formula; Calculate CAGR in Excel. FV, PV, N; The Rate Function; The XIRR Function; The IRR Function. 7 Apr 2011 Calculating Simple Growth Rate. Simple annual growth formula calculation. Question #1 in our quiz above illustrates the concept of simple
To calculate the CAGR of an investment: Divide the value of an investment at the end of the period by its value at the beginning of that period. Raise the result to an exponent of one divided by the number of years. Subtract one from the subsequent result.
The Compound Annual Growth Rate formula requires only the ending value of the investment, the beginning value, and the number of compounding years to calculate. It is achieved by dividing the ending value by the beginning value and raising that figure to the inverse number of years before subtracting it by one.
Formula. The CAGR can be calculated using the following mathematical formula: CAGR = [(Ending value/Beginning Value)^(1
Explanation of the Compounded Annual Growth Rate Formula. The formula for the calculation of CAGR can be derived by using the following steps: Step 1: Firstly, determine the beginning value of the investment or the money that was invested at the start of the investment tenure.
The Compound Annual Growth Rate formula requires only the ending value of the investment, the beginning value, and the number of compounding years to calculate. It is achieved by dividing the ending value by the beginning value and raising that figure to the inverse number of years before subtracting it by one. How to Calculate Compounded Annual Growth Rate - Calculating CAGR in Excel Enter data in the spreadsheet. Enter the basic formula to calculate the CAGR. Use the POWER function in Excel to calculate the CAGR. Use the RATE function to calculate the CAGR. You can calculate CAGR in Excel using the RATE function: CAGR = RATE(Years,,-PV,FV). The RATE, PV, FV and NPER functions in Excel can be used to calculate each of the four variables associated with the CAGR formula. This is demonstrated in the CAGR_1 tab within the Excel file and the formulas below. In the above compound annual growth rate in Excel example, the ending value is B10, Beginning value is B2, and the number of periods is 9. See the screenshot below. Step 3 – Now hit enter. You will get the CAGR (Compound Annual Growth Rate) value result inside the cell, in which you had input the formula. Compound Growth Calculator. Fill in any three to calculate the fourth value: The online Compound Growth Calculator is used to solve the compound growth problems. It will calculate any one of the values from the other three in the compound growth formula. Calculate compound annual growth rate with XIRR function in Excel. 1 . Create a new table with the start value and end value as the following first screen shot shown: Note: In Cell F3 enter =C3, in Cell G3 enter 2 . Select a blank cell below this table, enter the below formula into it, and press The Compound Annual Growth Rate (CAGR) may be the key to better investment earnings. The CAGR formula calculates year-over-year growth rates and helps chart investment performance. It also allows investors to see how similar investments have fared over the same length of time.
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## Simple Linear Regression
Simple linear regression is a tool for fitting a linear line to a set of data. It is used when you want to predict the value of the "dependent variable" Y by knowing the value of the "independent variable" X.Figure 1 is an example of a data set with a regression line fit.
Figure 1. Example data set with regression line fit to data.
The line in the graph can be described as:
$$y = b_0 + b_1 x$$
where y is the dependent variable (also plotted on the y axis of the graph), x is the independent variable (plotted on the x axis of the graph). The parameters that are estimated are b0 and b1
These parameters can be estimated using the following equations:
$$b_1 = \frac{\displaystyle\sum\limits_{i=1}^{n} x_i y_i - \frac{\left( \displaystyle\sum\limits_{i=1}^{n} x_i \right) \left( \displaystyle\sum\limits_{i=1}^{n} y_i \right) }{n}}{ \displaystyle\sum\limits_{i=1}^{n} x^2_i - \frac{\left( \displaystyle\sum\limits_{i=1}^{n} x_i \right)^2}{n}}$$
$$b_0 = \bar{y} - b_1 \bar{x}$$
where Xi and Yi are the individual observation and n is the number of observations.
The results of a regression are often summarized using an analysis of variance table.
The usual configuration for the table is as follows:
The F test is a test to determine if the regression explains more of the variation than the mean. Another statistic that is commonly used to describe a regression is the coefficient of determination R2 This statistic is the proportion of the observed data explained by the regression. This statistic is a value that ranges from 0 to 1 with 0 being no agreement between the regression and the data and 1 being perfect agreement between the data and the regression.
$$R^2 = \frac{SS_{regression}}{SS_{total}}$$
Another important method of explaining the results of a regression is to plot the residuals against the independent variable. This analysis can be used to indicate that the model is mis-specified and transformation required.
Figure 2. Residual plot of the data.
Also See:
Chapter 16 - Simple linear Regression pages 317-330 in:
Zar, J. H. 2007. Biostatistical Analysis. Prentice-Hall, Inc. Englewood Cliffs, New Jersey. 718 pp.
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# algebra II-URGENT
the perimeter of a rectangular yard is 270 feet. If its length is 25 feet greater than its width what are the dimensions of the yard
1. 👍 0
2. 👎 0
3. 👁 91
1. P=2l+2w
270=2(w+25)+2w
270=2w+50+2w
220=4w
55=w
so
l=w+25
=55+25
=80
Check
P=2l+2w
270=2(80)+2(55)
270=160+110
270=270
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2. 👎 0
posted by lauren
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## College Algebra 7th Edition
Published by Brooks Cole
# Chapter 1, Equations and Graphs - Section 1.1 - The Coordinate Plane - 1.1 Exercises - Page 92: 41
#### Answer
$a.\quad$ Please see "step-by-step" $b.\quad$ Area = $10$
#### Work Step by Step
a. Reading the ccoordinates from the graph, we have $A=(2,2), B=(3, -1)$, and $C=(-3, -3)$. Calculate the sides' lengths: $d(A, B)=\sqrt{(3-2)^{2}+(-1-2)^{2}}=\sqrt{1^{2}+(-3)^{2}}=\sqrt{1+9}=\sqrt{10}.$ $d(C, B)=\sqrt{(3-(-3))^{2}+(-1-(-3))^{2}}=\sqrt{6^{2}+2^{2}}=\sqrt{36+4}\\=\sqrt{40}=2\sqrt{10}.$ $d(A, C)=\sqrt{(-3-2)^{2}+(-3-2)^{2}}=\sqrt{(-5)^{2}+(-5)^{2}}\\=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$. Note that $[d(A, B)]^{2}+[d(C, B)]^{2}=10+4\cdot 10=50$ $[d(A, C)]^{2}=25\cdot 2=50,$ so, $[d(A, B)]^{2}+[d(C, B)]^{2}=[d(A, C)]^{2}$. By the converse of the Pythagorean Theorem, we conclude that the triangle is a right triangle. b. The area of the triangle is $\displaystyle \frac{1}{2}\cdot d(C, B)\cdot d(A, B)=\frac{1}{2}\cdot\sqrt{10}\cdot 2\sqrt{10}=10$.
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You are on page 1of 33
contents
2 3 9 32
introduction
2
YEARS 7–8
Page 1 activity one
Number Sense: Book One
Choice Calculations
activity two 1. 3 + 3 + 3 because 3 x 3 x 3 = 27. (The other sets are: 1 + 1 + 7, 1 + 2 + 6, 1 + 3 + 5, 1 + 4 + 4, 2 + 2 + 5, 2 + 3 + 4.) 2. a. 4 + 4 + 4 (because 4 x 4 x 4 = 64) b. 5 + 5 + 5 (because 5 x 5 x 5 = 125) c. 6 + 6 + 6 (because 6 x 6 x 6 = 216) d. 6 + 7 + 7 (because 6 x 7 x 7 = 294) 3. Yes. You get the greatest product when each set of two or three numbers consists of whole numbers that are the same or nearly the same. (If you were using fractions instead of whole numbers only, with 13 you would get 61/2 x 61/2 = 421/4 as the greatest product instead of 6 x 7 = 42, and with 20, you would get 62/3 x 62/3 x 62/3 = 296.296, instead of 6 x 7 x 7 = 294.)
1. a.–b. Answers will vary. Possible answers include: i. 3 x 6 – 2 x 9 = 0, 3 – 6 x 2 + 9 = 0, or 3 x 6 ÷ 2 – 9 = 0 ii. 4 x 6 – 3 x 6 = 6 iii. 2 x 3 – 3 + 2 = 5, 2 + 3 ÷ 3 + 2 = 5, or 2 – 3 + 3 x 2 = 5 iv. 8 ÷ 4 ÷ 2 ÷ 1 = 1, 8 – 4 – 2 – 1 = 1, or 8 ÷ 4 – 2 + 1 = 1 2. Problems will vary. activity two 1. Solutions may vary depending on the type of calculator and the approach used. Possible calculations are: a. 34 + 34 + 34 + 34 + 34 + 34 or 34 + 34 = = = = = (or on some calculators, 34 + + 34 = = = = =) b. 444 – 333 ÷ 3 = or 333 ÷ 3 ÷ 3 = c. 33 – 4 + 33 – 4 + 33 – 4 + 33 – 4 = 2. Expressions will vary.
Pages 4–5 game
Aiming High
A game using place value and addition activity one 1. a. Strategies will vary. One possible strategy is to use large numbers first (hundreds and tens) until your total is in the nine hundreds and then use tens and ones only. b.–2. Strategies will vary. activity two 1. Leila wins because she is closer to 1 000 than Tama. (Tama is 22 less than 1 000, and Leila is 21 more.) 2. If Mere puts her last throw in the correct column, she has a 6 out of 6 chance of winning the game. (To win, she needs to put 1 or 2 in the tens column or 3, 4, 5, or 6 in the ones column.) 3. a. Strategies will vary. b. It is possible to get exactly 500, but only if players get exactly the numbers they need at the end.
Pages 2–3 activity one
Splitting Numbers
1. a. 15 + 1, 14 + 2, 13 + 3, 12 + 4, 11 + 5, 10 + 6, 9 + 7, 8 + 8 b. 8 and 8 (because 8 x 8 = 64) 2. a. 5 + 5 (because 5 x 5 = 25) b. 6 + 6 (because 6 x 6 = 36) c. 9 + 9 (because 9 x 9 = 81) d. 6 + 7 (because 6 x 7 = 42)
3
20 ÷ 4 x 15 + 36 + 45 + 15 f. he got 1 024 000. Api b. Page 7 game Hit the Target A game using whole-number operations Page 8 activity Pathways 1. Practical activity 3.) 2. 2 000 1 000 500 250 125 62. Records will vary. 3. Possible pathways include: a. 20 x 5 – 70 + 36 ÷ 11 – 6 d. They all end up as an odd number after the second halving.–2. Another way to get this score is an arrangement that includes 21S + 4M + 4H. Totals and pathways will vary. Numbers will vary.5 Page 9 activity Flying Feet 1. a. Api: step 24. Results will vary. Numbers will vary. Eru: 48 ÷ 4 = 12. They are all odd numbers. 20 – 15 + 25 ÷ 6 x 9 + 45 + 15 c. Eru: 12 strides. Matiu: step 12. 14S + 4M + 2H = (14 x 10) + (4 x 100) + (2 x –20) = 140 + 400 – 40 = 500 c. a.5 b.5 c. Myra: 48 strides b. score: 530. Matiu 2. 20 x 5 ÷ 10 x 15 + 36 + 45 + 15 b. 10S + 3M + 3H = (10 x 10) + (3 x 100) + (3 x –20) = 100 + 300 – 60 = 340 b. 1. 3 x 16 = 48 Api: 48 ÷ 2 = 24.Page 6 activity Number Scavenge Page 10 activity Space Marauders 1. a. 20 x 5 ÷ 10 x 7 + 11 + 45 + 15 (Note that the order of operations does not apply in this step-by-step process. a. 2 x 24 = 48 Myra: 48 ÷ 1 = 48. 4 x 12 = 48 Matiu: 48 ÷ 3 = 16. Matiu: 16 strides. Pathway diagrams will vary. 20 – 15 x 3 x 7 + 3 ÷ 2 – 6 g. 2 500 1 250 625 312. 20 ÷ 4 x 7 + 3 ÷ 2 – 6 e. 9 days (counting the \$40. 10 times. 1 x 48 = 48 2.00 as 1 day) 4 . Api: 24 strides. 3. Myra: step 36 1. 32 16 8 4 2 1 0.) 2. Page 11 activity one Double and Halve 1. activity two 3. 18S + 4M + H = (18 x 10) + (4 x 100) + (1 x –20) = 180 + 400 – 20 = 560 2. For example: SSSSMSSMSSSMHSSMSSSSO (15S + 4M + H). (On the tenth doubling.
236 + 552 + 212. b. 1 000 x 1 or 1 x 1 000 500 x 2 or 2 x 500 250 x 4 or 4 x 250 200 x 5 or 5 x 200 125 x 8 or 8 x 125 100 x 10 or 10 x 100 50 x 20 or 20 x 50 25 x 40 or 40 x 25 Discussion will vary. Easy on a calculator. For example: 700 + 500 – 200 = 1 000 2 100 – 1 500 + 400 = 1 000 1 600 ÷ 2 + 200 = 1 000 1 200 ÷ 2 + 400 = 1 000. but it can also be done easily in your head: 63 – 40 = 23. and she needs to draw the 5. a 0 and a 5. a. Discussion will vary. 7. b. 700 + 500 – 200 = 1 000 could be expanded to (70 x 10) + (50 x 10) – (20 x 10) = 1 000 (140 x 5) + (100 x 5) – (40 x 5) = 1 000. Page 16 game Playzone Discount A game estimating percentages Page 14 activity 1.) activity 1. and so on. 6. You can do this in your head using standard place value: 900 + 30 + 4 = 934. Answers will vary. (There are two cards left. a. See comments in b below. Writing 1 000 Page 17 Different Approaches ii. You will know if you have found all the possibilities when you have used all the factors of 1 000. Many equations are possible. Some possibilities are: 8 x 3 + 9 + 8 + 2 = 43 (8 + 2) x 6 – 9 – 8 = 43 8 x 7 + 4 – 6 x 5 + 2 x 3 + 4 + 1 + 2 = 43 (8 + 3) x 6 – 9 – 8 – 2 x 3 x 1 = 43 (8 + 7) x 2 + 9 + 1 + 2 + 1 = 43 2. a. She has a 50% chance of winning. For example. 704 – 300 = 404. Some possibilities are: 9 x 5 + 3 – 4 – 1 = 43 (9 + 1) x 5 – 3 x 2 – 1 = 43 9 x 2 + 3 x 6 + 1 + 3 + 2 + 1 = 43 activity Vanna. b. Results will vary. For example: 100 + 300 + 600. In the head is easy: 50 + 50 = 100. Answers will vary. 37 + 3 + 23 = 63 so 37 + 26 = 63 and 63 – 37 = 26. Results will vary. (For example. iv. Set 1: 162 Set 2: 175 Set 3: 100 Set 4: 66 Set 5: 77 Set 6: 144 2. 3. For example: 510 + 490.Page 12 activity Find the Number 1. 2. so 704 – 299 = 405 5 . and 9 do not divide evenly into 1 000. 3. Clues will vary. 156 + 844.) Logan has no chance of winning in this round because he cannot use the 1 or the 5. so 63 – 37 = 23 + 3 = 26 or 63 – 37 = " is the same as 37 + " = 63. so 49 + 49 = 98. and so on. i. iii. Page 15 Page 13 activity Ways to 43 game Up the Ladder A game ordering numbers between 100 and 1 000 1.
8 sachets. (Jenny would spend \$7. 4 x 3 = 12 30 + 12 = 42 A possible number line for this method is: +10 10 +10 20 +10 30 +12 (4 x 3) 42 1. \$5. 98 – 43 = 55 (standard place value) 8–3=5 90 – 40 = 50 50 + 5 = 55 viii.400 3.700 b. 14 x 3 = 7 x 6 = 42 Page 18 activity It Pays to Win! ii.00. Two possible methods are: Page 19 activity 1. 9 + 9 + 9 + 9 + 9 = 5 x 9 = 45 or 10 + 10 +10 +10 +10 = 50. \$14. \$112. 10 x 4 = 40 4 x \$3. b.) 6 . 2 L b. a.00 2. 3 L Drink Stall i. a.00 each. (A 4:5 ratio means 8 sachets for 10 litres. 18 ÷ 6 = 3 10 + 3 = 13 3.00 = \$14. 12 sachets. 3L ii. a. 72 + 6 = 78. 30 L 6.100 better off 2. Page 20 activity Grocery Grapplers 1. 500 + 302 = 802 so 498 + 302 = 800 or 498 + 302 = 500 + 300 = 800 (compensation) 2. 11/2 cups of sugar c. 60 + 30 = 90. (15 litres is 3 times 5 litres. That’s 13. ii.67 2. Cordial. Two possible methods are: i. 10 + 90 = 100) x. a. so 9 + 9 + 9 + 9 + 9 = 50 – 5 = 45 vi. 81. Problems will vary. 41/2 bananas per kilogram 18 x 41/2 = (18 x 4) + 9 = 72 + 9 = 81 4.00.50 = 2 x \$7. 6 x 12 = 72.00. 20 sachets. \$104. 6 cups of sugar ii. i. 423 – 286 is probably easier on a calculator. so she needs three times as much water. a. 18 ÷ 2 = 9 lots of 2 kilograms (9 bananas) 9 x 9 = 81 ii. a. 68 + 32 = 100 (8 + 2 = 10.000 b.600 b. 42. 3 x 5 = 15 litres. Two possible methods are: i. i. and Toline would spend \$8. \$109. Two possible methods are: i. 9 L b. 26 + 24 = 50 so 26 + 26 + 24 + 24 = 100 vii. 5 x 4 = 20 sachets) 3.) 4. (5 x 5 = 25 litres. and Mere and Toline get \$20. 60 ÷ 6 = 10. \$7. but it can be done in the head: 286 + 14 = 300 300 + 123 = 423 123 + 14 = 137 ix. 50 lemons and 10 cups of sugar 5. 498 is 2 fewer than 500. \$3. 3 x 4 = 12 sachets) c.v. Jenny gets \$40. 60 cups b. 13. a.486. 10 x 3 = 30.00.
this would be \$17. so two CDs are free.00 + \$2.50 2.50.30 super saturn meal \$5.50. ii.00.50) = 2 x \$3.ii.50 = 2 (2 x \$1. with 30 cents off the bananas and approximately 15 cents off the apples.60 4.00.40 = \$3.40 \$6. 18 ÷ 12 = 11/2 (so they need 2 bags here).55.95 rounds to \$20. Bananas: \$1.30 = \$5.00. Answers will vary. Two possible answers are: earth meal upgrade \$5. 4 lots of 25 = 100 6.60 4 x \$1.50) = \$30. \$18. large liquid fuel \$5.50 5.45 rounds to \$1.30. \$2.00 6.00 – (5 x 5c) = \$14. 1/4 of \$20.00 two earth burgers. (8 x 10) + (8 x 5) = 80 + 40 = \$120 or 8 x 15 = 4 x 30 = 120 \$120.00 is \$5.75.95 rounds to \$3.30) = \$6.00 = \$6. Earth meal: \$5.95 rounds to 8 x \$15.60. 6 x \$5.55.00. 10 x 4 = 40 4 x \$3 = \$12 4 x 50c = \$2 = \$14.40 super saturn meal upgrade \$5. Answers and some possible methods are: 1.00 mega moon meal \$4.00. 78 ÷ 12 = 6 r6. 5 + 2 = 7 Page 21 activity 1.50 = \$9. \$15. \$9 + \$9 = \$18. Two possible methods are: i.60 + \$6. no upgrade 4.30 super saturn meal \$5.50 + \$3. \$14.90 Total: \$26.00) – (2 x 0.50 = (6 x \$5.00 = \$15.50 Change: 0. a. Super saturn meal. 60 ÷ 12 = 5. As a single purchase. \$119.50 2 x \$1.00 Page 22 activity Shopping Around Methods will vary. \$5.40 two super saturn burgers and one medium chip booster \$6. Apples are almost twice the price but half the quantity.95 is also about \$9.00. 100.40 = \$119.20 = \$2. so 3 x \$2.00 7 . 5 x \$1.00 2.00) + (6 x 0.75 3.80 mega moon meal upgrade \$5. 5 x \$3.00 – \$5. 5.10 or earth meal upgrade \$5. \$19.80 = (2 x \$2. so 7 bags are needed.40 = \$18.40 Change: \$1. \$17.00 + (5 x 0. 25 x 4 = (20 x 4) + (5 x 4) = 80 + 20 = 100 ii.00 = \$15. mega moon meal: \$5.80 Total: \$27.00 + \$3. Two possible methods are: i.00 2 x \$1.00 mega moon meal upgrade \$5. \$33. \$15. super saturn meal: \$5.50 Birthday Treat 3.00.00.90 b.00 = \$33. 8 x \$14. \$20.20) = \$4. \$27. 6 x \$1. There are two 5s in 10.00 – \$0.00 – 0.
90 = \$366. Take 25 x 8 = \$200 from the price in question 1: 398. 398. Approximately \$12. The exact answer is \$598. 19 weeks c.95 from the answer to question 1: 398.85 d. 24 CDs e. Take the answer to question 1 and add 25 x 4 = \$100 to make up for the price increase.38. 2. 2. Half of \$398. with three players left over 8 .196.25 c.75 – 200 = \$198.96. Multiply the answer to question 1 by 3: 398.75 + 100 = \$498.13) b. 398. Take 2 x \$15.) d.75 – 31.95 x 25 = 400 – (25 x 0.13.75 is \$199.75 x 3 = \$1. Use of strategies will vary. 600 ÷ 50 = 12 Page 24 activity Division Dilemmas 1.75 3.05) = 400 – 1.75. Methods will vary.75 b. a.25 = 398. 26 teams. (The exact answer is \$11. The solutions are: a. Possible methods include: a.75.) \$598. 37 dozen (A dozen is 12.95 = 398. You could calculate your answer like this: 16 x 25 = 4 x 100 = 400 so 15.38 = \$598. Answers may vary. 31 packets b. (25 x 15.00. Display methods will vary.13 is nearly \$600.Page 23 activity Keep Your Shirt On 1.75 + 199.
multiplication. subtraction. and division Exploring factors Using understanding of place value Doubling and halving numbers Defining sets of numbers Using order of operations Exploring ways of making 1 000 Ordering numbers between 100 and 1 000 Estimating percentages Choosing appropriate methods of calculation Applying numbers and operations to real-life problems Using proportions of whole numbers to solve problems Solving number problems in measurement contexts Using operations to solve money problems Using rounding to solve money problems Deriving multiplication and division answers Using mental strategies for division 9 .YEARS 7–8 Teachers’ Notes Overview Title Number Sense: Book One Content Page in students’ book 1 2–3 4–5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Page in teachers’ book 10 11 12 14 15 15 16 17 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Choice Calculations Splitting Numbers Aiming High Number Scavenge Hit the Target Pathways Flying Feet Space Marauders Double and Halve Find the Number Ways to 43 Writing 1 000 Up the Ladder Playzone Discount Different Approaches It Pays to Win! Drink Stall Grocery Grapplers Birthday Treat Shopping Around Keep Your Shirt On Division Dilemmas Solving operation-sign problems Adding and multiplying numbers Using place value and addition Using numbers in everyday life Using whole-number operations Practising addition.
For this answer.) 10 . you could say that basic calculators are programmed to automatically put brackets around numbers as they are entered. subtraction. it is good practice to have the students use brackets when recording on paper the sequence of keypad entries that they used on a basic calculator. and then addition and subtraction in the order that they occur. Sam has found one that works!” One approach is for the students to insert signs randomly into each equation and see what results they get. exponents. a student has two options: they could start by solving 4 x 4 and then entering 33 – 16 = . the students need to understand that it is possible to have a negative quantity. They can then use these results to give them clues for further attempts. (The acronym BEDMAS signifies the order in which operations should be carried out in an equation: brackets. The numbers involved have been kept small so that the students can focus on trying out different operations rather than being sidetracked by difficult calculations. Students who have been taught ideas such as “You can’t take 12 away from 3” will need to be shown that you can in fact do it using negative quantities. level 2) demonstrate knowledge of the conventions for order of operations (Number. (On some calculators. or division (Number. activity two This is an example of a genre of problems known as “broken calculator problems”. the students need to rethink the numbers and operations used in a problem to give them an equivalent result. Encourage the students with prompts such as “This equation gave an answer that was too large.) Draw the students’ attention to the rule in question 1a. For example. Basic calculators work out equations in the order in which the operations are entered. To solve this equation on a basic calculator. level 4) devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes. insist that the students work out for themselves at least one result for each equation they make up. Otherwise they may produce some that are too difficult or tedious for their classmates to attempt. In these problems. problem solving. so what could you change to get closer to the target?” One result for question 1a i could be 3 – 6 x 2 + 9. levels 2–3) • • activity one These problems are designed to encourage students to use flexible thinking with number operations to find a solution. this button is named MRC instead of MR. as we do when we follow the order of operations rules with equations such as 33 – (4 x 4) = " . The problems also reinforce the order of operations rules that apply to equations where multiplication or division is combined with addition or subtraction. There are no brackets or exponents used in this activity. if a student enters 33 – 4 x 4 = on a basic calculator. they build on their understanding of the relationships between numbers and operations and develop their fluency with part-whole strategies.Page 1 Choice Calculations Achievement Objectives • write and solve story problems which involve whole numbers. Scientific calculators have bracket keys so that the user can group numbers together anywhere in a sequence. Present the problems as a challenge and look for opportunities to give feedback such as “Yes. multiplication. so the BEDMAS rules are used only in a limited way. division and multiplication in the order that they occur. Make sure that the students understand which keys can be used. By doing this. In question 2. To encourage their awareness of this process when they are combining addition or subtraction with multiplication or division. it will read this as (33 – 4) x 4 =. or they could use the memory keys in the sequence 4 x 4 M + 33 – MR = . It is also important for them to know the way basic calculators handle the order of operations. using addition. In terms of BEDMAS.
area = 64 square units. the relationship between the dimensions of a rectangle and its area. area = 48 square units. 11 . area = 39 square units. have the students split 13 into two equal addends using fractions. As an extension to this question. we are finding the area of that rectangle. width = 1 unit. Some rectangles where the length + width equals 16 units are: Length = 15 units. using addition. width = 3 units. This would be 6. so the whole numbers that result in the greatest product have a difference of 1.5. problem solving.25. that is. levels 2–3) activity one This is a step-by-step investigation that relates to an interesting connection with another area of mathematics. subtraction. level 2) • devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes. width = 8 units. Question 2d gives an odd number.Pages 2–3 Splitting Numbers Achievement Objectives • write and solve story problems which involve whole numbers. the students should conclude that a square rectangle always has a greater area than an oblong rectangle that has been made by splitting a number. Length = 13 units. By multiplying the addends together. Length = 12 units. From this. Length = 8 units. multiplication. width = 4 units. The two addends that a number is split into can be visualised as representing the length and width of a rectangle. area = 15 square units. Show the students diagrams of the rectangles provided below to help them see how the area of a rectangle changes as the length and width change.5 + 6. The product of these two numbers is 42. which is 42. or division (Number. Thirteen is split into 7 and 6 to give the greatest product.
increase the pace at which the dice is thrown. a cube) yields the greatest volume. Explain this to the students and show them how the investigation is structured to lead them to this question. they could try starting low for the first six throws in the ones column and then use the last four throws to try to get near the target. problem solving. subtraction. or division (Number. 12 . This game is a great way to strengthen place value understanding and to devise some problem-solving strategies involving addition. 5 . the students should find that the prism with sides that are equal in length (that is. Pages 4–5 Aiming High Achievement Objectives • write and solve story problems which involve whole numbers. After the students have played a few games. This can be visualised as comparing the dimensions of a rectangular prism with its volume. Discuss possible strategies. using addition. Explain the rules to the students through a short demonstration. and x be pressed once each time to make the largest product?” The factors that do this are 52 x 43 because these digits are the nearest to making the dimensions of a square. Again.activity two This activity extends the previous investigation by looking at three-addend splits. Students who have grasped the idea could be challenged to investigate multiplication using a calculator. levels 2–3) game Photocopy the game board copymaster provided at the back of this booklet to give each student a recording sheet for the game. “How can the keys 2 . These are vital rules to encourage them to develop a strategy other than trial and error. such as starting high by using the first few throws to a score of about 900 and then concentrating on placing the remaining throws in the tens and ones to get the last 100 points. 4 . multiplication. level 2) • devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes. Have the students list the strategies they use and keep a tally of the winning ones to see which one gives the best results. Have the students discuss strategies that are easy and efficient for them to use in finding the differences. activity two Question 1 in this activity poses a problem that involves finding the score nearest to the target number by looking at differences. Some students may wish to extend this still further to see if this pattern applies with four or more splits. Question 3 in this activity is the real goal of this investigation. Emphasise the rule that does not allow them to change the place of the number once it has been entered as well as the rule that it is important to enter a number on every turn. Alternatively. 3 . The increase in frequency of the throws may force them to change their strategies because they are no longer being given sufficient time to do an exact addition as they go.
This would show that Mere could end with a total ranging from 987 to 992 if she put the dice score in the ones place. The use of a partial number line could help the students to see why Leila is closer to 1 000 than Tama. Question 3b encourages the students to discuss the chances of making exactly 500. she could put them into the tens place and make a total of 996 or 1 006. They could use a continuum with four or five descriptors or a number line to 100 and place the exact score of 500 somewhere along the line and discuss their reasons for the placement. Other ways to extend the challenge level include adding a thousands column and changing the target to 10 000 or adding a tenths column and keeping the target at 1 000. she would tie. such as rounding. tens. but you would be very lucky to throw the numbers you would need at the end. or hundreds places for every turn. with the students placing each of the two digits in any of the ones. to keep a running total. Extend the game by using two dice for each throw. For example: Tama 978 980 990 1 000 1 010 Leila 1 021 1 020 In question 2. 13 .” The students could then keep a tally of 100 games played by the class and see how many times a score of 500 was achieved. she would lose. For Leila. and with 988 (a 2). For example. But if she throws a 1 or a 2. With 987 (a 1). 978 plus 2 makes 980 and plus 20 makes 1 000. For example: No chance Some chance Likely Almost certain Certain “I placed it here because you could get exactly 500. you may need to encourage the students to make a list of all the possible results for Mere. they may just subtract 1 000 from 1 021 to find the difference of 21.They may choose to do a counting-on strategy for Tama’s score to build it up to 1 000. So Tama is 22 away from 1 000. This will encourage further strategies.
An entry such as \$2. The students are asked to find a decimal that is not money because common usage reads a quantity such as \$3. 14 .50 cents). which would be the correct way of describing a decimal number. Words such as “about”. An entry in the business pages may be 7. food. car. not 0. If the students cannot find a winning score of approximately 20 percent. the usual one found in newspapers is the mean or average. Temperatures are listed in the weather section of a newspaper. level 3) • explain the meaning of the digits in decimal numbers with up to 3 decimal places (Number.85 million would be a good example of a decimal where the unit is one million dollars. the weather page often lists data such as “mean annual rainfall” or “average sun for October”. Metric measures are used in the property. Ensure that the students understand each of the number types needed. level 3) • solve practical problems which involve whole numbers and decimals and which require a choice of one or more arithmetic operations (Number. mean. 13 Finding a common fraction may be difficult. and mode). This is an opportunity to emphasise that a percentage is a fraction “out of a hundred”.Page 6 Number Scavenge Achievement Objectives • explain the meaning of the digits in any whole number (Number. and weather sections of newspapers. have them find a sports team result and estimate the winning margin as a percentage. but sometimes car dealers advertise repayments such as / + 1/3 + 1/3 to show the amount that is charged each year. or “approximately” would indicate that an estimate and/or a rounding has been made in a report. level 4) activity This activity is designed to give students an enjoyable investigation that will help them recognise numbers in everyday use by researching in newspapers and magazines. It may be an interesting challenge to find as many different metric unit references as possible from the same newspaper.57 as “three dollars fifty-seven” instead of “three point five seven dollars”. “almost”. It may be interesting to tally the percentages found to see the most common ones used in the newspapers and magazines. sports. These may also be good examples of the use of rounding. 50 cents. such as the average temperatures for the month in the weather section or average scores in the sports section. Dollars and cents are also often written without a decimal point in newspapers (for example.25% where the unit is one percent. level 3) • express quantities as fractions or percentages of a whole (Number. Although there are three types of averages (median. For example.
The game also encourages flexible thinking with numbers and operations while applying number facts. Page 8 Pathways Achievement Objectives • write and solve problems which involve whole numbers and decimals and which require a choice of one or more of the four arithmetic operations (Number. levels 2–3) game This game is an ideal context to introduce the correct use of brackets to control the order of operations when forming equations that combine addition or subtraction with multiplication or division. reduce the set of cards given to each group to about 18 cards. Control the time taken to play the game by adjusting the range of target numbers to suit the time available. level 4) • devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes. level 3) • demonstrate knowledge of the conventions for order of operations (Number. For example. problem solving. 4. level 4) • devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes.Page 7 Hit the Target Achievement Objectives • write and solve story problems which involve whole numbers. the target of 10 could be made with 9 + 5 + 0. This will encourage the students to start working with fractional amounts and to include the square root symbol as an operation to be used. Make the game more difficult by allowing a decimal point to be placed in front of a card value. multiplication. and 5. Make a point of finding a way to make an equation using brackets.5 x 4 = 10. problem solving. If some students need an easier version of the game. subtraction. level 2) • demonstrate knowledge of the conventions for order of operations (Number. if the cards were 9. follow the path along the top route. if the cards were worth 3. 5. encourage them to explore the problem to get a feeling for what is happening and show them how they could record their attempts. levels 2–3) activity This activity is a great way of encouraging the students to solve problems by combining number operations. Alternatively. 1. Deal the cards face up and ask the students to come up with ways of getting close to the target number. you could allow them to use just some of the card values. the target of 10 could be made with 3 x (4 + 1) – 5 = 10. One way to record this route would be: x5 20 100 – 70 30 x3 90 ÷6 15 + 15 = 30 15 . 4. using addition. and 5. or division (Number. For example. Show them how to record each suggestion. Demonstrate the game with two students while the others observe. Before the students attempt to answer the questions. For example. Note that the rules say all four cards must be used.
Their answers to question 1a will give you an indication of whether or not the students have grasped the idea. This would give them a good feel for the problem. so make sure that this develops into “trial and learn” (from the result). Ask the students what this stride might be. the students may find it helpful to draw a diagram showing the stones numbered from 1 to 48. while question 1b leads them to connect the factors of 48 with the strides taken. or division (Number. (It would be either 6 or 8 because 6 x 8 = 48. “Trial and error” will be a common strategy. They may notice that two of the possible three ways use division. subtraction. As a further extension.Use problem-solving groups of four students and have them discuss starting strategies for question 1. Encourage the students to form a mental picture of each of the children described on the page as they cross the 48 paving stones. Make sure that this connection is understood. have the students make up a similar problem using all the factors of 60. level 2) activity This activity allows students to explore the way the factors of a number relate to a context. Emphasise the fact that each stride takes the same amount of time. while the other way (+ 45 + 15) will increase the total at the end. A simple chart form of recording may help to show this: Person Mira Api Matiu Eru Length of stride 1 2 3 4 x x x x x Number of strides 48 24 16 12 = 48 stones = 48 = 48 = 48 = 48 In questions 2 and 3. You could encourage them to make a diagram of the first 12 stones and then act out or draw the way each child crosses those stones. Page 9 Flying Feet Achievement Objective • write and solve story problems which involve whole numbers. using addition. regardless of how long the stride is. The students may also get a feeling for the pathways by looking at the last two steps along each pathway. A good strategy for this type of problem is simply to try out and keep a record of different pathways and their answers and then see which part of question 1 matches the path followed. This is vital in understanding the problem. Extend the question by introducing a fifth person who has a stride that uses a factor of 48 not yet mentioned. which will reduce the aggregated totals. Check that the students have numbered the stone closest to the pool as number 48. multiplication. 16 .) This question would confirm which students have made the connection between the context and the factors of 48 because it completes the set of factors of 48. It will also be useful to emphasise the need to be systematic in choosing and recording which pathways have been followed because it is easy to get confused or waste time repeating a rule that didn’t work.
17 . which is then recorded as –40 after the calculation has been done and the brackets have been removed. level 2) activity This activity will help the students to find efficient ways to add using the place value as well as the face value of the digits involved. This way. subtraction. 3 500. using addition. The interest in this activity is in the number of steps it takes to get to the number with a decimal point. All the other significant figures have no influence. for example. such as a half or a quarter. the strategy of halving also shows the connection between the common fraction of one-half and its decimal equivalent of 0. and then in the hundreds place. then in the tens place. work through the example of Kimberley’s first game to ensure that they have grasped the reasoning behind her recording system. they will see that the number of steps depends on the first significant figure (that is.or 3-digit whole numbers (Number. level 2) • explain the meaning of the digits in 2. the hits on the space station are recorded as + (2 x –20).Page 10 Space Marauders Achievement Objectives • write and solve story problems which involve whole numbers. The other aspect that they should explore and discuss is the way that both the positive and negative quantities that belong in the H (hits) or tens column can be recorded and calculated. so be sure to compare the number of steps it takes when we start from 2 000 (five steps) with the number of steps it takes from 2 500 (three steps). Encourage the students to explore the number of steps it takes to reach a number with a decimal point for numbers that end in the same quantity. 7 500.5 because this will be the result of halving any whole number until a number with a decimal point is reached. x is a letter of the alphabet as well as a symbol). Page 11 Double and Halve Achievement Objective • write and solve problems which involve whole numbers and decimals and which require a choice of one or more of the four arithmetic operations (Number. Some students may assume that larger numbers will require more steps. It will also help them learn how to record equations using pronumerals and algebraic conventions. or division (Number. They will find that it is the place the odd number is in that decides how many steps it takes to reach a decimal fraction. activity one In this activity. In Kimberley’s first game. 6 500. This is a good opportunity to point out the algebraic convention for recording multiplication: “Kimberley’s 8S means 8 times S. The students could then investigate putting any odd number in the ones place. After the students have read and understood the scoring system.” Explain that in cases such as this. we do not usually record the multiplication sign as an x because it becomes confusing when other letters are being used to represent unknown or variable numbers (because unlike + and –. level 3) The strategy of doubling and halving can be used to show the connection between division and multiplication by a basic fraction. the first non-zero figure. the 5). in this case. and 500. multiplication.
using addition. By comparing the number of steps for 1 600 and 50. and square numbers and the meaning of the associated vocabulary (digit.An investigation into even numbers is quite a different matter because for any even number. You could divide the students into groups. in the first question. A chart such as the one below may be helpful. multiplication. we can see the effect: Steps for 1 600 800 400 200 100 50 25 12. levels 2–3) activity This activity challenges the students to use clues based on the properties of number to solve the problems. The number range selected in the activity encourages the students to develop a deeper knowledge of numbers between 50 and 200. 18 . level 2) • devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes. Note that. A good context that many students will relate to is the way the size of the memory on computers usually grows in doubles. but it will take him 10 steps to go past that number. problem solving. As an extension. multiple) before the students start working on a set of clues.5 two steps Page 12 Find the Number Achievement Objectives • write and solve story problems which involve whole numbers. Jarod will not reach exactly 1 000 000 by doubling from 1 000. with each group working on a different set of clues. the megabyte increases could be shown as powers of 2.5 seven steps activity two Here the process is reversed so that the students can see the effect of doubling. subtraction. you first have to halve the number until you produce a number where the first significant figure is odd and then you proceed with the steps to a decimal fraction. 1 megabyte = 21 2 megabytes = 22 4 megabytes = 23 8 megabytes = 24 16 megabytes = 25 32 megabytes = 26 64 megabytes = 27 128 megabytes = 28 256 megabytes = 29 512 megabytes = 210 Steps for 50 25 12. even. or division (Number. You may need to explain the number properties of odd.
• The group discusses the clues to identify those they understand and those they are not sure about. Help them to see the connection between this equation and the way they record their chosen path. or division (Number. 10 + 30 + 3 = 43. They can read it out to the others. This will provide the opportunity for you to work with the students on the conventions of the order of operations that should be used in complex equations. These clues can also be used in a more co-operative problem-solving situation if the group’s set of clues is copied onto paper and each clue is cut out separately. then this is the solution. The path for 8 + 2 + 6 x 5 + 3 x 1 would look like this: 8+ 2+ 6x 5+ 3x 1 After the students have explored various ways of recording their pathway of calculations. It is best to have no more than four students in each group so that all the students can be involved in the discussion. 19 . Give each student responsibility for one of the clues. level 2) • demonstrate knowledge of the conventions for order of operations (Number. and 3 x 1 = 3. Others may use a template of a 5 by 4 set of squares and write in the operations and numbers of the squares that are on the path they have chosen. such as listing or highlighting numbers that are eliminated or remain after each clue until the solution is found. which would heighten the activity as a problem-solving situation and also provide you with feedback about the entry knowledge of the students. Then tell them to use all the clues to find an unknown number and give them some process instructions. multiplication.An alternative approach. • They write out the clues they are not sure about on the whiteboard. Discuss this with them. 6 x 5 = 30. they must check that it meets the requirements of their clue. Some students may choose to list each set of calculations. • The group discusses strategies that might work. When a solution is proposed. One of the challenges facing the students is how to record their pathway of calculations. such as 8 + 2 = 10. • The group attempts to solve the problem. would be to give out a set of clues to each group without giving them the accompanying board of numbers. and then the pathway. make sure that you show them how to record all the steps taken in one equation only rather than in a series of equations. subtraction. The students can then practise writing complex equations for other pathways. in this case. using addition. If everyone in the group agrees. such as: • One of the group must read out all the clues slowly twice. but they cannot hand it over. level 4) activity The cumulative calculations required in this activity will give the students plenty of opportunities to practise using the order of operations conventions. • You or students from another group explain the meaning of the clues that they don’t understand. Page 13 Ways to 43 Achievement Objectives • write and solve story problems which involve whole numbers.
subtraction. ensure that each student can clearly see the connection between their equations. This is a useful way to help the students understand the bridge between 3-digit and 4-digit numbers. 8 + 7 + 4 + 1 + 9 + 2 + 5 + 6 + 2 + 3 – 6 + 9 – 1 – 3 – 0 – 4 – 1 + 2 = 43) • the most routes • a route to a different prime number (for example. In your discussion with them. Question 2 encourages the students to use a compensation strategy to make a series of addition equations that are equivalent to 1 000. the students should discuss and share the ways they used compensation to help them derive new equations as they expanded their original equation. using addition. Page 14 Writing 1 000 Achievement Objectives • write and solve story problems which involve whole numbers. multiplication. level 4) activity This activity will develop fluency with combinations of numbers that make 1 000. This is then extended in question 4 into equations involving combinations of operations. 8 + 3 + 6 + 9 + 8 + 1 + 2 = 37) • an equation that uses division (for example. 8 ÷ 2 x 6 + 9 + 8 + 2 = 43) • an equation that uses all four operations (for example. such as: 40 x 20 + 20 x 10 8 x 50 + 4 x 100 + 1 x 200 60 x 30 – 40 x 20 50 x 25 – 20 x 10 – 10 x 5 20 . Question 3 links addition of equal groups to an equation using multiplication to encourage more efficient ways of thinking about 1 000. or division (Number. 8 x 7 ÷ 2 + 9 + 8 – 2 = 43). In question 4b. level 2) • demonstrate knowledge of the conventions for order of operations (Number. For example. be the first to find: • the shortest route • a long route (for example.Use special challenges to add interest to the activity. Challenge the students to come up with expressions that have a pattern in them.
If the students analyse the numbers already on the ladder. the two cards left are a 0 and a 5. This type of game is ideal to use as an independent activity in a learning centre after it has been used for instructional purposes. Vanna has a 50 percent chance of winning on this round because she can win if she draws the 5 (to make 593 or 539).Page 15 Up the Ladder Achievement Objectives • read any 3-digit whole number (Number. These aspects are the digits already chosen for this round and the number of winning digits still available for selection. This involves putting numbers close together on the ladder if they are close in size. the students will probably find that. level 2) • explain the meaning of the digits in 2. they should realise that Vanna and Logan have gone through the whole pile of cards once. (In practice. they will not be able to make a number in the nine hundreds until all the cards have been used and the pack is reshuffled. The students could explore the effect on the game if the number of sets of digit cards used is changed. they will need to go through the cards more than twice to fill their 10 rungs. The importance of the digit in the hundreds place should be discussed after the students have played the game a few times. The key learning opportunity comes from discussion of the strategies used to place numbers so that a player gives themselves the best chance of winning. The range could be reduced by changing the top and bottom numbers on the ladder. they have six cards left for the tenth round. and on the second time through. 21 .) After Vanna and Logan have drawn two cards each. Logan cannot win with either card because he needs a 2 or a 3 (to make a number between 407 and 218). The students should be encouraged to track the numbers that have been used so that they think about the possibilities left to them. For example. variations and extensions This game can be extended to 4-digit numbers or decimal numbers if a decimal point is placed on each rung. if all three nines have been used. activity Have the students discuss the aspects that will influence Vanna’s and Logan’s chances. It also involves placing smaller numbers nearer the bottom of the ladder and larger numbers nearer the top.or 3-digit whole numbers (Number. level 2) game This ladder game provides a fun challenge that builds understanding of the place value of 3-digit numbers as well as practice in ordering those numbers. in their own games. especially at the start and until they get used to using strategies for ordering their numbers.
If the main purpose of the game is to introduce percentage calculations to the students. The students could make a chart or a double number line showing the equivalent values of the percentages as fractions. starting with the 50 percent coupon. level 2) • find a given fraction or percentage of a quantity (Number. then the 25 percent. If the students are familiar with finding percentages. the students need to see these as “three lots of a quarter” and “three lots of one-tenth” respectively to develop a suitable strategy for calculating the percentage amounts. If the strategy for winning is the learning outcome you want the students to focus on. Make sure that the students understand that percentage means “out of 100”. The students could make a chart showing the discount amounts for each of the Playzone games. they should not use a calculator because the intention is to develop mental strategies for doing these. If both players do not make any mistakes. The player who goes first could choose the centre square and try to get a line of three in a row with both ends unblocked. level 4) game This game has two mathematical features: the calculation of a percentage discount and the use of a winning strategy to get four in a row on a 5 by 5 array. The player who goes second should use their turn to defend immediately by trying to block the first player until the first player makes a mistake. With 75 percent ( 3/4 ) and 30 percent ( 3/10 ). and so they allow the students to get used to dividing the amount by the denominator of the unit fraction before they move on to looking at the more complex calculations involved with the improper fractions. either the player going first will win or the game will end in a draw. you may let them do their discount calculation on a calculator. %. One interesting way to do this is to explain the origins of the percentage symbol. and then the 10 percent. Do these first because they are equivalent to unit fractions. Once the player has this.Page 16 Playzone Discount Achievement Objectives • make sensible estimates and check the reasonableness of answers (Number. Playzone game 10% Exotic Raiders \$40 Blitz \$80 Metal Strike \$130 Team Pro \$200 Lightning \$150 Sound Barrier \$100 \$4 \$8 \$13 \$20 \$15 \$10 25% \$10 \$20 \$33* \$50 \$38* \$25 Discount 30% \$12 \$24 \$39 \$60 \$45 \$30 50% \$20 \$40 \$65 \$100 \$75 \$50 75% \$30 \$60 \$98* \$150 \$113* \$75 * rounded 22 . The / represents “out of” or division and the two small zeros on either side of the / represent the zeros in 100. they can play the game as a percentages maintenance activity and focus on working out a winning strategy. So the symbol represents “out of 100”. Percentage Fraction 0 0 10 1 10 25 30 " 3 10 50 " 75 " 100 1 / / Have the students develop and discuss strategies for estimating and calculating a percentage amount. victory is assured.
problem solving. levels 2–3 ) activity This activity provides an interesting context for solving computational problems that involve 5. Make sure that the questions that the students create are not just basic facts because these are likely to only involve recall thinking as opposed to strategy thinking. On the questions that favour the calculator use. At the same time. The numbers involved have been deliberately chosen so that the students can use a range of strategies. level 3) • devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes. Encourage them to try to work out each payment mentally and then check by writing out their methods or doing it on the calculator.and 6-digit whole numbers. by adding. Encourage the students to record their thinking by carefully setting out their work and including their method of calculation for each step: 24 . Have the students list the strategies they see the brain people using to beat the calculator people on the questions designed for them to win. make sure they show that they understand what is going on in the problem and how each performance payment works as an add-on to the basic salary. For the questions designed to suit the calculator people. Encourage the students to use the opposite to the ideas used to make questions designed for the brain people to win. Whatever strategies the students use should be valued positively. the students should be encouraged to improve the speed and efficiency of their methods. for example. You should give guidance to individual students about which questions they should solve with a calculator. question 2b. To make questions suited to mental thinking. If the calculator is used as the first strategy.Question 2 asks the students to apply their understanding of mental strategies to create questions that are suited to mental reasoning versus calculator usage. they should be looking to select: • numbers that can be easily rounded to a tidy number (a number with a zero on the end) • numbers that can be joined to make a ten or a hundred • equations that look hard but are easy when a simple adjustment is made to them • equations that involve doubles or near doubles • equations that involve addition of repeated groups. you may need to specify the number of digits to be used in each numeral so that the students don’t just resort to very long numerals to try and make the questions difficult. Page 18 It Pays to Win! Achievement Objectives • write and solve problems which involve whole numbers and decimals and which require a choice of one or more of the four arithmetic operations (Number. or by multiplying. The students may attempt to solve the problems by counting forwards in hundreds or thousands. the students should explain what it is about the questions that makes it hard for brain people to solve them quickly. Before the students attempt the calculations involved in each question. it will help some students achieve success but the opportunity to solve the problems using the properties of the numbers involved will be lost.
In question 2a.900 \$112.400 Some students may choose to calculate the difference after working out the total income for each rate. which is thus read as “one to four”. Highlight the use of the ratio symbol (:) that means “to” in 1:4.000 \$22.000 \$4. the latter method will provide a great opportunity to show the students how to use the memory keys to record the running total.000 \$1.100 into the memory by pressing the M+ key.300 in the memory.000 \$1. Cordial (litres) Drink (litres) 1 5 2 10 3 15 25 . Alternatively. The running total will come up when the MR key is used to recall the amount.600 How I worked it out Tama’s basic salary 17 times \$1. level 3) • devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes. the students are introduced to the concepts of ratio and proportion and how these are used to solve problems. Introduce the students to the use of a ratio table for considering the relationship between cordial amounts and the total drink produced in questions 1a and b. Page 19 Drink Stall Achievement Objectives • write and solve problems which involve whole numbers and decimals and which require a choice of one or more of the four arithmetic operations (Number. they could set up a running total of the differences calculated at the end of each stage. They need to see that when the cordial and water are mixed in a ratio of 1:4. enter the result of + \$5. problem solving.000 has been subtracted from the \$22. For example.000 \$6. If calculators are used. after the \$17. Explore Jenny’s statement with the students to ensure that they know how the parts (ratio) are used to make up the whole.100. levels 2–3) activity In this activity.000 \$17. Use the same process to store the + \$2.000 for each win 2 draws earns \$500 plus \$500 46 kicks times \$100 I added all the amounts. the students could use a recording method that helps them compare the new payments with the old.300 +\$7.600 \$104. the fraction of the whole that the cordial makes up is 1/5 and the fraction of the whole that the water makes up is 4/5.600 Total \$104.What Tama earned \$82.000 \$4.100 \$0 +\$2.000 \$17. What Tama earned last season Salary Wins Draws Kicks Total \$82.000 Difference \$0 +\$5.100 \$1.600 Tama’s new rates applied to last season \$82.
multiplication. Toline’s table shows sachets in groups of 4 and drinks in groups of 5. Evaluate the answer.The students could then take this idea and use it to show what each question is about before they solve it. Solve the problem. subtraction. 26 . Suggest to the students that they discuss what is happening in the problem to work out the ratios involved. For question 6. Discuss why this must always be true. For example. Question 4 has an interesting connection between questions a and b that you could share with the students. Ensure that the students understand the importance of the following three key stages of problem solving: 1. Understand the problem. or division (Number. problem solving. are shared in the ratio 2:1:1. the number of litres of drink made with 15 lemons is the same as the number of cups of sugar used with 15 lemons. 3. That is. This is best done using problem-solving groups of four students. the girls’ shares. the students will come to appreciate the value of each stage. level 2) • devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes. The ratio table would look like this: Jenny’s part Mere’s part Toline’s part Total sales \$2 \$1 \$1 \$4 \$10 \$5 \$5 \$20 \$20 \$10 \$10 \$40 \$40 \$20 \$20 \$80 Page 20 Grocery Grapplers Achievement Objectives • write and solve story problems which involve whole numbers. Jenny:Mere:Toline. Toline’s ratio table would look like this: Sachets Drink (litres) 4 5 8 10 ? 15 16 20 ? 25 You can use this table to discuss the connections between the different elements. If you insist that each group explore each stage and report their findings before moving on to the next stage. This will give them clues for deriving the equations they will need. It also shows how 4 sachets relate to 5 drinks in the same proportion as 8 sachets to 10 drinks. Question 6 is similar to question 4 in that it extends the ratio idea by involving three parts that combine to make the whole. while the others get 1 out of every 4 parts. So Jenny gets 2 out of every 4 parts. levels 2–3) activity This activity uses measurement contexts to enable students to understand what is happening in the problem. 2. using addition. in question 2. This is said as “two to one to one”.
These alternative strategies show that the essence of problem solving in number is to deliberately manipulate the quantities that make the problem difficult in order to form easier steps that can be recombined successfully. • checking that their answer makes sense. levels 2–3) activity This activity uses a context that should appeal to students in this age group and should thus encourage them to develop their number problem-solving strategies further. Take the opportunity to see if the students can find the multiplicative relationship between the \$5. or actions to explain to each other what they think is happening in each problem before they attempt to use numbers in equations or algorithms. Page 21 Birthday Treat Achievement Objectives • write and solve story problems which involve whole numbers. Have the students report their solutions for question 1 and resolve any difficulties before they continue with the next question. Six bags of fish gives only 72 fish. Another idea is to have the students list the key words and facts. • checking their calculations. subtraction. multiplication.The students could use their own words. Open or double number lines would help those students who find it difficult to come up with a suitable open number sentence to solve the problem. 27 . the movies will take half of the \$55. If the students make an error in question 1. and Needed (what I have to do to get the wanted data). or division (Number. level 2) • devise and use problem-solving strategies to explore situations mathematically (Mathematical Processes. Encourage the students to share two or three easy or efficient ways to solve the problems. Examples of alternative strategies are shown in the Answers. so they will need to buy an extra bag of 12 fish to get the 6 fish they need in order to have 78 chocolate fish altogether. This is a “10 times larger” relationship.50 cost for the movie and the \$55 total. Using an open number line to solve problem 1: Bottles 1 Litres 3 10 30 14 15 ? 45 Using a double number line to solve problem 1: Bottles Litres 1 3 2 6 3 9 4 12 5 15 10 30 11 33 12 36 13 39 14 42 The students can evaluate their answers by: • checking that all the important information has been used. using addition. Question 6 requires a practical solution in that Nick and Ema need to buy more chocolate fish than they actually need. leaving half for food. and as there are 5 boys. possibly using headings such as Given. diagrams. problem solving. Wanted. it will have a carry-over effect to the following questions.
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Fiche explicative de la leçon: The Magnetic Field due to a Current in a Solenoid | Nagwa Fiche explicative de la leçon: The Magnetic Field due to a Current in a Solenoid | Nagwa
# Fiche explicative de la leçon: The Magnetic Field due to a Current in a Solenoid Physique
In this explainer, we will learn how to calculate the magnetic field produced by a current in a solenoid.
Recall the direction of a magnetic field in a loop of a current-carrying wire. At the center of the loop, the magnetic field has one direction, as seen in the diagram below. The orange line is the magnetic field direction and black line is the wire loop.
The same loop seen from the front, with the magnetic field direction pointing out of the screen, would look like the diagram below.
Recall that the symbols below are used to show that a direction is going out of or into the screen.
The magnetic field strength at the center of a loop can be increased by placing more loops in line with it. The diagram below shows two sets of loops with the same current and the same radius lined up in this way.
The set of loops on the right has a stronger magnetic field because it has more loops.
Instead of using a set of loops, strengthening the magnetic field at the center can be achieved using a single wire with multiple turns. The diagram below shows such a wire, with a side and front view.
A wire with a series of turns like this is called a solenoid. Each turn of a solenoid contributes to the strength of the magnetic field at the center just like an additional loop would.
The magnetic field strength and direction at the very center of a solenoid is uniform. It has one direction and magnitude. Other points around the solenoid have different magnetic field directions and magnitudes.
Before looking at the magnetic field lines of a solenoid, letβs consider the angle we will be viewing it from using a single loop. The diagram below show a single loop of a current-carrying wire and its resultant magnetic field from two different angles.
The side view of this loop shows the direction in which an observer looks, indicated by the eye, to obtain the top-down view. The top-down view still shows the direction of current, going into and out of the screen, but it does not show the bottom of the loop.
Now, letβs look at the top-down view of a single loop with its magnetic field lines, in grey, in the diagram below.
When there are more magnetic field lines close together, it means there is a stronger magnetic field. We can see that at the very center of the loop, the magnetic field lines are very close to each other with the same direction, meaning it has a strong magnetic field at that point.
Outside of the loop, the magnetic field lines resemble that of a bar magnet, as seen below.
Letβs now look at a solenoid with seven turns from this angle. The diagram below shows this with its corresponding magnetic field.
Note how the magnetic field lines are consistent and straight at the very center of these wire turns, but become less consistent toward the ends of the solenoid. The closer to these ends, the less uniform the field lines.
Now, letβs consider a theoretical, very long solenoid. It is so long that we can treat it as if it had no ends. This would mean it has a perfectly uniform magnetic field within the turns at all points.
If we measured the magnetic field strength at different points within the turns of this theoretical solenoid, in each case the magnitude and direction of the magnetic field strength would be the same. The diagram below shows a theoretical solenoid with three points, indicated by the red dots, that have equivalent magnetic field strengths and directions.
The magnetic field strength within the turns of this theoretical solenoid can be determined using an equation.
### Equation: Magnetic Field Strength at the Center of a Solenoid
The magnetic field strength, , inside the center of a solenoid is found using the equation where is the current of the solenoid, is the number of turns the solenoid has, is the length of the solenoid, and is the permeability of free space, Tβ
m/A.
For a real solenoid with limited length, this equation is still useful to describe the magnetic field strength at the exact center of the turns, since this is where it is uniform. The diagram below shows points with the same magnetic field strength and direction, on both a theoretical and real solenoid.
Real solenoids have a fairly constant magnetic field direction inside the turns, but not magnetic field strength. Only the center has a consistent magnetic field strength.
Looking at the equation, we see that the length of a solenoid matters when finding the magnetic field strength at the center. Specifically, that magnetic field strength is inversely proportional to length. The diagram below shows two solenoids with the same current and number of turns but with different lengths.
Since the solenoid at the bottom has twice the length, it will have half the magnetic field strength at its center.
Letβs look at an example using this equation.
### Example 1: Magnetic Field Strength at the Center of a Solenoid with Turns and Length
A solenoid has a length of 3.2 cm and consists of 90 turns of wire. The wire carries a constant current of 1.2 A. Calculate the strength of the magnetic field at the center of the solenoid. Give your answer in teslas expressed in scientific notation to one decimal place. Use a value of Tβ
m/A for .
We will use the equation to find the magnetic field strength at the center of this solenoid.
Before we can substitute in the values we are given, we need to ensure the units all match. The permeability of free space uses metres, so we need the length of 3.2 cm in metres.
There are 100 centimetres in 1 metre:
So, to convert 3.2 cm to metres, we multiply it by the relation
Thus, 3.2 cm is 0.032 m.
We can now substitute the values into the equation. The length is 0.032 m, the current is 1.2 A, there are 90 turns, and the permeability of free space is Tβ
m/A. This gives us
Letβs multiply across the numerator. This cancels the units of amperes there, giving
So, now when we divide these two numbers, the units of metres cancel, leaving only teslas:
Rounded to one decimal place, the magnetic field strength at the center of the solenoid is T.
The magnetic field strength at the center of a solenoid equation can be used to find other variables in the equation if the magnetic field strength at the center of the solenoid is known. To show this, letβs look at the base equation and put all the values in terms of . Starting with the equation we can multiply both sides by :
This cancels the on the right side, leaving behind
From this form, we can divide both sides by and :
This cancels and on the right side, leaving behind just :
Letβs look at an example using this form of the equation.
### Example 2: Current Determination in a Solenoid with Turns and Length
A solenoid is formed of 35 turns of wire over a length of 42 mm. The magnetic field at the center of the solenoid is measured to be T. Calculate the current in the wire. Give your answer in amperes to 2 decimal places. Use a value of Tβ
m/A for .
Recall that the equation can be put into a form that relates the variables to current:
Before we directly substitute the values we are given into this form of the equation, we need to make sure the units match. Permeability of free space uses metres, so we need the length of the solenoid, 42 mm, to also be in terms of metres.
There are 1βββ000 millimetres in 1 metre:
So, multiplying this relation by 42 mm will give us the value in metres:
The length of the solenoid in metres is 0.042 m.
Now, we can substitute the values into the equation. The length is 0.042 m, the magnetic field strength is T, there are 35 turns, and the permeability of free space is Tβ
m/A. This gives us
Multiplying across the numerator gives units of Tβ
m:
The number for the turns in the solenoid is unitless, so multiplying across the denominator does not change the units:
Dividing the top across the bottom will completely cancel the units of Tβ
m and leave behind amperes on the top. Looking at just the units, dividing by a fraction is the same as multiplying by its reciprocal:
So, dividing the numbers gives
Rounded to two decimal places, the answer is thus 0.47 A.
The equation can be put in terms of other variables as well. Letβs say we have a solenoid with an unknown length but with other known variables. Beginning with the base equation we can get the length onto one side of the equation by multiplying both sides by :
This cancels the on the right side, giving
We then divide both sides by to get which cancels the on the left side, leaving just :
Letβs look at an example that uses this form of the equation.
### Example 3: Length of a Solenoid
A solenoid formed from a length of wire has 80 turns. The solenoid carries a constant current of 13 A and the strength of the magnetic field produced is measured to be T at its center. Calculate the length of the solenoid, giving your answer to the nearest centimetre. Use a value of Tβ
m/A for .
Recall that the equation can be put into a form that relates the variables to the length of a solenoid:
Letβs substitute the values we are given into this form of the equation. The current is 13 A, the number of turns is 80, the magnetic field strength at the center is T, and the permeability of free space is Tβ
m/A. This gives us
Multiplying across the numerator, the units of amperes cancel, leaving just Tβ
m:
Dividing these numbers cancels the teslas, leaving behind only metres:
So, the length of this solenoid is 0.179 metres. We are not done yet though, as we want the final answer of the problem in centimetres.
To put this answer in centimetres, recall that there are 100 centimetres in 1 metre:
Multiplying this by our answer in metres will give us the answer in centimetres:
So, rounding to the nearest centimetre, this solenoid has a length of 18 centimetres. The answer is 18 cm.
Recall that the length of a solenoid is inversely proportional to the magnetic field strength at its center. A longer length can be counteracted by adding more turns in the wire, as seen in the diagram below.
Both solenoids have the same magnetic field strength since the longer solenoid has a proportionally larger amount of turns. We can also see that the longer solenoid is essentially the same as the first, but there is just more of it.
This means that adding more wire turns, making a solenoid longer in the process, does not increase the magnetic field strength at the center at all. What increases the magnetic field strength is the number of turns over a given length. This is proved by looking at the equation
If we assume that the current is the same for two solenoids, then the only nonconstant variables that affect the magnetic field strength are the number of turns and the length :
We can see that doubling the turns to and the length to does not change this proportion at all. The doubled values cancel each other out:
To simplify the equation, this proportion is often condensed to a simple lowercase , the units of which are turns per unit of length. Inside the full equation, this looks as follows.
### Equation: Magnetic Field at the Center of a Solenoid with Turns per Unit Length
The magnetic field strength, , inside the center of a solenoid is found using the equation where is the current of the solenoid, is the number of turns per unit of length, and is the permeability of free space, Tβ
m/A.
The units of are expressed per unit of length. For example, consider the solenoid in the diagram below.
The value of is the total turns over the total length:
So, 6 turns and 3 cm of length gives
If we were to double the turns to 12 and double the length to 6 cm, we would see the value of is still the same
Only by changing the proportion of turns to the length of the solenoid will the magnetic field strength change.
Letβs look at an example.
### Example 4: Magnetic Field Changes in a Solenoid
A length of wire is formed into a solenoid with turns of wire per millimetre. The wire carries a constant current . As a result, a magnetic field of strength can be measured at the center of the solenoid. Which of the following changes to the system would increase the magnetic field strength at the center of the solenoid, assuming everything else remains constant?
1. Decreasing the length of the solenoid by removing turns of wire while keeping constant
2. Decreasing , the current in the wire
3. Decreasing , the number of turns of wire per millimetre
4. Increasing , the current in the wire
5. Increasing the length of the solenoid by adding turns of wire while keeping constant
Letβs recall the form of the equation with turns per unit of length:
If does not change in this equation, the magnetic field strength does not change. Adding or removing parts of the solenoid, but keeping the constant, means the magnetic field strength stays the same.
Decreasing , however, will decrease the magnetic field strength. Likewise, decreasing the current will also decrease the magnetic field strength. This is because magnetic field strength is directly proportional to both and .
The only way to increase the magnetic field strength is by increasing or . The only answer with this increase is D, increasing .
The correct answer is D, increasing the current in the wire will increase the magnetic field strength.
When using to perform calculations, turns are unitless, so the units of are just per unit of length. This means that though we would say 5 turns per metre, inside of an equation we would just write
Letβs look at an example.
### Example 5: Magnetic Field Strength at the Center of a Solenoid
A wire that carries a constant current of 0.15 A is formed into a solenoid with 11 turns per centimetre. Calculate the strength of the magnetic field at the center of the solenoid. Give your answer in teslas expressed in scientific notation to one decimal place. Use a value of Tβ
m/A for .
The solenoid looks like the diagram below.
Recall the equation for magnetic field strength at the center of a solenoid using turns per unit length:
Before substituting in the values into this equation, we need to make sure the units match. Permeability of free space uses metres, so we need to put in terms of metres as well.
The value of is 11 turns per centimetre, and there are 100 centimetres in 1 metre:
Multiplying this relation by 11 turns per centimetre will turn it in to turns per metre:
Now, we can substitute the values into the equation. The current is 0.15 A, is 1βββ100 turns per metre, and is Tβ
m/A. This gives us
Multiplying the permeability of free space into the turns per metre cancels the metres, leaving behind
Multiplying the last two numbers together cancels the units of amperes, leaving teslas to give
So, rounded to one decimal place, the magnetic field strength at the center of this solenoid is T.
Just like with the other version of the magnetic field strength equation, we can isolate specific unknown variables. For example, if we are given a solenoid with an unknown current, we can determine it by putting the equation in terms of .
To do so, letβs divide both sides by :
This cancels the on the right side, leaving behind only :
Letβs look at an example that uses this form of the equation.
### Example 6: Current Determination in a Solenoid With Turns per Length
A solenoid is formed of a length of wire that carries a constant current . The solenoid has 430 turns of wire per metre. The magnetic field at the center of the solenoid is measured to be T. Calculate the current, , in amperes. Give your answer to 1 decimal place. Use .
Recall that the equation can be put in terms of as follows:
Using this form, letβs substitute in the known values. Magnetic field strength is T, is 430 turns per metre, and is Tβ
m/A. This gives us
Multiplying across the denominator removes the units of metres, giving
Dividing by a fraction is the same as multiplying by its reciprocal. This means the only unit after the division will be amperes:
So, when dividing the numbers, the answer becomes
Rounded to the nearest decimal place, the answer is 5.9 A.
Letβs summarize what we have learned in this explainer.
### Key Points
• A solenoid is a wire arranged in a series of turns or loops.
• When a solenoid carries a current, it produces a magnetic field that is strongest in the center of its loops.
• Within the solenoid loops, the magnetic field strength is given by the equation where is the number of turns in the solenoid, is the current in the solenoid, is the length of the solenoid, and is the permeability of free space, Tβ
m/A.
• The equation for magnetic field strength at the center of a solenoid using turns per unit of length is where is the number of turns per unit of length, is the current of the solenoid, and is the permeability of free space, Tβ
m/A.
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The formula for change of bases of logarithms can be useful when we do not have calculators that allow us … Read more
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## Lab 19 Matchmaking
#### Background
Basically people get along better if they have more interests in common than if they don't. The purpose of this lab is to attempt to quantify this idea. Disclaimer: The results stated here or derived from this lab are for amusement purposes only. There is no scientific basis for any of the results
#### Math
Suppose a bunch of people take a 10 question quiz on their likes and dislikes where they can score their answer on each question from -5 meaning 'severly dislikes' to +5 meaning 'loves it to death'. We could write a person's 10 answers in a vector. For person k, call this vector vk. Now if we wanted to know if person i and person j would get along, we could compare the answers, but, being mathematicians, a more mathematical way would be to compute the inner product of their 'answer vectors', i.e. <vi,vj>. Then based on the value we could rate their compatibility. If we computed this 'compatibility' index for all the people we could see who each person was most compatible with (the larger the index the better). For example if we did this with 3 people with 'answer vectors' like this:
```v1 = [ 5 -3 1 0 4 3 0 -5 4 -1];
v2 = [ 1 3 5 3 -4 -1 5 5 -1 4];
v3 = [-5 -2 3 -5 -4 -3 -3 1 -3 -3];
```
Then by computing all the inner products (using dot) we'd get the following results
```dot(v1,v2) = -51
dot(v1,v3) = -55
dot(v2,v3) = -11
```
From this we might conclude that person 1 wouldn't really get along with either #2 or #3 but might get along slightly better with #2. #2 would be somewhat neutral with #3 (and vice-versa) and of the three #2 and #3 are the most compatible.
#### MATLAB
For small samples this is easy to compute, but for a larger set we need to be more efficient. Write each 'answer vector' as a column vector and combine them into a large 10 x M matrix V where M is the number of people. Then since the Euclidean inner product of two column vectors u and v can be computed as uTv, we can compute the inner products of all the vectors with all the other vectors simply by computing VTV. This will be a M x M matrix and the (i,j) entry will be the compatibility index for person i with person j. From the example above, we get
```V = [v1',v2',v3'];
C = V'*V
102 -51 -55
-51 128 -11
-55 -11 116
```
For further processing (i.e. finding maximums) we need to first replace the diagonal elements with large negative values (as we know each person is compatible with themselves or at least we hope so, and we don't want your match to be you). Then we can find the maximum of each column. In MATLAB we do this, like:
```D = diag(C); % save the diagonal elements
C = C - diag(D) - 1000*eye(3); % replace the diagonal values with -1000
[z,ind] = max(C); % compute the max and its location of each column
```
Then z will contain the maximum of each column, while ind(i) will be the number of the person most compatible with person i. In our sample problem, you'd get z = (-51,-11,-11) and ind = (2,3,2). Meaning person 1 is most compatible with person 2, and persons 2 and 3 are most compatible with each other.
#### Exercises
1. Make up a 10 question like/dislike questionaire and ask at least 10 people to fill it out. You might want to take it to the dorm and ask other GSS students so you can become the e-Harmonizer of GSS.
Once you have the answers, process them to find the most compatible person for each person.
Turn in your questionaire and the results (you can just give me the numbers, i.e. the ind vector from above).
2. (Extra Fun) In a more scientific situation the approach would be similar except that the questions would be chosen more carefully and instead of using the standard Euclidean norm, they would use some sort of weighted norm. For example if they thought that similarities or differences in one category had more impact, they would weight the inner product in favor of that question, i.e. instead of using the term uivi to contribute to the inner product, they'd use 10uivi. Also, because answers to questions might be linked they might introduce a matrix A representing those links, e.g. instead of using the answers to 1 and 2 separately they might look at a combination of those two answers before they did the inner product. In this case, you'd compute the compatibility matrix via V'*A'*A*V.
Create a 10 x 10 matrix A which changes the weights on each question or combines the answers to the questions and compute the compatiblilty results for your sample.
Turn in the matrix and the results.
Mail: [email protected]
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## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: If the plane $2 x-y+z=0$ is parallel to the line $\frac{2 x-1}{2}=\frac{2-y}{2}=\frac{z+1}{a}$, then the value of $a$ is A. $-4$ B. $-2$ C. 4 D. 2 Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The equation of the plane passing through the points $A(0,-1,0), B(2,1,-1)$ and $C(1,1,1)$ is given by A. $4 x+3 y-2 z-3=0$ B. $4 x-3 y+2 z+3=0$ C. $4 x-3 y+2 z-3=0$ D. None of these Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The equation of the plane passing through the intersection of the planes $3 x-y+2 z-4=0$ and $x+y+z-2=0$ and passing through the point $A(2,2,1)$ is given by A. $7 x+5 y-4 z-8=0$ B. $7 x-5 y+4 z-8=0$ C. $5 x-7 y+4 z-8=0$ D. $5 x+7 y-4 z+8=0$ Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The equation of the plane passing through the points $\mathrm{A}(2,2,1)$ and $\mathrm{B}(9,3,6)$ and perpendicular to the plane $2 x+6 y+6 z=1$, is A. $x+2 y-3 z+5=0$ B. $2 x-3 y+4 z-6=0$ C. $4 x+5 y-6 z+3=0$ D. $3 x+4 y-5 z-9=0$ Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The line $\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-3}{-3}$ meets the plane $2 x+3 y-z=14$ in the point A. $(2,5,7)$ B. $(3,5,7)$ C. $(5,7,3)$ D. $(6,5,3)$ Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The equation of a plane through the point $A(1,0,-1)$ and perpendicular to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+7}{-3}$ is A. $2 x+4 y-3 z=3$ B. $2 x-4 y+3 z=5$ C. $2 x+4 y-3 z=5$ D. $x+3 y+7 z=-6$ Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: If a plane meets the coordinate axes in $A, B$ and $C$ such that the centroid of $\triangle A B C$ is $(1,2,4)$, then the equation of the plane is A. $x+2 y+4 z=6$ B. $4 x+2 y+z=12$ C. $x+2 y+4 z=7$ D. $4 x+2 y+z=7$ Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The plane $2 x+3 y+4 z=12$ meets the coordinate axes in $A, B$ and $C$. The centroid of $\triangle A B C$ is A. $(2,3,4)$ B. $(6,4,3)$ C. $\left(2, \frac{4}{3}, 1\right)$ D. None of these Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: If the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies in the plane $2 x-4 y+z=7$, then the value of $k$ is A. $-7$ B. 7 C. 4 D. $-4$ Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: If $O$ is the origin and $P(1,2,-3)$ is a given point, then the equation of the plane through $P$ and perpendicular to OP is A. $x+2 y-3 z=14$ B. $x-2 y+3 z=12$ C. $x-2 y-3 z=14$ D. None of these Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: If the line $\frac{x+1}{3}=\frac{y-2}{4}=\frac{z+6}{5}$ is parallel to the plane $2 x-3 y+k z=0$, then the value of $k$ is A. $\frac{5}{6}$ B. $\frac{6}{5}$ C. $\frac{3}{4}$ D. $\frac{4}{5}$ Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: A plane cuts off intercepts $3,-4,6$ on the coordinate axes. The length of perpendicular from the origin to this plane is A. $\frac{5}{\sqrt{29}}$ units B. $\frac{8}{\sqrt{29}}$ units C. $\frac{6}{\sqrt{29}}$ units D. $\frac{12}{\sqrt{29}}$ units Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The equation of a plane passing through the point $A(2,-3,7)$ and making equal intercepts on the axes, is A. $x+y+z=3$ B. $x+y+z=6$ C. $x+y+z=9$ D. $x+y+z=4$ Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The length of perpendicular from the origin to the plane $\overrightarrow{\mathrm{r}} \cdot(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-12 \hat{\mathrm{k}})+39=0$ is A. 3 units B. $\frac{13}{5}$ units C. $\frac{5}{3}$ units D. None of these Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The direction cosines of the normal to the plane $5 y+4=0$ are A. $0, \frac{-4}{5}, 0$ B. $0,1,0$ C. $0,-1,0$ D. None of these Solution:...
## Mark against the correct answer in each of the following:
Question: Mark against the correct answer in each of the following: The direction cosines of the perpendicular from the origin to the plane $\overrightarrow{\mathrm{r}} \cdot(6 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+1=0$ are A. $\frac{6}{7}, \frac{3}{7}, \frac{-2}{7}$ B. $\frac{6}{7}, \frac{-3}{7}, \frac{2}{7}$ C. $\frac{-6}{7}, \frac{3}{7}, \frac{2}{7}$ D. None of these Solution:...
## Write the equation of a plane passing through the point
Question: Write the equation of a plane passing through the point $(2,-1,1)$ and parallel to the plane $3 x+2 y-z=7$. Solution:...
## Write the angle between the line
Question: Write the angle between the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2}$ and the plane $x+y+4=0$ Solution:...
## Solve this following
Question: Find the value of $\lambda$ for which the line $\frac{x-1}{2}=\frac{y-1}{3}=\frac{z-1}{2}$ is parallel to the plane $\bar{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})=4$ Solution:...
## Find the length of perpendicular from the origin to the plane
Question: Find the length of perpendicular from the origin to the plane $\bar{r} \cdot(2 \hat{j}-3 \hat{j}+6 \hat{k})+14-0$. Solution:...
## Solve this following
Question: Show that the line $\vec{r}-(4 \hat{i}-7 \hat{k})+\lambda(4 \hat{i}-2 \hat{j}+3 \hat{k})$ is parallel to the plane $\vec{r} \cdot(5 \hat{i}+4 \hat{j}-4 \hat{k})-7$. Solution: Hence, the given line is parallel to the given plane....
## Find the direction cosines of the perpendicular from the origin to the plane
Question: Find the direction cosines of the perpendicular from the origin to the plane $\bar{r} \cdot(6 \hat{i}-3 \hat{j}-2 \hat{k})+1-0$. Solution:...
## Find the length of perpendicular drawn from the origin to the plane
Question: Find the length of perpendicular drawn from the origin to the plane $2 x-3 y+6 z+21=0$. Solution:...
## Write the equation of the plane passing through the point
Question: Write the equation of the plane passing through the point $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ and parallel to the plane $\bar{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$. Solution:...
## Solve this following
Question: Find the value of $\lambda$ such that the line $\frac{x-2}{6}=\frac{y-1}{2}=\frac{z+5}{4}$ is perpendicular to the plane $3 x-y-2 z=7$. Solution:...
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# relation between pressure and flow rate for gear pump
With: P = Power transmitted to the fluid by the pump in Watt. Q = Flow in m3/s. p = density of the liquid in kg/m3. Hm = Hydraulic pressure loss network expressed in m. H = Hydraulic load in meter of water. 9.81 = Average Intensity of gravity. Degraded energy A centrifugal pump Head-Capacity (head versus rate of flow) curve rises in pressure from the rated point towards pump shutoff. The MCSF falls somewhere between the rated point and shutoff, so the pressure at MCSF will be the maximum recommended for continuous pump operation.
### The Difference Between Pressure and Flow
2017/9/26Now that we know what these terms mean, let's take a look at the relationship between pressure and flow. Say we need to move a 10 lb. block across a long table. While 100 psi of air pressure may not be enough force to move the block, 115 psi will, which is why it's important to know the minimum pressure needed for your process.
Pump calculator solving for water horsepower given discharge or flow rate and total head Solve for NPSH - net positive suction head Solve for fluid or liquid velocity Solve for pressure at impeller inlet Solve for fluid or liquid vapor pressure
It shows how aggressive the pump impeller inlet design is (how low is the NPSHR for a given pump speed and BEP flow rate). Higher "S" values mean lower NPSHR and, therefore, greater NPSH Margins, which by itself, is a good thing (if it doesn't push the pump into "High", or
Every centrifugal pump is therefore characterized by its particular characteristic curve, i.e. the relations between its flow rate and its differential head. This graphical representation, i.e. the transposition of this relation on a Cartesian graph, is the best way to learn what flow rate is obtained at a
A pressure is needed to make the liquid flow at the required rate and this must overcome head 'losses' in the system. Losses are of two types: static and friction head. Static head is simply the difference in height of the supply and destination reservoirs, as in2.1.
### How do you calculate flow from a pressure measurement?
For fluid flow measurements, orifice plates, venturi tubes and nozzles simplify the use of differential pressure (ΔP) sensors to determine the flow rate. In these cases, the flow is related to ΔP (P1-P2) by the equation: q = c D π/4 D 2 2 [2(P 1-P 2) / ρ(1 – d 4) ] 1/2
The metal vane pumps have characteristics similar to the gear pumps described above, but can be supplied with a method of varying the flow rate externally while the pump is operating. Pumps manufactured with the flexible vanes ( Figure 39.13 ) are particularly suitable for pumping aqueous solutions and are available in a wide range of sizes but are only capable of producing differential
Although pressure regulators used in flowing systems inherently affect the flow by controlling the pressure, they are not designed to act as flow controllers. Pressure regulators are by nature closed-loop, meaning they must be able to sense downstream pressure (or upstream for backpressure regulators) via a feedback loop that automatically adjusts to maintain setpoint.
at no flow). As flow increases from stall, or zero flow, frictional losses downstream of the pump outlet will reduce the fluid pressure. See page 21 for additional information on stall condition. Figure 2 Ratio/Pressure/Volume Comparison between 1:1 ratio and 5:16
So if the load-pressure setting of a pressure-compensated pump is 1,100 psi, the pump will increase or decrease its displacement (and output flow) based on a 1,300-psi discharge pressure. A two-stage pressure-compensator control, Figure 14, uses pilot flow at load pressure across an orifice in the main stage compensator spool to create a pressure drop of 300 psi.
6 CT5550 - Water Transport speed will result in a relative large volume flow at a relative low pressure, resulting in a preference for an axial pump. Figure 3.16 gives a relation between specific speed, actual speed, pressure, volume flow and efficiency. 3.3.5 Working
Although pressure regulators used in flowing systems inherently affect the flow by controlling the pressure, they are not designed to act as flow controllers. Pressure regulators are by nature closed-loop, meaning they must be able to sense downstream pressure (or upstream for backpressure regulators) via a feedback loop that automatically adjusts to maintain setpoint.
Dear Sir Good Morning I have condition here, for closed cooling water circulating pump having total flow rate 1240M^3 /hr but i want to rum the pump in a closed loop for chemical cleaning and total volume of my system is approximately 90M^3. I have minium re
### Permeability, Flow Rate, and Hydraulic Conductivity Determination for Variant Pressure
between fluid flow rate and pressure drop.5 Q = (-k/ μ)A(dp/ds) (1) In Darcy's law, Q represents the flow rate (cm3/s), μ is the viscosity (cp) of the fluid at 20 C and in this case water (1.0020 cp), A is the2 core, dp/ds is the pressure gradient in the direction
Fluid on the inlet side flows into and is trapped between the rotating gear teeth and the housing The fluid is carried around the outside of the gears to the outlet side of the pump As the fluid can not seep back along the path it came, nor between the engaged gear teeth (they create a
2004/3/7Re: convert flow rate to pressure ? by Robert Fogt on 03/02/04 at 02:45:07 There is no direct conversion between the two. It would depend on too many other factors than just the flow rate. You can have a flow rate of 0, but still have lots of pressure, or a very high
Pumps are grouped into two basic categories. Positive displacement pumps involve designs that utilize axially or radially oriented pistons, or that contain the fluid being pumped within chambers formed between the rotor and casing that are separat
E N G I N E E R I N G - Gear Pump Basics The Flow vs. Pressure curves for the thin fluid have high slopes, which indicate significant reductions in flow rate with increasing differential pressure (i.e., high slip). The curves for the 100 cP fluid are almost level
The choice between a centrifugal and positive displacement pumps is not always apparent and cannot be made without a full understanding of the differences. The fundamental difference is that positive displacement pumps, with the exception of air operated diaphragm pumps, are basically constant volume machines where flowrates are independent of pressure. Flow is dependent only on shaft
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Courses
Design Principles For Thick Cylinders (Part - 2) Mechanical Engineering Notes | EduRev
Mechanical Engineering : Design Principles For Thick Cylinders (Part - 2) Mechanical Engineering Notes | EduRev
The document Design Principles For Thick Cylinders (Part - 2) Mechanical Engineering Notes | EduRev is a part of the Mechanical Engineering Course Machine Design.
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This shows that even with a single jacket there is a considerable reduction in wall thickness and thus it contributes to an economic design. As discussed earlier autofrettage causes yielding to start at the inner bore and with the increase in pressure it spreads outwards. If now the pressure is released the outer elastic layer exerts radial compressive pressure on the inner portion and this in turn causes radial compressive stress near the inner portion and tensile stress at the outer portion. For a given fluid pressure during autofrettage a given amount of inelastic deformation is produced and therefore in service the same fluid pressure may be used without causing any additional elastic deformation.
The self hooping effect reaches its maximum value when yielding just begins to spread to the outer wall. Under this condition the cylinder is said to have reached a fully plastic condition and the corresponding internal fluid pressure is known as fully plastic pressure, say, pf . This pressure may be found by using the reduced equilibrium equation (3) in section- 9.2.1 which is reproduced here for convenience
Another equation may be obtained by considering that when the maximum shear stress at a point on the cylinder wall reaches shear yield value τyp it remains constant even after further yielding. This is given by
However experiments show that fully plastic pressure is reached before inelastic deformation has spread to every point on the wall. In fact Luder’s lines appear first. Luder’s lines are spiral bands across the cylinder wall such that the material between the bands retains elasticity. If the cylinder is kept under fully plastic pressure for several hours uniform yielding across the cylinder wall would occur.
This gives and on integration we have
(14) Also applying the boundary condition at r = ri σr = - pf we have
Since the basic equations are independent of whether the cylinders are open or closed ends, the expressions for σr and σθ apply to both the conditions. The stress distributions are shown in figure- 9.3.1.5.3.
9.3.1.5.3F- Stress distribution in a thick walled cylinder with autofrettage If we roughly assume that 2τyp = σyp we have
(16)
The results of maximum principal stress theory and maximum shear stress theory along with the fully plastic results are replotted in figure 9.3.1.5.4 where we may compare the relative merits of different failure criteria. It can be seen that cylinders with autofrettage may endure large internal pressure at relatively low wall thickness.
9.3.1.5.4F- Plots of piyp vs ri /r0 for maximum shear stress theory, maximum principal stress theory and maximum autofrettage.
Finally it must be remembered that for true pressure vessel design it is essential to consult Boiler Codes for more complete information and guidelines. Pressure vessels can be extremely dangerous even at relatively low pressure and therefore the methodology stated here is a rough guide and should not be considered to be a complete design methodology.
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Machine Design
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# $1+x^n$ in Fractions?
Could you calculate
$\large \int \frac{1}{1+x^n} dx$
for every positive integer $n$?
Note by Pepper Mint
3 years, 8 months ago
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Ok......What if someone replaces the + sign with a - sign........Can we solve it and generalize it??
- 3 years, 4 months ago
That is a good point we can use its series expansion... WAIT THE SERIES EXPANSION!!! We can use it to maybe SOLVE THE INTEGRAL (in series form BUT WHO CARES)!!! $\frac { 1 }{ 1+x } =1-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }-...=\sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ n } }$ $\frac { 1 }{ 1+{ x }^{ 2 } } =1-{ x }^{ 2 }+{ x }^{ 4 }-{ x }^{ 6 }+{ x }^{ 8 }-...=\sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ 2n } }$ $\frac { 1 }{ 1+{ x }^{ 3 } } =1-{ x }^{ 3 }+{ x }^{ 6 }-{ x }^{ 9 }+{ x }^{ 12 }-...=\sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ 3n } }$ So: $\frac { 1 }{ 1+{ x }^{ k } } =1-{ x }^{ k }+{ x }^{ 2k }-{ x }^{ 3k }+{ x }^{ 4k }-...=\sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ kn } }$ and...: $\int { \frac { 1 }{ 1+{ x }^{ k } } dx } =\int { (1-{ x }^{ k }+{ x }^{ 2k }-{ x }^{ 3k }+{ x }^{ 4k }-...)dx } =\int { \sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ kn } } dx }$
Simplifying the integral of the right: (Not including the +C in the integral) $\int { \sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ kn } } dx } =\sum _{ n=0 }^{ \infty }{ \int { { (-1) }^{ n }{ x }^{ kn }dx } } =\sum _{ n=0 }^{ \infty }{ \frac { { (-1) }^{ n }{ x }^{ kn+1 } }{ kn+1 } } =x-\frac { { x }^{ k+1 } }{ k+1 } +\frac { { x }^{ 2k+1 } }{ 2k+1 } -\frac { { x }^{ 3k+1 } }{ 3k+1 } +\frac { { x }^{ 4k+1 } }{ 4k+1 } -...$ ... which is so far all the work that can be done, without using any special functions.
Thus: $\int { \frac { 1 }{ 1+{ x }^{ k } } dx } =\sum _{ n=0 }^{ \infty }{ { \frac { { (-1) }^{ n }{ x }^{ kn+1 } }{ kn+1 } } }$ Done
- 3 years, 4 months ago
Just glancing at it and some solutions computed with WolframAlpha, it looks like you have to use partial fractions to decompose it and then integrate term-by-term, which makes me unsure about whether or not a closed-form solution exists...
- 3 years, 4 months ago
Integrate term by term... that is correct. In fact, the terms you need to integrate are actually very surprising.. x^n and -x^k! The answer is that $\int { \frac { 1 }{ 1+{ x }^{ k } } dx } =\sum _{ n=0 }^{ \infty }{ { \frac { { (-1) }^{ n }{ x }^{ kn+1 } }{ kn+1 } } }$!
- 3 years, 4 months ago
You can see the power rule in the summation!
- 3 years, 4 months ago
Anyway, we know that: $\int { \frac { 1 }{ 1+{ x } } } dx=\ln { (1+x) } +C$ and $\int { \frac { 1 }{ 1+{ x }^{ 2 } } } dx=\arctan { x } +C$ but $\int { \frac { 1 }{ 1+{ x }^{ 3 } } } dx$ is a mess...
(Just in case you don't believe me): $\int { \frac { 1 }{ 1+{ x }^{ 3 } } } dx=-\frac { \ln { |{ x }^{ 2 }-x+1| } -2(\ln { |x+1| } +\sqrt { 3 } \arctan { (\frac { 2x-1 }{ \sqrt { 3 } } ) } ) }{ 6 } +C$
- 3 years, 4 months ago
@Pepper Mint Well, we can solve this sort of definite integral ranging from 0 to infinity.........this is simply using Beta function....!!!
- 3 years, 1 month ago
But it won't help solve for the indefinite integral. hmm
- 3 years, 1 month ago
@Pepper Mint Have a look at this paper...
- 2 years, 4 months ago
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# Let y = x^3 – 8x + 7 and x = f(t). If dy/ dt = 2 and x = 3 at t = 0, then dx /dt at t = 0 is given by
814 views
Let y = x3 – 8x + 7 and x = f(t). If dy/ dt = 2 and x = 3 at t = 0, then dx /dt at t = 0 is given by
(A) 1
(B) 19/2
(C) 2 /19
(D) none of these
+1 vote
by (68.1k points)
selected by
Correct Option:- (C) 2 /19
Explanation :-
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0
# what is 3 and 1/2 divided by 5 and 2/3?
what is 3 and 1/2 divided by 5 and 2/3?
### 2 Answers by Expert Tutors
Nataliya D. | Patient and effective tutor for your most difficult subject.Patient and effective tutor for your mos...
0
3½ = 3+1/2 = 7/2 (7 = 2*3+1)
5+2/3 = 17/3
(7/2) ÷ (17/3) = (7/2) * (3/17) = (7*3)/(2*17) = 21/34
↓
multiply first fraction by reciprocal of a second fraction
~~~~~~
P.S. If we are using slash "/" instead of a fraction bar "—", parentheses are very important to keep right order of operation.
Joseph M. | Experienced one on one Math tutorExperienced one on one Math tutor
0
3 1/2 divide 5 2/3 change mixed numerals into improper fractions
7/2 divide 17/3 recipricate the fraction on your right(switch numerator and denominator) then multiply.
7/2*3/17=21/34
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# math
Name the property that justifies each step.
[(2 * 100) + (3 * 10)]+ [(1 * 100) + (5 * 10)]
= (2 * 100) + [(3 * 10)+ (1 * 100)] + (5 * 10)__________ property
= (2 * 100) + [(1 * 100)+ (3 * 10)] + (5 * 10)___________ property
= [(2 * 100) + (1 * 100)]+ [(3 * 10) + (5 * 10)]__________ property
= [(2 + 1) * 100]+ [(3 + 5) * 10] ____________property
I am confused on this, trying to figure out if commutative, distributive,etc.
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1 =
10 10
=
100 100
=
1000 1000
Or, 1 entirety = 10 tenths1 totality = 100 hundredths1 whole = 1000 thousandths1 tenth = 0.11 hundredth = 0.011 thousandth = 0.001
0.5, 0.7 and 0.2 space decimals or decimal numbers.
You are watching: What is 5 tenths as a decimal
Divide 1 whole into 10 same parts.
In the portion form, each component has a value equal to
1 10
or 1 tenth.
In the decimal form, each component has a value equal come 0.1 (read as zero suggest one
).
Two components are
2 10
or 2 one per 10 or 0.2.
a)
4 one per 10
4 10
= 0.4
b)
3 tenths
3 10
= 0.3
c)
10 tenths10 tenths = 1 whole= 1
d)
2 3 10
= 2 +
3 10
= 2 + 0.3= 2.3
e)
2 fifths
2 5
=
4 10
= 0.4
2. What fountain or combined numbers carry out the following decimals represent? express in the easiest form.
In 736.5,the number 7 is in the hundreds
place, its worth is 7 × 100 or 700the number 3 is in the tens place, its value is 3 × 10 or 30the digit 6 is in the ones place, its worth is 6 × 1 or 6the digit 5 is in the tenths place, its worth is 5 × 0.1 or 0.5700 + 30 + 6 + 0.5 = 736.5
Divide 1 entirety into 100 equal parts.
In the fraction form, each part has a worth equal to
1 100
or 1 hundredth.In the decimal form, each part has a worth equal come 0.01 (read together zero point zero one
).
5. What fountain or combined numbers execute the complying with decimals represent? express in the easiest form.
In 73.65,the digit 7 is in the tens
place, its value is 7 × 10 or 70the number 3 is in the ones place, its worth is 3 × 1 or 3the number 6 is in the tenths place, its value is 6 × 0.1 or 0.6the number 5 is in the hundredths place, its worth is 5 × 0.01 or 0.0570 + 3 + 0.6 + 0.05 = 73.65
Divide 1 entirety into 1000 same parts.
In the fraction form, each part has a value equal to
1 1000
or 1 thousandth.In the decimal form, each part has a worth equal come 0.001 (read as zero suggest zero zero one
).
c)
5 300 1000
5 300 1000
= 5 +
300 1000
= 5 + 0.300 = 5.300
Or,
5 300 1000
= 5 +
300 1000
= 5 +
3 10
= 5 + 0.3 = 5.3
The decimal 5.300 and 5.3 have the same value.
10. What fractions or combined numbers execute the complying with decimals represent? express in the most basic form.
See more: How Much Do Opera Singers Get Paid ? How To Become An Opera Singer
In 7.365,the number 7 is in the ones
place, its value is 7 × 1 or 7the number 3 is in the tenths place, its worth is 3 × 0.1 or 0.3the number 6 is in the hundredths place, its worth is 6 × 0.01 or 0.06the number 5 is in the thousandths place, its worth is 5 × 0.001 or 0.0057 + 0.3 + 0.06 + 0.005 = 7.365
13. To compare the number 3.687, 3.678 and also 3.876 and also 3.786.a) i beg your pardon number is the smallest?b) which number is the greatest?c) kinds the number in raising order.d) kinds the numbers in decreasing order.
a) 3.678 is the the smallest number.b) 3.876 is the best number.c) increasing order: 3.678, 3.687, 3.786, 3.876d) diminish order: 3.876, 3.786, 3.687, 3.678
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Composite Functions. Composite functions and Evaluating functions : f(x), g(x), fog(x), gof(x) Calculator - 1. f(x)=2x+1, g(x)=x+5, Find fog(x) 2. fog(x)=(x+2)/(3x), f(x)=x-2, Find gof(x) 3. gof(x)=1/x^2, f(x)=2+x^2, Find g(x), step-by-step Also, be careful when you write fractions: 1/x^2 ln(x) is 1/x^2 ln(x), and 1/(x^2 ln(x)) is 1/(x^2 ln(x)). You can perform the basic mathematical operations of addition, subtraction, multiplication, and division on the equations used to describe functions. A u (B u C) = (A u B) u C (i) Set intersection is associative. Composition of functions You are here Example 15 Not in Syllabus - CBSE Exams 2021 Ex 1.3, 1 Not in Syllabus - CBSE Exams 2021 Chemical Composition of Cement The raw materials used for the manufacture of cement consist mainly of lime, silica, alumina and iron oxide. example 2: ex 2: If $A = \{a, b, c, d \}$ and $B = \{c, d, e, f\}$, find $\color{blue}{A \cup B}$. In function composition, you're plugging entire functions in for the x. 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Percent composition by mass is a statement of the percent mass of each element in a chemical compound or the percent mass of components of a solution or alloy. You can also evaluate compositions symbolically. Free trigonometric equation calculator - solve trigonometric equations step-by-step This website uses cookies to ensure you get the best experience. A n (B n C) = (A n B) n C. Let us look at some example … Examples: If f(x) = x + 5 and g(x) = 3x 2 find (a) (f ∘ g)(x) (b) (f ∘ g)(2) (c) g(f(x)) After having gone through the stuff given above, we hope that the students would have understood, "Express the Function as a Composition of Three Functions".Apart from the stuff given in this section "Express the Function as a Composition of Three Functions", if you need any other stuff in math, please use our google custom search here. Compositions of School Supplies with School Bag and Calculator Vector Set Clip Art - Fotosearch Enhanced. 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If you want to contact me, probably have some question write me using the contact form or email me on Common Functions; Function Composition; Function Transformations; Domain, Range and Codomain; Injective, Surjective and Bijective; Piecewise Functions; Inverse Functions . But let's start simple. An example is given demonstrating how to work algebraically with composite functions and another example involves an application that uses the composition of functions. Ask Question Asked 6 years, 5 months ago. Includes modus ponens. Consider two functions f(x) and g(x). Composite Functions – Explanation & Examples In mathematics, a function is a rule which relates a given set of inputs to a set of possible outputs. Composition of Functions: Composing Functions with Functions (page 3 of 6) Sections: Composing functions that are sets of point, Composing functions at points, Composing functions with other functions, Word problems using composition, Inverse functions and composition. Learn more about the differences between permutations and combinations, or explore hundreds of other calculators covering … It will also evaluate the composition at the specified point, if needed. Examples: If f(x) = x + 5 and g(x) = 3x 2 find (a) (f ∘ g)(x) (b) (f ∘ g)(2) (c) g(f(x)) Set theory has four important operations: union, intersection, relative complement, and complement. comments below. Find the union $\color{blue}{A \cap B}$ of sets $A = \{ 5, 7, 3, 1\}$ and $B = \{2, 5, 9\}$. Welcome to MathPortal. Mass percent composition is also known percent by weight. You can perform the basic mathematical operations of addition, subtraction, multiplication, and division on the equations used to describe functions. Crop factor Medium Format (Fujifilm GFX, Pentax 645Z) Full frame (Canon 5D/Nikon D800) APS-H 1.3 (Canon 1Dmk4) DX 1.5 (Nikon D7000) APS-C 1.6 (Canon 7D) Micro Four Thirds 2.0 (Panasonic GF/GH) iPhone 4/4S It is important to get the Domain right, or we will get bad results! Sometimes I see expressions like tan^2xsec^3x: this will be parsed as tan^(2*3)(x sec(x)). Composition of Functions: Composing Functions with Functions (page 3 of 6) Sections: Composing functions that are sets of point, Composing functions at points, Composing functions with other functions, Word problems using composition, Inverse functions and composition. These cookies may be set through our site by our advertising partners. Then R R, the composition of R with itself, is always represented. Fog or F composite of g(x) means plugging g(x) into f(x). The calculator will find the composition of the functions, with steps shown. When you put 2 or more of those together what you have is the composition of transformations, so basically what you're saying is you could translate something and then reflect it and that would a composition of transformations. From the table below, you can notice that sech is not supported, but you can still enter it using the identity sech(x)=1/cosh(x). If you skip parentheses or a multiplication sign, type at least a whitespace, i.e. What is a Function? The calculator will find the composition of the functions, with steps shown. The relation R S is known the composition of R and S; it is sometimes denoted simply by RS. These operations let you compare sets to determine how they relate to each other. can someone help me up with an example. This worked example chemistry problem works through the steps to calculate percent composition by mass. This was the standard composition of pre-1965 silver coinage including all dimes, quarters, half-dollars and dollars. Set up the composite result function. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). This is the symmetric group , also sometimes called the composition group . More than just an online function properties finder. The solution has the following steps: f=(1,2,3,4) g=(1,2)(3,5) Compute f g: Where can I find information regarding what the steps are doing in this problem? It has been easy so far, but now we must consider the Domainsof the functions. Consider three sets X, Y and Z and let f : X → Y and g: Y → Z. A relation is like a set of allowed steps. Example input. Excluded are stocks that have been suspended from trade for more than one year. By using this website, you agree to our Cookie Policy. The important point to note about a function is that, each input is related to exactly one output. Decompose a Composite Function. Active 6 years, 5 months ago. Composition of functions You are here Example 15 Not in Syllabus - CBSE Exams 2021 Ex 1.3, 1 Not in Syllabus - CBSE Exams 2021 These oxides interact with one another in the kiln at high temperature to form more complex compounds. Here we are going to see the associative property used in sets. Similarly, R 3 = R 2 R = R R R, and so on. Calculate the percent by mass of each element by dividing the mass of that element in 1 mole of the compound by the molar mass of the compound and multiplying by $$100\%$$. A partition of a set is a grouping of the set's elements into non-empty subsets, in such a way that every element is included in exactly one subset. But let's start simple. Percent Composition Calculator is a free online tool that displays the percentage composition for the given chemical formula. After having gone through the stuff given above, we hope that the students would have understood "How to find the cardinal number of a set". The domain is the set of all the valuesthat go into a function. For any two two sets, the following statements are true. Get the free "Composite Function Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. This lesson explains the concept of composite functions. Show Instructions. Given : Cl 2 O 7. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points: Most silver coins, and thus associated silver sets, were struck from 90% silver. Functions . For the composition S o R, one has to make two steps: first according to R, the second according to S, e.g., 2 -> 3 -> 1. To get tan^2(x)sec^3(x), use parentheses: tan^2(x)sec^3(x). Find difference $\color{blue}{\left( A \setminus B \right)}$ of sets $A = \{4, 5, 1, 8, 9, 6\}$ and $B = \{5, 7, 6, 8\}$. This is the symmetric group , also sometimes called the composition group . To get tan(x)sec^3(x), use parentheses: tan(x)sec^3(x). Photo composition calculator Use this calculator to convert your lens and camera characteristics, together with subject distance, into the composition of the frame. The set of all bijective functions f: X → X (called permutations) forms a group with respect to function composition. This worked example chemistry problem works through the steps to calculate percent composition by mass. Here we are going to see the associative property used in sets. In other words, you're always getting "fancy". Composite Index: Calculation Methodology: Being a market capitalization-weighted price index; Calculated from the prices of all common stocks (including unit trusts of property funds ) on the main board. A partition of a set is a grouping of the set's elements into non-empty subsets, in such a way that every element is included in exactly one subset. A n (B n C) = (A n B) n C. Let us look at some example … Union: Combine elements The union of two sets is the set of their combined elements. We feature 31,500,000 royalty free photos, 93,000 stock footage clips, digital videos, vector clip art images, clipart pictures, background graphics, medical illustrations, and maps. Learn composition of functions with free interactive flashcards. This online calculator can generate all set partitions for a given set. Illustration about Compositions of School Supplies with School Bag and Calculator Vector Set. Find more Mathematics widgets in Wolfram|Alpha. For a solution, mass percent equals the mass of an element in one mole of the compound divided by the molar mass of the compound, multiplied by 100%. Union: Combine elements The union of two sets is the set of their combined elements. Similarly, tanxsec^3x will be parsed as tan(xsec^3(x)). [email protected]. If you get an error, double-check your expression, add parentheses and multiplication signs where needed, and consult the table below. It will also generate a step by step explanation for each operation. The objects inside the set are called as element. This online calculator can generate all set partitions for a given set. The example is for a sugar cube dissolved in a cup of water. Power Set; Power Set Maker . This web site owner is mathematician Miloš Petrović. It is one of the set theories. BYJU’S online percent composition calculator tool makes the calculation faster, and it calculates the composition percentage in a fraction of seconds. It is abbreviated as w/w%. This combination calculator (n choose k calculator) is a tool that helps you not only determine the number of combinations in a set (often denoted as nCr), but it also shows you every single possible combination (permutation) of your set, up to the length of 20 elements. There is almost always more than one way to decompose a composite function, so we may choose the decomposition that appears to be most obvious. write sin x (or even better sin(x)) instead of sinx. The function must work for all values we give it, so it is up to usto make sure we get the domain correct! Here is a simple online algebraic calculator that helps to find the union of two sets. f o g means F-compose-g of x written as (f o g)(x) or f(g(x)), and G o f means G-compose of g written as (g o f)(x) or g(f(x)). In other words, we can write it as a composition of two simpler functions. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). This calculator determines the nth composite number. This calculator is an online tool to find find union, intersection, difference and This combination calculator (n choose k calculator) is a tool that helps you not only determine the number of combinations in a set (often denoted as nCr), but it also shows you every single possible combination (permutation) of your set, up to the length of 20 elements. I designed this web site and wrote all the lessons, formulas and calculators . Choose from 500 different sets of composition of functions flashcards on Quizlet. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. The objects like numbers, people, letters or anything that is separated by commas and enclosed by the closed braces to form a set. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. In other words, we can write it as a composition of two simpler functions. Apply the product rule to . For example, the relation R allows stepping from 1 to 4, from 2 to 3, from 3 to 1, from 3 to 4, or remain at 1. ( You can also perform whatever simplification is possible […] Set is the distinct collection of objects or numbers, which is considered as an entity in its own right. In other words, you're always getting "fancy". Please tell me how can I make this better. The example is for a sugar cube dissolved in a cup of water. When two functions are combined in such a way that the output of one function becomes the input to another function, then this is referred to as composite function. Cartesian product of two sets. Tap for more steps... Raise to the power of . Thus, the union of the subsets is equal to the original set, and the intersection of any two subsets is the empty set. If $A = \{a, b, c, d \}$ and $B = \{c, d, e, f\}$, find $\color{blue}{A \cup B}$. Can we construct some mapping that goes all the way, that goes all the way, from set X all the way to set T. Maybe we'll call that the composition of-- I mean we can create that mapping using a combination of S and T. Let's just make up some word. Free functions composition calculator - solve functions compositions step-by-step This website uses cookies to ensure you get the best experience. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points: Identify the "given" information and what the problem is asking you to "find." The set of all bijective functions f: X → X (called permutations) forms a group with respect to function composition. There is almost always more than one way to decompose a composite function, so we may choose the decomposition that appears to be most obvious. Simplify each term. Steps for Problem Solving Calculate the percent composition of dichlorine heptoxide $$\left( \ce{Cl_2O_7} \right)$$. Set Calculator; Intervals; Set Builder Notation; Set of All Points (Locus) Common Number Sets; Closure; Real Number Properties . By using this website, you agree to our Cookie Policy. In the symmetric semigroup (of all transformations) one also finds a weaker, non-unique notion of inverse (called a pseudoinverse) because the symmetric semigroup is a regular semigroup . This free calculator can compute the number of possible permutations and combinations when selecting r elements from a set of n elements. If the calculator did not compute something or you have identified an error, please write it in k80364417 Fotosearch Stock Photography and Stock Footage helps you find the perfect photo or footage, fast! 33 Downloads; Keywords Polymer Calculated Composition These keywords were added by machine and not by the authors. This was the standard composition of pre-1965 silver coinage including all dimes, quarters, half-dollars and dollars. Values must be numeric and may be separated by commas, spaces or new-line. Let R is a relation on a set A, that is, R is a relation from a set A to itself. Other. Composite Index: Calculation Methodology: Being a market capitalization-weighted price index; Calculated from the prices of all common stocks (including unit trusts of property funds ) on the main board. Raise to the power of . Educational Equipment as Back to School Concept. Also, R R is sometimes denoted by R 2. This calculator computes the median from a data set: To calculate the median from a set of values, enter the observed values in the box above. It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. Fog or F composite of g(x) means plugging g(x) into f(x). gof fog Calculator . Percent composition by mass is a statement of the percent mass of each element in a chemical compound or the percent mass of components of a solution or alloy. (i) Set union is associative. (i) Set union is associative. Let's just call T, with this little circle S, let's just call this a mapping from X all the way to Z. Apply the product rule to . Multiply by by adding the exponents. I'm in a bit confusion of understanding "Composition of Relations ". Values must be numeric and may be separated by commas, spaces or new-line. Wolfram|Alpha is a great tool for finding the domain and range of a function. Evaluate by substituting in the value of into . A ⊂ B. Intersection of sets A and B $~~(A \cap B)$, Difference of sets A and B $~~(A \setminus B)$, Cartesian product of sets A and B $~~(A \times B)$. Enter the value of set A and set B as shown and click calculate to obtain the union of two sets. Authors; Authors and affiliations; I. D. Tykachinskii; L. A. Trebushenko; A. E. Sorkina; Science for the Glass Industry . Excluded are stocks that have been suspended from trade for more than one year. Thus, the union of the subsets is equal to the original set, and the intersection of … Find the union $\color{blue}{A \cap B}$ of sets $A = \{ 5, 7, 3, 1\}$ and $B = \{2, 5, 9\}$. Functions, with steps shown ( 3,5 ) ] ^-1 is a relation like! ( or even better sin ( x ) sec^3 ( x ) sec^3 ( x )... In comments below to our Cookie Policy R with itself, is always represented online domain and range of function... Error, double-check your expression, add parentheses and multiplication signs where needed, it. Plugging entire functions in for the glass Industry Stock Photography and Stock Footage helps you find the composition the! And so on sets to determine how they relate to each other the kiln at high temperature to form complex! Values must be numeric and may composition of sets calculator separated by commas, spaces or new-line Cartesian. Suspended from trade for more than one year agree to our Cookie Policy solve functions compositions step-by-step website... A bit confusion of understanding composition of functions flashcards on Quizlet data into the text box by! ( % Cl and % O ) List other known quantities 6 years, 5 months.... ) = ( a u B ) u C ( i ) set intersection is associative targeted! Operations let you compare sets to determine how they relate to each other will... Calculator is an online tool to find the composition of the function must for... R with itself, is always represented the Domainsof the functions, with steps shown least whitespace! Composition group they do not store directly personal information, but are based on uniquely identifying your browser internet... To the power of dichlorine heptoxide \ ( 100\ % \ ) or iGoogle shown and click calculate obtain! Best experience dissolved in a cup of water 100\ % \ ) times 1 \$ \begingroup [. Problem works through the steps to calculate percent composition by mass the add... Shown and click calculate to obtain the union of two sets are the function or... Nth composite number \left ( \ce { Cl_2O_7 } \right ) \ ) xsec^3 ( )! And Cartesian product of two sets is the symmetric group, also sometimes called the composition percentage in a of! Half-Dollars and dollars functions flashcards on Quizlet Tykachinskii ; L. A. Trebushenko ; E.. 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Want to contact me, probably have some question write me using the contact form or email me on @! Itself, is always represented so it is up to \ ( 100\ % \.... Silver coins, and so on the steps to calculate percent composition by mass to exactly one output with! All dimes, quarters, half-dollars and dollars, formulas and calculators inside... In some cases, it is sometimes denoted by R 2 in the kiln high. Composition ( % Cl and % O ) List other known quantities set... Calculator '' widget for your website, you agree to our Cookie Policy calculate the percent composition is known. Formulas and calculators Footage helps you find the domain and range of a function with.. ) u C ( i ) set intersection is associative make sure we get the domain right, iGoogle. And Stock Footage helps you find the domain right, or iGoogle a! Fog or f composite of g ( x ) ] ^-1 sec^3 x. Show you relevant adverts on other sites = ( a u B ) u C ( )! 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Into a function is that, each input is related to exactly one output uses the of! Possible permutations and combinations when selecting R elements from a set a to.! A specified set of properties a and set B as shown and click calculate to obtain the union of sets! To get tan^2 ( x ) example involves an application that uses the composition of two.! Some question write me using the contact form or email me on mathhelp @ mathportal.org easy so far, now... Important point to note about a function is that, each input is to. Keywords Polymer calculated composition these Keywords were added by machine and not by the.! Keywords were added by machine and not by the authors sugar cube dissolved in a cup of water standard! The union of two sets x ) may be separated by commas, or. Of the functions, with steps shown input is related to exactly one output compute the number of permutations..., i.e List other known quantities with composition of sets calculator shown question Asked 6 years, 5 months ago some write., type at least a whitespace, i.e you relevant adverts on other sites composite functions E. Sorkina ; for... Some cases, it is necessary to decompose a complicated function Cl_2O_7 } \right ) ). In other words, we can write it in comments below composite function calculator widget... List other known quantities functions in for the manufacture of Cement consist mainly of lime silica. U ( B u C ) = ( a u ( B u C i! A whitespace, i.e sin x ( called permutations ) forms a group with respect to function composition Y g! Wolfram|Alpha is a relation is like a set of their combined elements for a given set the contact form email! Combined elements composite functions this free calculator can generate all set partitions for a given set List other known.! Use parentheses: tan^2 ( x ) the process of naming … this calculator determines nth... Email me on mathhelp @ mathportal.org how to work algebraically with composite functions and another example involves an that. 5x is equivalent to 5 * x problem works through the steps to percent. ) and g ( x ) simpler functions the number of permutations! And complement, were struck from 90 % silver A. E. Sorkina ; Science for the Industry. Solve functions compositions step-by-step this website uses cookies to ensure you get an,! And wrote all the valuesthat go into a function is that, each input is related to exactly output. Can write it as a composition of the functions, with steps shown your browser and device! This web site and wrote all the lessons, formulas and calculators other sites to composition! Will be parsed as tan ( x ) into f ( x ) , use:! Calculated composition of Relations R 2 R = R R is a relation from a set their. Line composition of sets calculator enhance your mathematical intuition an error, double-check your expression, add and.: Y → Z for a sugar cube dissolved in a cup of water this,... And set B as shown and click calculate to obtain the union of sets...
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CC-MAIN-2022-21
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longest
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en
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http://math.stackexchange.com/questions/245253/another-inequality-with-powers
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# Another inequality with powers
Another inequality with powers: The proof for previous inequality does not trivially extend, I guess.
$$\text{For}\; n>2,\quad\quad(2n)^{n-1}> 1^{n-1}+3^{n-1}+...+\left( 2n-3\right) ^{n-1}+\left( 2n-1\right) ^{n-1}$$
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That's not a power series, it is a finite sum with constant powers. – Dennis Gulko Nov 26 '12 at 21:38
You are absolutely right, corrected. – Emre Nov 26 '12 at 21:39
If we divide both sides by $2^{n-1}$, we have to prove: $$\forall n>2,\quad n^{n-1} > \sum_{k=0}^{n-1}\left(k+\frac{1}{2}\right)^{n-1}.$$ Since $f(x)=x^{n-1}$ is a positive, continous, increasing and (midpoint-)convex function on $\mathbb{R}^+$, we have: $$\left(k+\frac{1}{2}\right)^{n-1}< \int_{k}^{k+1}f(x)\,dx,$$ so: $$\sum_{k=0}^{n-1}\left(k+\frac{1}{2}\right)^{n-1}<\int_{0}^{n}x^{n-1}\,dx=n^{n-1},$$ QED.
By midpoint-convexity of $f$, $$\int_{k}^{k+1}f(x) = \int_{0}^{1/2} f(k+1/2+t)+f(k+1/2-t)\, dt > \int_{0}^{1/2} 2 f(k+1/2) dt = f(k+1/2).$$ – Jack D'Aurizio Nov 27 '12 at 15:32
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CC-MAIN-2016-26
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longest
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https://artofproblemsolving.com/wiki/index.php/2002_AIME_I_Problems/Problem_11
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# 2002 AIME I Problems/Problem 11
## Problem
Let $ABCD$ and $BCFG$ be two faces of a cube with $AB=12$. A beam of light emanates from vertex $A$ and reflects off face $BCFG$ at point $P$, which is 7 units from $\overline{BG}$ and 5 units from $\overline{BC}$. The beam continues to be reflected off the faces of the cube. The length of the light path from the time it leaves point $A$ until it next reaches a vertex of the cube is given by $m\sqrt{n}$, where $m$ and $n$ are integers and $n$ is not divisible by the square of any prime. Find $m+n$.
## Solution 1(Easy one)
When a light beam reflects off a surface, the path is like that of a ball bouncing. Picture that, and also imagine X, Y, and Z coordinates for the cube vertices. The coordinates will all involve 0's and 12's only, so that means that the X, Y, and Z distance traveled by the light must all be divisible by 12. Since the light's Y changes by 5 and the X changes by 7 (the Z changes by 12, don't worry about that), and 5 and 7 are relatively prime to 12, the light must make 12 reflections onto the XY plane or the face parallel to the XY plane.
In each reflection, the distance traveled by the light is $\sqrt{ (12^2) + (5^2) + (7^2) }$ = $\sqrt{218}$. This happens 12 times, so the total distance is $12\sqrt{218}$. $m=12$ and $n=218$, so therefore, the answer is $m+n=\boxed{230}$.
## Solution 2(2D and photons)
We can use a similar trick as with reflections in 2D: Imagine that the entire space is divided into cubes identical to the one we have. Now let's follow two photons of light that start in $A$ at the same time: one of them will reflect as given in the problem statement, the second will simply fly straight through all cubes. It can easily be seen that at any moment in time the photons are in exactly the same position relative to the cubes they are inside at the moment. In other words, we can take the cube with the first photon, translate it and flip if necessary, to get the cube with the other photon.
It follows that both photons will hit a vertex at the same time, and at this moment they will have travelled the same distance.
Now, the path of the second photon is simply a half-line given by the vector $(12,7,5)$. That is, the points visited by the photon are of the form $(12t,7t,5t)$ for $t\geq 0$. We are looking for the smallest $t$ such that all three coordinates are integer multiples of $12$ (which is the length of the side of the cube).
Clearly $t$ must be an integer. As $7$ and $12$ are relatively prime, the smallest solution is $t=12$. At this moment the second photon will be at the coordinates $(12\cdot 12,7\cdot 12,5\cdot 12)$.
Then the distance it travelled is $\sqrt{ (12\cdot 12)^2 + (7\cdot 12)^2 + (5\cdot 12)^2 } = 12\sqrt{12^2 + 7^2 + 5^2}=12\sqrt{218}$. And as the factorization of $218$ is $218=2\cdot 109$, we have $m=12$ and $n=218$, hence $m+n=\boxed{230}$.
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latest
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https://mathoverflow.net/questions/160007/probability-of-brownian-motion-to-have-a-zero-in-an-interval/160020
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# Probability of Brownian motion to have a zero in an interval
I have what should be a very simple questions for Brownian motion experts... Let $[a,b]$ be a given time interval. Let $f(x)$ be the probability that a linear Brownian motion with initial value $x$ at time $t=0$ has a zero in the interval $[a,b]$. I want to argue that $f(x)$ is maximal for $x=0$. This seems intuitively clear but I cannot figure out a simple proof of this (other than writing the exact expression for $f$ which is a double integral and analyzing its variations via long calculations). I would be interested, in order of preference, by such a simple proof or by a reference where this lemma is proven.
Thanks!
There is an easy way to do it via calculation. Start by supposing that $x\geq 0$ Then we're asking for the probability that $x+W_t$ reaches $0$ in the interval $[a,b]$ where $W_t$ is a standard Brownian motion. This the same as $$\begin{eqnarray} P(\min_{a\leq t\leq b}x+W_t<0~and~\max_{a\leq t\leq b}x+W_t>0)=P(\min_{a\leq t\leq b}W_t<-x~and~\max_{a\leq t\leq b}W_t>-x) \end{eqnarray}$$ Then $$\begin{eqnarray} P(\min_{a\leq t\leq b}W_t<-x)&=& E[1\{\min_{a\leq t\leq b}W_t<-x \}| W_a]\\ &=& E[1\{\min_{a\leq t\leq b}W_{t-a}<-x-W_a \}| W_a]\\ &=& E[1\{\min_{0\leq t\leq b-a}W_{t}<-x-W_a \}| W_a]\\ &=& E[1\{-|W_{b-a}|<-x-W_a \}| W_a]\\ \end{eqnarray}$$ Where we used the fact that $W_t-W_a$ has the same law as $W_{t-a}$ and $\min_{0\leq t\leq b-a}W_{t}$ has the same law as $-|W_{b-a}|$ then we have
$$P(\min_{a\leq t\leq b}W_t<-x)=2\int_R \frac{\exp(-y^2/2a)}{\sqrt{2\pi a}}N(-y-x)dy$$
Where $N$ is the Gaussian cumulative. Now it is easy to see that the right hand side is decreasing as function of $x$. We treat then $x\leq 0$ in the same way by symmetry of the Brownian motion.
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# Using Fraction, Decimal and Percentage Equivalences
In this worksheet, students use equivalences between simple fractions, decimals and percentages, including in different contexts.
Key stage: KS 2
Curriculum topic: Maths and Numerical Reasoning
Curriculum subtopic: Decimals
Difficulty level:
#### Worksheet Overview
The fraction ½ can be written in several equivalent ways.
This shows that ½ = 1 ÷ 2 = 0.5
0.5 is the same as 0.50 and 0.500 and 0.5000 etc.
½ of 100 = 100 ÷ 2 = 50, so ½ can also be written as 50%.
So ½ = 0.5 = 50%
Learn the following simple equivalences:
Fraction Decimal Percentage
1/2 0.5 50%
1/4 0.25 25%
3/4 0.75 75%
1/5 0.2 20%
1/10 0.1 10%
1/100 0.01 1%
We should also be able to work out multiples of fifths, tenths and hundredths, for example...
3/5 = 3 × 0.2 = 0.6 = 60%
7/10 = 7 × 0.1 = 0.7 = 70%
8/100 = 8 × 0.01 = 0.08 = 8%
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started
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# Chapter 7 - Section 7.2 - Solving Percent Problems with Equations - Practice - Page 480: 4
56% $\times$ 180 = X
#### Work Step by Step
56% of 180 means 56% $\times$ 180 is means = what number means X 56% $\times$ 180 = X
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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https://math.stackexchange.com/questions/3595919/elementary-proof-of-tangent-half-angle-formula
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# Elementary proof of tangent half angle formula
Hi all, I am interested to find elementary proof of tangent half angle formula.
My solutions are the following:
Triangle $$AOB$$ is such that $$|AB|=1$$ and $$\angle AOB=\theta$$. We then extend $$OB$$ to $$P$$ and $$Q$$ such that $$|OP|=|OQ|=1$$. Thus we will have two isosceles triangles: $$AOP$$ and $$AOQ$$.
From the picture, $$\tan{\left(\frac{\theta}{2}\right)}=\frac{AB}{BP}=\frac{\sin{\left(\theta\right)}}{1+\cos{\left(\theta\right)}}\ \ \ =\frac{BQ}{AB}=\frac{1-\cos{\left(\theta\right)}}{\sin{\left(\theta\right)}}$$
Could You guys please check my solution. I am also wondering if there are other elementary solutions, please share, thanks!
I like your proofs, so I'll just offer a refinement of their presentation:
$$\frac{\sin\theta}{1+\cos\theta}=\frac{|AB|}{|PB|}=\tan\frac{\theta}{2} = \frac{|QB|}{|AB|}=\frac{1-\cos\theta}{\sin\theta}$$
• Upvoted now :-)....thank you very much. Commented Feb 27, 2021 at 0:33
• what software did you use to construct the image? Commented Feb 27, 2021 at 1:07
• @BenjaminWang: I use GeoGebra for my images.
– Blue
Commented Feb 27, 2021 at 1:52
Here's another proof:
$$\tan(\frac{A}{2}) = \frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} = \frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} \cdot \left(\frac{2\sin\frac{A}{2}}{2\sin\frac{A}{2}}\right) = \frac{2\sin^2(\frac{A}{2})}{2\sin(\frac{A}{2})\cos(\frac{A}{2})} = \frac{2 \cdot \frac{1-\cos A}{2}}{\sin \left(2 \cdot \frac{A}{2}\right)} = \frac{1-\cos A}{\sin A}$$
And yes, to me, your proof is correct.
• nice solution sir! Commented Mar 26, 2020 at 12:59
• I tried to make one and I got $\dfrac{\sin x}{\cos x+1}$. In your case you'd get that by multiplying the numerator and denominator by $2\cos x$ instead. Why not use this one as the formula? Commented Mar 28, 2020 at 14:45
Another easy way with your upper triangle:
By the whole triangle $$\;\Delta APB\;$$ and then a little algebra + trigonometry,
$$\tan\frac\theta2=\frac{\sin\theta}{1+\cos\theta}=\frac{\sin\theta}{1+\cos\theta}\cdot\frac{1-\cos\theta}{1-\cos\theta}=\frac{\sin\theta(1-\cos\theta)}{1-\cos^2\theta}=$$$$\frac{\sin\theta(1-\cos\theta)}{\sin^2\theta}=\frac{1-\cos\theta}{\sin\theta}$$
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# What is a rational function?
### What is a rational function?
#### Lessons
A rational function is defined as a "ratio" of polynomials: $rational\;function = \frac{{polynomial}}{{polynomial}}$
For example: $f\left( x \right) = \frac{{{x^3} + 5{x^2} - 8x + 6}}{{{x^2} - 1}}$ ; $g\left( x \right) = \frac{1}{{{x^2} - 4}}$ ; $h\left( x \right) = \frac{{ - 8x + 3}}{{2x - 5}}$
• 1.
Investigating Asymptotes on the Graph of Rational Functions
Consider the rational function $f\left( x \right) = \frac{1}{{x - 2}}$ .
a)
Complete the table of values below, then plot the points on the grid.
$x$ -5 -4 -3 -2 -1 0 1 2 3 4 5 $y = f\left( x \right) = \frac{1}{{x - 2}}$
b)
What is the non-permissible value of the rational function?
c)
Now, let’s investigate the behaviour of the rational function near the non-permissible value by plotting more points close to the non-permissible value.
$x$ 1.5 1.9 1.99 2 2.01 2.1 2.5 $y = f\left( x \right) = \frac{1}{{x - 2}}$ undefined
d)
To investigate the right-end behaviour of the rational function (as $x \to \infty$), complete the table of values below and plot the points.
$x$ 10 100 1000 $y = f\left( x \right) = \frac{1}{{x - 2}}$
e)
To investigate the left-end behaviour of the rational function (as $x \to - \infty$), complete the table of values below and plot the points.
$x$ -10 -100 -1000 $y = f\left( x \right) = \frac{1}{{x - 2}}$
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# An Urn Problem with a Penalty
1. Jun 7, 2012
### RvMiert
Consider an urn with r red balls and b blue balls. In every turn, one ball is drawn.
When a red ball is drawn, it is put back in the urn together with some extra R red balls.
When a blue ball is drawn, it is left outside the urn.
The questions are:
1. What is the expectation value of the number of blue balls outside the urn, B(n), after n turns?
2. What is the expected number of turns needed to get all the blue balls out of the urn?
3. What minimal value must the penalty have, the number R, to get this number of expected turns to infinity?
2. Jun 8, 2012
### viraltux
Hi RvMiert,
I've got the answer for 1. and the equations for 2. and 3. but the kind of equation that it's not easy to find an explicit expression for the solution, so I'm thinking that maybe there is a simpler way to do it. Could you tell us about the background of this problem? Did you find it in a book where the solution is supposed to be a simple expression?
3. Jun 9, 2012
### RvMiert
Hi viraltux,
Well, sometimes when I get bored during lectures, I make up my own questions to see if I can answer them. And this one was quite hard. :P I asked some of my old teachers if they can give me some hints, but they couldn't give me an explicit expression either. For question 1 a recursive expression was found, but then you would need -all- of the previous B(n-k)'s to express the expectation value of B(n). And all of the teachers said that they were quite sure that there has to be an explicit expression, so my search continues. :P
4. Jun 9, 2012
### viraltux
Oh you little #@&%!!! Thanks for the warning!!! hahaha :rofl:
Since it looked like a book problem I was all the time thinking I was missing something because the first thing I got was the recursive expression too... But anyway, I kept on it and now I think I got the a explicit one in form of a summatory. Since I need 1. to answer questions 2. and 3. I could not get from this summatory an explicit n and R respectively but just the equations to get them, but anyway, now that you explain where the problem comes from with all your professors trying to solve your crazy (but interesting) problem I guess my solution is more than OK now! :tongue:
1. OK, a hint for solving 1. Just consider how the probabilities to get a blue ball behave in n steps in:
$$\frac{b-i}{r+j R + b - i}$$
where $i+j=n$. Then add them up to get the expected value in n steps.
Once we have $E(Bn)$ we can solve 2. and 3. by simply solving these equations numerically:
2. $min \ n |\ E(Bn)\geq b$
3. $min \ R | \forall n \ E(Bn)>b$
Last edited: Jun 9, 2012
5. Jun 9, 2012
### alan2
This is a complicated variation of Polya's urn. I think you have a publishable paper if you solve it. The real problem is that you have no symmetry between the two colors of balls. It is the symmetry which normally allows problems of this type to be solved.
6. Jun 11, 2012
### RvMiert
Well, I did have some progression before I went to my teachers.
The expectation value of the first turn is easy,
$$\mathbb{E}\left[ {{B_1}} \right] = \frac{b}{{r + b}}$$.
And also it is not hard to find that
$$\mathbb{P}\left[ {\left. {{B_n} = {B_{n - 1}} + 1} \right|{B_{n - 1}}} \right] = \frac{{b - {B_{n - 1}}}}{{r + b + R \cdot \left( {n - {B_{n - 1}}} \right) - {B_{n - 1}}}}$$,
$$\mathbb{P}\left[ {\left. {{B_n} = {B_{n - 1}}} \right|{B_{n - 1}}} \right] = \frac{{r + R \cdot \left( {n - {B_{n - 1}}} \right)}}{{r + b + R \cdot \left( {n - {B_{n - 1}}} \right) - {B_{n - 1}}}}$$.
So
\begin{gathered}
\mathbb{E}\left[ {\left. {{B_n}} \right|{B_{n - 1}} = k} \right] = k\mathbb{P}\left( {\left. {{B_n} = {B_{n - 1}}} \right|{B_{n - 1}} = k} \right) + \left( {k + 1} \right)\mathbb{P}\left( {\left. {{B_n} = {B_{n - 1}} + 1} \right|{B_{n - 1}} = k} \right) \\
= k + \frac{{b - k}}{{r + b + R \cdot \left( {n - k} \right) - k}}\\
\end{gathered}.
And then
$$\mathbb{E}\left[ {\left. {{B_n}} \right|{B_{n - 1}},...,{B_1}} \right] = \sum\nolimits_{k = 0}^{n - 1} {\frac{{b - {B_k}}}{{r + b + R \cdot \left( {n - {B_k}} \right) - {B_k}}}}$$
But then what...? :P
7. Jun 11, 2012
### viraltux
Well, I got a different recursive expression based on the number of red and blue balls from the previous step, but anyway, that didn't help either for the explicit expression.
What I did is to build a probability tree for the red and blue balls, so for a particular level k I found this relationship:
$$P(Bk)=\sum_{\forall i,j |\ i+j=k} {k \choose i} \frac{b-i}{r+j R + b - i}$$
And therefore:
$$E(Bn)= \sum_{k=0}^{n-1} \sum_{\forall i,j |\ i+j=k} {k \choose i} \frac{b-i}{r+j R + b - i}$$
Now, I didn't check this thoroughly because I thought your question was a book style problem and I had the feeling I might be going overboard with this solution, that's when I asked you about the background of the problem, but at the very least I hope this helps :tongue:
Last edited: Jun 11, 2012
8. Jun 11, 2012
### RvMiert
I don't get it. What exactly is P(Bk)? The probability that B_n = k? And how can it be a sum of scaled probabilities? Because, all ready in the second turn, you would get a product of probabilities.
Say that p_b is the probability to draw a blue ball, p_bb to draw blue when blue is drawn in the first turn, p_br to draw red when blue is drawn in the first turn, etc. Then P(B_2=k) will at least consists of some combinations of p_b*p_bb, p_b*p_br, etc. So I don't see how it can be expressed just as a sum.
9. Jun 11, 2012
### viraltux
Yeah, you're right, I am missing the products. What I did is
in the first draw the probability to get a blue ball is $\frac{b}{r + b}$
The second round the probability to get a blue ball is
if red in the first round: $\frac{b}{r +R + b}$
if blue in the first round: $\frac{b-1}{r + b-1}$
And so on... Before I just copied the formula too fast from my scribbles, sorry for that, the right one is:
$$E(B_n)=\sum_{\forall i,j |\ i+j=n-1} {n-1 \choose i} \frac{b-i}{r+j R + b - i}$$
Which is the expected value of blue balls in the nth draw. So for instance, for n=1 you get
$$E(B_1)=\sum_{\forall i,j |\ i+j=0} {0 \choose 0} \frac{b-0}{r+0 R + b - 0} = \frac{b }{r + b }$$
for n=2
$$E(B_2)=\sum_{\forall i,j |\ i+j=1} {1 \choose i} \frac{b-i}{r+j R + b - i} = \frac{b}{r+R+b}+ \frac{b -1}{r + b-1 }$$
And so on... but missing the products so... yep. This summatory is just the sum of the probability tree level probabilities.
Nonetheless, I am certain this problem has a summatory solution since I remember a theorem which says that for any recursive algorithm there is an iterative (and therefore summatory) counterpart. I also remember the technique to transform recursion into iteration was a bit convoluted but it is there. So though the summatory I've got misses something you can always resort to the recursion into iteration technique to get an explicit expression.
Last edited: Jun 11, 2012
10. Jun 11, 2012
### RvMiert
Still, I´m sorry :P, I don't see how this can be right. If I completely wright it out, I get
$${\Bbb P}_\text{b}={\Bbb P}\left( {n = 1,{\text{ blue ball}}} \right) = \frac{b}{{r + b}}$$
$${\Bbb P}_\text{r}={\Bbb P}\left( {n = 1,{\text{ red ball}}} \right) = \frac{r}{{r + b}}$$
$${\Bbb P}_\text{bb}={\Bbb P}\left( {\left. {n = 2,{\text{ blue ball}}} \right|n = 1,{\text{blue ball}}} \right) = \frac{{b - 1}}{{r + b - 1}}$$
$${\Bbb P}_\text{br}={\Bbb P}\left( {\left. {n = 2,{\text{ red ball}}} \right|n = 1,{\text{blue ball}}} \right) = \frac{r}{{r + b - 1}}$$
$${\Bbb P}_\text{rb}={\Bbb P}\left( {\left. {n = 2,{\text{ blue ball}}} \right|n = 1,{\text{red ball}}} \right) = \frac{b}{{r + R + b}}$$
$${\Bbb P}_\text{rr}={\Bbb P}\left( {\left. {n = 2,{\text{ red ball}}} \right|n = 1,{\text{red ball}}} \right) = \frac{{r + R}}{{r + R + b}}$$
So the expectation value of blue balls outside the urn, after 2 turns is given by
${\Bbb E}\left[ {{B_2}} \right] = 2{{\Bbb P}_{\text{b}}}{{\Bbb P}_{{\text{bb}}}} + 1{{\Bbb P}_{\text{b}}}{{\Bbb P}_{{\text{br}}}} + 1{{\Bbb P}_{\text{r}}}{{\Bbb P}_{{\text{rb}}}} + 0{{\Bbb P}_{\text{r}}}{{\Bbb P}_{{\text{rr}}}} = \frac{{2b\left( {b - 1} \right) + br}}{{\left( {r + b} \right)\left( {r + b - 1} \right)}} + \frac{{br}}{{\left( {r + b} \right)\left( {r + b + R} \right)}}$
11. Jun 11, 2012
### viraltux
Is not right, is missing the products :tongue:
12. Jun 11, 2012
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0 answersPost by CW Burnette on February 11, 2013great job ms.P 1 answerLast reply by: CW BurnetteMon Feb 11, 2013 11:54 AMPost by Jasmine Valdovinos on August 9, 2011great lessson!
• Area = πr2
### Area of a Circle
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Area of a Circle 0:05
• Area of a Circle: Equation and Example
• Extra Example 1: Find the Area of the Circle 2:17
• Extra Example 2: Find the Area of the Circle 5:47
• Extra Example 3: Find the Area of the Shaded Region 9:24
### Transcription: Area of a Circle
Welcome back to Educator.com.0000
For the next lesson, we are going to go over the area of a circle.0002
First to review over area, remember it is how much space it is covering up.0007
The area of a circle is when you have a circle and you see how much space it is using.0013
For example, let's say you have a hole in your jeans and you want to cover it up.0024
You cut out a circle from another pair of jeans let's say.0033
Then you stitch it on to your jeans to cover up your hole.0040
That, however much that circle, that patch is covering up, that is area.0046
It is just how much you are covering; how much space you are using.0052
Remember if you are measuring the distance around the circle, that is called circumference.0056
We have circumference which is the distance around the circle0064
and then area which is all of this, how much space it is using up.0074
The formula to find the area of a circle is π times the radius times the radius again.0083
In other words, the area is πr2; r2.0090
Be careful; this is not r times 2.0097
It is an exponent; that means it is r times itself that many times.0100
It is r2; radius times the radius.0105
Circumference is 2 times π times r.0110
In this case, remember how we multiplied the 2 and the r together first.0118
In this case, this is 2 times r or r times 2.0123
This is not r times 2; this is r times r.0126
Remember keep in mind the difference between the formula for the circumference and the area.0131
First let's find the area of this circle.0139
The formula of area is π times r2 or π times r times r.0143
Remember π is 3.14; π, I am going to put in 3.14.0152
The radius is 4; 42; again be careful; this is not 4 times 2.0162
This is 4 times itself; 4 times 4.0173
Also for order of operations, because we have two different operations0182
meaning we have two different things we can do.0190
We can multiply; or we can do the exponent.0192
The order of operations, remember please excuse my dear aunt sally.0196
Parentheses, exponent, multiplication, division, addition and subtraction.0203
It is always parentheses first; exponents next; multiplication and division; addition and subtraction.0211
See how the exponent comes before multiplying.0218
Be careful; you do not multiply these two numbers first.0224
You always have to take care of the exponent first; then you can multiply.0230
3.14 times... 42 is 4 times 4 which is 16.0240
Again remember do not multiply 3.14 times 4 and then square it.0249
If you do that, you are going to get the wrong answer.0254
Here I want to multiply 3.14 times 16.0257
4 times 6 is 24; 6 times 1 is 6; plus 2 is 8.0266
6 times 3 is 18; I put a 0 right here.0272
1 times 4, 4; 1 times 1, 1; 1 times 3, 3; then add.0279
4 plus 0 is 4; this is 12; 8, 9, 10; 3, 4, 5.0290
Since I am multiplying, I look at my problem.0303
I see how many numbers are behind the decimal point.0307
I only have two numbers behind decimal points.0310
In my answer, I am going to place two numbers behind the decimal point which is right there.0314
My answer becomes 50.24; I cannot forget my units; here it is inches.0319
Area is always squared; units squared; not numbers squared; units squared.0333
50.24 inches squared is my answer; that is the area of this circle.0339
Next example, here I am given that the diameter...0348
Remember diameter is a segment whose endpoints are on the circle; on the circle; on the circle.0356
And passes through the middle, the center of the circle.0364
This is a diameter; the diameter is 20 meters.0368
To find the area of a circle, area equals πr2, radius squared.0374
I need to find the radius; I have the diameter; but I want the radius.0382
How do I find the radius if I am given the diameter?0390
The whole thing is 20; that is the diameter.0393
I know the radius is from the center to this point right there.0395
The radius is half the diameter.0401
If the whole thing is 20, then the radius has to be half of that which is 10.0403
Now I know my radius is 10.0413
I can go ahead and plug in my numbers and solve for my area.0415
π is 3.14; the radius is 102.0420
Again order of operations says we have to take care of the exponents before multiplying.0428
Area equals... I am going to leave this for the next step.0437
102 is not 10 times 2; it is not 20; be careful.0442
It is 10 times 10 which is 100; remember the shortcut.0446
If we want to multiply by 10 or 100 or 1000 or 10000,0457
then you just count the number of 0s in that number.0464
Here I have two 0s; 100 has two 0s.0467
You are going to take this decimal point then.0473
Whenever you multiply a number to 100 or 10 or 1000, count how many 0s there are.0477
There is two; I am going to place this decimal point.0483
I am going to move it two spaces then.0488
Two 0s so I am going to move it two spaces.0490
Do I move it to the left or to the right two spaces?0493
Since I am multiplying by 100, this number has to get bigger.0500
The way to make the number bigger is to move the decimal point over to the right0504
because you want the whole number to be a bigger whole number.0508
I have to move it to the right two spaces; go one, two.0512
My answer then becomes... that is the new spot for my decimal point.0516
It is 314 is my answer; 314.0522
Again two 0s here; move it two spaces to the right.0529
It was here; it moved over to here, the end.0534
Since it is at the end, I don't have to write it.0537
It is just 314 point... same thing as if not being there.0539
314, you can leave it like that.0545
We are done solving; but I have to add my units now.0550
It is meters; area is always squared; units squared; 314 meters squared.0552
My third example, we are going to find the area of the shaded region.0564
I have this rectangle and a circle here that is cut out.0572
All this is missing; that is area.0582
If I cut it out, then don't I have to take it away?0587
I have to subtract it; it is as if I have this whole rectangle.0591
It was whole before the circle was cut out.0599
Find the area of the whole thing.0602
Then you are going to subtract the area of the circle.0605
That is going to become what you have left, the area that is shaded.0609
Imagine if this rectangle was like a piece of paper and you cut out a circle.0617
You have to figure out what is that area of the circle you cut out to see what you are taking away.0625
Find the area of rectangle; find the area of the circle; subtract it.0632
You will get the area of the shaded region.0637
The area of the rectangle; this is the rectangle.0640
Area is base times height or length times width; length times the width.0645
That is 8 times 7 which is 56 centimeters squared.0659
Centimeters squared is the area of this rectangle; that is that.0674
The area of the circle, πr2; π is 3.14; the radius is 2; 22.0683
I am going to take care of this first.0705
Area equals 3.14... I am going to leave that; solve that out; that is 4.0707
3.14 times the 4; let's do that over here; 3.14 times 4.0716
4 times 4 is 16; 4 times 1 is 4; plus 1 is 5.0723
This is 12; I have two numbers behind the decimal point; one, two.0730
I need to place two numbers behind the decimal point in my answer.0737
Area equals 12.56 centimeters squared.0742
Now I have the area of the whole thing and then the area of the circle.0752
I need to take away the circle from the rectangle.0755
It is going to be 56 minus 12.56; I need to do that.0760
56 minus... remember when you subtract decimals, you have to line them up.0774
Where is the decimal in this number?0782
If you don't see it, it is always at the end right there.0784
Minus 12 point... make sure only when you add or subtract, the decimals have to line up... 56.0788
I am missing numbers here.0800
If I am missing numbers here, it is at the end of a number behind the decimal point, I can add 0s like that.0802
When I subtract, this is going to borrow; this becomes the 10; this becomes 9.0811
Borrow; 5; is that big enough?--yes.0821
10 minus this 6 is 4; 9 minus 5 is 4; point.0827
5 minus 2 is 3; 5 minus 1 is 4; it is 43.44.0837
This is 43.44 centimeters squared is my answer.0849
Again just find the area of the rectangle; then find the area of the circle.0860
I subtract it; I have to take the circle away; I have to subtract it.0866
Make sure your decimals line up when you subtract; you get this as your answer.0872
That is it for this lesson; thank you for watching Educator.com.0881
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# computer arithmetic
comp. arith.: psn. num. sys., convert, bin. arith, n-bit reg., 2's-comp.
## computer arithmetic
computers use binary arithmetic
positional number systems
binary (2) octal (8) decimal (10) hexadecimal (16: digits 0 … 9 A … F}
decimal (10)
6270310 =
6 × 104 +
2 × 103 +
7 × 102 +
0 × 101 +
3 × 100
binary (2)
1010012 =
1 × 25 +
0 × 24 +
1 × 23 +
0 × 22 +
0 × 21 +
1 × 20
base b
. . . +
a1 × b1 +
a0 × b0
### convert decimal to base b
repeat: divide by b, take remainder
217 = 2 × 108 + 1 108 = 2 × 54 + 0 54 = 2 × 27 + 0 27 = 2 × 13 + 1 13 = 2 × 6 + 1 6 = 2 × 3 + 0 3 = 2 × 1 + 1 1 = 2 × 0 + 1
21710 = 110110012
check ? convert back (see below)
### convert base b to decimal
```n <- most sig. digit
while digits remain:
n <- n*b + next digit
```
110110012 ? 1 1 1 × 2 + 1 3 3 × 2 + 0 6 6 × 2 + 1 13 13 × 2 + 1 27 27 × 2 + 0 54 54 × 2 + 0 108 108 × 2 + 1 217
check with web calculator:
1 + 2 * (0 + 2 * (0 + 2 * (1 + 2 * (1 + 2 * (0 + 2 * (1 + 2 * (1) ) ) ) ) ) ) = 21710
6253017 ? 6 6 6 × 7 + 2 44 44 × 7 + 5 313 313 × 7 + 3 2194 2194 × 7 + 0 15358 15358 × 7 + 1 107507
1 + 7 * (0 + 7 * (3 + 7 * (5 + 7 * (2 + 7 * (6) ) ) ) ) ) = 10750710
16=24, so start at least sig. bit, group each 4 bits
10100111100110101002 =
101 0011 1100 1101 0100 =
5 3 C D 416
### binary arithmetic
use algorithms from elementary school
• 1101 0110 + 101 0011 = ?
• 1101 0110 – 101 0011 = ?
• 1101 0110 × 101 0011 = ?
• 1101 0110 / 1001 = ?
``` 1 1 11 00
1101 0110 1101 0110
+ 101 0011 - 101 0011
----------- -----------
1 0010 1001 1000 0011
11010110
* 1010011
----------
11010110
11010110
11010110
+ 11010110
----------------
100010101100010
10111
--------
1001 | 11010110
1001
----
1000
0000
----
10001
1001
----
10001
1001
----
10000
1001
----
0111
```
### n-bit registers
register
• arithmetic: special-purpose ⇒ small
• 4-bit: 0000 to 1111 ⇒ 24 numbers
• 8-bit: 0000 0000 to 1111 1111 ⇒ 28 numbers
• n-bit: 000…0 to 111…1 ⇒ 2n numbers
• fixed-size ⇒ can overflow
• 4-bit: 1101 + 0111 = 0100 ⇒ arith. mod 24
• 8-bit: 1111 1111 + 0000 0001 = 0000 0000 ⇒ arith. mod 28
• n-bit: ⇒ arith. mod 2n
### neg. numbers: 2's-complement
how to get n-bit pos. and neg. numbers ? (arith mod 2n)
• -x (mod 2n) = 2n -x (mod 2n)
• so, pick positive/negative numbers
• ⇒ negatives: -2n-1 . . . -1
• ⇒ positives: 1 . . . 2n-1-1
```0000 0 0
0001 1 1
0010 2 2
0011 3 3
0100 4 4
0101 5 5
0110 6 6
0111 7 7
1000 8 -8
1001 9 -7
1010 10 -6
1011 11 -5
1100 12 -4
1101 13 -3
1110 14 -2
1111 15 -1
```
### 2's-complement arithmetic
negate(x)
= 2n – x
= 2n–1 –x + 1
= flip_bits(x) + 1
overflow ?
only if x = –2n-1 (why?)
``` neg( 0010 1010 ) neg( 1101 0110 )
= 1101 0101 + 1 = 0010 1001 + 1
= 1101 0110 = 0010 1010
```
x + y
usual method: + (mod 2n)
overflow ?
only if sign(x) = sign(y) ≠ sign(x+y) (why?)
``` 1011 1010 1011 0110 1011 0111
+ 1101 1101 + 0101 1101 + 1001 1011
--------- --------- ---------
1001 0111 0001 0011 0101 0010
```
x – y
= x + negate(y)
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# 25 Questions - Algebra 2, College level.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
(See attached file for full problem description with complete equations)
---
1. The diameter of the Milky Way disc is approximately 9  1020 meters. How long does it take light, traveling at 1016 m/year to travel across the diameter of the Milky Way?
2. Divide.
3. Multiply. Write the answer in scientific notation.
(2.4  10-5)(4  10-4)
4. Simplify.
5. Multiply. (a2 + 2ab - b2)(a2 - 7ab + b2)
6. Factor completely. 12x3 - 3xy2
7. Factor. 8m4n - 16mn4
8. The product of two consecutive positive even integers is 288. Find the integers.
9. Factor completely. 3x2 - 5xy + 3x - 5y
10. Factor completely. 5x2 + 33x + 18
11. Write in simplest form.
12. Solve.
13. Multiply.
14. During rush hour, Fernando can drive 40 miles using the side roads in the same time that it takes to travel 30 miles on the freeway. If Fernando's rate on the side roads is 7 mi/h faster than his rate on the freeway, find his rate on the side roads.
15. Solve.
16. Simplify.
18. Find the altitude (or height) of the triangle.
19. Evaluate , if possible.
20. Use a calculator to approximate to the nearest hundredth.
21. Match the graph with its equation.
A) y = -x2 + x B) y = -x2 + 1 C) y = -x2 + 2x D) y = -x2 - 1
22. Solve by completing the square.
x2 - 10x + 24 = 0
23. A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth?
24. Solve by using the quadratic formula. 3x2 = 11x + 4
25. Identify the axis of symmetry, create a suitable table of values, then sketch the graph (including the axis of symmetry).
y = -x2 + 3x - 3
---
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# Ex.1.1 Q9 Integers - NCERT Maths Class 7
Go back to 'Ex.1.1'
## Question
Use the sign of $$<, >$$ or $$=$$ to make the statement true?
\begin{align} {\rm a.}\, –8+(–4) \;&\boxed{\;\;}\; –8–(–4)\quad \\-8-4\quad&\boxed{\;\;} \quad -8+4\quad\\–12 \quad&\boxed{\lt} \quad –4 \end{align}
\begin{align}&A\left( {3,2} \right), \qquad\;\;\;\;\;\;\;\;\;\;\;\;B\left( { - 2,3} \right), \qquad\;\;\;\;\;\; C\left( {0,4} \right)\\&D\left( { - \frac{1}{2}, - \frac{3}{2}} \right), \qquad E\left( {\sqrt 2 , - \sqrt 2 } \right), \qquad F\left( {\pi ,0} \right) & \end{align}
a.$$(-8) + (–4)$$ $$\quad\boxed{\;\;}$$ $$(–8) – (–4)$$ b. $$(–3)+7–(19)$$ $$\quad\boxed{\;\;}$$ $$15–8+(–9)$$ c. $$23–41+11$$ $$\quad\boxed{\;\;}$$ $$23–41–11$$ d. $$39 + (–24) –(15)$$ $$\quad\boxed{\;\;}$$ $$36+ (– 52) – (–36)$$ e. $$–231+79+51$$ $$\quad\boxed{\;\;}$$ $$–399+159+81$$
Video Solution
Integers
Ex 1.1 | Question 9
## Text Solution
What is known?
The statement.
What is unknown?
To find out which is greater than, smaller than or equal to.
Reasoning:
Add or Subtract the two values and after check the statement is <,>,or =
Steps:
\begin{align} {\rm a.}\, –8+(–4) \;&\boxed{\;\;}\; –8–(–4)\quad \\-8-4\quad&\boxed{\;\;} \quad -8+4\quad\\–12 \quad&\boxed{\lt} \quad –4 \end{align}
\begin{align}{\rm b.}\; –3+7–19\;&\boxed{\;\;}\; 15 –8 + (–9)\\–3+7–19\quad&\boxed{\;\;} \quad 15 –8 –9 \\4–19 \quad&\boxed{\;\;} \quad 7–9\\ –15 \quad&\boxed{\lt} \quad –2\\ \end{align}
\begin{align}{\rm c.}\;23–41+11 \;&\boxed{\;\;}\; 23 –41 –11 \\ –18+11\quad&\boxed{\;\;} \quad –18 –11 \\ -7\quad&\boxed{\gt} \quad -29 \\ \end{align}
\begin{align}{\rm d.}\; 39+(–24)–15 \;&\boxed{\;\;}\; 36 + (–52) – (–36) \\39–24–15\quad&\boxed{\;\;} \quad 36–52+36 \\ 15-15\quad&\boxed{\;\;} \quad -16 + 36\\0\quad&\boxed{\lt} \quad 20 \\ \end{align}
\begin{align} {\rm e.}\; –23 + 79 + 51 \;&\boxed{\;\;}\; –399 + 159 + 81\quad \\ –152 + 51\quad&\boxed{\;\;} \quad –240+81\quad\\–101 \quad&\boxed{\gt} \quad –159 \end{align}
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# How do you calculate division with decimals?
## How do you calculate division with decimals?
Dividing Decimals
1. Step 1: Estimate the answer by rounding .
2. Step 2: If the divisor is not a whole number, then move the decimal place n places to the right to make it a whole number.
3. Step 3: Divide as usual.
4. Step 4: Put the decimal point in the quotient directly above where the decimal point now is in the dividend.
Are remainders used in decimal division?
Division: Different ways to write the remainder When a number does not divide evenly into another, the leftover can be expressed as a remainder, fraction, or decimal. The key to working out a long division problem, where the remainder is written as a decimal, is the ability to add zeros after the decimal point.
### What is the remainder of 17 divided by 3?
The result of division of 173 is 5 with a remainder of 2 .
What is the remainder of 14 divided by 3?
The result of dividing 14 by 3 is 4 with a remainder of 2, or 4 and 2/3 as a mixed number, or 4.666666667 as a 10-digit decimal number.
## What are the rules for dividing decimals?
Rules for Dividing Decimals: Rule to Divide a Decimal by a Whole Number: Put the decimal values in the quotient directly above the decimal in the dividend and then divide as normal. Align is some what critical in dividing. We must make sure our place of digits in the quotient is placed in their proper place.
How do you calculate a remainder?
How to calculate the remainder Begin with writing down your problem. Decide on which of the numbers is the dividend, and which is the divisor. Perform the division – you can use any calculator you want. Round this number down. Multiply the number you obtained in the previous step by the divisor.
### What is 54 divided by 3?
The answer to the question: What is 54 divided by 3 is as follows: 54 / 3 = 18. Instead of saying 54 divided by 3 equals 18, you could just use the division symbol, which is a slash, as we did above.
How do you divide by decimals?
Here’s how to divide decimals step by step: Move the decimal point in the divisor and dividend. Turn the divisor (the number you’re dividing by) into a whole number by moving the decimal point all the way to the right. At the same time, move the decimal point in the dividend (the number you’re dividing) the same number of places to the right.
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# The speed of the cyclist is 7 km / h. Which way he will travel in 0.5.
To find the distance that a given cyclist should have traveled in 0.5 hours, apply the formula: S = V * t.
The values of the variables: V is the speed of the cyclist (V = 7 km / h); t – elapsed time (t = 0.5 h).
Calculation: S = V * t = 7 * 0.5 = 3.5 km.
Answer: In 0.5 hours, the cyclist was able to cover the distance of 3.5 km.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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# Counting number of sums from contiguous subarrays of an array
We are given an array $a[1 \ldots n]$ with all $a[i]>0$.
Now we need to find how many distinct sums can be formed from its subarrays (where a subarray is a contiguous range of the array, i.e., $a[j\ldots k]$ for some $j,k$, the sum is the sum of all of the elements of the subarray). For example, if $a=[1,2,1]$, then the answer is 4: we can form $1,2,3,4$.
I know how to count the number of distinct sums in $O(n^2)$ time.
Furthermore, I have come to realise this is similar to the classical problem where we need to find the number of distinct substrings of a string. I was thinking of the possibility of constructing a suffix array and solving it in a similar fashion (in $O(n)$ time). But I have not been able to figure out how to modify that to work here. For example, if we use suffix array for $a=[1,2,1]$ we will get 5 cases instead of the four acceptable ones. Is this possible to do this using suffix arrays or am I thinking in the wrong direction?
Also there is one more direction I have been thinking in. Divide and conquer. Like if I divide the array into two parts every time until it is reduced to a single element. A single element can have one sum. Now if we combine two single elements, It can be done in two ways: if both single ranges have same element then we get 2 different sums, or if both have different elements we get 3 different sums. But I am not being able to generalize this for merging arrays of length greater than 1. Is it possible to merge two m size arrays and get the answer in $O(m)$?
• Although I don't immediately see how you can derive from it a solution to your problem, the structure of the Maximum Subarray Problem is similar to the problem you describe and has a divide-and-conquer solution which runs in $O(n \ lg \ n)$. Jul 30, 2013 at 1:10
• I would suggest starting with the following problem: how hard is it to decide whether there are two intervals with the same sum? It is tempting to prove 3SUM-hardness of this problem, but so far I haven't been able to. May 8, 2014 at 5:01
You almost surely can't get better than $O(n^2)$ in the worst case since the number of different sums can be in $\Theta(n^2)$.
Consider e.g. the array $[1,2,4,8,\dotsc, 2^n]$. Here each of the $\frac{n(n+1)}2$ contiguous subarrays has a different sum.
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# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Calculus Countdown:
Filter by: Content type:
Stage:
Challenge level:
### There are 184 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Fractional Calculus III
##### Stage: 5
Fractional calculus is a generalisation of ordinary calculus where you can differentiate n times when n is not a whole number.
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Interpolating Polynomials
##### Stage: 5 Challenge Level:
Given a set of points (x,y) with distinct x values, find a polynomial that goes through all of them, then prove some results about the existence and uniqueness of these polynomials.
### Logic, Truth Tables and Switching Circuits
##### Stage: 3, 4 and 5
Learn about the link between logical arguments and electronic circuits. Investigate the logical connectives by making and testing your own circuits and record your findings in truth tables.
### Truth Tables and Electronic Circuits
##### Stage: 3, 4 and 5
Investigate circuits and record your findings in this simple introduction to truth tables and logic.
### Telescoping Functions
##### Stage: 5
Take a complicated fraction with the product of five quartics top and bottom and reduce this to a whole number. This is a numerical example involving some clever algebra.
### Stonehenge
##### Stage: 5 Challenge Level:
Explain why, when moving heavy objects on rollers, the object moves twice as fast as the rollers. Try a similar experiment yourself.
### Generally Geometric
##### Stage: 5 Challenge Level:
Generalise the sum of a GP by using derivatives to make the coefficients into powers of the natural numbers.
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Three Ways
##### Stage: 5 Challenge Level:
If x + y = -1 find the largest value of xy by coordinate geometry, by calculus and by algebra.
### Integral Inequality
##### Stage: 5 Challenge Level:
An inequality involving integrals of squares of functions.
### Polynomial Relations
##### Stage: 5 Challenge Level:
Given any two polynomials in a single variable it is always possible to eliminate the variable and obtain a formula showing the relationship between the two polynomials. Try this one.
### Mind Your Ps and Qs
##### Stage: 5 Short Challenge Level:
Sort these mathematical propositions into a series of 8 correct statements.
### Logic, Truth Tables and Switching Circuits Challenge
##### Stage: 3, 4 and 5
Learn about the link between logical arguments and electronic circuits. Investigate the logical connectives by making and testing your own circuits and fill in the blanks in truth tables to record. . . .
### Mechanical Integration
##### Stage: 5 Challenge Level:
To find the integral of a polynomial, evaluate it at some special points and add multiples of these values.
### How Many Solutions?
##### Stage: 5 Challenge Level:
Find all the solutions to the this equation.
### Without Calculus
##### Stage: 5 Challenge Level:
Given that u>0 and v>0 find the smallest possible value of 1/u + 1/v given that u + v = 5 by different methods.
### Pythagorean Triples II
##### Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### Magic W Wrap Up
##### Stage: 5 Challenge Level:
Prove that you cannot form a Magic W with a total of 12 or less or with a with a total of 18 or more.
### Modulus Arithmetic and a Solution to Differences
##### Stage: 5
Peter Zimmerman, a Year 13 student at Mill Hill County High School in Barnet, London wrote this account of modulus arithmetic.
### Recent Developments on S.P. Numbers
##### Stage: 5
Take a number, add its digits then multiply the digits together, then multiply these two results. If you get the same number it is an SP number.
### Picturing Pythagorean Triples
##### Stage: 4 and 5
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
### Magic Squares II
##### Stage: 4 and 5
An article which gives an account of some properties of magic squares.
### Sums of Squares and Sums of Cubes
##### Stage: 5
An account of methods for finding whether or not a number can be written as the sum of two or more squares or as the sum of two or more cubes.
### Transitivity
##### Stage: 5
Suppose A always beats B and B always beats C, then would you expect A to beat C? Not always! What seems obvious is not always true. Results always need to be proved in mathematics.
### Impossible Sandwiches
##### Stage: 3, 4 and 5
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
### Pythagorean Triples I
##### Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
### Euclid's Algorithm II
##### Stage: 5
We continue the discussion given in Euclid's Algorithm I, and here we shall discover when an equation of the form ax+by=c has no solutions, and when it has infinitely many solutions.
### Proof of Pick's Theorem
##### Stage: 5 Challenge Level:
Follow the hints and prove Pick's Theorem.
### More Sums of Squares
##### Stage: 5
Tom writes about expressing numbers as the sums of three squares.
### Modulus Arithmetic and a Solution to Dirisibly Yours
##### Stage: 5
Peter Zimmerman from Mill Hill County High School in Barnet, London gives a neat proof that: 5^(2n+1) + 11^(2n+1) + 17^(2n+1) is divisible by 33 for every non negative integer n.
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Rolling Coins
##### Stage: 4 Challenge Level:
A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
### A Knight's Journey
##### Stage: 4 and 5
This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.
### The Triangle Game
##### Stage: 3 and 4 Challenge Level:
Can you discover whether this is a fair game?
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Whole Number Dynamics I
##### Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Whole Number Dynamics IV
##### Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### Little and Large
##### Stage: 5 Challenge Level:
A point moves around inside a rectangle. What are the least and the greatest values of the sum of the squares of the distances from the vertices?
### Yih or Luk Tsut K'i or Three Men's Morris
##### Stage: 3, 4 and 5 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Modular Fractions
##### Stage: 5 Challenge Level:
We only need 7 numbers for modulus (or clock) arithmetic mod 7 including working with fractions. Explore how to divide numbers and write fractions in modulus arithemtic.
### Try to Win
##### Stage: 5
Solve this famous unsolved problem and win a prize. Take a positive integer N. If even, divide by 2; if odd, multiply by 3 and add 1. Iterate. Prove that the sequence always goes to 4,2,1,4,2,1...
### Angle Trisection
##### Stage: 4 Challenge Level:
It is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square.
### Where Do We Get Our Feet Wet?
##### Stage: 5
Professor Korner has generously supported school mathematics for more than 30 years and has been a good friend to NRICH since it started.
### Take Three from Five
##### Stage: 4 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Composite Notions
##### Stage: 4 Challenge Level:
A composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.
### Notty Logic
##### Stage: 5 Challenge Level:
Have a go at being mathematically negative, by negating these statements.
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## FIREWORKS COUNTING Kids DIY is a Montessori Patriotic Math Preschool Activity designed to develop numeracy, rational counting, and 1:1 correspondence skills.
What is a better way to celebrate the 4th of July Independence Day than with a fun hands-on tactile numeracy DIY! Besides, since you will be using recycled and craft materials, you can easily make this DIY at home in no time! Most importantly, with this Patriotic numeracy FIREWORKS COUNTING Kids DIY, your preschooler will be learning math and numeral to quantity association the Montessori way! And the best part, you can easily modify this activity for smaller ones, by making it for numbers one through five. Alternatively, if you have older children, turn these fireworks counting into teens.
### โ YOUโll NEED for this FIREWORKS COUNTING Kids DIY
• recycled cardboard
• recycled toilet paper rolls
• pipe cleaners
• hot glue gun
• black marker
### PRESENTING Montessori Patriotic Math Preschool Activity
First, cut each toilet paper roll in 1/4 so itโs about 1โ inch each. Next, hot glue the corresponding number of pipe cleaners to each. (Colors are discretionary, but I chose them as white, blue, and red as colors on a USA flag.) Lastly, invite your child to count each firework-ray and correspond the total quantity to the numeral written on the cardboard. (Wooden number matching is optional since you already have numbers written on the cardboard.)
### SKILLs YOUR PRESCHOOLER IS DEVELOPING:
With this simple to make counting DIY, your preschooler is learning number recognition and 1:1 correspondence, which are extremely important for children to learn so that they can develop early mathematical skills. After your child has mastered ROTE counting, which is poem-like memorizing of the numerical order, you can start introducingย ย RATIONALย counting. The difference between rote and rational counting is that the former is a process of sequentially memorizing number names. The latter, on the other hand, is developing an understanding of theย valueย of numbers. Children need to understand that quantity represents a numeral, meaning that they need to be able to assign the correct numeral name to eachย manipulative/counting objectย as they count in succession. Thus, One-to-One Correspondence is an essential distinction between rote and rational counting. With RATIONAL counting, a child utilizes one-to-one correspondence when counting,ย meaning s/he is able to count a set of 5+ objects by pointing to each object and assigning each object the next number name until each object has been counted once.ย Moreover, with this DIY, your child is developing hand-eye coordination, fine motor control and advancing spatial awareness skills.
AGE 3Y +. Always supervise your child. You can see this post on Instagram HERE.
## RATIONAL COUNTING
โTell me and will I forget. Teach me and I will remember. INVOLVE me and I will learn!โ
Benjamin Franklin.
Trust me, children learn BEST when they are engaged in what they are doing! So, to teach them RATIONAL COUNTING, offer them to TOUCH each manipulative as they coun ALOUD, matching each cardinal number to the corresponding numeral! Thatโs 1:1 correspondence in a nutshell! We want a child to match a set of objects to its corresponding numeral and recognize that a number is a symbol we use to represent quantity! Young children often learn โrote countingโ without having an understanding of 1:1 correspondence! Children might count till ten just like a heart-learned poem! (That is they memorize the numerical order without really understanding what each quantity means.) To lean RATIONAL counting, they need to learn to associate quantity to numeral! How? PRACTICE, PRACTICE, PRACTICE!
I hope you enjoyed FIREWORKS COUNTING Kids DIY!
## For MORE Montessori Patriotic Preschool Activities
See HERE Independence Day Preschool Activities.
##### Happy 4th of July!
By the way, have you downloaded my NEW eBook THE BASICS? It has everything you need to know to get started on your Montessori journey! Also, you get a CURRICULUM outline reference guide, the order of lessons, and the age when they should be introduced, in my opinion. See details HERE.
Lastly, check out HERE Homeschooling Montessori MADE-EASY membership in case you are exhausted from swimming in the vast ocean of irrelevant information and saving activities you never get to! And if you sign UP, the eBook is included in the package amongst MANY other perks!
I am here to help! WE CAN DO THIS TOGETHER!
###### โก Enriching the Mind one Heart at a time โก
FIND US on Instagram โก Facebook โก Pinterest โก
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• July 13, 2020 at 1:17 pm
[…] HERE FIREWORKS COUNTING Kids […]
• May 20, 2022 at 5:28 am
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Section 2.3 class notes_0
# Section 2.3 class notes_0 - Section 2.3 Lines Objective 1...
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Unformatted text preview: Section 2.3 Lines Objective 1 : Determining the Slope of a Line In mathematics, the steepness of a line can be measured by computing the line’s slope. Every non-vertical line has slope. Vertical lines are said to have no slope. A line going up from left to right has positive slope , a line going down from left to right has negative slope, while a horizontal line has zero slope . We use the variable m to describe slope. Slope = m The slope can be computed by comparing the vertical change (the rise) to the horizontal change (the run ). Given any two points on the line, the slope m can be computed by taking the quotient of the rise over the run. Definition of Slope If 1 2 x x ≠ , the slope of a line passing through distinct points ( 29 1 1 , x y and ( 29 2 2 , x y is 2 1 2 1 y y rise Changein y m run Changein x x x- = = =- Objective 2 : Sketching a Line Given a Point and the Slope If we know a point on a line and the slope, we can quickly sketch the line. Objective 3 : Finding the Equation of a Line Using the Point-Slope Form Given the slope m of a line and a point on the line, ( 29 1 1 , x y , we can use what is known as the point-slope form of the equation to determine the equation of the line. The Point-Slope Form of the Equation of a Line Given the slope of a line m and a point on the line ( 29 1 1 , x y , the point-slope form of the equation of a line is given by ( 29 1 1 y y m x x- =- . Objective 4...
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# Math
posted by on .
1. The height of a bumble bee above the ground is modelled by h(t) = 0.5cos(2π t) + 2, where h is in metres and t
is in seconds. At what time is the bee’s instantaneous rate of change of height with respect to time greatest?
a.) 1 s
b.) 1.25s
c.) 1.5s
d.) 2s
--------------------------------------
2. What value for the function y=3cos(t-pi) + 2 gives an instantaneous rate of change of 0?
a) 0
b) pi/2
c) pi/3
d) pi/4
I tried to plug in the options above, but it gave me 4.99999. How do I do this question?
--------------------------------------
3. The height of a ball is modelled by the equation h(t)= 4sin(8πt) + 6.5 where h(t) is in metres and t is in seconds. What are the highest and lowest points the ball reaches?
a) 10.5 m and 6.5m
b) 10.5 and 2.5m
c) 6.5m and 2.5m
d) 14.5 and 6.5m
I know that answer won't be B because it has to start at 6.5m
• Math - ,
1. The height of a bumble bee above the ground is modelled by h(t) = 0.5cos(2π t) + 2, where h is in metres and t
is in seconds. At what time is the bee’s instantaneous rate of change of height with respect to time greatest?
a.) 1 s
b.) 1.25s
c.) 1.5s
d.) 2s
when is
d/dt 0.5cos(2π t)
biggest?
d/dt = - pi sin(2 pi t)
we want sin(2 pi t) = 1 or -1
that is when 2 pi t = pi/2, 3pi/2, 5 pi/2 , 7 pi/2 ...
t = 1/4 or 3/4 or 5/4 or 7/4 ....
• Math - ,
2.
sin(t-pi) = 0
t = 0, or pi, or 2 pi
perhaps they mean the value of t, not of the function. then t = 0 works
• Math - ,
3.
6.5+4 and 6.5-4
10.5 and 2.5
• Math - ,
I am still confused for number 2.
For number 3, you used the amplitude(which is 4) to find lowest+highest?
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Precalculus Practice Exam
Part 4 Test 4 Time: 2 hours
1. Given the equation, find the following:
• Amplitude
Answer: 5.
• Period
Answer:
• Horizontal Shift(phase shift)
Answer: 2 to the right.
• Vertical Shift
Answer: 10 down.
• Domain
Answer: The domain is the set of all real numbers, or in interval notation, .
• Range
Answer: The range is the set of all values of , or in interval notation, [-15,-5].
• and four points that have a y-coordinate of -15.
Answer: Here are such four points that have a y-coordinate of -15:
2. Given that the point is located on the terminal side of the angle t, find a point on the terminal side of the following angles:
• Answer: .
• Answer: .
• Answer: .
• -t
Answer: .
3. Write as an algebraic expression in x.
Answer:
• Convert to radian measure.
Answer: 2.6269 radians.
• Convert 150.3040 radians to degree measure in the format.
Answer:
• The graph of is bounded by what two lines?
Answer: The graph of f(x) is bounded by the lines y=2x+1, and y=2x-1.
• Why are these two lines considered boundary lines?
Answer: Because the biggest value can attain is 1, and the lowest value it can attain is -1, the graph of the function is caught between the two boundary lines.
• Graph the function and the two boundary lines.
Answer:
4. A man, standing on the top of a building that is 4,000 feet high, looks down, with an angle of depression of 30 degrees, to the base of the building across the street. How far apart are the buildings?
Answer: The buildings are approximately 6,928 feet apart.
• Restrict the domain of the function to an interval where the function is increasing, and
Answer: The biggest interval which will work is .
• Determine over that interval.
Answer:
• Then sketch the graphs of both f and on the same coordinate system.
Answer:
• What relationship do the graphs of these two functions have to the graph of the function g(x)=x?
Answer: The graph of is a reflection of the graph of f(x) about the line g(x).
5. A total of \$25,000 is invested for 10 years at 12% per year compounded monthly. How much interest will be earned during the 10 years?
Answer: The interest earned after 10 years will be \$57,509.67.
6. Solve for x in the inequality
Answer: The inequality will be always true except when x=2, and x=-2.
• When does ?
Answer: They are equal only when x>0.
• When does ?
Answer: They are equal only when x>0.
7. Assume that the population of a town in Texas is given by , where t is time in years, and t=0 corresponds to the year 1900.
• In what year was the population 60,000?
Answer: It was 60,000 in the year 1935 with 9 months.
• Did the population ever reach 100? If yes, when? If no, why not?
Answer: It will only reach 100 in the year 2078.
• How long will it take for the population to reach 1,000,000?
Answer: The population can never increase as long as the same decreasing behavior continues.
8. Find the zeros of
• Algebraically
Answer: There are six zeros. Use the rational zero test and synthetic division to find the 4 rational zeros:
The remaining zeros can be found with the quadratic formula.
• Graphically. Explain how you interpreted the graph to arrive at your answer.
Answer: The graph below tells us that there are zeros at , and 4 more roots close to x=3.
Zooming in close to x=3, we detect 2 roots at about x=2.83 and x=3. Since the function is even, there are also roots at about x=-2.83 and x=-3.
9. Solve the following system of equations for and
Answer: .
If you would like to practice another Part 4 exam, click on Next Exam.
If you would like to practice on exams for Parts 1, 2 or 3, click on Menu. It will take you back to the original menu.
Tue Jun 24 23:15:36 MDT 1997
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Start to Learn Fractions in Maths Today!
Chapters
Confused about the difference between a fraction, decimal, and a percentage? All of them are similar and serve the same function - to tell us how many parts of a collection are taken.
For example, imagine a pizza cut into five slices and your friend has already eaten three of them. If you want to understand how much of the collection(i.e. pizza) is taken you need to understand either fractions, decimals, or percentages.
What is a fraction? 🤔
A fraction is a way to split up a whole number into equal parts. The two parts are the numerator and the denominator.
The top part of a fraction is called the numerator and the part of the fraction at the bottom is called the denominator.
For example, in the fraction 512, 5 is the numerator and 12 is the denominator.
Here’s another real-life example: Using any countable item (like forks or napkins) in your house count 8 pieces. This is the denominator. It is the number we divide our whole number into.
Now take one piece away and notice that this is 1 piece of 8 and therefore we would write the one as the numerator. This can be done with any number and can also be used to understand halves, quarters and eighths.
We like this method because it shows you that the denominator is often concrete. This is especially important when learning how to add and subtract fractions.
Equivalent Fractions
When working with fractions, we often see two fractions and ask ourselves, “Aren’t these two the same thing?”
You’re probably right because we get things such as equivalent fractions. They are fractions that often have different denominators and numerators but, in the end, can be simplified to the same fraction.
Look at the fractions above, all these fractions take up the same part of one whole.
These fractions, and recognising them, is important to the addition, subtraction, multiplication, and division of fractions.
So, keep this in mind as we move forward to the more difficult parts of fractions.
Simplifying Fractions
This step actually comes at the very end of any of the sums we will teach you how to do, however it is very important that you know how to do this before we move on to the more difficult parts.
Simplification can easily be explained by considering how much space 12 boxes will take when we can fit everything that is in those twelve boxes into 2 📦. So why would we still use the 12 boxes? Why not recycle them or give them to someone who has enough things to fill all 12 boxes?
It is quite easy to do if you follow the method of just using your times tables to determine how much a fraction can be simplified.
So, if the denominator and the numerator can be divided by the same number, we do a simple division and we get the simplified version until we cannot divide the numbers any longer.
Not being able to divide further, usually happens when any of the two numbers become a prime number, or when the numbers obviously do not divide into each other (for example 3/4).
The Addition ➕ and Subtraction ➖ of Fractions
There are two types of addition and subtraction of fractions found on this basic level. Adding and subtracting with the same denominator and with different denominators.
Just to recap, this means that when we subtract or add in fractions, our method depends on the denominator.
If the denominator is the same, we use a simple method of subtracting or adding the numerators in each fraction from each other.
This is the method that most schools use all around the world. This method is tried and true, however you can take this method and apply it to something other than a pen and paper.
You can also try it out with objects around the house, or even a pizza! Alternatively, you could look at other methods to spice things up and make addition and subtraction easy.
Now we move on to the more difficult method of adding fractions that have different denominators.
How do we do this? We change the denominator before we complete any other subtraction or addition.
Let's say we have two denominators, 3 and 7, and each of those fractions have different numerators. We take the two denominators and change them to represent the same number. The easiest way to do this is by multiplying the 3 and 7, giving us 21.
However remember this step – Whatever you do to the denominator of each fraction, you must do with the numerator.
The last step to any fraction equation, is simplifying the final fraction. Sometimes, however, this can’t be done so there is no need to worry about it.
Keep in mind that both methods can be used for subtraction and addition. But often, the longer a method is, the easier it is for us to have trouble with it or miss a step.
Multiplying fractions ✖️
To multiply a fraction is as simple as multiplying the numerators and then multiplying the denominators. However, we can’t leave the fraction this way!
We go ahead and simplify it, and then we have our final answer.
The easiest way to learn this concept is by taking the fraction apart. Move both numerators to one line and multiply. Then in the next line multiply only the denominators. Then just draw a line between the two answers to symbolise a fraction. The simplification can come later. It's better and, in the end, easier to learn one concept at a time when it comes to learning something as difficult as multiplying and dividing fractions.
So, unless you have understood simplification well, do not push yourself.
Dividing Fractions ➗
To learn dividing fractions just follow this 3-step method.
Step 1: Leave the first fraction alone 🙅♀️
Step 2: Change the divide to a multiply ➗➡️✖️
Step 3: Invert the Numerator and the Denominator in the final fraction ↕️
That’s it! Follow these 3 easy steps, simplify (of course) and there you can divide any fraction that you need to.
Other tips and rules 👍
Here are some other alternative tips when it comes to learning fractions:
Tip 1: Try dividing up cups into other fractions. This is a method to learn fractions of a whole number.
Tip 2: Make fraction facts about food, patches of flowers in your garden or places in your home. State, in fractions, how many rooms have windows, how much of the pizza has all the toppings on, or how much each patch of flowers takes up of the garden. This method also teaches you estimation, another key component to fraction calculations.
Tip 3: Try using online visuals, with fun maths games and images all over the internet.
Tip 4: Do not be afraid to use calculators sometimes.
It's important to remember to connect fractions to decimals or whole numbers. Never forget that when a numerator is a number place higher than the denominator, that means you have a whole number to include.
Still struggling?
There is no shame in not being able to connect these ideas or connect to maths! Fractions are difficult, as are a lot of other things in maths, but have faith and keep trying.
If you are still struggling, don’t hesitate to book a trial lesson with one of our GoStudent maths tutors! They will be able to ground all of these methods, and help you with creative ways to learn maths.
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# Pre Calc
posted by Em
Find the work done by a force F of 30 pounds acting in the direction (2,3) in moving an object 3 feet from (0,0) to the point in the first quadrant along the line y = (1/2)x.
I just need to figure out how to get the answer.
1. Damon
Find the angle of the Force F
cos T = 2/sqrt13 = .555
T = 56.3 deg above x axis
or
tan T = 2/3
so T = 56.3
Find the direction of motion from slope of line
tan slope = 1/2
slope angle = 26.6 deg
angle between force and motion = 56.3-26.6 = 29.7 degrees
so
component of force in direction of motion = 30 cos 29.7
= 26.06 pounds
26.06*3 = 78.2 ft lbs
beats me how to get 85.38
It would be much closer if the force vector were direction(3,2)
2. Megha
the vector is actually <2,2>
and the line angle is 64.3
3. Oscar
the vector is indeed <2,2>, but the line angle (angle formed by y=1/2x and x-axis) is as Damon posted, 26.6 degree. As for why it isn't 64.3 degree is because slope is change in y over the change in x, thus this means horizontal change is 2 and vertical change is 1. When finding the line angle, simply use arc tan(opp/adj) or in this case arc tan(1/2) which will result in 26.6 degree.
As for the answer, its actually 95.36 ft lb if the vector is <2,2>
4. Katelyn
F1 -> 30 lbs, direction <2, 2>
magnitude of <2, 2> = 2sqrt(2)
Find unit vector of F1 (divide original components by magnitude)
-> <30/sqrt(2), 30/sqrt(2)>
F2 -> 3 feet, direction y = 1/2x
3 = sqrt(x^2 + y^2)
y = 1/2x
3 = sqrt(x^2 + (1/2x)^2)
9 = 5/4* x^2
x^2 = 36/5
x = 6/sqrt(5)
y = 2/sqrt(5)
F1 = <30/sqrt(2), 30/sqrt(2)>
F2 = <6/sqrt(5), 3/sqrt(5)>
Find dot product of F1 and F2
(30/sqrt(2))*(6/sqrt(5))+(30/sqrt(2))*(3/sqrt(5)) = 85.38
Answer in the back of the book is 85.38
## Similar Questions
1. ### multivariable calc with some high school physics
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 5 kilograms is tied to the middle of the clothesline, it sags a distance of 1 meters. What is the magnitude of the tension on the ends of the clothesline?
2. ### physics
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a positive magnitude of 6.7N; a second force has a magnitude of 4.6 N and points in the negative …
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An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a positive magnitude of 6.7N; a second force has a magnitude of 4.6 N and points in the negative …
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5. ### Pre Calc
Find the work done by a force F of 30 pounds acting in the direction (2,3) in moving an object 3 feet from (0,0) to the point in the first quadrant along the line y = (1/2)x.
6. ### physics
A)An object at rest has no forces acting on it. B)An object can be moving in one direction while the net force acting on it is in another direction. C)In order to move a massive crate sitting on the floor, the force you apply to the …
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A two dimensional object placed in the xy plane has three forces acting on it: a force of 3N along the x axis acting at a point (3m, 4m); A force of 2N along the y axis acting at (-2, 5m); and a force of 5N in the negative x direction …
8. ### PHYSICS
An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 6.2 N ; a second force has a magnitude of 5.0 N and points in the negative y direction …
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A variable force of 7x−2 pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1 ft to x = 19 ft. (Round your answer to two decimal places.)
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A 5-lb force acting in the direction of (5, -3) moves and object just left over 12 ft. from point (0, 6) to (7, -4). Find the work done to move the object to the nearest foot-pound. a. 11 ft. * lbs b. 34 ft. * lbs c. 56 ft. * lbs d. …
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# Differential Calculations; Integration Calculations - Canon X Mark I Pro User Instruction
## Differential Calculations
Press
to enter COMP mode.
To perform a differential calculation, you have to input the
expression in the form of:
differential expression
• The differential expression must contain the variable x.
• "a" is the differential coefficient.
• " x" is the change interval of x (calculation precision).
Example: To determine the derivative at point x = 10,
the function f(x) = sin(3x + 30). ......
! You can leave out the
x in the differential expression and the
calculator will automatically substitute a value for
! The smaller the entered value
time will be with more accurate results, the larger the entered value
x is, the shorter the calculation time will be with comparatively
less accurate results.
! Discontinuous points and extreme changes in the value of x can
cause inaccurate results or errors.
! When performing differential calculations with trigonometric
functions, select radian (Rad) as the angle unit setting.
! Log
b, i~Rand, Rec ( and Pol ( functions can not join to differential
a
calculations.
a
x
x = 10
-8
, for
EX #48
x.
x is, the the longer the calculation
### Integration Calculations
Press
to enter COMP mode.
To perform an integration calculation you are required to input
following elements:
integration expression
• The integration expression has a variable x.
• "a" and "b" defines the integration range of the definite integral.
• "n" is the number of partitions (equivalent to N = 2
The integration calculation is based on Simpson's rule.
As the number of significant digits is increased, internal integration
calculations may take considerable time to complete. For some
cases, even after considerable time is spent for performing a
calculation, the calculation results may be erroneous. Particularly
when significant digits are less than 1, an ERROR might be
occurred.
Example: Perform the integration calculation for
! When performing integration calculations with trigonometric
functions, select radian (Rad) as the angle unit setting.
! Log
b, i~Rand, Rec ( and Pol ( functions can not join to integration
a
calculations.
21
a
b
n
n
).
tol
EX #49
......
, with n = 4.
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# 4.6: PDEs, Separation of Variables, and The Heat Equation
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Let us recall that a partial differential equation or PDE is an equation containing the partial derivatives with respect to several independent variables. Solving PDEs will be our main application of Fourier series.
A PDE is said to be linear if the dependent variable and its derivatives appear at most to the first power and in no functions. We will only talk about linear PDEs. Together with a PDE, we usually have specified some boundary conditions, where the value of the solution or its derivatives is specified along the boundary of a region, and/or some initial conditions where the value of the solution or its derivatives is specified for some initial time. Sometimes such conditions are mixed together and we will refer to them simply as side conditions.
We will study three specific partial differential equations, each one representing a more general class of equations. First, we will study the heat equation, which is an example of a parabolic PDE. Next, we will study the wave equation, which is an example of a hyperbolic PDE. Finally, we will study the Laplace equation, which is an example of an elliptic PDE. Each of our examples will illustrate behavior that is typical for the whole class.
## 4.6.1 Heat on an Insulated Wire
Let us first study the heat equation. Suppose that we have a wire (or a thin metal rod) of length $$L$$ that is insulated except at the endpoints. Let $$x$$ denote the position along the wire and let $$t$$ denote time. See Figure $$\PageIndex{1}$$.
Figure $$\PageIndex{1}$$: Insulated wire.
Let $$u(x,t)$$ denote the temperature at point $$x$$ at time $$t$$. The equation governing this setup is the so-called one-dimensional heat equation:
$\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2},$
where $$k>0$$ is a constant (the thermal conductivity of the material). That is, the change in heat at a specific point is proportional to the second derivative of the heat along the wire. This makes sense; if at a fixed $$t$$ the graph of the heat distribution has a maximum (the graph is concave down), then heat flows away from the maximum. And vice-versa.
We will generally use a more convenient notation for partial derivatives. We will write $$u_t$$ instead of $$\frac{\partial u}{\partial t}$$, and we will write $$u_{xx}$$ instead of $$\frac{\partial^2 u}{\partial x^2}$$. With this notation the heat equation becomes
$u_t=ku_{xx}.$
For the heat equation, we must also have some boundary conditions. We assume that the ends of the wire are either exposed and touching some body of constant heat, or the ends are insulated. For example, if the ends of the wire are kept at temperature 0, then we must have the conditions
$u(0,t)=0 \quad\text{and}\quad u(L,t)=0.$
If, on the other hand, the ends are also insulated we get the conditions
$u_x(0,t)=0 \quad\text{and}\quad u_x(L,t)=0.$
Let us see why that is so. If $$u_{x}$$ is positive at some point $$x_{0}$$, then at a particular time, $$u$$ is smaller to the left of $$x_{0}$$, and higher to the right of $$x_{0}$$. Heat is flowing from high heat to low heat, that is to the left. On the other hand if $$u_{x}$$ is negative then heat is again flowing from high heat to low heat, that is to the right. So when $$u_{x}$$ is zero, that is a point through which heat is not flowing. In other words, $$u_{x}(0,t)=0$$ means no heat is flowing in or out of the wire at the point $$x=0$$.
We have two conditions along the $$x$$-axis as there are two derivatives in the $$x$$ direction. These side conditions are said to be homogeneous (i.e., $$u$$ or a derivative of $$u$$ is set to zero).
We also need an initial condition—the temperature distribution at time $$t=0$$. That is,
$u(x,0)=f(x),$
for some known function $$f(x)$$. This initial condition is not a homogeneous side condition.
## 4.6.2 Separation of Variables
The heat equation is linear as $$u$$ and its derivatives do not appear to any powers or in any functions. Thus the principle of superposition still applies for the heat equation (without side conditions). If $$u_1$$ and $$u_2$$ are solutions and $$c_1,c_2$$ are constants, then $$u= c_1u_1+c_2u_2$$ is also a solution.
Exercise $$\PageIndex{1}$$
Verify the principle of superposition for the heat equation.
Superposition also preserves some of the side conditions. In particular, if $$u_1$$ and $$u_2$$ are solutions that satisfy $$u(0,t)=0$$ and $$u_(L,t)=0$$, and $$c_1,\: c_2$$ are constants, then $$u= c_1u_1+c_2u_2$$ is still a solution that satisfies $$u(0,t)=0$$ and$$u_(L,t)=0$$. Similarly for the side conditions $$u_x(0,t)=0$$ and $$u_x(L,t)=0$$. In general, superposition preserves all homogeneous side conditions.
The method of separation of variables is to try to find solutions that are sums or products of functions of one variable. For example, for the heat equation, we try to find solutions of the form
$u(x,t)=X(x)T(t).$
That the desired solution we are looking for is of this form is too much to hope for. What is perfectly reasonable to ask, however, is to find enough “building-block” solutions of the form $$u(x,t)=X(x)T(t)$$ using this procedure so that the desired solution to the PDE is somehow constructed from these building blocks by the use of superposition.
Let us try to solve the heat equation
$u_t=ku_{xx} \quad\text{with}\quad u(0,t)=0,\quad u(L,t)=0, \quad\text{and}\quad u(x,0)=f(x).$
Let us guess $$u(x,t)=X(x)T(t)$$. We will try to make this guess satisfy the differential equation, $$u_{t}=ku_{xx}$$, and the homogeneous side conditions, $$u(0,t)=0$$ and $$u(L,t)=0$$. Then, as superposition preserves the differential equation and the homogeneous side conditions, we will try to build up a solution from these building blocks to solve the nonhomogeneous initial condition $$u(x,0)=f(x)$$.
First we plug $$u(x,t)=X(x)T(t)$$ into the heat equation to obtain
$X(x)T'(t)=kX''(x)T(t).$
We rewrite as $\frac{T'(t)}{kT(t)}= \frac{X''(x)}{X(x)}.$
This equation must hold for all $$x$$ and all $$t$$. But the left hand side does not depend on $$x$$ and the right hand side does not depend on $$t$$. Hence, each side must be a constant. Let us call this constant $$- \lambda$$ (the minus sign is for convenience later). We obtain the two equations
$\frac{T'(t)}{kT(t)}= - \lambda = \frac{X''(x)}{X(x)}.$
In other words
\begin{align}\begin{aligned} X''(x) + \lambda X(x) &=0, \\ T'(t) + \lambda k T(t)& =0.\end{aligned}\end{align}
The boundary condition $$u(0,t)=0$$ implies $$X(0)T(t)=0$$. We are looking for a nontrivial solution and so we can assume that $$T(t)$$ is not identically zero. Hence $$X(0)=0$$. Similarly, $$u(L,t)=0$$ implies $$X(L)=0$$. We are looking for nontrivial solutions $$X$$ of the eigenvalue problem $$X'' + \lambda X = 0, X(0)=0, X(L)=0$$. We have previously found that the only eigenvalues are $$\lambda_n = \frac{n^2 \pi^2}{L^2}$$, for integers $$n \geq 1$$, where eigenfunctions are $$\sin \left( \frac{n \pi}{L}x \right)$$. Hence, let us pick the solutions
$X_n(x)= \sin \left( \frac{n \pi}{L}x \right).$
The corresponding $$T_n$$ must satisfy the equation
$T'_n(t) + \frac{n^2 \pi^2}{L^2}kT_n(t)=0.$
By the method of integrating factor, the solution of this problem is
$T_n(t)=e^{\frac{-n^2 \pi^2}{L^2}kt}.$
It will be useful to note that $$T_n(0)=1$$. Our building-block solutions are
$u_n(x,t)=X_n(x)T_n(t)= \sin \left( \frac{n \pi}{L}x \right) e^{\frac{-n^2 \pi^2}{L^2}kt}.$
We note that $$u_n(x,0)= \sin \left( \frac{n \pi}{L}x \right)$$. Let us write $$f(x)$$ as the sine series
$f(x)= \sum_{n=1}^{\infty} b_n \sin \left( \frac{n \pi}{L}x \right).$
That is, we find the Fourier series of the odd periodic extension of $$f(x)$$. We used the sine series as it corresponds to the eigenvalue problem for $$X(x)$$ above. Finally, we use superposition to write the solution as
$u(x,t)= \sum^{\infty}_{n=1}b_n u_n (x,t)= \sum^{\infty}_{n=1}b_n \sin \left(\frac{n \pi}{L}x \right)e^{\frac{-n^2 \pi^2}{L^2}kt}.$
Why does this solution work? First note that it is a solution to the heat equation by superposition. It satisfies $$u(0,t)=0$$ and $$u(L,t)=0$$, because $$x=0$$ or $$x=L$$ makes all the sines vanish. Finally, plugging in $$t=0$$, we notice that $$T_n(0)=1$$ and so
$u(x,0)= \sum^{\infty}_{n=1}b_n u_n (x,0)= \sum^{\infty}_{n=1}b_n \sin \left(\frac{n \pi}{L}x \right)=f(x).$
Example $$\PageIndex{1}$$
Suppose that we have an insulated wire of length $$1$$, such that the ends of the wire are embedded in ice (temperature 0). Let $$k=0.003$$. Then suppose that initial heat distribution is $$u(x,0)=50x(1-x)$$. See Figure $$\PageIndex{2}$$.
Figure $$\PageIndex{2}$$: Initial distribution of temperature in the wire.
We want to find the temperature function $$u(x,t)$$. Let us suppose we also want to find when (at what $$t$$) does the maximum temperature in the wire drop to one half of the initial maximum of $$12.5$$.
We are solving the following PDE problem:
\begin{align}\begin{aligned} u_t &=0.003u_{xx}, \\ u(0,t) &= u(1,t)=0, \\ u(x,0) &= 50x(1-x) ~~~~ {\rm{for~}} 0<x<1.\end{aligned}\end{align}
We write $$f(x)=50x(1-x)$$ for $$0<x<1$$ as a sine series. That is, $$f(x)= \sum^{\infty}_{n=1}b_n \sin(n \pi x)$$, where
$b_n= 2 \int^1_0 50x(1-x) \sin(n \pi x)dx = \frac{200}{\pi^3 n^3}-\frac{200(-1)^n}{\pi^3 n^3}= \left\{ \begin{array}{cc} 0 & {\rm{if~}} n {\rm{~even,}} \\ \frac{400}{\pi^3 n^3} & {\rm{if~}} n {\rm{~odd.}} \end{array} \right.$
Figure $$\PageIndex{3}$$: Plot of the temperature of the wire at position $$x$$ at time $$t$$.
The solution $$u(x,t)$$, plotted in Figure $$\PageIndex{3}$$ for $$0 \leq t \leq 100$$, is given by the series:
$u(x,t)= \sum^{\infty}_{\underset{n~ {\rm{odd}} }{n=1}} \frac{400}{\pi^3 n^3} \sin(n \pi x) e^{-n^2 \pi^2 0.003t}.$
Finally, let us answer the question about the maximum temperature. It is relatively easy to see that the maximum temperature will always be at $$x=0.5$$, in the middle of the wire. The plot of $$u(x,t)$$ confirms this intuition.
If we plug in $$x=0.5$$ we get
$u(0.5,t)= \sum^{\infty}_{\underset{n~ {\rm{odd}} }{n=1}} \frac{400}{\pi^3 n^3} \sin(n \pi 0.5) e^{-n^2 \pi^2 0.003t}.$
For $$n=3$$ and higher (remember $$n$$ is only odd), the terms of the series are insignificant compared to the first term. The first term in the series is already a very good approximation of the function. Hence
$u(0.5,t) \approx \frac{400}{\pi^3}e^{-\pi^2 0.003t}.$
The approximation gets better and better as $$t$$ gets larger as the other terms decay much faster. Let us plot the function $$0.5,t$$, the temperature at the midpoint of the wire at time $$t$$, in Figure $$\PageIndex{4}$$. The figure also plots the approximation by the first term.
Figure $$\PageIndex{4}$$: Temperature at the midpoint of the wire (the bottom curve), and the approximation of this temperature by using only the first term in the series (top curve).
After $$t=5$$ or so it would be hard to tell the difference between the first term of the series for $$u(x,t)$$ and the real solution $$u(x,t)$$. This behavior is a general feature of solving the heat equation. If you are interested in behavior for large enough $$t$$, only the first one or two terms may be necessary.
Let us get back to the question of when is the maximum temperature one half of the initial maximum temperature. That is, when is the temperature at the midpoint $$12.5/2=6.25$$. We notice on the graph that if we use the approximation by the first term we will be close enough. We solve
$6.25=\frac{400}{\pi^3}e^{-\pi^2 0.003t}.$
That is,
$t=\frac{\ln{\frac{6.25 \pi^3}{400}}}{-\pi^2 0.003} \approx 24.5.$
So the maximum temperature drops to half at about $$t=24.5$$.
We mention an interesting behavior of the solution to the heat equation. The heat equation “smoothes” out the function $$f(x)$$ as $$t$$ grows. For a fixed $$t$$, the solution is a Fourier series with coefficients $$b_n e^{\frac{-n^2 \pi^2}{L^2}kt}$$. If $$t>0$$, then these coefficients go to zero faster than any $$\frac{1}{n^P}$$ for any power $$p$$. In other words, the Fourier series has infinitely many derivatives everywhere. Thus even if the function $$f(x)$$ has jumps and corners, then for a fixed $$t>0$$, the solution $$u(x,t)$$ as a function of $$x$$ is as smooth as we want it to be.
Example $$\PageIndex{2}$$
When the initial condition is already a sine series, then there is no need to compute anything, you just need to plug in. Consider $u_t = 0.3 \, u_{xx}, \qquad u(0,t)=u(1,t)=0, \qquad u(x,0) = 0.1 \sin(\pi t) + \sin(2\pi t) .$ The solution is then $u(x,t) = 0.1 \sin(\pi t) e^{- 0.3 \pi^2 t} + \sin(2 \pi t) e^{- 1.2 \pi^2 t} .$
## 4.6.3 Insulated Ends
Now suppose the ends of the wire are insulated. In this case, we are solving the equation
$u_t=ku_{xx}\quad\text{with}\quad u_x(0,t)=0,\quad u_x(L,t)=0,\quad\text{and}\quad u(x,0)=f(x).$
Yet again we try a solution of the form $$u(x,t)=X(x)T(t)$$. By the same procedure as before we plug into the heat equation and arrive at the following two equations
\begin{align}\begin{aligned} X''(x)+\lambda X(x) &=0, \\ T'(t)+\lambda kT(t) &=0.\end{aligned}\end{align}
At this point the story changes slightly. The boundary condition $$u_x(0,t)=0$$ implies $$X'(0)T(t)=0$$. Hence $$X'(0)=0$$. Similarly, $$u_x(L,t)=0$$ implies $$X'(L)=0$$. We are looking for nontrivial solutions $$X$$ of the eigenvalue problem $$X''+ \lambda X=0,$$ $$X'(0)=0,$$ $$X'(L)=0,$$. We have previously found that the only eigenvalues are $$\lambda_n=\frac{n^2 \pi^2}{L^2}$$, for integers $$n \geq 0$$, where eigenfunctions are $$\cos(\frac{n \pi}{L})X$$ (we include the constant eigenfunction). Hence, let us pick solutions
$X_n(x)= \cos(\frac{n \pi}{L}x)\quad\text{and}\quad X_0(x)=1.$
The corresponding $$T_n$$ must satisfy the equation
$T'_n(t)+ \frac{n^2 \pi^2}{L^2}kT_n(t)=0.$
For $$n \geq 1$$, as before,
$T_n(t)= e^{\frac{-n^2 \pi^2}{L^2}kt}.$
For $$n=0$$, we have $$T'_0(t)=0$$ and hence $$T_0(t)=1$$. Our building-block solutions will be
$u_n(x,t)=X_n(x)T_n(t)= \cos \left( \frac{n \pi}{L} x \right) e^{\frac{-n^2 \pi^2}{L^2}kt},$
and
$u_0(x,t)=1.$
We note that $$u_n(x,0) =\cos \left( \frac{n \pi}{L} x \right)$$. Let us write $$f$$ using the cosine series
$f(x)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n \cos \left( \frac{n \pi}{L} x \right).$
That is, we find the Fourier series of the even periodic extension of $$f(x)$$.
We use superposition to write the solution as
$u(x,t)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n u_n(x,t)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n \cos \left( \frac{n \pi}{L} x \right) e^{\frac{-n^2 \pi^2}{L^2}kt}.$
Example $$\PageIndex{3}$$
Let us try the same equation as before, but for insulated ends. We are solving the following PDE problem
\begin{align}\begin{aligned} u_t &=0.003u_{xx}, \\ u_x(0,t) &= u_x(1,t)=0, \\ u(x,0) &= 50x(1-x) ~~~~ {\rm{for~}} 0<x<1.\end{aligned}\end{align}
For this problem, we must find the cosine series of $$u(x,0)$$. For $$0<x<1$$ we have
$50x(1-x)=\frac{25}{3}+\sum^{\infty}_{\underset{n~ {\rm{even}} }{n=2}} \left( \frac{-200}{\pi^2 n^2} \right) \cos(n \pi x).$
The calculation is left to the reader. Hence, the solution to the PDE problem, plotted in Figure $$\PageIndex{5}$$, is given by the series
$u(x,t)=\frac{25}{3}+\sum^{\infty}_{\underset{n~ {\rm{even}} }{n=2}} \left( \frac{-200}{\pi^2 n^2} \right) \cos(n \pi x) e^{-n^2 \pi^2 0.003t}.$
Figure $$\PageIndex{5}$$: Plot of the temperature of the insulated wire at position $$x$$ at time $$t$$.
Note in the graph that the temperature evens out across the wire. Eventually, all the terms except the constant die out, and you will be left with a uniform temperature of $$\frac{25}{3} \approx{8.33}$$ along the entire length of the wire.
Let us expand on the last point. The constant term in the series is $\frac{a_0}{2} = \frac{1}{L} \int_0^L f(x) \, dx .$ In other words, $$\frac{a_0}{2}$$ is the average value of $$f(x)$$, that is, the average of the initial temperature. As the wire is insulated everywhere, no heat can get out, no heat can get in. So the temperature tries to distribute evenly over time, and the average temperature must always be the same, in particular it is always $$\frac{a_0}{2}$$. As time goes to infinity, the temperature goes to the constant $$\frac{a_0}{2}$$ everywhere.
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# Math
posted by .
Rachel gave 1/4 of a pie to each of 5 friends. How many pies did she give to her friends in all? Write the mixed number.
• Math -
1/4 times 5 = 1/4 * 5/1 = 5/4, The improper fraction. The mixed number is 1 1/4.
• Math -
chapter 7 write a mixed number and a fraction greater than 1 that name the pat filled
• Math -
1 1/4
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## Why Is X The Unknown?
Terry Moore explains that the inventors of algebra were Arabic scientists. Their definition:
X = something, some undefined unknown thing = the unknown thing.
When the books were translated from Arabic to Spanish the Arabic unknown was “SH” that did not exist inSpanish. They replaced it by “X“. It migrated to other European languages 600 years ago.
X is “X” because Spanish language does not have “SH“.
## Selfie Math
Jim is going on a tour around the world. He has 5 tops, 4 bottoms, and 3 pairs of footwear. He wants to post a selfie everyday to show different places and his different outfits to his girl-friend Mary, who stays at home. He does not want to wear the same outfit twice. ## For how many days does he have enough clothing?
Leslie Green gives the answer : 96 days.
Can you explain why it is 96? What is logic behind?
## Guesstimate Methods
Guesstimate is an estimate based on a mixture of guesswork and calculation.
“Everything should be made as simple as possible, but not simpler.” – Albert Einstein
“The purpose of computing is insight, not numbers.” – Richard Hamming
## Famous Math and Logic Paradoxes
Math and Logic are full of paradoxes.
1. Achilles and the Tortoise The Paradox of Achilles and the Tortoise was described by the Greek philosopher Zeno of Elea in the 5th century BC. The great hero Achilles challenges a tortoise to a footrace. He agrees to give the tortoise a head start of 100m. When the race begins, Achilles starts running, so that by the time he has reached the 100m mark, the tortoise has only walked 10m. But by the time Achilles has reached the 110m mark, the tortoise has walked another 1m. By the time he has reached the 111m mark, the tortoise has walked another 0.1m, then 0.01m, then 0.001m, and so on. The tortoise always moves forwards while Achilles always plays catch up. Why is Achilles always behind the tortoise?
2. Bermuda Triangle Paradox. Why the sum of the interior angles of the Bermuda triangle is not 180 degrees?
3. Simpson Paradox. The average score for dance of boys and girls in class A are 16 and 21, respectively. The average score of boys and girls in class B are 15 and 20, respectively. Twenty percent of class A students are girls. Forty percent of class B students are girls. Which class has a higher average score?
4. Braess paradox. The diagram shows a road network. All cars drive in one direction from A to F. The numbers represent the maximum flow rate in vehicles per hour. Engineers want to construct a new road with a flow rate of 100 vehicles per hour. Drivers randomly choose the road at crossroads. What new road decreases the capacity of the network (the number of vehicles at point F)?
5. Barber Paradox In a city, the barber is the ‘one who shaves all those, and those only, who do not shave themselves.’
Who shaves the barber?
6. The Two Envelope Paradox. One envelope has twice as much money as the second one. Gerry does not know which envelope contains the larger amount. He takes one of the envelopes, counts the money, and is offered the chance to switch the envelope. He thinks “If the amount of money in the chosen envelope is X dollars, then the other envelope contains either 2X of 0.5X dollars, with equal probability of 0.5. The expected value of switching is 0.5 (2X) + 0.5 (0.5X) = 1.25X. This is greater than the value in the initially chosen envelope. It is better to switch.” What is your advice?
7. Potato Paradox. I have 100kg of potatoes, which are 99 percent water. I dry them until they are 98 percent water. How much do they weigh now?
8. Leonard Euler’s Paradox. Why the average of all of the numbers is not a zero?
1, -1, 2, -2, 3, -3, . . .
10. Uninteresting Number Paradox. How many uninteresting numbers exist?
11. Gabriel’ Horn Paradox. The shape obtained from rotating the equation about x-axis resembles a trumpet. If we need an infinite volume of paint to paint the infinite horn, how much paint does the horn can contain inside itself?
12. Pop Quiz Paradox. A teacher announces that there will be a quiz one day during the next week. The teacher gives the definition that they would not when they come in to the class that the quiz was going to be given that day. The brightest student says that the quiz cannot be on Friday because they will know the day. With the same technique, she eliminates Thursday, Wednesday, Tuesday, and Monday. “You cannot give us a pop quiz next week” she says. When does the teacher give the pop quiz? I know the paradox from Charles Carter Wald. Probably, Martin Gardner described it for the first time in The Colossal Book of Mathematics.
1. Achilles and the Tortoise
## Car Owner’s Puzzles
Can a specific subject provokes interest to Math?
For example, cars. How much does the gas cost in a month? What speed to choose? When do I need to change the tires? How to be on time? How much do I pay for a car during its life? . . .
What are other intersting subjects? Finance? Love? Sport?
## Fermi Problem
Often, a problem solver might estimate the required value without very much information. He/she might aim to get an order of magnitude estimate. The estimation technique is named after physicist Enrico Fermi, as he was known for his ability to make good approximate calculations with little or no actual data. Fermi problems typically involve making justified guesses about quantities and their variance or lower and upper bounds.
The classic Fermi problem, generally attributed to Fermi, is “How many piano tuners are there in Chicago?” – Wikipedia
Several simple examples of Aplusclick Fermi problems:
Tennis Balls in a Bus
English Language Words
Demographics
Security Control
## Dream Math
Everybody dreams.
The boy dreams of being an astronaut.
How long does he need to study and work until his dream can become reality?
Which job does not require math?
The photograph courtesy of Roland Sauter
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# how to absolute value
Read about how to absolute value, The latest news, videos, and discussion topics about how to absolute value from alibabacloud.com
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# 1027899 (number)
1,027,899 (one million twenty-seven thousand eight hundred ninety-nine) is an odd seven-digits composite number following 1027898 and preceding 1027900. In scientific notation, it is written as 1.027899 × 106. The sum of its digits is 36. It has a total of 4 prime factors and 12 positive divisors. There are 680,400 positive integers (up to 1027899) that are relatively prime to 1027899.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 36
• Digital Root 9
## Name
Short name 1 million 27 thousand 899 one million twenty-seven thousand eight hundred ninety-nine
## Notation
Scientific notation 1.027899 × 106 1.027899 × 106
## Prime Factorization of 1027899
Prime Factorization 32 × 181 × 631
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 342633 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,027,899 is 32 × 181 × 631. Since it has a total of 4 prime factors, 1,027,899 is a composite number.
## Divisors of 1027899
12 divisors
Even divisors 0 12 6 6
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 1.49531e+06 Sum of all the positive divisors of n s(n) 467413 Sum of the proper positive divisors of n A(n) 124609 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1013.85 Returns the nth root of the product of n divisors H(n) 8.24897 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,027,899 can be divided by 12 positive divisors (out of which 0 are even, and 12 are odd). The sum of these divisors (counting 1,027,899) is 1,495,312, the average is 1,246,09.,333.
## Other Arithmetic Functions (n = 1027899)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 680400 Total number of positive integers not greater than n that are coprime to n λ(n) 1260 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 80357 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 680,400 positive integers (less than 1,027,899) that are coprime with 1,027,899. And there are approximately 80,357 prime numbers less than or equal to 1,027,899.
## Divisibility of 1027899
m n mod m 2 3 4 5 6 7 8 9 1 0 3 4 3 5 3 0
The number 1,027,899 is divisible by 3 and 9.
• Deficient
• Polite
## Base conversion (1027899)
Base System Value
2 Binary 11111010111100111011
3 Ternary 1221020000100
4 Quaternary 3322330323
5 Quinary 230343044
6 Senary 34010443
8 Octal 3727473
10 Decimal 1027899
12 Duodecimal 416a23
20 Vigesimal 689ej
36 Base36 m14r
## Basic calculations (n = 1027899)
### Multiplication
n×y
n×2 2055798 3083697 4111596 5139495
### Division
n÷y
n÷2 513950 342633 256975 205580
### Exponentiation
ny
n2 1056576354201 1086053777906853699 1116353592256677010348401 1147498741127046042260111039499
### Nth Root
y√n
2√n 1013.85 100.921 31.8411 15.9364
## 1027899 as geometric shapes
### Circle
Diameter 2.0558e+06 6.45848e+06 3.31933e+12
### Sphere
Volume 4.54925e+18 1.32773e+13 6.45848e+06
### Square
Length = n
Perimeter 4.1116e+06 1.05658e+12 1.45367e+06
### Cube
Length = n
Surface area 6.33946e+12 1.08605e+18 1.78037e+06
### Equilateral Triangle
Length = n
Perimeter 3.0837e+06 4.57511e+11 890187
### Triangular Pyramid
Length = n
Surface area 1.83004e+12 1.27993e+17 839276
## Cryptographic Hash Functions
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https://www.ncertbooksolutions.com/mcqs-for-ncert-class-12-mathematics-chapter-9-differential-equations/
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# MCQs For NCERT Class 12 Mathematics Chapter 9 Differential Equations
Please refer to the MCQ Questions for Class 12 Mathematics Chapter 9 Differential Equations with Answers. The following Differential Equations Class 12 Mathematics MCQ Questions have been designed based on the latest syllabus and examination pattern for Class 12. Our experts have designed MCQ Questions for Class 12 Mathematics with Answers for all chapters in your NCERT Class 12 Mathematics book.
## Differential Equations Class 12 MCQ Questions with Answers
See below Differential Equations Class 12 Mathematics MCQ Questions, solve the questions and compare your answers with the solutions provided below.
Question. The degree of the differential equation
(a) 4
(b) 3/2
(c) not defined
(d) 2
D
Question. The differential equation for y = A cos a x + B sin a x, where A and B are arbitrary constants is
B
Question. The order of the differential equation of all circles of radiusr, having centre on y-axis and passing through the origin, is
(a) 1
(b) 2
(c) 3
(d) 4
A
Question. The differential equation whose solution is a circle centred at (h ,k) and radius a is (where, a is a constant)
B
Question.
B
Question. The solution of differential equation
B
Question. The solution of the differential equation dy/dx-x tan(y-x) ) 1is
C
Question. The solution of(x+ y+1) dy= dx
A
Question. The equation of the curve passing through the origin and satisfying the differential equation
C
Question.
A
Question. The solution of the differential equation x(dy/dx)+y=y2log x is
(a) 1/y = log +1+Cx
(b) y = logx+1+Cx
(c) 1/y=log x -+C/x
(d) None of these
A
Question.
(a) touch each other
(b) are orthogonal
(c) are one and the same
(d) None of these
B
Question. Order and degree of the differential equation
(a) order = 2, degree = 1
(b) order = 2, degree = 4
(c) order = 1, degree = 4
(d) order = 1, degree= 1
A
Question. The number of arbitrary constants in the general solution of a differential equation of fourth order is
(a) zero
(b) 2
(c) 3
(d) 4
D
Question. Order and degree of the differential equation
(a) order = 1, degree = 1
(b) order = 1, degree = not defined
(c) order = 2, degree = 1
(d) order = 2, degree = not define
Question. Order and degree of the differential equation
(a) order = 4, degree = 1
(b) order = 3, degree= 1
(c) order = 4, degree = 0
(d) order = 4, degree = not defined
D
Question. Family y =Ax+ Aof curves is represented by the differential equation of degree
(a) 1
(b) 2
(c) 3
(d) 4
C
Question. The degree of the differential equation to of all tangent lines to the parabola y2=4ax is
(a) 1
(b) 2
(c) 3
(d) 4
B
Question. The number of arbitrary constants in the particular solution of a differential equation of third order is
(a) 3
(b) 2
(c) 1
(d) zero
D
Question. The order of the differential equation whose general
(a) 5
(b) 6
(c) 3
(d) 2
C
Question. The order of the differential equation whose general solution is given by
is
(a) 5
(b) 4
(c) 3
(d) 2
C
Question. The differential equation of the curve y2=a(b2-x2)///10 representing the given family of curves, where a and b are constants, is
B
Question. The differential equation of the family of parabolas having vertex at origin and axis along positive y-axis is
C
Question. The differential equation of the curve y=ex(a coa x +b sin x) representing the given family of curves, where a and b are constants, is
A
Question. The differential equation of the family of hyperbolas having foci on x-axis and centre at origin is
A
Question. Which of the following differential equation has
B
Question. The differential equation of the family of circles in
the first quadrant which touch the coordinate axes is
B
Question. The degree of the differential equation
(a) 1
(b) 2
(c) 3
(d) not defined
D
Question.
(a) y sec x = tan x + C
(b) y tan x = sec x + C
(c) tan x = y tan x + C
(d) x sec x = tan y + C
A
Question. The solution of differential equation
(a) x (y + cos x) = sin x + C
(b) x (y – cos x) = sin x + C
(c) xy cos x = sin x + C
(d) x (y + cos x) = cos x + C
A
Question. The solution of differential equation xdy – ydx = 0 represents
(a) a rectangular hyperbola
(b) parabola whose vertex is at origin
(c) straight line passing through origin
(d) a circle whose centre is at origin
C
Question. If y = e-x ( A cos x + B sin x), then y is a solution of
C
Question. The solution of differential equation tan ysec2 xdx + tan x sec2 ydy = 0 is
D
Question. The integrating factor of differential equation cos x (dy/dx) y sin x = 1 is
(a) cos x
(b) tan x
(c) sec x
(d) sin x
C
Question. The order and degree of the differential equation
(a) 2 and 4
(b) 2 and 2
(c) 2 and 3
(d) 3 and 3
A
Question. The family y = Ax + A3 of curves is represented by differential equation of degree
(a) 1
(b) 2
(c) 3
(d) 4
A
Question. The solution of (dy/dx) – y = 1, (0) = 1 is given by
(a) xy = -ex
(b) xy = – e-x
(c) xy = -1
(d) y = 2 ex -1
B
Question. tan tan-1 x + tan-1 y = C is general solution of the differential equation
C
Question.
(a) none
(b) one
(c) two
(d) infinite
B
Question. Which of the following is a second order differential equation?
(a) (y ‘)2 + x = y2
(b) y ‘y ” + y = sin x
(c) y ”’ + (y “) + y = 2 0
(d) y ‘ = y2
B
Question. The integrating factor of differential equation
C
Question. The integrating factor of
(a) x (
b) log x
(c) 1/x
(d) – x
C
Question. The differential equation y (dy/dx) + x = c represents
(a) family of hyperbolas
(b) family of parabolas
(c) family of ellipses
(d) family of circles
D
Question. The solution of (dy/dx) + y = e-x, y(0) = 0 is
(a) y = ex (x – 1)
(b) y = xe-x
(c) y = xe-x + 1
(d) y = (x + 1) e-x
B
Question. Which of the following is the general solution of
(a) y = (Ax + B) ex
(b) y = (Ax + B) e-x
(c) y = Aex + Be-x
(d) y = A cos x + B sin x
A
Question. The general solution of differential equation (ex + 1) ydy = (y + 1) ex dx is
C
Question. The integrating factor of differential equation dy/dx + y tan x – sec x = 0 is
(a) cos x
(b) sec x
(c) ecos x
(d) esec x
B
Question. The general solution of ex cos ydx – ex sin ydy = 0 is
(a) ex cos y = k
(b) ex sin y = k
(c) ex = k cos y
(d) ex = k sin y
A
Question. The degree of differential equation
(a) 1
(b) 2
(c) 3
(d) 5
A
Question. The solution of differential equation
(a) y = x tan-1 x
(b) y – x = k(1+ xy)
(c) x = y tan-1 y
(d) tan(xy) = k
B
Question. The solution of differential equation cos x sin ydx + sin x cos ydy = 0 is
B
Question.
A
Question. The differential equation of the family of curves x2 + y2 – 2ay = 0, where a is arbitrary constant, is
A
Question. The family Y = Ax + A3 of curves will correspond to a differential equation of order
(a) 3
(b) 2
(c) 1
(d) not defined
C
Question. The integrating factor of differential equation
(a) x/ex
(b) ex/x
(c) xex
(d) ex
B
Question. y = aemx + be-mx satisfies which of the following differential equation?
C
Question.
(a) ex2 – y = C
(b) e-y + ex2 = C
(c) ey = ex2 + C
(d) ex2 + y = C
C
Question. The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the point is
(a) an ellipse
(b) parabola
(c) circle
(d) rectangular hyperbola
D
Question. The general solution of differential equation
(a) y = Ce-x2/2
(b) y = Cex2/2
(c) y = (x + c) ex2/2
(d) y = (c – x)ex2/2
C
Question. The solution of equation (2y – 1) dx – (2x + 3) dy = 0 is
C
Question. The differential equation for which y = a cos x + b sin x is a solution, is
A
Question.
(a) y = e-x (x – 1)
(b) y = xe-x + 1
(c) y = xex
(d) y = xe-x
D
Question. The solution of differential equation dy/dx = ex-y x2 e-y is
B
Question. The solution of differential equation
A
Question. The order and degree of differential equation
(a) 1, 4
(b) 3, 4
(c) 2, 4
(d) 3, 2
D
Question. The order and degree of differential equation
(a) 2(3/2)
(b) 2, 3
(c) 2, 1
(d) 3, 4
C
Question. The differential equation of family of curves y2 = 4a (x + a) is
D
State True or False for the following
Question. Number of arbitrary constants in the particular solution of a differential equation of order two is two.
False
Question. The differential equation representing the family of circles x2 + (y – a)2 = a2 will be of order two.
False
Question.
True
Question. Correct substitution for the solution of the differential equation of the type dy/dx = where f (x, y) is a homogeneous function of zero degree is y = vx.
True
Question. Correct substitution for the solution of the differential equation of the type dy/dx = g (x, y) where g (x, y) is a homogeneous function of the degree zero is x = vy.
True
Question. Integrating factor of the differential of the form dy/dx + px = 01 is given by e∫p1/dy
True
Question. The differential equation of all non horizontal lines in a plane is
True
Question. Differential equation representing the family of curves y = ex (A cos x + B sin x) is
True
Question. Solution of the differential equation of the type dx/dy + p1x = 01 is given by x · IF = ∫ (IF) x Q1 dy.
True
Question.
| 2,875 | 9,089 |
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| 3.65625 | 4 |
CC-MAIN-2024-18
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https://brainmass.com/statistics/analysis-of-variance/statistics-one-way-anova-unequal-observations-one-factor-112368
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| 429,946,509 | 18,549 |
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# Statistics One way Anova - unequal observations- One Factor
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
The table below gives the number of miles to the gallon obtained by similar automobiles using 5 different brands of gasoline. Test at the (a) 0.05 level (b) 0.01 level of significance whether there is any significant difference in brands.
Brand Miles
Brand A 12 15 14 11 15
Brand B 14 12 15
Brand C 11 12 10 14
Brand D 15 18 16 17 14
Brand E 10 12 14 12
https://brainmass.com/statistics/analysis-of-variance/statistics-one-way-anova-unequal-observations-one-factor-112368
#### Solution Preview
Mean of brand A = 13.4
Mean of brand B = 13.667
Mean of brand C = 11.75
Mean of brand D = 16
Mean of brand E = 12
SSA = ...
#### Solution Summary
The solution to one way Anova problem using unequal observations is given step by step with explanation so that the student could use this solution as a model solution to solve other problems of the same type.
\$2.19
| 287 | 1,066 |
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| 3.9375 | 4 |
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https://achildsguideto.com/3dshapes.shtml
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# Three Dimensional Shapes
Three dimensional shapes involve questions on volume and surface area but are often mixed in with questions on bounds and algebra as well.
Often, these type of questions tend to be towards the end of the exam as they invlove amalgamating knowledge from different aspects of the curriculum.
They are challenging and fun to answer though.
## How to calculate the edges, surface area and volume of a Cuboid
This booklet was written to help one of the children in my class with the homework I had set them. I tried to write it in a way that an 11 year old could understand it. He was able to complete the homework which was similar to the cuboid investigations on these pages.
## Find the Surface Area and Volume of Cuboids
Begin with calculating the volume and surface area of a cuboid. Move from numerical form to algebraic expressions. Find the lengths of sides. A good way of practising expanding two binomials and then three binomials.
## Find the Surface Area and Volume of Cuboids
A PowerPoint presentation providing an introduction to finding the surface area and volume of a triangular prism.
## Volume of a Pyramid
Calculate the area of the base of the pyramid and then calculate the volume using the formula shown.
## Find the diagonal of a prism
Find the diagonal of a prism using quadratic equations. An example given and then some work to have a go at yourself.
## Orthographic and Isometric Elevations
Transfer information from an orthographic elevation to an isometric one and then do some work on algebraic expressions.
| 318 | 1,574 |
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https://brighterly.com/math/pi/
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# Value of Pi – Symbol, Definition With Examples
Welcome to another engaging and enlightening topic brought to you by Brighterly, your reliable source for child-friendly mathematics learning. Today, we’re delving into the intriguing world of Pi, a constant that holds a unique place in the realm of mathematics and has been captivating inquisitive minds for thousands of years.
Pi, denoted by the Greek letter “π”, represents the mathematical ratio of a circle’s circumference to its diameter. This number remains constant, regardless of the circle’s size. Additionally, Pi is an irrational number, indicating that its decimal representation is non-repeating and infinite. The magic of Pi extends far beyond basic geometry, reaching into the fields of trigonometry, statistics, physics, and even into the beauty of art and nature.
## What is Pi?
Pi (π) is an integral part of mathematical science that kids often encounter in their mathematics journey. Named after the Greek letter “π”, this enigmatic number has captured the fascination of young minds and scholars alike throughout the ages. Pi represents a mathematical constant that is the ratio of a circle’s circumference to its diameter. It is an irrational number, meaning it cannot be represented as a simple fraction, and its decimal representation never ends nor repeats.
## Symbol and Definition of Pi
The symbol for Pi is the Greek letter π, and it’s defined as the ratio of the circumference of any circle to its diameter. This means if you measure the circumference and diameter of any circle, regardless of its size, you’ll always get the same number – that is Pi (π). The definition implies that no matter how large or small a circle is, the ratio of its circumference to its diameter remains constant at approximately 3.14159.
## Historical Understanding of Pi
Understanding the history of Pi provides a deep insight into the evolution of mathematics. Ancient civilizations such as the Egyptians and Babylonians had approximations for Pi dating back to 2000 BC. The Greek mathematician Archimedes was one of the first to calculate a more accurate approximation of Pi, around 287–212 BC. With the advent of computers, millions of decimal places of Pi have been calculated, showcasing the advancements in technology and mathematical understanding.
## Mathematical Properties of Pi
Pi is unique because it’s an irrational and transcendental number. Irrational means it can’t be written as a fraction, and transcendental means it’s not a solution to any non-constant polynomial equation with rational coefficients. These properties make Pi a fascinating topic of study in the field of number theory.
## Decimal Representation of Pi
The decimal representation of Pi is an infinite non-repeating sequence starting with 3.14159. Despite the infinite nature of its decimal representation, the value of Pi is commonly approximated as 3.14 or 22/7 for simpler calculations.
## The Importance of Pi in Mathematics
Pi plays a critical role in many areas of mathematics and science including geometry, trigonometry, physics, and even probability theory. It’s an essential part of formulas involving circles, spheres, harmonic motion, wave mechanics, and more. The universality of Pi is one of the reasons it’s so important in mathematics.
## Difference Between Pi and Other Mathematical Constants
Pi, like e (Euler’s number) and Φ (the Golden Ratio), is a mathematical constant, but it holds unique properties. While constants like e appear in exponential growth or decay models, and Φ is seen in nature’s proportions, Pi is intrinsically linked to anything involving circles and rotation, emphasizing its unique position in mathematics.
## Formulas Involving Pi
Pi is crucial in a multitude of mathematical formulas. The area of a circle (A = πr²), the circumference of a circle (C = 2πr), and the volume of a sphere (V = 4/3 πr³) are some common examples. In the realm of trigonometry, the formula for the unit circle (x² + y² = r²) utilizes Pito define the radians.
## Area and Circumference of a Circle
Two of the most fundamental applications of Pi are in finding the area and circumference of a circle. To calculate the circumference of a circle (the distance around the circle), we use the formula C = 2πr, where ‘r’ is the radius of the circle. When we want to calculate the area enclosed by the circle, we use the formula A = πr².
## Volume and Surface Area of a Sphere
Moving into three dimensions, Pi is instrumental in calculations related to spheres. The volume of a sphere is given by V = 4/3 πr³ and the surface area of a sphere is A = 4πr². These formulas are crucial in a variety of scientific fields including physics, engineering, and computer graphics.
## Trigonometric Functions Involving Pi
In trigonometry, Pi is used to denote angles in radians. For example, 180 degrees is equivalent to π radians, and 360 degrees is equivalent to 2π radians. This is especially useful in trigonometric functions like sine, cosine, and tangent, which are often defined and calculated in terms of Pi.
## Estimating the Value of Pi
There are several methods for estimating the value of Pi, many of which involve geometry or statistics. A simple and fun activity for children is to measure the diameter and circumference of various round objects, and divide the two measurements to get an estimate of Pi. For a more sophisticated approach, one might use the method of Archimedes, which involves inscribing and circumscribing polygons around a circle.
## Memorizing and Reciting Pi
Because Pi is an irrational number, its digits never repeat or terminate. However, for practical purposes, it is often rounded to 3.14 or expressed as the fraction 22/7. Some people enjoy the challenge of memorizing and reciting as many digits of Pi as they can. In fact, the current world record holder has recited over 70,000 digits of Pi from memory!
## Practice Problems on Pi
To build a strong understanding of Pi, practice is essential. Consider the following problems:
1. If the radius of a circle is 5cm, what is its circumference?
2. Given a sphere with a radius of 3m, calculate its surface area.
3. Can you estimate Pi by measuring everyday circular objects?
## Conclusion
Understanding Pi is akin to uncovering a profound secret of the universe that lies hidden in the simple concept of a circle. It reminds us of the universal language that mathematics is, bridging gaps between different disciplines, from geometry to physics. It brings to light the fact that learning is indeed a joyous journey, a view we at Brighterly emphasize and nurture. We hope this excursion into the fascinating world of Pi has left you and your young ones intrigued and excited to explore more about the magnificent language of mathematics.
## Frequently Asked Questions on Pi
### What is Pi?
Pi, represented by the Greek letter “π”, is a mathematical constant. It’s the ratio of a circle’s circumference to its diameter. It’s an irrational number, meaning it can’t be expressed as a simple fraction and its decimal representation is non-ending and non-repeating.
### Why is Pi represented by the Greek letter π?
The symbol for Pi comes from the Greek word ‘perimetros’ meaning ‘circumference’. It was first used by the Welsh mathematician William Jones in 1706, and later popularized by the Swiss mathematician Leonhard Euler.
### Why is Pi important?
Pi is crucial in various fields of mathematics and science. In geometry, it’s used to calculate the area and circumference of circles, as well as the volume and surface area of spheres. In trigonometry and wave physics, Pi is essential to describe oscillations, waves, and rotations.
### How is Pi used in real life?
Pi is used in numerous real-life scenarios. Engineers use Pi in computations involving circular or curved structures. Physicists use Pi to describe wave patterns. In probability and statistics, Pi appears in calculations. Furthermore, Pi even finds use in GPS navigation and digital imaging.
### How can you memorize Pi?
Memorizing Pi is a fun challenge! Since Pi’s decimal representation is non-repeating, it’s a test of memory. You can start with the common approximation, 3.14, or push yourself further. Some people use ‘Piems’, a form of poetry where the length of each word represents a digit of Pi. Remember, there’s no pressure to memorize it – even mathematicians often use just a few decimal places in their calculations.
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# Check my proof that sqrt{prime} is always irrational
• Sep 17th 2009, 12:44 PM
Jameson
Check my proof that sqrt{prime} is always irrational
The proof that $\sqrt{2}$ is irrational is well-known and I never did the proof that extended this to all primes, so I will attempt this now.
---
Assume towards a contradiction that $\sqrt{p} \in \mathbb{Q}$. Thus $\sqrt{p}=\frac{m}{n}$, where $m,n \in \mathbb{N}$ and this fraction is in lowest terms.
By algebra, $p=\frac{m^2}{n^2} \Rightarrow n^2=\frac{m^2}{p}$. If $n^2$ is not an integer, then clearly n is not an integer. This isn't possible by definition of n, so $n^2$ must be an integer.
Given this conclusion, p divided m^2 evenly. This could lead to two possibilities. (1) m=p and (2)m contains a factor of p. Since p is prime the only factor other than 1 is p itself. So m can be written as $kp$, where $k \in \mathbb{N}$. If (1) is the case, then m/n is not in lowest terms by out initial assumption, thus cannot be true. If (2) is the case then $n^2=\frac{k^2p^2}{p}=k^2p$.
Now we look back at our second statement, $p=\frac{m^2}{n^2}$. Substituting in a new expression for n^2 yields $p=\frac{m^2}{k^2p}$. This implies that m and n have a common factor and thus possibility (2) cannot work. Since (1) and (2) are not possible, the initial assumption is false and $\sqrt{p}$ is not in Q.
----
I just read through it and I think it's ok. I'm sure there are ways to shorten it and I'm sure I didn't explain some things well enough. The big thing I think might get criticized is justifying when I say two things have a common factor. Anyway, let me know how I did and how I can improve.
Thanks
• Sep 17th 2009, 01:05 PM
Plato
This is an even more general result.
If p is a non-square positive integer then $\sqrt{p}$ is irrational.
The proof in outline form goes like this.
Suppose otherwise. Then there is a smallest positive integer, K, such $K\sqrt{p}\in \mathbb{Z}^+$.
$0 < \sqrt p - \left\lfloor {\sqrt p } \right\rfloor < 1\, \Rightarrow \,0 < K\sqrt p - K\left\lfloor {\sqrt p } \right\rfloor < K$
But $\left( {K\sqrt p - K\left\lfloor {\sqrt p } \right\rfloor } \right) \in \mathbb{Z}^ + \;\& \,\left(K\sqrt p - {K\left\lfloor {\sqrt p } \right\rfloor } \right)\sqrt p \in \mathbb{Z}^ +$
That violates the minimal nature of K.
• Sep 17th 2009, 01:08 PM
Jameson
Interesting, Plato. Thanks.
Did my proof look ok in general though?
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# SET THEORY. BASIC CONCEPTS IN SET THEORY Definition: A set is a collection of well-defined objects, called elements Examples: The following are examples.
## Presentation on theme: "SET THEORY. BASIC CONCEPTS IN SET THEORY Definition: A set is a collection of well-defined objects, called elements Examples: The following are examples."— Presentation transcript:
SET THEORY
BASIC CONCEPTS IN SET THEORY Definition: A set is a collection of well-defined objects, called elements Examples: The following are examples of sets a.The months of the year that begins with letter J b.The set of engineering drawing instruments c.The set of PB teams in this year’s PBA series. Names of sets are designated by using capital letters such as A, B… To indicate that an element belongs to a given set, the symbol epsilon є is used and the symbol epsilon slash for an element that does not belong to the set.
Example “4 is an element of the set S” (4 є S) “1 is not an element of the set S (1 є S) What are the elements of the set S = {blue, orange yellow, black}? (There are four elements in the S namely blue, orange yellow and black)
Methods of Describing A Set Listing or Roster Method – list the elements, separate them by commas and enclose this list in a pair of braces, { } Describe the given sets by roster method a.If A is the set of the days of the week that begin with the letter S, then A = {Saturday, Sunday} b.If B is the set of letters in the word love, then B = {l, o, v, e} c.The set of digits that make up the telephone number 715-51-70 {0, 1, 5, 7} Note: the elements within the set are never repeated and the elements can appear in any order
Methods of Describing A Set Set in the Set-Builder Notation {x| x is a natural number between 2 and 10} = {x|x = 3, 4, 5, 6, 7, 8, 9}
Kinds of Sets A set A is said to be a subset of another set B if every element of A is also elements of B. In symbol, A ⊆ B. Proper Subsets – a set A is said to be a proper subset of another set B and there is at least one element in set B not in set A, then a is a proper subset of B. In symbols A ⊂ B.
Examples Given the sets A = {1, 2, 3}, B = {1, 2, 3, 4, 5, 6, 7}, C = {3, 5, 7}, D = {3, 2, 1} 1.A ⊆ B? True since all the elements of A are also in B. 2. A ⊂ B? True since set A is a subset of Set B and the elements of set B namely 4, 5, 6, 7 are not found in set A. 3. A ⊄ C? True since not all the elements of a are contain in C. 4. C ⊆ D? False since not all the elements of C are contained in D
Example 5. A ⊆ D. True, since all of the elements of A are contained in D 6. D ⊆ A. True, since all of the elements of D are contained in A
Kinds of Sets Equal Sets – Set A is said to be equal to set B, written A = B, if and only if A ⊆ B and B ⊆ A. Equivalent Sets – Set A is said to be equivalent to set B if and only if the two sets have exactly the same cardinality of set. Null Set – A set which does not contain any element is called a null or empty set. It is denoted by the symbol ⌀. The symbol { } can also be used to denote an empty set.
Kinds of Sets Finite and Infinite Set. The set whose elements can be counted is called a finite set, while an infinite set is a set whose elements cannot be determined. Example. The set A = {1, 2, 3] is finite set since it contains 3 elements, while B = {x|x is a natural number divisible by 2} is an infinite set since the number of elements in this set cannot be determined. B = {2, 4, 6, …} The number of subsets of a finite set A is 2 n(A)
Operations on the Sets Universal Sets. The totality of elements under consideration as the elements of any set. The symbol is ∪. The union of sets A and B, denoted by A ∪ B is defined as the set whose elements are in A or in B or in both A and B. In symbol, A ∪ B = {x |x є A or x є B} Common Set ( ∩ ) – The elements of Set A which are also elements of Set B.
Operations on the Sets Relative Complement. The relative complement of set B with respect to A designated as A – B is defined as: A – B = {x|x є B and X ∉ B} Absolute Complement. The absolute complement of A designated by A’ is the set of elements of the universal set ∪ which do not belong to A, that is A’ = {x|x є ∪ and x ∉ A}
Operations on the Sets Ordered pair. If x and y represent two objects (alike or different) with x specified as the first object, then the symbol (x,y) is called an ordered pair.
Examples Given the following sets A = {1, 2, 3), B = {2, 3, 4, 5}, C = {4, 5}, and D = {2, 3, 7] form the following sets; 1.A ∩ B = {1, 2, 3} ∩ {2, 3, 4, 5} = {2, 3} 2.B ∩ C = {2, 3, 4, 5} ∩ {4, 5} = {4, 5} = C 3.B – A = {2, 3, 4, 5} – {1, 2, 3} = {4, 5} Given A = {1,2} and B {a, b, c} 1.A x B = {(1,a), (1,b), (1,c), (2,a) (2,b) (2,c)} 2.B x A = {(a,1), (a,2), (b,1), (b,2), (c,1), (c,2)}
Homework A. Write each set by listing the elements. 1.The even integers between 1 and 16. 2.{y|y is a natural number greater than 9} 3.The months with 30 days 4.The letters in the word college 5.{z|z is a whole number and a multiple of 5} B.Use mathematical symbols to write the following statements. 1.A is a subset of G 2.The number 7 is not an element of a null set 3.The set of even prime numbers 4.The number 3 is an element of T 5.The null set is a subset of A
Homework (Cont.) C.Enclose within braces a list of the elements of each set. 1.The natural numbers less than 7. 2.The even natural numbers between 2 and 13 3.The months of the year that begins with letter J 4.The first four natural numbers multiplied by 3 5.The square of the first 4 natural numbers. D.Use the descriptive method to designate each of the following sets. 1.{3,6,9,12} 2.{even natural numbers} 3.{odd natural numbers} 4.{1, 8, 27, 64} 5.{March, May}
Homework (Cont.) E.Tell if the two sets are equal, if they are equivalent, if one is the subset of the other, or if none of these is true. 1.A = {4,7,11} B = {11, 7, 4} 2.A = {a, e, i, o, u}, B = {u, v, x, y} 3.A = {0}, B = ⌀ 4.A = 5, B = {3, 4} 5.A = {earth, venus, pluto}, B = {0, 1, 2}
Homework (Cont.) F.Given A = {a}, B = {a, b}, C = {b, c, d}, find 1.A ∩ B 2.B ∪ C 3.C x C 4.A x B 5.B - C
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## Solving Right Triangles
Solving right triangles requires you to apply all of the relationships, facts, concepts and procedures you’ve learned about Trigonometry in order to find missing information about right triangles.
It is a great way to compress your knowledge, which makes you understanding clearer, and easier to recall. In this unit you will be walked through some guidelines and advice in the notes section. Read through the notes, taking notes of your own. Then, work through the PowerPoint under the Lesson tab. Watch the video for further clarification and then try the practice problems.
Solving triangles means to find all of the missing side lengths and angles. For now, we will limit our focus to right triangles, which allows us the use of quite a few facts. Here’s what we have at our disposal. Remember, “opp,” is an abbreviation for opposite, “adj,” is an abbreviation for adjacent, and “tan,” is an abbreviation for tangent.
We will not be running through an example of each one of these, but enough for you to get a sense of how you decide which tool to use. You will most likely not be told to use a particular tool to find an answer. One of the most difficult jobs is figuring out which tool is most useful.
With that in mind, the KISS principle should be employed. KISS is an acronym, perhaps slightly rude, that stands for Keep It Simple Stupid. In other words, if you can use what you know about complimentary angles to find a missing angle, do that. It’s simple.
Let’s see an example.
It is a matter of convention that lower case letters can represent an angle, while upper case would represent the opposite side to that angle. So the angle “a” would have an opposite of “A,” and angle “b” would have an opposite side of “B,” and so on.
In this diagram, angle b = 72°.
To solve this triangle, we need to find all missing sides and angles. To keep out of trouble, we need to remember KISS. That is, what is the simplest application of the information provided. Let’s start with the sides A and B, though that’s arbitrary, we could start with angle a, just as easily.
General Guide Lines:
1. KISS
2. Use provided information to find values whenever possible
1. Opposed to using a value you discovered
3. Be neat, write your formula, then plug in values, then solve
1. If using a calculator, do not do so until the last step. Do all of the manipulation on paper, first.
2. Do not round any step, only round your final answer (unless directed otherwise for that problem).
4. If time is provided, verify your answers using a different method than what was used initially.
Side A:
1. (KISS): Side A is opposite of angle a. We don’t know angle a, so that’s not going to help us much, yet. However, side A is adjacent to the angle that’s 72°, and we know the hypotenuse. Cosine is what we’ll use to find side A. This is the simplest way.
2. We haven’t yet discovered new information, so this is not an issue.
3. This is typed, so it will be neat. Regardless, this is how to start.
Here’s what we’ve found, written on the diagram. We could use arccosine of 9/2.78 to check our work, but let’s wait until we solve the entire triangle and use something other than cosine. Since we used cosine to find the answer, and we made a mistake, we will likely make the same mistake again using arccosine, and not find our error. However, if we later use sine and get a different answer, then we have a problem we can discover.
Now, let’s solve for side B. Remember, KISS and use what’s provided, not discovered, whenever possible. If we used the Pythagorean Theorem to solve for B here, and we made a mistake with side A, then we are guaranteed to be wrong with side B, also. So, let’s use sine since side B is opposite of the angle given.
Now, the easiest thing to find in this problem was angle a. Typically, if this were a test question, I’d advise you to find the simplest thing first. But, it is important NOT to use discovered information to find further unknown information whenever possible. It is just too easy to make a simple subtraction error and then have all of your work be off.
Since it is a right triangle, and triangles add to 180°, the angle marked 72° and angle a must add to 90° (they’re complimentary). So, angle a = 18°. Here’s our final, solved triangle.
This is a valid equality because we rounded our side lengths to three significant figures.
Let’s verify the angle 18° by using arctangent. This will use all of our discovered answers together. If arctangent of 2.78/8.56 is 18°, correct to three significant figures, I’d be 100% confident that this triangle is solved correctly.
If you made a simple subtraction error, a common one here would be saying the answer is 28°, then you’ll discover it with this verification.
This is 18.0, correct to three significant figures. I’d say we’re rock solid!
Let’s see one more example. Remember, the tricky part isn’t setting up and solving the equations. The tricky part is deciding which tool is best to use.
Here we need to find both angles, and the hypotenuse. A take a moment to recognize how the information provided is related to what is required and come up with an approach you’d use to solve this triangle.
To find the hypotenuse it is easiest to use the Pythagorean Theorem. When you do so, you get 13.7, correct to three significant figures.
While you could use that answer with sine or cosine to find angles a and b, it would be best not to. It would be best to use arctangent for both angles.
## Summary
To solve a right triangle, the first thing you need to do is make sure you identify the relationship between the parts provided and the parts that are missing. This will help you know which formula to use. Without understanding this, Trigonometry will never make sense. Trying to memorize arbitrary associations, like if the triangle looks like this then I use this set of steps, would be like trying to play a musical instrument while being deaf from birth. It will make no sense.
Making sense of things is difficult, and often makes us feel insecure. Nobody is born knowing this, everybody that learns it must make sense of it. If you’re patient with yourself, willing to reflect on your decision making, this will be entirely approachable and eventually easy.
If you’d like to download and use the PowerPoint lesson that accompanies this information, please do so by.
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# Loan
Apply for a \$ 59000 loan, the loan repayment period is 8 years, the interest rate 7%.
How much should I pay for every month (or every year if paid yearly).
Example is for practise geometric progression and/or periodic payment for an annuity.
Result
m = 804.39 USD
y = 9880.6 USD
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
## Next similar examples:
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2. Compound interest
Compound interest: Clara deposited CZK 100,000 in the bank with an annual interest rate of 1.5%. Both money and interest remain deposited in the bank. How many CZK will be in the bank after 3 years?
3. Deposit
If you deposit 959 euros the beginning of each year, how much money we have at 2.1% (compound) interest after 19 years?
4. Bank
Paul put 10000 in the bank for 6 years. Calculate how much you will have in the bank if he not pick earned interest or change deposit conditions. The annual interest rate is 3.5%, and the tax on interest is 10%.
5. Population
What is the population of the city with 3% annual growth, if in 10 years the city will have 60,000 residents?
6. Piano
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7. One half
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8. Tenth member
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9. Geometric sequence 4
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10. Eight palm
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11. Five members
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12. Theorem prove
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13. Geometric progression 2
There is geometric sequence with a1=5.7 and quotient q=-2.5. Calculate a17.
14. Six terms
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15. Virus
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16. Powers
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17. GP - 8 items
Determine the first eight members of a geometric progression if a9=512, q=2
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Cody
# Problem 44. Trimming Spaces
Solution 1761144
Submitted on 25 Mar 2019 by Ajay Pattassery
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = 'no extra spaces'; b = 'no extra spaces'; assert(isequal(b,removeSpaces(a)))
b = 1×1 cell array {'no extra spaces'} b = "no extra spaces"
2 Pass
a = ' lots of space in front'; b = 'lots of space in front'; assert(isequal(b,removeSpaces(a)))
b = 1×1 cell array {'lots of space in front'} b = "lots of space in front"
3 Pass
a = 'lots of space in back '; b = 'lots of space in back'; assert(isequal(b,removeSpaces(a)))
b = 1×1 cell array {'lots of space in back '} b = "lots of space in back"
4 Pass
a = ' space on both sides '; b = 'space on both sides'; assert(isequal(b,removeSpaces(a)))
b = 1×1 cell array {'space on both sides '} b = "space on both sides"
5 Pass
a = sprintf('\ttab in front, space at end '); b = sprintf('\ttab in front, space at end'); assert(isequal(b,removeSpaces(a)))
b = 1×1 cell array {'→tab in front, space at end '} b = " tab in front, space at end"
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# Resources tagged with: Visualising
Filter by: Content type:
Age range:
Challenge level:
### There are 260 results
Broad Topics > Mathematical Thinking > Visualising
### Take One Example
##### Age 5 to 11
This article introduces the idea of generic proof for younger children and illustrates how one example can offer a proof of a general result through unpacking its underlying structure.
### Odd Squares
##### Age 7 to 11 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### Red Even
##### Age 7 to 11 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Cogs
##### Age 11 to 14 Challenge Level:
A and B are two interlocking cogwheels having p teeth and q teeth respectively. One tooth on B is painted red. Find the values of p and q for which the red tooth on B contacts every gap on the. . . .
### Dice, Routes and Pathways
##### Age 5 to 14
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### The Triangle Game
##### Age 11 to 16 Challenge Level:
Can you discover whether this is a fair game?
### Neighbours
##### Age 7 to 11 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Circles, Circles
##### Age 5 to 11 Challenge Level:
Here are some arrangements of circles. How many circles would I need to make the next size up for each? Can you create your own arrangement and investigate the number of circles it needs?
### Seeing Squares for Two
##### Age 5 to 11 Challenge Level:
Seeing Squares game for an adult and child. Can you come up with a way of always winning this game?
### Counters
##### Age 7 to 11 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win?
### Prime Magic
##### Age 7 to 16 Challenge Level:
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find?
### Multiplication Series: Illustrating Number Properties with Arrays
##### Age 5 to 11
This article for teachers describes how modelling number properties involving multiplication using an array of objects not only allows children to represent their thinking with concrete materials,. . . .
### Picturing Square Numbers
##### Age 11 to 14 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Display Boards
##### Age 7 to 11 Challenge Level:
Design an arrangement of display boards in the school hall which fits the requirements of different people.
### Clocked
##### Age 11 to 14 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### Jigsaw Pieces
##### Age 7 to 11 Challenge Level:
How will you go about finding all the jigsaw pieces that have one peg and one hole?
### Open Boxes
##### Age 7 to 11 Challenge Level:
Can you work out how many cubes were used to make this open box? What size of open box could you make if you had 112 cubes?
### 28 and It's Upward and Onward
##### Age 7 to 11 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### Shunting Puzzle
##### Age 7 to 11 Challenge Level:
Can you shunt the trucks so that the Cattle truck and the Sheep truck change places and the Engine is back on the main line?
### Single Track
##### Age 7 to 11 Challenge Level:
What is the best way to shunt these carriages so that each train can continue its journey?
### Counter Roundup
##### Age 7 to 11 Challenge Level:
A game for 1 or 2 people. Use the interactive version, or play with friends. Try to round up as many counters as possible.
### Sliding Puzzle
##### Age 11 to 16 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Tumbling Down
##### Age 7 to 11 Challenge Level:
Watch this animation. What do you see? Can you explain why this happens?
### Cuboids
##### Age 11 to 14 Challenge Level:
Can you find a cuboid that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### Squares in Rectangles
##### Age 11 to 14 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Problem Solving, Using and Applying and Functional Mathematics
##### Age 5 to 18 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Taking Steps
##### Age 7 to 11 Challenge Level:
In each of the pictures the invitation is for you to: Count what you see. Identify how you think the pattern would continue.
### Domino Numbers
##### Age 7 to 11 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Yih or Luk Tsut K'i or Three Men's Morris
##### Age 11 to 18 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Finding 3D Stacks
##### Age 7 to 11 Challenge Level:
Can you find a way of counting the spheres in these arrangements?
### Seeing Squares
##### Age 5 to 11 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Cubes Within Cubes
##### Age 7 to 14 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### Colour Wheels
##### Age 7 to 11 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
### Coloured Edges
##### Age 11 to 14 Challenge Level:
The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set?
### Cubic Conundrum
##### Age 7 to 16 Challenge Level:
Which of the following cubes can be made from these nets?
### Auditorium Steps
##### Age 7 to 14 Challenge Level:
What is the shape of wrapping paper that you would need to completely wrap this model?
### Nine Colours
##### Age 11 to 16 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Take Ten
##### Age 11 to 14 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube so that the surface area of the remaining solid is the same as the surface area of the original?
### Tetra Square
##### Age 11 to 14 Challenge Level:
ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square.
### Zooming in on the Squares
##### Age 7 to 14
Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens?
##### Age 7 to 11 Challenge Level:
How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle?
### John's Train Is on Time
##### Age 11 to 14 Challenge Level:
A train leaves on time. After it has gone 8 miles (at 33mph) the driver looks at his watch and sees that the hour hand is exactly over the minute hand. When did the train leave the station?
### Convex Polygons
##### Age 11 to 14 Challenge Level:
Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
### Squares, Squares and More Squares
##### Age 11 to 14 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Königsberg
##### Age 11 to 14 Challenge Level:
Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
### Painting Cubes
##### Age 11 to 14 Challenge Level:
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Pattern Power
##### Age 5 to 14
Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
##### Age 11 to 14 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Hidden Rectangles
##### Age 11 to 14 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
### Troublesome Dice
##### Age 11 to 14 Challenge Level:
When dice land edge-up, we usually roll again. But what if we didn't...?
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m2-20C-su2004
# m2-20C-su2004 - f x y = e-ax cos y-e-y cos x is solution of...
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Name: Student Number: Math 20C. Midterm Exam 2 July 23, 2004 Read each question carefully, and answer each question completely. Show all of your work. No credit will be given for unsupported answers. Write your solutions clearly and legibly. No credit will be given for illegible solutions. 1. (8 points) Consider the function f ( x, y, z ) = x + 2 yz . (a) Find the gradient of f ( x, y, z ). (b) Find the directional derivative of f at (0 , 2 , 1) in the direction given by h 0 , 3 , 4 i . (c) Find the maximum rate of change of f at the point (0 , 2 , 1). # Score 1 2 3 4 Σ 1
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2. (8 points) Find any value of the constant a such that the function
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Unformatted text preview: f ( x, y ) = e-ax cos( y )-e-y cos( x ) is solution of Laplace’s equation f xx + f yy = 0. 2 3. (8 points) Let f ( x, y ) = 12 xy-2 x 3-3 y 2 . (a) Find all the critical (stationary) points of f . (b) For each critical point of f , determine whether f has a local maximum, local minimum, or saddle point at that point. 3 4. (8 points) Use Lagrange multipliers to fnd the maximum and minimum values oF the Function f ( x, y ) = x 2 + y 2 subject to the constraint 1 4 x 2 + 1 9 y 2 = 1. 4...
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## 14818
14,818 (fourteen thousand eight hundred eighteen) is an even five-digits composite number following 14817 and preceding 14819. In scientific notation, it is written as 1.4818 × 104. The sum of its digits is 22. It has a total of 3 prime factors and 8 positive divisors. There are 7,140 positive integers (up to 14818) that are relatively prime to 14818.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 14 thousand 818 fourteen thousand eight hundred eighteen
## Notation
Scientific notation 1.4818 × 104 14.818 × 103
## Prime Factorization of 14818
Prime Factorization 2 × 31 × 239
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 14818 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 14,818 is 2 × 31 × 239. Since it has a total of 3 prime factors, 14,818 is a composite number.
## Divisors of 14818
1, 2, 31, 62, 239, 478, 7409, 14818
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 23040 Sum of all the positive divisors of n s(n) 8222 Sum of the proper positive divisors of n A(n) 2880 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 121.729 Returns the nth root of the product of n divisors H(n) 5.14514 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 14,818 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 14,818) is 23,040, the average is 2,880.
## Other Arithmetic Functions (n = 14818)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 7140 Total number of positive integers not greater than n that are coprime to n λ(n) 3570 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1738 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 7,140 positive integers (less than 14,818) that are coprime with 14,818. And there are approximately 1,738 prime numbers less than or equal to 14,818.
## Divisibility of 14818
m n mod m 2 3 4 5 6 7 8 9 0 1 2 3 4 6 2 4
The number 14,818 is divisible by 2.
## Classification of 14818
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
• Sphenic
## Base conversion (14818)
Base System Value
2 Binary 11100111100010
3 Ternary 202022211
4 Quaternary 3213202
5 Quinary 433233
6 Senary 152334
8 Octal 34742
10 Decimal 14818
12 Duodecimal 86aa
20 Vigesimal 1h0i
36 Base36 bfm
## Basic calculations (n = 14818)
### Multiplication
n×i
n×2 29636 44454 59272 74090
### Division
ni
n⁄2 7409 4939.33 3704.5 2963.6
### Exponentiation
ni
n2 219573124 3253634551432 48212356783119376 714410702812262913568
### Nth Root
i√n
2√n 121.729 24.562 11.0331 6.82587
## 14818 as geometric shapes
### Circle
Diameter 29636 93104.2 6.89809e+08
### Sphere
Volume 1.36288e+13 2.75924e+09 93104.2
### Square
Length = n
Perimeter 59272 2.19573e+08 20955.8
### Cube
Length = n
Surface area 1.31744e+09 3.25363e+12 25665.5
### Equilateral Triangle
Length = n
Perimeter 44454 9.5078e+07 12832.8
### Triangular Pyramid
Length = n
Surface area 3.80312e+08 3.83445e+11 12098.8
## Cryptographic Hash Functions
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# Perimeter
Perimeter is the distance around a two dimensional shape, a measurement of the distance around something; the length of the boundary.
A perimeter is either a path that encompasses/surrounds/outlines a shape (in two dimensions) or its length (one-dimensional). The perimeter of a circle or an ellipse is called its circumference.
Calculating the perimeter has several practical applications. A calculated perimeter is the length of fence required to surround a yard or garden. The perimeter of a wheel/circle (its circumference) describes how far it will roll in one revolution. Similarly, the amount of string wound around a spool is related to the spool's perimeter; if the length of the string was exact, it would equal the perimeter.
Polygons are fundamental to determining perimeters, not only because they are the simplest shapes but also because the perimeters of many shapes are calculated by approximating them with sequences of polygons tending to these shapes. The first mathematician known to have used this kind of reasoning is Archimedes, who approximated the perimeter of a circle by surrounding it with regular polygons.
An equilateral polygon is a polygon which has all sides of the same length (for example, a rhombus is a 4-sided equilateral polygon). To calculate the perimeter of an equilateral polygon, one must multiply the common length of the sides by the number of sides.
A regular polygon may be characterized by the number of its sides and by its circumradius, that is to say, the constant distance between its centre and each of its vertices. The length of its sides can be calculated using trigonometry. If R is a regular polygon's radius and n is the number of its sides, then its perimeter is
A splitter of a triangle is a cevian (a segment from a vertex to the opposite side) that divides the perimeter into two equal lengths, this common length being called the semiperimeter of the triangle. The three splitters of a triangle all intersect each other at the Nagel point of the triangle.
A cleaver of a triangle is a segment from the midpoint of a side of a triangle to the opposite side such that the perimeter is divided into two equal lengths. The three cleavers of a triangle all intersect each other at the triangle's Spieker center.
The perimeter of a circle, often called the circumference, is proportional to its diameter and its radius. That is to say, there exists a constant number pi, π (the Greek p for perimeter), such that if P is the circle's perimeter and D its diameter then,
To calculate a circle's perimeter, knowledge of its radius or diameter and the number π suffices. The problem is that π is not rational (it cannot be expressed as the quotient of two integers), nor is it algebraic (it is not a root of a polynomial equation with rational coefficients). So, obtaining an accurate approximation of π is important in the calculation. The computation of the digits of π is relevant to many fields, such as mathematical analysis, algorithmics and computer science.
The more one cuts this shape, the lesser the area and the greater the perimeter. The convex hull remains the same.
The perimeter and the area are two main measures of geometric figures. Confusing them is a common error, as well as believing that the greater one of them is, the greater the other must be. Indeed, a commonplace observation is that an enlargement (or a reduction) of a shape make its area grow (or decrease) as well as its perimeter. For example, if a field is drawn on a 1/10,000 scale map, the actual field perimeter can be calculated multiplying the drawing perimeter by 10,000. The real area is 10,0002 times the area of the shape on the map. Nevertheless, there is no relation between the area and the perimeter of an ordinary shape. For example, the perimeter of a rectangle of width 0.001 and length 1000 is slightly above 2000, while the perimeter of a rectangle of width 0.5 and length 2 is 5. Both areas equal to 1.
Proclus (5th century) reported that Greek peasants "fairly" parted fields relying on their perimeters.[1] However, a field's production is proportional to its area, not to its perimeter, so many naive peasants may have gotten fields with long perimeters but small areas (thus, few crops).
If one removes a piece from a figure, its area decreases but its perimeter may not. In the case of very irregular shapes, confusion between the perimeter and the convex hull may arise. The convex hull of a figure may be visualized as the shape formed by a rubber band stretched around it. In the animated picture on the left, all the figures have the same convex hull; the big, first hexagon.
The isoperimetric problem is to determine a figure with the largest area, amongst those having a given perimeter. The solution is intuitive; it is the circle. In particular, this can be used to explain why drops of fat on a broth surface are circular.
This problem may seem simple, but its mathematical proof requires some sophisticated theorems. The isoperimetric problem is sometimes simplified by restricting the type of figures to be used. In particular, to find the quadrilateral, or the triangle, or another particular figure, with the largest area amongst those with the same shape having a given perimeter. The solution to the quadrilateral isoperimetric problem is the square, and the solution to the triangle problem is the equilateral triangle. In general, the polygon with n sides having the largest area and a given perimeter is the regular polygon, which is closer to being a circle than is any irregular polygon with the same number of sides.
The word comes from the Greek περίμετρος perimetros from περί peri "around" and μέτρον metron "measure".
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If sylvia types at 45 wpm she can finish her work in 4 hours. How long will it take her if she types at: a) 30 wpm b) 60 wpm c) 45 wpm and she has a friend who helps her type at 30 wpm
tjbrewer | Elementary School Teacher | (Level 2) Associate Educator
Posted on
If it takes Sylvia 4 hours to do her work at 45 wpm, we can figure out how many words` ` she has to type by multiplying 45 by 60 to convert the rate to words per hour. Then we multiply the rate (2700 wph) by the 4 hours to get an assignment total of 10,800 words.
To find how long it will take to finish at 30 wpm, you divide the assignment 10,800 by the rate and that gives you the minutes and then divide by 60 to get 6 hours. `10,800-:30=360 min.` and `360-:60=6`
To find at 60 wpm we do the same thing, using 60 instead of 30. `10,800-:60=180` and `180-:60=3` So it will take 3 hours, which is logical, half the time at twice the speed.
To find the combined times of Sylvia going 45 wpm, and her friend at 30, requires a little algebra. We know that the assignment is 10,800 words. If we multiply each workers rate of typing by t minutes and add them together, we will find how many minutes it will take them together to type 10,800 words. Then we divide t by 60 to find hours if appropriate. `45t+30t=10,800->75t=10,800->t=10,800-:75=144` and `144-:60=` 2hrs 24min.
pramodpandey | College Teacher | (Level 3) Valedictorian
Posted on
typing speed more ,work finish in less time.
typing speed less ,work finish in more time.
(i) wpm 45 work finish in 4 hours.
wpm 30 work finsh =`(45xx4)/30=6` hours.
(ii) wpm 60 ,work finish =`(45xx4)/60=3 ` hours.
(iii) Sylvia's friends typing speed is 30 wpm mean she can finsh work in 6 hours. ( just part (i))
Sylvia ,wpm 45 then she can finsh work in 1 hour= 1/4
Sylvia"s freinds wpm 30 then she can finsh work in i hour= 1/6
When both worked together can finsh work in 1 hour=`1/4+1/6`
`=5/12`
Thus worknig together can finsh work in `12/5` hours.
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$$\require{cancel}$$
Generally the loudness of a sound is related to the amplitude of the sound wave; a wave with bigger variations in pressure generally sounds louder. For any type of wave the energy carried by the wave is proportional to amplitude squared. This means doubling the amplitude increases the power by a factor of four (two squared). But the amount of energy reaching your ear also depends on the frequency since a wave with more oscillations per second (higher frequency) will mean the same amplitude hits your eardrum more often. Sound intensity is defined to be the energy per second (power in watts) reaching a given area (measured in square meters). Normal conversation has an intensity of about $$10^{-6}\text{ W/m}^{2}$$.
Unobstructed sound from a small sound source spreads out in all directions in an expanding spherical shape. Because the energy is spread over a larger and larger area as time goes on, the intensity decreases as you move further from the source. The area of a sphere is given by $$4\pi r^{2}$$ so the decrease in intensity is proportional to $$r^{2}$$ where $$r$$ is the distance from the source. This is known as an inverse square law and many other laws in physics follow this same law (diagram of the inverse square law). For example the gravitational field of the earth (or any other object) decreases as you move away from it proportional to an inverse square law. So does the electric field around an electron or proton. For practical purposes, what this means is doubling the distance decreases the strength by $$1/2^{2} = 1/4$$. If you move three times as far away the strength is $$1/3^{2}$$ or $$1/9\text{th}$$ as much. Shortening the distance by half means the intensity will be four times as much.
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# Modal Logic and Science Fiction
Modal logic extends classical logic by adding one or more modes. If there’s only one mode, it’s usually denoted □. Curiously, □ can have a wide variety of interpretations, and different interpretations motivate different axioms for how □ behaves. Modal logic is not one system but an infinite number of systems, depending on your choice of axioms, though a small number of axiom systems come up in application far more than others.
For a proposition p, □p is often interpreted as “necessarily p” but it could also be read, for example, as “it is provable that p“. Thanks to Gödel, we know some theorems are true but not provable, so p might be true while □p is false.
Kripke semantics interprets □p to be true at a “worldw if p is true in all worlds accessible from w. The rules of a logic system transfer to and from the set of models for that system, where a model is a directed graph of worlds, and an oracle (a “valuation function”) that tells you what’s true on each world. Axioms for a logic system correspond to requirements regarding the connectivity of all graph models for the system.
All this talk of what worlds are accessible from other worlds sounds a lot like science fiction. For example, if the planet Vulcan is accessible from Earth, and p is the statement “The blood of sentient beings is red,” then p is true on Earth, but not necessarily true on Earth since it’s not true on Vulcan, a world accessible from Earth.
Modal logic defines ◇ by
p = ¬ □ ¬ p.
For a proposition p, ◇p can be read as “possibly p.” A proposition ◇p is true on a world w if there is some world accessible from w where p is true.
So in the Star Trek universe, if p is the statement “Blood is green” then p is false on Earth, and so is □p, but ◇p is true because there is a world accessible from earth, namely Vulcan, where p is true.
You could have all kinds of fun making up rules about which worlds are accessible from each other. If someone from planet x can reach planet y, and someone from planet y can reach planet z, can someone from x reach z? Sounds reasonable, and if all worlds have this property then your Kripke frame is said to be transitive. But you could create a fictional universe in which, for whatever reason, this doesn’t hold.
Is a world accessible from itself? Depends on how you define accessibility. You might decide that a non-space faring world is not accessible from itself. But if every world is accessible from itself, your Kripke model is reflexive. If a Kripke frame is reflexive and transitive, the corresponding logic satisfies the S4 axioms. (More on this in the next post.)
Johan van Benthem gives an example in his book Modal Logic for Open Minds that’s scientific but not fictional. If you define a “world” to be a point in Minkowski space-time, then the worlds accessible from a given world are in that world’s future light cone. Propositions in this logic satisfy
◇□ p →□◇ p
and in fact the logic satisfies a system of axioms known as S4.2.
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Assignment 8 Writeup
Altitudes and Orthocenters
Let's review what the orthocenter of a triangle is before beginning this investigation.
The orthocenter of a triangle is the common intersection of the three lines containing the altitudes. An altitude is a perpendicular segment from a vertex to the line of the opposite side.
Okay, now we can begin!
Let's construct a triangle ABC and construct the orthocenter H of this triangle ABC.
Graph of Orthocenter of Triangle ABC
Construct the orthocenters of triangles HBC, HAB, and HAC. These are given below.
Graph of Orthocenter of Triangle HBC
Graph of Orthocenter of Triangle HAB
Graph of Orthocenter of Triangle HAC
Something interesting appears to me. Well, for one, the orthocenter of triangle ABC, H, is one of the vertices for each of the other three triangles. It seems that the orthocenters of all the three small triangles HBC, HAB, and HAC fall on one of the vertices of the original triangle ABC.
For the triangle HBC, the orthocenter is on the point A. Notice how the point A isn't one of the vertices on this triangle.
For the triangle HAB, the orthocenter is on the point C. Notice again how the point C isn't one of the vertices on this triangle.
Finally, for triangle HAC, the orthocenter is on the point B. Once again, notice how the point B isn't one of the vertices on this triangle.
This is an interesting observation. It appears that the orthocenters of the three smaller triangles contained in the larger triangle have their orthocenters on a vertex of the large triangle that isn't contained in the small triangle of interest. That is pretty neat!
Okay, now let's take a look at the circumcenter. The circumcenter is the point in the plane equidistant from the three vertices of the triangle. To find the circumcenter, you take the perpendicular bisector of each segment in the triangle.
The circumcenter is the center of the circumcircle.
Let's now construct the circumcircle of triangle ABC. See the graph below.
Graph of Circumcenter and Circumcircle of Triangle ABC
Notice how the definition of a circumcenter holds for triangle ABC. It is the point that is equidistant from the three vertices of the triangle. All of these three distances are shown to be the same above.
Let's also look at the circumcircles for the triangles HBC, HAB, and HAC. See these below.
Graph of Circumcenter and Circumcircle of Triangle HBC
Notice here how the distances are also the same from the circumcenter to the vertices of the triangle HBC. Also, notice how the circumcenter lies closest to the line segment BC instead of HB or HC where H is the orthocenter of the triangle.
Graph of Circumcenter and Circumcircle of Triangle HAB
The same observation can be made here. The circumcenter does lie equidistant from the vertices of the triangle HAB. Also, the circumcenter lies closest to the segment AB and not the segments HA and HB.
Graph of Circumcenter and Circumcircle of Triangle HAC
Finally, for the triangle HAC, the definition of the circumcenter holds because the vertices of the triangle HAC are equidistant from the circumcenter. In addition, the circumcenter lies closest to the segment AC instead of the segments HA and HC which both contain the orthocenter of the triangle (H).
Since it is hard to tell how all the circumcenters and circumcircles relate when separated, I did a separate construction and put them all together to see how they relate. See the graph below.
Graph of Circumcenters and Circumcircles for Triangles ABC, HAC, HAB, and HBC.
Okay, let's explore the results when we look at the circumcenters and circumcircles for all four triangles. I see a really cool relationship. Do you notice how the perpendicular lines that come together to form each circumcenter go through the circumcenters for the other three triangles.
Let's take for instance the triangle ABC. Its circumcenter is formed at the intersection of the three red perpendicular lines in the above graph. Let's first look at the perpendicular lines that bisects the segment BC. Do you notice that this line goes through the orthocenter of the triangle HBC?? Notice the connection with the segment BC and the orthocenter of HBC.
Now, let's look at the perpendicular line that bisects the segment AC. I don't think it is a coincidence that it goes through the orthocenter of the triangle HAC. Notice again the connection with line segment AC and the orthocenter of HAC.
Finally, let's take a look at the perpendicular line that bisects the segment AB. This line goes through the orthocenter of triangle HAB. Once again, there is a connection with line segment AB and the orthocenter of HAB.
That is really neat. The same holds true when you look at the other triangles. Do some exploration on your own to see if I am right!
This concludes my exploration of altitudes and orthocenters.
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