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App that helps with math problems
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Solving complex numbers can be a challenge, but it is doable with the right approach. To solve a complex number, one must first understand what a complex number is. A complex number is any number that can be expressed in the form a + bi, where a and b are real numbers, and i is the imaginary unit. The imaginary unit is equal to the square root of -1. With this in mind, one can tackle solving complex numbers. The first step is to identify
Negative exponents in fractions can be solved by using the reciprocal property. This states that when taking the reciprocal of a number, the exponent becomes the opposite sign. So, to solve a fraction with a negative exponent, you simply take the reciprocal of the base number and then change the sign of the exponent.
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To solve a piecewise function, you need to carefully examine the given equations and determine which one to use for which x-values. You also need to be careful when graphing piecewise functions, as the different equations can produce different shapes for the graph.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > nfceqdf Structured version Visualization version GIF version
Theorem nfceqdf 2757
Description: An equality theorem for effectively not free. (Contributed by Mario Carneiro, 14-Oct-2016.)
Hypotheses
Ref Expression
nfceqdf.1 𝑥𝜑
nfceqdf.2 (𝜑𝐴 = 𝐵)
Assertion
Ref Expression
nfceqdf (𝜑 → (𝑥𝐴𝑥𝐵))
Proof of Theorem nfceqdf
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 nfceqdf.1 . . . 4 𝑥𝜑
2 nfceqdf.2 . . . . 5 (𝜑𝐴 = 𝐵)
32eleq2d 2684 . . . 4 (𝜑 → (𝑦𝐴𝑦𝐵))
41, 3nfbidf 2090 . . 3 (𝜑 → (Ⅎ𝑥 𝑦𝐴 ↔ Ⅎ𝑥 𝑦𝐵))
54albidv 1846 . 2 (𝜑 → (∀𝑦𝑥 𝑦𝐴 ↔ ∀𝑦𝑥 𝑦𝐵))
6 df-nfc 2750 . 2 (𝑥𝐴 ↔ ∀𝑦𝑥 𝑦𝐴)
7 df-nfc 2750 . 2 (𝑥𝐵 ↔ ∀𝑦𝑥 𝑦𝐵)
85, 6, 73bitr4g 303 1 (𝜑 → (𝑥𝐴𝑥𝐵))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∀wal 1478 = wceq 1480 Ⅎwnf 1705 ∈ wcel 1987 Ⅎwnfc 2748 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-9 1996 ax-12 2044 ax-ext 2601 This theorem depends on definitions: df-bi 197 df-an 386 df-ex 1702 df-nf 1707 df-cleq 2614 df-clel 2617 df-nfc 2750 This theorem is referenced by: nfceqi 2758 nfopd 4394 dfnfc2 4427 dfnfc2OLD 4428 nfimad 5444 nffvd 6167 riotasv2d 33762 nfcxfrdf 33772 nfded 33773 nfded2 33774 nfunidALT2 33775
Copyright terms: Public domain W3C validator
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# Definition Perimeter of a square
WHAT IS A PERIMETER?
The length of the boundary or sides of any two-dimensional shape is called the perimeter. For polygons, perimeter can be calculated by adding the distances as you move around the shape.
WHAT IS A SQUARE?
A square is a plane closed figure with all the four sides equal and four right (90°) angles. A square has two diagonals that are equal. The diagonals of the square bisect each other at 90°. The diagonal of the square divide it into two isosceles right-angled triangles.
NOTE: 1. The diagonal of the square is greater than the sides of the square.
1. Perimeter of a regular polygon = Number of sides × Length of one side.
2. In circle, The distance around a circle is known as its circumference.
3. If a wire in the shape of a square is rebent into a rectangle, then the area of both the shapes remains the same, but perimeter may vary.
PERIMETER OF A SQUARE
The perimeter of a square is the total length of the four equal sides of the square.
FORMULA OF PERIMETER OF A SQUARE
= Sum of all the sides
= side + side + side + side
As we know, length of all the sides of a square is equal.
So, Perimeter(P) = 4 x side
Let the side of the square be a units
So, Perimeter of square = a + a + a + a
= 4 x a
= 4 x side
So, Perimeter of square = 4 x side
UNIT OF PERIMETER OF SQUARE:
The unit to determine the value of the square of the perimeter can be millimetre (mm), centimetre (cm), metre (m), etc.
METHODS OF FINDING THE PERIMETER OF SQUARE
• Using the Formula of Perimeter of a Square
The perimeter of the square can also be calculated by adding the value of all the four sides of a square.
From the formula of Perimeter of a square,
P = 4 x side
We can find the perimeter of a square.
• Perimeter of Square using Diagonal
The perimeter of the square can also be calculated if the length of its diagonal is known.
From the Pythagorean theorem, we know that the side of a square is equal to the diagonal divided by the square root of 2.
So, Side = Diagonal/√2
Then, Side = √2(Diagonal/2)
And Perimeter of square = 4 x side
= 4 x √2(Diagonal/2)
= 2√2 (Diagonal)
Perimeter of square = 2√2(Diagonal) units.
1. Perimeter of Square using Area
The area of a square is the space occupied by the square.
The formula for area of square is given by:
Area of a square = side × side = side² sq. units
where “s” is the length of one side.
Now, we can write, side of the square as,
side = √Area
Hence, the perimeter of square will be:
Perimeter = 4 x side
So, Perimeter of square = 4 x √Area units.
DIFFERENCE BETWEEN AREA AND PERIMETER
Area: The area is the space occupied or enclosed by a closed shape. The area of the square is obtained by the product of the length of any two sides of a square. The unit of area is in square units like sq.cm, sq. m, sq. km etc.
Perimeter: The perimeter is the total distance along the boundary of a closed figure. The perimeter of the square is obtained by adding all the sides of a square. The unit of the perimeter is in cm, mm, m, km etc.
## Solution:
Given: Side of a square is 4cm
We know that, Perimeter of a square = 4 x side
= 4 x 4 = 16cm.
Therefore, the perimeter of square = 16cm.
## Solution:
Given: Perimeter of a square is 96cm
We know that, Perimeter of a square = 4 x side
or, Side of a square = Perimeter of a square/4
= 96/4
= 24cm.
So, the side of square is 24cm.
## Solution:
Side of a square is 200cm
And, Perimeter of a square = 4 x side
= 4 x 200
= 800cm
We know that, 1m = 100cm or 100cm = 1m
So, 800 cm = 800/100m
= 8m
So, Perimeter of the square is 8m.
## Solution:
Perimeter of square = 4√Area
= 4√81
= 4 x 9
= 36cm.
## Solution:
Side of the square park is 45m.
We know, the perimeter of a square = 4 x side
= 4 x 45 metres
= 180 metres.
Therefore, the boundary of the square park measures 180 metres.
If Ram takes four rounds of the park, then the total distance travelled by Ram
= 4 x 180 metres
= 720 metres.
## Solution:
To find the cost of fencing a square park, we have to find out the length of the boundary or the perimeter of a square park.
We know that,
The perimeter of a square = 4 x side
If the side of the square park is 100 metres,
So, the perimeter of a square = 4 x 100
= 400 metres.
The cost of fencing 1 metre is Rs 10
So, the cost of fencing 400 metres will be 400 x 10 =Rs. 4,000.
Therefore, the cost of fencing a square park of sides 100 metres is Rs 4,000.
## Solution:
Let the breadth of the rectangle be x.
Given: The perimeter of the rectangle is equal to the perimeter of the square.
Perimeter of rectangle = 2(l + b)
Perimeter of square = 4 x side
Therefore, 2(80+x) = 4 × 60
80 + x = 240/2
80 + x = 120
x = 120 – 80 = 40
i.e., breadth of the rectangle = 40 m
Area of the square = Side x Side
= (60 × 60) m2
= 3600 m2
Area of the rectangle = length x breadth
= (40 × 80) m2
= 3200 m2
Hence, the square field is larger than the rectangular field.
## Perimeter of Square Practice Questions
17
Created on By jpci
Attend Perimeter of Square Quiz Questions
1 / 5
perimeter of a square equal to how many sides ?
2 / 5
Formula of Perimeter of Square is ?
3 / 5
Perimeter of a square can be calculated by adding ?
4 / 5
perimeter of a 4 cm square?
5 / 5
Perimeter of Square also know as
The average score is 85%
0%
### What is the perimeter of a square?
Perimeter is the total distance around a closed figure.
### What is the formula of the perimeter of a square?
The formula for the perimeter of square is equal to 4 times the length of each side of square.
Perimeter = 4 x length of a side of the square.
### What is the unit of perimeter of a square?
The unit of perimeter of a square is in m, cm, mm etc.
### How to find the perimeter of the square?
Add all the sides of the square to find the perimeter of the square.
### What is the area of a square?
The number of square units that are required to fill the total amount of space bounded by the four equal sides of the square or it can be referred as the total amount of region enclosed in the four sides of a square.
Area of a square = Side x Side = (Side)2
### What is the unit of area of a square?
Unit of area of a square is in square units like cm2, m2, km2 etc.
### How to calculate the area of square when diagonal is given?
Area of a square using diagonals = Diagonal²/2
For example: The diagonal of a square is 8m.
Area = 82/2 = 64/2 = 32 units.
### If the side of a square is 7 m, then find the perimeter of the square.
Given: Side of a square = 7m
As we know that, Perimeter of a square = 4 x side
So, Perimeter of a square = 4 x 7 = 28m.
### Find the perimeter of a square whose area is 100 m². Solution:
Given: Area = 100 m²
Perimeter of square = 4√Area
= 4√100
= 4 x 10 = 40m.
Therefore, the perimeter of a square is 40m.
Acute Angled Triangle Right Angled Triangle Obtuse Angled Triangle
### Videos(3)
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https://maqsad.io/classes/first-year/math/quadratic-equations/4-%20solve%20by%20using%20synthetic%20division%20if(ii)%203%20is%20the%20root%20of%20the%20equation%202%20xexp%203%20-3%20xexp%202%20-11%20x+6%200
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Classes
Class 9Class 10First YearSecond Year
$4. Solve by using synthetic division if(ii) 3 is the root of the equation 2 x^{3}-3 x^{2}-11 x+6=0$
$Solve the following equations.4. \sqrt{3 x+100}-x=4$
now playing
$4. Solve by using synthetic division if(ii) 3 is the root of the equation 2 x^{3}-3 x^{2}-11 x+6=0$
$1. Write the following ratic equations in the standard form and point out pure ratic equations.(vi) \frac{x+1}{x+2}+\frac{x+2}{x+3}=\frac{25}{12}$
$Q.33 Roots of ratic equation a x^{2}+b x+c=0 are imaginary and unequal if :(a) b^{2}-4 a c=0 (b) b^{2}-4 a c>0 (c) b^{2}-4 a c<0 (d) b^{2}-4 a c \geq 0$
$Q.13 Roots of ratic equation 2 x^{2}+3 x+1=0 are :(a) \frac{1}{2}-1 (b) -\frac{1}{2} 1 (c) -\frac{1}{2}-1 (d) 1-1 Faisalabad Board 2013$
$Q.42 If the sum of the roots of the equation a x^{2}-2 x+2 a=0 is equal to their product then the value of a is :(a) 1(b) 2(c) 3(d) 4$
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1. ## Factoring Polynomials
I don't understand factoring - here are some problewms I am looking to see complete step by step because no matter what I do I am not getting the right answers!!
6x^2y^2 + 12xy^2 + 12y^2
a^2 + 2a - 24
z^3 + 9z^2 +18z
a^4 - ab3
and then for review I am confused on these:
t^8 / t^2
8t^
____
2t^2
I have read the book and gone step by step but something just isn't clicking... I am desperate as I have a test next week and I can't even do my homework!
Thank you in advance to anyone who helps!
2. Ok let's start off with one. Notice anything similar with the 3 terms? They all have $6y^{2}$ in them so we can factor out $6y^{2}$ and get $6y^{2}(x^{2} + 2x + 2)$ I don't think that can be factored any further. Well without getting into complex roots.
Now #2-
Let's find factors of -24 that will give us 2. Hmmm....how about 6 and -4? That will give us $(x+6)(x-4)$ Expand and check your work if you need to.
#3- Hmm notice something in these three terms too? Yup z is in all 3 so take that out of there $z(z^{2}+9z+18)$ Now let's find factors of 18 that will sum to 9. Aha! 6 & 3 So that gives us $(x+6)(x+3)$ Expand and check if needed.
#4- Hmmm...there's an a in both terms there so get him out of there. $a(a^{3} - b^{3})$ Now we're going to factor out cubes (I don't know if you guys are doing this yet, but I'll show you anyway, sorry if this confuses you) We'll take the cubed root of a and b and get $(a-b)$ then we square the first term multiply the 1st and 2nd term once and switch the sign and then square the last term to get $a(a-b)(a^{2}+ab+b^{2})$ and we have factored the polynomial.
For the last to just simplify $\frac{a^{5}}{a^{3}} = a^{5-3}$ apply that do both of those. (Don't forget to simplify $\frac{8}{2}$ as well.)
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PROBLEM 44- 1251: The elongation, E, of a steel wire when a mass, m, is hung
from its free end varies jointly as m and the length, x, of the
wire and inversely as the cross sectional area A, of the wire.
Given that E = 0.001 inches when m = 20 pounds, x = 10
inches, and A = 0.01 square inches, find E when m = 40
pounds, x = 15.5 inches, and A = 0.015 square inches.
Solution: If E is directly proportional to m and x and inversely proportional to
A, our equation is
E = kmx / A
where k is a constant which we can determine from the given information.
Since, when m = 20, x = 10, and A = .01, E = .001, we can use this
information to solve for k. Substituting we obtain:
0.001 = {k(20)(10)} / .01, and multiplying both
sides by .01/{(20)(10)} we obtain:
Multiplying numerator and denominator by 100,000 (move the decimal 5
places to the right) we obtain:
Hence, E = (1 / 20,000,000) (mx/A)
and when m= 40, x= 15.5, and A = 0.015, we have:
E =[1 / 20,000,000] [{40(15.5)} / 0.015]
= 0.00207 inches (approximately).
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# Thread: Monotonically increasing
1. ## Monotonically increasing
Here´s the question.
Ok so the derivative of f
$\displaystyle {\frac {d}{dx}}f \left( x \right) =\cos \left( x \right) -1+3\,\alpha \,{x}^{2}$
The function is monotonically increasing if the derivative of f is bigger than zero in the interval. Or if $\displaystyle \alpha<1/3\,{\frac {1-\cos \left( x \right) }{{x}^{2}}}$
So my task should be to determine the lowest value of the right hand side of the inequality and say that alpha should be less than that? Is it so? I´ve tried that but ended up with zeros in the denominator and stuff. Any ideas?
2. min(cosx-1) in [0,pi/2] = -1
max(cosx-1) in [0,pi/2] = 0
3. Originally Posted by Also sprach Zarathustra
min(cosx-1) in [0,pi/2] = -1
max(cosx-1) in [0,pi/2] = 0
Yes. min(cosx-1) in [0,pi/2] =-1 and this happens as x --> pi/2 and therefore f´(0) larger than zero if
$\displaystyle 4/3\,{\pi }^{-2}<\alpha$
But what happens when x-->0 and how about all the other values between 0 and pi/2?
I´m still not sure how to choose alpha so that f´(x) is larger than zero in the interval.
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Polynomial derivative causes Prime to Crash? - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: Polynomial derivative causes Prime to Crash? (/thread-6549.html) Polynomial derivative causes Prime to Crash? - Spybot - 07-13-2016 05:05 PM Hello! Perhaps just trying to save some time while finding the derivative of this polynomial function: d/dx(5x^(3/5)+4x^(3/4)) ... using the Prime Virtual Calculator, it failed causing the Emulator to crash, then I tried to solve it separating each term to find out the symbolic results are no the ones I expected. [attachment=3753] [attachment=3754] I noticed that evaluating the derivative numerically it returns correct values... but symbolically I'm not sure, what do you think? Thank you. RE: Polynomial derivative causes Prime to Crash? - Arno K - 07-13-2016 05:16 PM Well, the answer is correct, you got to simplify yourself. Arno RE: Polynomial derivative causes Prime to Crash? - parisse - 07-13-2016 06:19 PM I don't think it crashes (at least it does not inside Xcas), but you would have to wait much too long, I will change some parameters to avoid that (maximum degree for common algebraic extension built for simplifications). RE: Polynomial derivative causes Prime to Crash? - roadrunner - 07-13-2016 10:56 PM In xcas diff(5*x^(3/5)+4*x^(3/4),x) returns: 3/(x^(1/5))^2+3*1/(x^(1/4))^3*sqrt(x) However, simplify(diff(5*x^(3/5)+4*x^(3/4),x)) returns: Code: `rootof([[1849290596989009920*x^3-144653156352000000*x^2+21432560400000000000*x+76904296875000,3178468213574860800*x^3+7094037381120000000*x^2+2440993500000000000*x-116729736328125,9635845027921920000*x^3-386546688000000000*x^2+229471875000000000*x,7879278565785600000*x^3-1874283840000000000*x^2-22113281250000000*x,(-9423782220272762880)*x^4+1713490624512000000*x^3-107266723200000000000*x^2+1757812500000000*x,(-23106632681898639360)*x^4-34563695837184000000*x^3-98018458200000000000*x^2+71411132812500*x,(-62818287878563430400)*x^4-25623209410560000000*x^3-10937965500000000000*x^2+487518310546875*x,(-72469455631810560000)*x^4+9139958784000000000*x^3-886802343750000000*x^2,19445417291078959104*x^5-40962633184051200000*x^4+223432480800000000000*x^3+86163574218750000*x^2,(-449515361885057187840)*x^5+101669760663552000000*x^4-5546714328000000000000*x^3-29462585449218750*x^2,(-708857389430907863040)*x^5-1855984556261376000000*x^4-502953623550000000000*x^3+31589813232421875*x^2,(-2431921742405015961600)*x^5+130928497781760000000*x^4-44452603406250000000*x^3-796508789062500*x^2,(-20610723694661074944)*x^6-1971674608293642240000*x^5+265932066816000000000*x^4+7454288671875000000*x^3,(-1129025475352763301888)*x^6-157730966234726400000*x^5-13308929464680000000000*x^4-767644409179687500*x^3,(-2769378177893825249280)*x^6-4396264306606080000000*x^5-8788632003900000000000*x^4+115511627197265625*x^3,(-6561699281264726507520)*x^6-2377594621145088000000*x^5-1056820966931250000000*x^4+32634887695312500*x^3,12198562540686409728*x^7-9550672312568433868800*x^6+1825856668567680000000*x^5-57416284212890625000*x^4+711822509765625*x^3,(-156421643493552685056)*x^7-2674904881032069120000*x^6-2496402776412000000000*x^5+2680500750732421875*x^4,(-138243318795485577216)*x^7-1439651046187008000000*x^6+2043538964595000000000*x^5-236938568115234375*x^4,(-298436070998016000000)*x^7+493404453398016000000*x^6-194979280837500000000*x^5+10898437500000000*x^4],[1,0,0,0,-5*x,-4*x,0,0,10*x^2,-260*x^2,6*x^2,0,-10*x^3,-620*x^3,-340*x^3,-4*x^3,5*x^4,-140*x^4,110*x^4,-20*x^4,-x^5+x^4]])` with these warnings: Warning, choice of an algebraic branch for root of a polynomial with parameters might be wrong. The choice is done for parameters value=0 if 0 is regular, otherwise randomly. Actual choice is Vector [92] Warning, replacing 92 by 92.0, a substitution variable should perhaps be purged. Warning, replacing 92 by 92.0, a substitution variable should perhaps be purged. Warning, replacing 92 by 92.0, a substitution variable should perhaps be purged. Warning, replacing 92 by 92.0, a substitution variable should perhaps be purged. Warning, replacing 92 by 92.0, a substitution variable should perhaps be purged. Warning, replacing 92 by 92.0, a substitution variable should perhaps be purged. Warning, replacing 92 by 92.0, a substitution variable should perhaps be purged. Warning, replacing 92 by 92.0, a substitution variable should perhaps be purged. Evaluation time: 66.129 What am I doing wrong? -road RE: Polynomial derivative causes Prime to Crash? - Spybot - 07-14-2016 12:21 AM Thank you Parisse, I appreciate you will do that for us. RE: Polynomial derivative causes Prime to Crash? - compsystems - 07-14-2016 01:57 AM 3*x^(-2/5) == diff(5*x^(3/5)) [enter] returns 1 (true) [up] [up] [copy] => 3*x^(-2/5) == 5*x^(3/5)' [enter] returns 1 (true) ok, but shifted the single quotation mark out of the entire expression, in this case not fails, because 5 is taken as constant should be [up] [up] [copy] =>3*x^(-2/5) == (5*x^(3/5))' in the following case fails diff(x/y) [enter] returns 1/y ok but [up] [up] [copy] => x/y' [enter] returns ±∞. BUG CONFIRMED should be [up] [up] [copy] => (x/y)' [enter] 1/y ok FOR HP-PRIME TEAM solution to bug diff(expr) => parser as (expr)' and not as expr' --------------------------------- for Bernard, we would see the solution as we make to naturally diff(5*x^(3/5) => 3*x^(-2/5) or 3/x^(2/5) and not 3*(x^(1/5))^3/x (is not so simplified superficially) as you can improve the simplifier? RE: Polynomial derivative causes Prime to Crash? - parisse - 07-14-2016 05:15 AM roadrunner: nothing is wrong. You get the result as rootof(P,Q) i.e. P(alpha) where Q(alpha)=0. This is because the simplifier rewrite all algebraic extensions in a unique one, in order to certify that it will not miss a simplification, the analog as when you add fractions like 1/2+1/3+1/6 you can get a simplification to an integer. However here it requires heavy computations. I would recommend not to run simplify on expressions with fractional powers, by the way the title of the topic is misleading, this is not a polynomial derivative. Regarding fractional powers, the CAS rewrites them as polynomials over simpler intermediate variables, here x^(3/5) as x^(1/5)^3.
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https://secularfrontier.infidels.org/2011/01/how-many-ways-to-analyze-the-word-god-part-3/
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# How Many Ways to Analyze the Word ‘God’? – Part 3
II. Only Three Attributes are Relevant
In this second case there will be far fewer possible definitions of ‘divine person’ because the assumption that one of the four attributes is irrelevant means that we only have three attributes to use in constructing sets of conditions that will in turn be used to generate definitions. Fewer elements means fewer combinations of elements can be formed.
The following table illustrates the various possibilities in which only three of the four attributes are relevant (R = relevant; I = irrelevant):
Since there are only three conditions in each set of conditions, each column in the above table represents 10 x 10 x 10 = 100 x 10 = 1,000 different sets of three conditions. Thus, the four columns of this table represent a total of 4,000 different sets of three conditions.
A. All Three Conditions are Criterial and None are Necessary Conditions
If all three conditions are taken as criterial, then each of the 4,000 sets of conditions is ambiguous. This is because two different strengths of requirement can be imposed on a set of three criterial conditions:
At least two conditions must be satisfied
At least one condition must be satisfied
Therefore, each of the 4,000 sets of three conditions can be used to make two criterial definitions. This means that there are 4,000 sets of conditions x 2 strengths of requirement = 8,000 definitions consisting of three criterial conditions.
B. All Three Conditions are Necessary Conditions and None are Criterial
Since there is just one strength of requirement for definitions that are composed of only necessary conditions (i.e. every condition must be satsified, without exception), the 4,000 sets of three conditions can be used to generate 4,000 definitions that are composed of three necessary conditions.
C. Two Conditions are Criterial and One is a Necessary Condition
When you mix criterial conditions and necessary conditions, they can occur in different orders. With sets of three conditions, there are three different possible permutations consisting of two criteria and one necessary condition (C = criterial condition; N = necessary condition):
This means that we can take the 4,000 different sets of three conditions, and generate a total of 12,000 sets of conditions that consist of two criteria and one necessary condition (4,000 sets of conditions x 3 different permutations = 12,000 sets consisting of two criteria and one necessary condition).
There is only one strength requirement that can be used for definitions involving just two criteria:
At least one of the (criterial) conditions must be satisfied
So, we can generate only one defintion from each set of conditions, and thus this subset generates 12,000 definitions that consist of two criteria and one necessary condition.
D. One Condition is Criterial and Two are Necessary Conditions
This subset of definitions is easy to enumerate. There are none, because any correct definition that includes a criterial condition must include at least two criteria.
E. Adding Up the Definitions from each Subset (A-D)
8,000 definitions (consisting of three criterial conditions)
4,000 definitions (consisting of three necessary conditions)
12,000 definitions (consisting of two criteria and one necessary condition)
0 definitions (consisting of one criterial condition and two necessary conditions)
The total number of different definitions of ‘divine person’ that can be generated in the case that only three of the four attributes are relevant is:
24,000 definitions
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https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-10-geometry-10-4-area-and-circumference-exercise-set-10-4-page-646/2
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## Thinking Mathematically (6th Edition)
Published by Pearson
# Chapter 10 - Geometry - 10.4 Area and Circumference - Exercise Set 10.4 - Page 646: 2
#### Answer
12$ft^{2}$
#### Work Step by Step
According to the figure, l = 4ft and w = 3ft. The formula for area of a rectangle is: A = l x w. A=4 x3= 12$ft^{2}$
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://answerdata.org/waht-is-the-next-number-in-this-sequence-100-52-28-16-10/
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# Waht is the next number in this sequence 100 52 28 16 10?
• (100/2) + 2 = 52
( 52/2 ) + 2 = 28
(28 / 2) +2 = 16
(16 / 2) + 2 = 10
(10/2 ) + 2 = 7 >>>> answer
• Seven
52 = (100 + 4)/2
28 = (52 + 4)/2
16 = (28 + 4)/2
10 = (16 + 4)/2
7 =(10+4)/2
• 7
100/2+2=52
52/2+2=28
28/2+2=16
16/2+2=10
10/2+2=7
Interesting the +4 approach noted above.
• 206. hmm it seems i worked from the other side.
the answer would be 7 then.
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# Quant Boosters - Nitin Gupta - Set 2
• Q20) What is the minimum value of the expression 6x^2 + 3y^2 - 4xy - 8x + 6y + 2?
(a) -13/7
(b) -2
(c) -8/5
(d) -7/3
(e) none of the foregoing
• The expression 6x^2 + 3y^2 - 4xy - 8x + 6y + 2 can be written as (14x^2 -12x - 3)/3 + 3(y - (4x-6)/6)^2
To minimize (y - (4x-6)/6) = 0 and 14x^2 -12x - 3 should be minimum.
14x^2 -12x - 3 is minimum for x = 3/7 => y = -5/7.
Hence, option (a) is the right answer
• Q21) Let P(x) = nx + a where n and a are integers with n > 0. If the solution to 2^P(x) = 5 is x = log8 (10 ) , then 4n - 3a is
(a) 11
(b) 12
(c) 13
(d) 14
(e) None of the foregoing
• Taking log on both the sides of 2^P(x) = 5 we get P(x) = log5/log2.
x = log10/log8 = 1/3(1 + log5/log2) => n = -3 and a = 1.
Hence, option (e) is the right answer
• Q22) he number of permutations of the set {1, 2, 3, 4} in which no two adjacent positions are filled by consecutive integers (increasing order)
(a) is a prime
(b) is a composite number divisible by 3
(c) does not exceed 17
(d) is at most 19
(e) None of the foregoing
• The permutations where there are consecutive numbers are 1234, 1243, 1342, 1423, 2134, 2341, 2314, 3124, 3412, 3421, 4123, 4312, 4231
Hence, choice (a) is the right answer
• Q23) Let f(x) be an algebric expression of odd degree ( > 1). If the degree of f(x) + f(1-x) is atleast two less than the degree of f(x) and the coefficients of x and x^2 are equal in magnitude but opposite in sign in the given expression, then the highest possible degree of f(x) is
(a) 9
(b) 13
(c) 5
(d) 7
(e) none of the foregoing
• Taking f(x) = ax^n + bx^n-1 + cx^n-2 + .... , where the third last coefficient and the second last coefficient are equal in magnitude but opposite in sign and n is odd.
f(x) + f(1-x) has atleast first 2 terms as 0 => 2b + na = 0. Thus, we can have such f(x) for any n, thus n-> Infinity.
Hence, choice (e) is the right answer.
• Q24) A car gives the mileage of 60km/L, 50km/L and 40km/L when driven at the speeds of 40km/hr, 50km/hr and 60km/hr respectively. Assume that each car is driven only at the three speeds mentioned above. The car is driven for 3 hours using 2 Litres of petrol. The distance covered by the car
(a) > 120 km
(b) < 120 km
(c) = 120 km
(d) can not be determined
(e) none of the foregoing
• Q25) One day Vikram was out bicycling. After entering a one-way tunnel and after having ridden one-fourth of the distance through it, he looked back over his shoulder and saw a bus approaching the tunnel entrance at a speed of 80 miles/hr. Doing a quick mental exercise, Vikram realized that if he accelerated immediately to his top speed, he could just escape with his life, whichever direction he rode. What is Vikram's top biking speed in miles/hr?
a) 32
(b) 36
(c) 40
(d) 48
(e) none of the foregoing
• Let d = distance truck is in front of tunnel entrance, L= length of tunnel, x = Vikram's speed.
Case 1: Vikram turns around and heads for entrance, a distance of L/4.
Vikram and truck get to entrance at same time T1 = d/80 = (L/4)/x.
Case 2: Vikram streaks for exit, a distance of 3L/4. Vikram and truck get to exit at same time
T2 = (L+d)/80 = 3L/4)/x.
Solving both equations for x and setting them equal, x = (L/4) * 80/d = (3L/4) * 80/(L+d)
After simplifying, d = L/2, hence x = 40.
Hence, choice (c) is the right answer.
• Q26) Four persons go to a birthday party. They leave their top-coats and hats in the lounge and pick them while returning back.The number of ways in which none of them picks up his own top-coat as well as his own hat is p. The number of ways in which exactly one of them picks up his own top-coat as well as his own hat is q. The number of ways in which a person picks up someone else’s top-coat and yet someone else’s hat is r. Then p+q+r is
(a) 117
(b) 104
(c) 113
(d) 108
(e) none of the foregoing
• The no. of ways in which the 1st person doesn't pick up his own top-coat is 3. The 2nd person (whose top-coat the 1st one picked)can pick someone else's top-coat in 3 more ways. For the 3rd and 4rth person we have just 1 choice. Thus, in all when none among the 4 pick his/her own top-coat is 9. Thus p = 9 * 9 = 81.
A person among the 4 who picks his own top-coat as well as hat can be selected in 4 ways. The other 3 falls in the category [not own top-coat or not own hat] = not own top-coat + not-own hat - (not won top-coat and not own hat) = 2 * 6 + 2 * 6 - 2 * 2 = 20. Thus, q = 80.
As explained in the 1st case, each person can pick someone else's top-coat in 9 ways. To each of these 9 ways we have 2 ways where he picks yet someone else's hat. Thus, r = 9 * 2 = 18
Hence, choice (e) is the right answer.
• Let the distance travelled with 40km/hr, 50km/hr, 60km/hr is A, B and C respectively.
=> A/40 + B/50 + C/60 = 3 and A/60 + B/50 + C/40 = 2.
Eliminating A, we get 2B/5 + 5C/6 = 0 => B = C = 0 => A = 120 km
Hence, choice (c) is the right answer.
• Q27) Five containers each have one litre of 10%, 20%, 30%, 40%, 50% milk in them. By mixing the contents of just two of the containers, one litres each of w%, x%, y%, z% and 42% (22 < w < = x < = y < = z < 42) milk is prepared, contents of no two containers are mixed twice in different proportions to get different concentrations. Also each original container is used exactly twice in making new mixtures. If the containers are mixed only in the multiples of 100 ml then z + y - x - w = ?
a) can not be determined
(b) 4
(c) 16
(d) 20
(e) none of the foregoing
10% milk can be mixed with 50% milk to produce 42% milk but on doing so we are left with 800 ml of 10% milk which cant give any possible combination with others to produce milk more than 22% milk. Only 30% milk in quantity 400 ml with 50% milk in 600 ml gives us a 42% combination.
Now we are left with 600 ml of 30% and 400ml of 50%.Now since we are combining 600 and 400 ml of 2 hence all the solution have to be broken in two parts each of 400 ml and 600 ml.
Proceeding further from here, we find only 1 valid combination so that all the values are satisfied in the given range.
400ml of C + 600 ml of E ---> [120 + 300] ---> 420 out of 1000 ---> 42%
400ml of E + 600 ml of A ---> [200 + 60 ] ---> 260 out of 1000 ---> 26%
400ml of A + 600 ml of D ---> [40 + 240] ---> 280 out of 1000 ---> 28%
600ml of B + 400 ml of D ---> [120+ 160] ---> 280 out of 1000 ---> 28%
400ml of B + 600 ml of C ---> [80 + 180] ---> 260 out of 1000 ---> 26%
Hence, z + y - x - w = 28 + 28 - 26 - 26 = 4
Hence, choice (b) is the right answer.
• Q28) Let a, b, c be distinct non-zero integers such that -5 < = a, b, c < = 5. How many solutions (a, b, c) does the equation 1/a + 1/b + 1/c = 1/(a+b+c) have?
a) 72
(b) 120
(c) 192
(d) 240
(e) none of the foregoing
• 1/a + 1/b + 1/c = 1/(a+b+c) reduces to (a+b)(b+c)(c+a) = 0;
If a + b = 0 then we can chose b in 10 ways (integers from -5 to 5 excluding 0) and c in 10 - 2 = 8 ways (since a, b and c are different). So total 80 ways
Similarly, b + c = 0, and c + a = 0 gives 80 solutions each.
So total 80 + 80 + 80 = 240
Hence, choice (d) is the right answer.
• Q29) A biologist catches a random sample of 60 fish from a lake, tags them and releases them. Six months later she catches a random sample of 70 fish and finds 3 are tagged. She assumes 25% of the fish in the lake on the earlier date have died or moved away and that 40% of the fish on the later date have arrived (or been born) since. What does she estimate as the number of fish in the lake on the earlier date?
(a) 420
(b) 560
(c) 630
(d) 720
(e) 840
• Let no. of Fishes on earlier day = x; 60 fishes are tagged, out of which 25% have died/moved remaining fishes that are tagged = 45.
No. of fishes on day of second observation = 5/4x
No. of fishes caught = 70
No. of Fishes tagged = 3 = 4.2857% (3/70)
Total No. fishes on Second day of Observation = 45/.042857 (45 * 70/3) = 1050
No of Fishes on earlier day = x = 4/5 * 1050 = 840
Hence, choice (e) is the correct option
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https://www.shaalaa.com/question-bank-solutions/if-v-volume-cuboid-dimensions-a-b-c-s-its-surface-area-then-prove-that-surface-area-of-a-cuboid_61584
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# If V Is the Volume of a Cuboid of Dimensions A, B, C And S is Its Surface Area, Then Prove that - Mathematics
If V is the volume of a cuboid of dimensions a, b, c and S is its surface area, then prove that $\frac{1}{V} = \frac{2}{S}\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$
#### Solution
$\text { It is given that V is the volume of a cuboid of length = a, breadth = b and height = c . Also, S is surface area of cuboid . }$
$\text { Then, V = a } \times b \times c$
$\text { Surface area of the cuboid } = 2 \times \text { (length } \times \text { breadth + breadth }\times \text { height + length } \times \text { height) }$
$\Rightarrow S = 2 \times (a \times b + b \times c + a \times c)$
$\text { Let us take the right - hand side of the equation to be proven } .$
$\frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) = \frac{2}{2 \times (a \times b + b \times c + a \times c)} \times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
$=\frac{1}{(a \times b + b \times c + a \times c)} \times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
$\text { Now, multiplying the numerator and the denominator with a } \times b \times c, \text { we get: }$
$\frac{1}{(a \times b + b \times c + a \times c)} \times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \times \frac{a \times b \times c}{a \times b \times c}$
$=\frac{1}{(a \times b + b \times c + a \times c)} \times (\frac{a \times b \times c}{a}+\frac{a \times b \times c}{b}+\frac{a \times b \times c}{c}) \times \frac{1}{a \times b \times c}$
$=\frac{1}{(a \times b + b \times c + a \times c)} \times (b\times c+a\times c+a\times b) \times \frac{1}{a \times b \times c}$
$=\frac{1}{(a \times b + b \times c + a \times c)}\times(a\times b+b\times c+a\times c) \times \frac{1}{a \times b \times c}$
$=\frac{1}{a \times b \times c}$
$=\frac{1}{V}$
$\therefore \frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) = \frac{1}{V}$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 8 Maths
Chapter 21 Mensuration - II (Volumes and Surface Areas of a Cuboid and a Cube)
Exercise 21.4 | Q 2 | Page 30
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Algebra: A Combined Approach (4th Edition)
$\frac{4(-3) + (-8)}{2 + (-2)}$ = $\frac{-12 + (-8)}{2 + (-2)}$ = $\frac{-20}{0}$ = undefined, you cannot divide by zero
| 62 | 164 |
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# Crossroads
The rectangular crossroads comes passenger car and an ambulance, the ambulance left. Passenger car is at 43 km/h and ambulance 52 km/h.
Calculate such a relative speed of the ambulance moves to the car.
Result
Relative speed of ambulance to car: 67.48 km/h
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Two vectors given by its magnitudes and by included angle can be added by our vector sum calculator. Pythagorean theorem is the base for the right triangle calculator.
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# If Machine X can make 40 widgets in 8 minutes, how many widgets can it
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If Machine X can make 40 widgets in 8 minutes, how many widgets can it [#permalink]
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19 Oct 2017, 23:17
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If Machine X can make 40 widgets in 8 minutes, how many widgets can it make in 1 hour?
(A) 90
(B) 180
(C) 240
(D) 300
(E) 320
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Re: If Machine X can make 40 widgets in 8 minutes, how many widgets can it [#permalink]
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19 Oct 2017, 23:30
Bunuel wrote:
If Machine X can make 40 widgets in 8 minutes, how many widgets can it make in 1 hour?
(A) 90
(B) 180
(C) 240
(D) 300
(E) 320
If the machine makes 40 widgets in 8 minutes, it would make 5 widgets in a minute!
In an hour(60 minutes long),
the machine would make 60*5 = 300 widgets(Option D)
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Re: If Machine X can make 40 widgets in 8 minutes, how many widgets can it [#permalink]
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20 Oct 2017, 03:44
X can make =40/8 = 5 widgets per min.
No. of widgets by X in 60 min (1 hr) = 5*60 = 300
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Re: If Machine X can make 40 widgets in 8 minutes, how many widgets can it [#permalink]
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20 Oct 2017, 05:59
Bunuel wrote:
If Machine X can make 40 widgets in 8 minutes, how many widgets can it make in 1 hour?
(A) 90
(B) 180
(C) 240
(D) 300
(E) 320
Number of widgets that can be made by Machine X in 1 hour = 40/8 = 5
Number of widgets it can make in 1 hour = 5*60 = 300
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Re: If Machine X can make 40 widgets in 8 minutes, how many widgets can it [#permalink]
### Show Tags
24 Oct 2017, 05:15
Bunuel wrote:
If Machine X can make 40 widgets in 8 minutes, how many widgets can it make in 1 hour?
(A) 90
(B) 180
(C) 240
(D) 300
(E) 320
Since 1 hour is 60 minutes, Machine X can make 40/8 x 60 = 5 x 60 = 300 widgets.
Alternate solution:
We can set up a proportion in which x is the number of widgets that can be made in 1 hour, or 60 minutes:
40/8 = x/60
5 = x/60
300 = x.
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Re: If Machine X can make 40 widgets in 8 minutes, how many widgets can it &nbs [#permalink] 24 Oct 2017, 05:15
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## Elementary Statistics (12th Edition)
a) In each selection there are 10 ways to select, threfore the number of selections: $10\cdot 10 \cdot 10 \cdot 10=10000$ ways. b) Out of the 1000 possibilities 1 possibility would win us the prize therefore the probability is:$\frac{1}{10000}$. c)net profit=winning-bet=5000-1=4999 d) Expected value: $5000\cdot\frac{1}{10000}+(-1)\cdot \frac{10000}{10000}=0.5-1=-0.5=-50 cents$ e)Both have the expected value of -50 cents, therefore betting on neither of them is better.
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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## GMATPrep 2 tagged by: Brent@GMATPrepNow ##### This topic has 5 expert replies and 1 member reply ## GMATPrep 2 Attached. _________________ Please press "thanks" if you think my post has helped you.. Cheers!! ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15388 messages Followed by: 1872 members Upvotes: 13060 GMAT Score: 790 Quote: Is xy > 0 ? 1. x - y > -2 2. x - 2y < -6 Statement 1: x > y-2 If y=2 and x=1, is 1*2 > 0? Yes. If y=-1 and x=1, is 1*(-1) > 0? No. Since the answer is YES in the first case but NO in the second case, INSUFFICIENT. Statement 2: x < 2y-6 If y=1 and x=-10, is 1*(-10) > 0? No. If y=10 and x=1, is 1*10 > 0? Yes. Since the answer is NO in the first case but YES in the second case, INSUFFICIENT. Statements combined: Linking together the two statements, we get: y-2 < x < 2y-6 y-2 < 2y-6 y > 4. Since y > 4 and x > y-2, we know that x > 2. Thus, x and y are both positive, with the result that xy > 0. SUFFICIENT. The correct answer is C. _________________ Mitch Hunt Private Tutor for the GMAT and GRE [email protected] If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at [email protected]. Student Review #1 Student Review #2 Student Review #3 Last edited by GMATGuruNY on Tue Feb 03, 2015 6:57 am; edited 2 times in total Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 13046 messages Followed by: 1253 members Upvotes: 5254 GMAT Score: 770 Quote: Is xy > 0 ? 1. x - y > -2 2. x - 2y < -6 Target question: Is xy>0? Statement 1: x-y > -2 There are several pairs of numbers that meet this condition. Here are two: Case a: x = 5 and y = 1, in which case xy is greater than 0 Case b: x = 5 and y = -1, in which case xy is not greater than 0 Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT Statement 2: x - 2y < -6 There are several pairs of numbers that meet this condition. Here are two: Case a: x = 1 and y = 5, in which case xy is greater than 0 Case b: x = -1 and y = 5, in which case xy is not greater than 0 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Statements 1 and 2 combined: Here's what we know: x-y > -2 x-2y < -6 Since both inequalities have an x, let's isolate x in both of them to get: y-2 < x x < 2y-6 Aside: Notice that I rewrote them so that the 2 inequality symbols are pointing in the same direction. Now we can combine these inequalities to get: y-2 < x < 2y-6 Next, remove the x to get: y-2 < 2y-6 Then subtract y from both sides and add 6 to both sides to get: 4 < y Great, we now know that y is positive. Also, if y-2 < x (and y>4), then we know that x must also be positive Since we now know that x and y are positive, we can be certain that xy is greater than 0 So, the answer is C Cheers, Brent _________________ Brent Hanneson – Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 04 Dec 2012 Posted: 2094 messages Followed by: 240 members Upvotes: 1443 Mitch, I think you accidentally posted the solution to the wrong question! I'll add 2 things to Brent's great explanation: a) As I'm sure you know, any time you see an inequality relative to 0, frame your question in terms of positives and negatives. Is xy > 0? Rephrased question: do x and y have the same sign? Brent gave good examples of testing numbers to prove that x and y could have the same sign or different signs with each statement individually. b) If you'd prefer not to isolate x first, you can combine the inequalities by simply flipping the second one around: x - 2y < -6 --> -6 > x - 2y Now, line up the inequalities and you can add them together: x - y > -2 -6 > x - 2y _______________ x - y - 6 > x - 2y - 2 Simplify: -y - 6 > -2y - 2 y - 6 > -2 y > 4 Substitute: x - (grt 4) > -2 x > 2 x and y are both positive, so together the statements are sufficient. _________________ Ceilidh Erickson Manhattan Prep GMAT & GRE instructor EdM in Mind, Brain, and Education Harvard Graduate School of Education Manhattan Prep instructors all have 99th+ percentile scores and expert teaching experience. Sign up for a FREE TRIAL, and learn why we have the highest ratings in the GMAT industry! Free Manhattan Prep online events - The first class of every online Manhattan Prep course is free. Classes start every week. ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15388 messages Followed by: 1872 members Upvotes: 13060 GMAT Score: 790 ceilidh.erickson wrote: Mitch, I think you accidentally posted the solution to the wrong question! Thanks for pointing this out, Ceilidh. My post above now offers a solution to the correct question. _________________ Mitch Hunt Private Tutor for the GMAT and GRE [email protected] If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at [email protected]. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. Senior | Next Rank: 100 Posts Joined 12 Mar 2013 Posted: 44 messages Upvotes: 3 but if substitute in 2nd equation does not hold? y>4 -6 > x - 2y -6> x - 2(grt4) -6> x -grt8 grt 2 > x x could be 0 or neg ? ### GMAT/MBA Expert GMAT Instructor Joined 12 Sep 2012 Posted: 2635 messages Followed by: 117 members Upvotes: 625 Target GMAT Score: V51 GMAT Score: 780 J N, I'd think of it this way. x - y > -2, or x > y - 2. Since y > 4, y - 2 > 2. We know that x > y - 2, so x > 2. Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now! • FREE GMAT Exam Know how you'd score today for$0
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# 89.16 kg to lbs - 89.16 kilograms to pounds
Do you want to know how much is 89.16 kg equal to lbs and how to convert 89.16 kg to lbs? Here it is. In this article you will find everything about kilogram to pound conversion - theoretical and practical too. It is also needed/We also want to emphasize that whole this article is devoted to only one amount of kilograms - that is one kilogram. So if you need to know more about 89.16 kg to pound conversion - keep reading.
Before we go to the practice - it means 89.16 kg how much lbs conversion - we are going to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s move on.
How to convert 89.16 kg to lbs? 89.16 kilograms it is equal 196.5641527992 pounds, so 89.16 kg is equal 196.5641527992 lbs.
## 89.16 kgs in pounds
We will start with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, formally known as International System of Units (in abbreviated form SI).
Sometimes the kilogram could be written as kilogramme. The symbol of this unit is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. This definition was simply but hard to use.
Later, in 1889 the kilogram was described by the International Prototype of the Kilogram (in abbreviated 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 substituted by a new definition.
Today the definition of the kilogram is build 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 can be also divided into 100 decagrams and 1000 grams.
## 89.16 kilogram to pounds
You learned something about kilogram, so now we can go to the pound. The pound is also a unit of mass. It is needed to point out that there are more than one kind of pound. What are we talking about? For example, 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. Naturally, 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 exactly 0.45359237 kilograms. One avoirdupois pound is divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was placed 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 89.16 kg?
89.16 kilogram is equal to 196.5641527992 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 89.16 kg in lbs
Theoretical part is already behind us. In next part we are going to tell you how much is 89.16 kg to lbs. Now you learned that 89.16 kg = x lbs. So it is time to know the answer. Just see:
89.16 kilogram = 196.5641527992 pounds.
That is an exact result of how much 89.16 kg to pound. It is possible to also round off this result. After rounding off your outcome is as following: 89.16 kg = 196.152 lbs.
You learned 89.16 kg is how many lbs, so see how many kg 89.16 lbs: 89.16 pound = 0.45359237 kilograms.
Obviously, this time you can also round it off. After rounding off your outcome is exactly: 89.16 lb = 0.45 kgs.
We are also going to show you 89.16 kg to how many pounds and 89.16 pound how many kg results in tables. Have a look:
We are going to start with a chart for how much is 89.16 kg equal to pound.
### 89.16 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
89.16 196.5641527992 196.1520
Now see a table for how many kilograms 89.16 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
89.16 0.45359237 0.45
Now you know how many 89.16 kg to lbs and how many kilograms 89.16 pound, so we can go to the 89.16 kg to lbs formula.
### 89.16 kg to pounds
To convert 89.16 kg to us lbs a formula is needed. We are going to show you two versions of a formula. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 196.5641527992 outcome in pounds
The first formula will give you the most accurate result. In some cases even the smallest difference could be significant. So if you want to get a correct outcome - first formula will be the best solution to know how many pounds are equivalent to 89.16 kilogram.
So let’s go to the another version of a formula, which also enables calculations to know how much 89.16 kilogram in pounds.
The another formula is down below, see:
Number of kilograms * 2.2 = the outcome in pounds
As you see, the second formula is simpler. It could be the best solution if you need to make a conversion of 89.16 kilogram to pounds in fast way, for example, during shopping. You only need to remember that your result will be not so exact.
Now we are going to learn you how to use these two formulas in practice. But before we will make a conversion of 89.16 kg to lbs we are going to show you another way to know 89.16 kg to how many lbs totally effortless.
### 89.16 kg to lbs converter
An easier way to learn what is 89.16 kilogram equal to in pounds is to use 89.16 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Calculator is based on first version of a formula which we showed you in the previous part of this article. Due to 89.16 kg pound calculator you can easily convert 89.16 kg to lbs. You only need to enter amount of kilograms which you want to convert and click ‘convert’ button. You will get the result in a second.
So let’s try to convert 89.16 kg into lbs using 89.16 kg vs pound converter. We entered 89.16 as a number of kilograms. Here is the outcome: 89.16 kilogram = 196.5641527992 pounds.
As you see, our 89.16 kg vs lbs calculator is easy to use.
Now let’s move on to our chief issue - how to convert 89.16 kilograms to pounds on your own.
#### 89.16 kg to lbs conversion
We are going to start 89.16 kilogram equals to how many pounds conversion with the first formula to get the most correct result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 196.5641527992 the result in pounds
So what have you do to check how many pounds equal to 89.16 kilogram? Just multiply amount of kilograms, this time 89.16, by 2.20462262. It is equal 196.5641527992. So 89.16 kilogram is 196.5641527992.
You can also round off this result, for instance, to two decimal places. It is equal 2.20. So 89.16 kilogram = 196.1520 pounds.
It is time for an example from everyday life. Let’s calculate 89.16 kg gold in pounds. So 89.16 kg equal to how many lbs? And again - multiply 89.16 by 2.20462262. It is exactly 196.5641527992. So equivalent of 89.16 kilograms to pounds, if it comes to gold, is equal 196.5641527992.
In this example it is also possible to round off the result. This is the outcome after rounding off, in this case to one decimal place - 89.16 kilogram 196.152 pounds.
Now let’s move on to examples converted with short formula.
#### How many 89.16 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 196.152 the result in pounds
So 89.16 kg equal to how much lbs? As in the previous example you have to multiply number of kilogram, this time 89.16, by 2.2. See: 89.16 * 2.2 = 196.152. So 89.16 kilogram is equal 2.2 pounds.
Do another conversion with use of this formula. Now convert something from everyday life, for instance, 89.16 kg to lbs weight of strawberries.
So calculate - 89.16 kilogram of strawberries * 2.2 = 196.152 pounds of strawberries. So 89.16 kg to pound mass is equal 196.152.
If you know how much is 89.16 kilogram weight in pounds and are able to convert it using two different versions of a formula, let’s move on. Now we are going to show you all outcomes in tables.
#### Convert 89.16 kilogram to pounds
We know that outcomes presented in charts are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Thanks to this you can easily make a comparison 89.16 kg equivalent to lbs outcomes.
Let’s start with a 89.16 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
89.16 196.5641527992 196.1520
And now see 89.16 kg equal pound chart for the second formula:
Kilograms Pounds
89.16 196.152
As you see, after rounding off, when it comes to how much 89.16 kilogram equals pounds, the results are not different. The bigger number the more significant difference. Remember it when you want to do bigger amount than 89.16 kilograms pounds conversion.
#### How many kilograms 89.16 pound
Now you learned how to calculate 89.16 kilograms how much pounds but we will show you something more. Are you curious what it is? What about 89.16 kilogram to pounds and ounces conversion?
We are going to show you how you can calculate it step by step. Let’s begin. How much is 89.16 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, in this case 89.16, by 2.20462262. So 89.16 * 2.20462262 = 196.5641527992. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To know how much 89.16 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 final result is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then your result is exactly 2 pounds and 33 ounces.
As you see, conversion 89.16 kilogram in pounds and ounces simply.
The last conversion which we want to show you is conversion of 89.16 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To calculate foot pounds to kilogram meters it is needed another formula. Before we give you it, let’s see:
• 89.16 kilograms meters = 7.23301385 foot pounds,
• 89.16 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 89.16 foot pounds to kilograms meters you need to multiply 89.16 by 0.13825495. It is exactly 0.13825495. So 89.16 foot pounds is exactly 0.13825495 kilogram meters.
It is also possible to round off this result, for example, to two decimal places. Then 89.16 foot pounds is 0.14 kilogram meters.
We hope that this conversion was as easy as 89.16 kilogram into pounds conversions.
This article was a big compendium about kilogram, pound and 89.16 kg to lbs in conversion. Due to this calculation you know 89.16 kilogram is equivalent to how many pounds.
We showed you not only how to make a calculation 89.16 kilogram to metric pounds but also two other conversions - to know how many 89.16 kg in pounds and ounces and how many 89.16 foot pounds to kilograms meters.
We showed you also other solution to do 89.16 kilogram how many pounds conversions, it is with use of 89.16 kg en pound converter. It is 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 can do 89.16 kilogram equal to how many pounds calculation - on your own or using our 89.16 kgs to pounds converter.
Don’t wait! Convert 89.16 kilogram mass to pounds in the best way for you.
Do you need to make other than 89.16 kilogram as pounds calculation? For example, for 10 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so easy as for 89.16 kilogram equal many pounds.
### How much is 89.16 kg in pounds
We want to sum up this topic, that is how much is 89.16 kg in pounds , we prepared one more section. Here we have for you all you need to remember about how much is 89.16 kg equal to lbs and how to convert 89.16 kg to lbs . It is down below.
How does the kilogram to pound conversion look? The conversion kg to lb is just multiplying 2 numbers. How does 89.16 kg to pound conversion formula look? . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
Now you can see the result of the conversion of 89.16 kilogram to pounds. The accurate result is 196.5641527992 pounds.
You can also calculate how much 89.16 kilogram is equal to pounds with another, shortened version of the formula. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So in this case, 89.16 kg equal to how much lbs ? The answer is 196.5641527992 lbs.
How to convert 89.16 kg to lbs quicker and easier? You can also use the 89.16 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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## Geometry: Common Core (15th Edition)
$x = 5\sqrt 2$
In a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, the hypotenuse is $\sqrt 2$ times each leg. Let's write an equation to solve for $x$, one of the legs: $10 = \sqrt 2(x)$ Divide each side of the equation by $\sqrt 2$ to solve for $x$: $x = \frac{10}{\sqrt 2}$ To simplify this fraction, we multiply both the numerator and denominator by the denominator: $x = \frac{10}{\sqrt 2} • \frac{\sqrt 2}{\sqrt 2}$ Multiply to simplify: $x = \frac{10\sqrt 2}{2}$ Divide the numerator and denominator by their greatest common factor, $2$: $x = 5\sqrt 2$
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https://www.jiskha.com/questions/106054/find-de-if-d-2-4-6-and-e-1-2-5-7-1-0-4-3-4-i-followed-the-example-in-the
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Find DE if D=[-2 4 6] and E=[1 -2]
[5 -7 1] [0 4]
[-3 4]
I followed the example in the book on how to multiply matrices and one I found on the Internet and keep coming up with the same answer which is not one of the solutions given to choose from.
WORK: [(-2)(1)+(4)(0)+(6)(-3)]=-20
[(-2)(-2)+(4)(4)+(6)(4)]=44
[(5)(1)+(-7)(0)+(1)(-3)]=2
[(5)(-2)+(-7)(4)+(1)(4)]=-34
I keep getting [-20 44]
[2 -34]
A) [20 44] B) [-20 8]
[8 42] [44 42]
I don't know where I am going wrong, so hopefully this will stay aligned and you can see where I am making my mistake!!
Thanks.
1. 👍
2. 👎
3. 👁
1. did not stay aligned so I will try something else.
1. 👍
2. 👎
-20__44
__2_-34
1. 👍
2. 👎
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# Math 22 Lagrange Multipliers
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Method of Lagrange Multipliers
If ${\displaystyle f(x,y)}$ has a maximum or minimum subject to the constraint ${\displaystyle g(x,y)=0}$, then it will occur at one of the critical numbers of the function defined by
${\displaystyle F(x,y,\lambda )=f(x,y)-\lambda g(x,y)}$.
In this section, we need to set up the system of equations:
${\displaystyle F_{x}(x,y,\lambda )=0}$
${\displaystyle F_{y}(x,y,\lambda )=0}$
${\displaystyle F_{\lambda }(x,y,\lambda )=0}$
Example: Set up the Lagrange Multipliers:
1) Maximum: ${\displaystyle f(x,y)=xy}$ and Constraint ${\displaystyle x+y-14=0}$
Solution:
So, ${\displaystyle F(x,y,\lambda )=f(x,y)-\lambda g(x,y)=xy-\lambda (x+y-14)=xy-\lambda x-\lambda y+14\lambda }$
${\displaystyle F_{x}(x,y,\lambda )=y-\lambda }$
${\displaystyle F_{y}(x,y,\lambda )=x-\lambda }$
${\displaystyle F_{\lambda }(x,y,\lambda )=-x-y+14}$
2) Maximum: ${\displaystyle f(x,y)=xy}$ and Constraint ${\displaystyle x+3y-6=0}$
Solution:
So, ${\displaystyle F(x,y,\lambda )=f(x,y)-\lambda g(x,y)=xy-\lambda (x+3y-6)=xy-\lambda x-3\lambda y+6\lambda }$
${\displaystyle F_{x}(x,y,\lambda )=y-\lambda }$
${\displaystyle F_{y}(x,y,\lambda )=x-3\lambda }$
${\displaystyle F_{\lambda }(x,y,\lambda )=-x-3y+6}$
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| 651,146,757 | 35,285 |
# File
Warm Up #1
• What are some various forms of energy (what
do we use to power stuff)?
• Of the ones you listed in #1, which ones are
better? Why are they better?
• What is horsepower?
Warm Up #2
• Where have you seen the word “watts” being
used before? What do you think it means?
• How can you tell a car, like a Acura NSX, is
powerful? (i.e. what quality about the car
makes it powerful?)
• How many meters are in a kilometer? What
about a megameter (yes, that’s a real thing)?
Geosphere
Energy: Renewable and NonRenewable
Energy Math
Part I
Laws of Energy (Thermodynamics)
• Law 1: Energy cannot be created or destroyed
– You can’t make something out of nothing
• Law 2: When energy is converted from one
form to another, some of that energy is lost
– Moving up food chain, some energy lost as HEAT
Other Forms of Energy: Part 1
• Horsepower (HP) - measures the power
generated by horses (original definition)
– Ex. 1 horsepower = power of 1 horse
Conversion:
• 1 HP = 746 watts
– Random, I know. Sorry about it.
• Ferrari vs. Prius
Other Forms of Energy: Part 2
• British Thermal Units (Btu) – Amount of
energy needed to raise 1 lb. of water by 1oF
– Think of it like a calorie (food)
Random Conversions:
• 1 watt = 3.4 Btu
• 1 HP = 2,540 Btu
Measuring Energy: Watts
• Watt – standard unit of measurement for
energy
• Two forms you will see:
• Kilowatts (1 kW = 1000 watts)
– 1x103 watts
• Megawatts (1 MW = 1,000,000 watts)
– 1x106 watts
Kilowatt Hours & Your Electric Bill
• Your electric bill – measures amount of power used
over a period of time
• You will see: Kilowatt Hours (kWh) – amount of watts
generated in 1 hour
– Kilowatts x hours = kWh
For these conversions:
• Know what you are starting with
• Know what they are asking for
• Convert accordingly
Warm Up #6
A resident of Beverly Hills, a community of 10,000
homes, uses on average about 5,000 kilowatt hours
of electricity per year. The cost of electricity in this
community is \$0.20 per kWh.
• How much electricity would a resident use in a 20
year period?
• How many kWh of electricity does all of Beverly
Hills use in one year?
• What would their electric bill be as a community for
twenty years?
Warm Up #7
The Cobb family’s current electric hot-water heater requires
0.20kWh/gallon of energy, costing \$0.10/kWh. Each person in
the family of four showers once a day, 10 min. each. Each
shower is approx. 5.0 gallons per min. Calculate:
• Total amount of water the family uses per year taking showers.
• The annual cost of electricity for family showers, assuming 2.5
gallons per min. of water is used from the water heater.
• The family wants to replace their hot-water heater with a new,
\$1,000 more efficient model. It requires only 0.10 kWh/gallon
of energy. How many days would it take to recover the initial
costs?
• Describe two practical measures the family could take to
reduce overall water use.
• Describe two practical measures that the Cobb family can
lower their overall energy consumption (other than hot water).
Practice Problem
West Fremont is a community of 3,000 homes. A
small coal-burning plant supplies the electricity of
the town, with a capacity of 12 megawatts (MW).
The average household consumes 8,000 kWh of
electrical energy every year, at a cost of \$0.10 per
kWh. The town leaders suggest switching to wind
power. They want to install 10 wind turbines, each
with a 1.2 MW capacity, and each costing \$3
million. Their lifespan is 25 years.
Practice Problem Questions
(a) Assuming the power plant is operating at full capacity
for 8,000 hours/yr. How many kWh of electricity can
be produced by the plant each year?
(b) At the current rate of energy per household, how
many kWh of electrical energy does the community
consume in 1 year?
you would not expect their answers to match.
(d) Assuming the energy needs do not change during the
25-year lifespan of the wind turbines, what would the
cost of electricity be over 25 years (in \$/kWh)?
(e) Identify and explain TWO environmental benefits of
switching to wind power, and TWO environmental
costs for switching to wind power from coal.
Warm Up # 3
A wind turbine in Palm Springs, a community with
800 turbines, uses on average about 10,000
kilowatt hours of electricity per year. The cost of
electricity in this community is \$0.20 per kWh.
• Calculate the amount of energy generated from 1
turbine in 15 years.
• Calculate the amount of energy generated by all
of the wind turbines in 1 day
• For this day, how much would it cost the people
of Palm Springs
Practice Problem
For a certain dairy farm with 500 cows, the cost of installing a
digester is approximately \$400,000. Assume that the farm uses
800,000 kilowatt-hours (kWh) of electricity each year at a cost of
\$0.10 per kWh. The waste from a single cow can produce 3.0
kWh of electricity each day.
(c) Assuming that the cost of electricity remains constant and the
farmer starts using the manure from the cows in an anaerobic
digester to produce electricity on the farm, calculate:
• (i) The number of kWh of electricity that can be produced in one
year
• (ii) The amount of money the farmer can save in one year, NOT
counting the installation cost of the digester. (You may round your
• (iii) The amount of time, in years, that it will take to recover the
cost of installing an anaerobic digester on the farm. (You may
(d) Calculate the minimum number of cows the farm would need to
produce 800,000 kWh of electricity per year.
Warm Up #4
You live in a community of 3,000 homes that rely on coal as a source of fuel.
The average home uses 8,000 kwh of energy per year at the cost of
\$.10/kwh.
Instead, town members want to switch to wind power; installing 10 wind
turbines (\$3 million each) which will last 25 years.
• List two pros and two cons for switching to wind power.
• How much energy does your entire community use per year?
• If the energy needs stay the same over the 25 year lifetime of
the wind turbines, what would be the cost for the community
for wind power (in \$/kwh)? How does it compare to the cost
of coal?
13 cards
28 cards
15 cards
30 cards
23 cards
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How do you determine the limit of (x+2) / (x+3) as x approaches 3-?
Aug 1, 2017
As $x$ approaches $3$ from the left,
the numerator approaches $5$ and
the denominator approaches $6$.
Since the denominator does not approach $0$, we know that the quotient approaches $\frac{5}{6}$ and so the limit is $\frac{5}{6}$.
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# draft
Time：2022-5-1
Hello, children. I’m your teacher Yi. Let’s learn the double insertion problem together today. [1.9 seconds] let’s take a look at the topic first. After moving the decimal point of a number to the right by one digit, the number obtained is 108 more than the original number. What is the original number? After reading it, don’t worry about the formula. Let’s analyze the topic first. [1.9 seconds] what does it mean to move the decimal point of a number one digit to the right? Is to expand this number by 10 times. [1.9 seconds] how is it represented by a graph? We can use a circle to represent the original number, so how many circles to represent the expanded number? [1.9 seconds] Yes, we use 10 circles. The number obtained is 108 more than the original number, [1.9 seconds]. In the figure, we can see that this is the extra part, that is, this part is 108. And this part has 10 minus 1 circle, that is to say, the number represented by these 9 circles is 108, so what is the number represented by a circle? We can list the formula: [1.1 seconds] 108 divided by 10 minus 1, and the final result is equal to 12 In other words, the original number was 12 Is it very simple? Let’s try a problem! [1.1 seconds] look at the title. [1.1 seconds] the sum of a and B is 803. After moving the decimal point of a to the left, the number obtained is equal to B. What are the numbers of a and B respectively? [1.1 seconds] similarly, through the drawing method, we can get the following formula: [1.1 seconds] number B: 803 divided by 10 plus 1 equals 73, [1.1 seconds] number A: 73 times 10 equals 730 or 803 minus 73 equals 730. Children, have you learned? [1.1 seconds] that’s all for today. Bye.
## Fastems.Mms.Client.BaseData.Service.BaseDataService之FixtureSummaries
in Fastems.Common.dll, Fastems.Mms.Client.BaseData.Views， In Fixtures/Views/FixtureLibraryView.xaml, the data source of Fixtures has been unclear. The records are as follows. The data source of the View is the Fixtures in Fastems.Mms.Client.Fixtures.ViewModel.FixtureLibraryViewModel, And the source of Fixtures is Fixtures = new PagingCollection(_fixtures, FixtureFilterSources); BatchingObservableCollection _fixtures; _fixtures creates a collection object BatchingObservableCollection in the constructor public class BatchingObservableCollection : […]
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# Material thickness to withstand Pressure
Let's say I have a Sphere made out of Acrylic, with a diameter d meters.
And I want to submerge it down in the ocean. The depth is such that it reaches up to p PSI.
How thick does the Sphere have to be, in order to withstand pressure of up to p PSI?
(I used the variables d and p so I can be given a formula that I can learn/understand instead of a raw answer.)
• You need the strength of the material to start with and there are questions similar to this on this site... – Solar Mike Sep 8 '17 at 21:07
take note if you are talking about dropping to the deepest parts of the ocean the pressure can be as high as 15,000 psi. the Yield strength of acrylic is maybe 10,000 psi.
Your wall will be very thick! You can not use most of the thin shell based formulae you might look up.
You can take an extreme view and imagine a large block of material with a tiny void. The wall hoop stress at the void wall is approximately 3/2*pressure, or over 20,000 psi. Which is to say your cavity is liable to collapse no matter how thick you make it.
The basic formula used in analyzing thin-walled spherical pressure vessel is:
stress = PD / 4t
where P is the pressure inside or outside, D is the inner diameter, and t is the wall thickness.
therefore:
t = PD / [(allowable stress of the material) * 4]
Note that this formula is only applicable if D/t >= 10
Note also that allowable stress in acrylic is 10,000 psi @ room temperature. This changes when you submerge this in water, so might want to consider this too.
Edit: The unit of measurement in the formula is:
For English system: stresses should be in pounds per sq.inch (psi) while dimensions should be in inches (in.)
For SI system: stresses should be in megaPascal (MPa), and dimensions should be in millimeters (mm)
• Why does temperature have to be involved in this when it's not involved in the Formula ^? On datasheets of materials (found online), when they mention how much pressure (PSI) the material can hold, at what conditions do they mean it? What temperature, thickness, shape etc.? Also, what is the unit measurement for each variable in the formula ^? – Coto Sep 9 '17 at 0:33
• Material tend to loose strength when they are heated or cooled. Imagine a steel bar being heated, it will lose its elasticity. Same goes when you are going to submerge in deep ocean, where temperature starts to vary @ very large depths. I am only saying this because room temperature is absolutely different from sea temperature. Thus, he needs to consider the case in determining the allowable stress of material. :) – Jem Eripol Sep 9 '17 at 0:38
• Where can I find how much a material gets affected under x Degrees? – Coto Sep 9 '17 at 0:41
• @JemEripol why will thin wall expressions be suitable in this case? – Solar Mike Sep 9 '17 at 21:00
• @JemEripol so in short, you have provided a "solution" that will not give correct results in the situation described. Why do you think the different formulae for thin and thick pressure vessels were developed ? Boilers exploded due to this... – Solar Mike Sep 11 '17 at 6:07
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# According to the table, the median acreage managed by the Federal Land
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Re: According to the table, the median acreage managed by the Federal Land [#permalink]
1
Kudos
rishi02 wrote:
According to the table, the median acreage managed by the Federal Land Management agencies is how many million acres?
A 87
B 92
C 133
D 191
E 270
Attachment:
Median.png
Solution:
Arranging the numbers from the smallest to the largest, we have:
25, 87, 92, 191, 270
Therefore, the middle value of 92 is the median.
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Re: According to the table, the median acreage managed by the Federal Land [#permalink]
Solution
Given
In this question, we are given that
• A table containing the data related to the acreage managed by different agencies, and the average millions of acres that is managed by each of the agencies
To find
We need to determine
• The median acreage managed by the Federal Land Management agencies
Approach and Working out
If we arrange the numbers either in ascending order or descending order, the median will be the middle-most number
(note that, as the number of elements here is an odd number, there will be only one middle element here)
• Ascending order = 25, 87, 92, 191, 270
• Descending order = 270, 191, 92, 87, 25
We can see in either of the cases, the middle element is 92, which is the median here
Thus, option B is the correct answer.
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Re: According to the table, the median acreage managed by the Federal Land [#permalink]
What if ascending and descending orders differ? What would be in that case?
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Re: According to the table, the median acreage managed by the Federal Land [#permalink]
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Re: According to the table, the median acreage managed by the Federal Land [#permalink]
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# Witch 96: Root Con All
This is the fourth of our three WitCHes on VCAA’s Specialist Mathematics Exam 1 Sample Questions. Yeah, yeah it’s Douglas Adams-ish, but it wasn’t deliberate. We were undecided on this last one, and it’s not as compelling as the others. But, for similar reasons for the other WitCHes, we changed our mind and decided to post it. (There were other close calls, and we’ll soon update this post with brief comments on all the questions.)
### UPDATE (14/02/23)
This one is interesting. As noted above, it was an afterthought to WitCH it; it is not nearly as bad as the two other contradiction questions considered in the webinar; although there are simple direct proofs, it is not totally absurd to ask this as a proof by contradiction. Still, as commenters have noted, there are definitely aspects to criticise, and there’s plenty to discuss and to clarify.
First of all, let’s give some direct proofs.
Most simply, we have Anonymous’s one-liner:
√3 + √5 > √3 + √3 = 2√3 = √12 > √11.
A clunkier version, which Anonymous also proposed, is to bluntly estimate the roots. For instance,
√3 + √5 > 1.5 + 2 = 7/2,
and then just note that (7/2)2 > 11.
If nothing else, these two proofs point out the importance of choosing numbers in problems with proper care: if VCAA had gone with √13 or similar, such a direct proof would have required either more cleverness or more work.
Our immediate reaction to the problem was a third direct proof, which begins with first squaring the LHS:
(√3 + √5)2 = 3 + 5 + 2√15 = 8 + 2√15.
We then want to see if this is greater than the square of the RHS, and of course that’s easy:
8 + 2√15 > 8 + (2 x 3) = 14 > 11.
Since LHS2 > RHS2, and since both sides are positive, we also have LHS > RHS. Done.
The expected contradiction proof is, in fundamental working, very similar to this third direct proof. Assuming the conclusion is false, we have
√3 + √5 ≤ √11.
Squaring gives
8 + 2√15 ≤ 11,
and we easily get the desired contradiction.
Now, a couple questions. First of all, which is preferable: the third direct proof or the proof by contradiction? It is perhaps a matter of taste, but for us the direct proof is clearly preferable, for two reasons. First of all, direct proofs are always preferable: why enter the fantasyland of contradiction if it is not required? Secondly, the direct proof begins with a quantity and then manipulates and estimates that quantity; this seems to us much more natural than manipulating statements, than manufacturing a chain of inequalities.
The second question concerns the following equivalence, for positive x and y:
All the above proofs use one or the other direction of this inequality, and necessarily so: in the end, we only know something about a square root through squaring it. Can both directions be assumed by students? Is one direction “easier” than the other? Note that the reverse implication is standardly used in solving questions such as Q16 of the sample questions.
As with the induction WitCH, this seems a case where VCAA has not stated the ground rules for the students, and so it is simply impossible for a student to be sure what may or may not be assumed.
## 50 Replies to “Witch 96: Root Con All”
1. Anonymous says:
It’s very easy to prove this directly in a single line:
√3 + √5 > √3 + √3 = 2√3 = √12 > √11
What the hell does VCAA actually want us to use proof by contradiction for? I can’t actually think of any useful way to use proof by contradiction to solve this problem.
In fact, if I were to do this question in an exam, I’d probably write the following:
Line 1: √3 + √5 > √3 + √3 = 2√3 = √12 > √11
Line 2: Let us assume that √3 + √5 ≤ √11.
Line 3: By line 1, this is false.
Line 4: Therefore, by proof of contradiction, √3 + √5 > √11.
I wonder how VCAA would mark that proof “by contradiction.”
1. marty says:
Now that’s funny.
Thanks very much, A. To be honest I didn’t think of directly estimating the roots as you did. But even without that, proof by contradiction seems a silly route.
1. Anonymous says:
Yeah, my first reaction to this question was just to directly compute the roots (to be more precise upper and lower bounds for the roots). But I then realized you could just do the trick I did above.
Never would I ever have thought of using proof by contradiction.
2. Tungsten says:
Assume √3 + √5 ≤ √11
√3 + √5 ≤ √11
3 + 2√15 + 5 ≤ 11
√60 ≤ 3
60 ≤ 9
Contradiction therefore √3 + √5 > √11.
Correct me if there are any mistakes but I think that I would do it that way if I were asked that question. Anyway, I have no particular issue with asking for a specific proof technique although the fact that VCAA seems to do so I think is a separate problem.
1. marty says:
Yes, that’s fine, and clearly what VCAA expected. And it’s kind of ok to ask for this contradiction proof, even if there are easier direct methods. That’s why I hesitated to post this one. But there’s something else that bugs me about the proof you’ve given.
1. Amber says:
Is it something to do with squaring and square rooting both sides of an inequality?
1. marty says:
Thanks, Amber, and yes. Would that be permitted or not? Can it be used in a direct proof?
1. Amber says:
I’ve been thinking about that – it seems a clear potential trap for students – yes if the proof is over natural numbers, or R+? Here no number set is defined but everything is positive so it works but is the issue with only taking the positive square root in that step? There was an example given at the MAV conference recently with an inequality geometric proof involving triangles and pythagoras where the lecturer explicitly stated that lengths needed to be positive so I assumed we needed that line to make the proof work.
1. Red Five says:
What if we first make a zero right hand side? Does that then fix the issue, because we can assume the LHS is positive/negative and hence squaring both sides either will or will not reverse the inequality.
2. Amber says:
oops just realised there is no square rooting only squaring…
2. Tungsten says:
Squaring both sides of the inequality would be permitted I think, but clearly requires a high degree of care. For example, squaring any inequality involving a negative on either side, and the fact that does not imply . In my proof above we do assume that everything we need is positive as Amber mentioned and then the argument is valid. I see no reason why a direct proof would have different permissions but maybe somebody else can give some insight on that.
2. marty says:
Thanks, Amber, and Tungsten and RF. You’ve caught onto the reason I posted this question. I will try to focus your puzzling.
First of all, the negative numbers thing is mostly a red herring. We’re dealing with explicit positives here. So, yes, it may be reasonable or VCAA-mandated to declare the positivity, but that’s just bookkeeping.
Now, sticking to positives,
(a) Does x2 > y2 imply that x > y?
(b) If so, can that form part of a direct proof of VCAA’s sample question?
(c) If so, would that inequality in (a) first have to be proven?
If the answer to all three is “yes”, then i think we have a problem.
1. Red Five says:
implies .
So .
Now, if then so we can conclude
Which means .
1. marty says:
That answers (a).
1. Red Five says:
(b) is beyond my ability to comprehend.
Or rather: I say it can. What VCAA says might be a different matter.
1. marty says:
No it isn’t. Try to give a direct proof. How might you go about it, if you want to avoid directly estimating the roots?
1. marty says:
No. Forget about what VCAA might want or might do. How could you use (a) to give a direct proof of the sample question?
2. marty says:
I got rid of your 0, but there seem to be other typos. Not sure what “assume” is assuming. (And why assume anything? It’s meant to be a direct proof.)
If you tell me the corrections (by email if you wish), I’ll correct.
2. marty says:
Sorry, my (b) was unclear, and probably unmeaningful.
What I’m asking for is an easy direct calculation (no contradiction) that establishes THIS > THAT. Then, that inequality combined with (a) constitutes a proof of the sample question.
1. Red Five says:
I think such an argument IS possible, but the one I sketched out looks really ugly.
The argument Anonymous proposes at the top of this thread is so much nicer to look at…
2. aps says:
are you looking for a direct version of Tungsten’s proof? as in:
(using your (a) and its converse)
(again using (a) and its converse)
True
or do you mean something completely different?
1. Red Five says:
Something like that except starting from the final line and working backwards.
Otherwise, it may be argued that you are starting from the assumption that the inequality is True which VCAA may not like.
1. marty says:
They’re equivalences, so it’s ok. (Which may not stop VCAA being stupid enough to object.)
2. marty says:
Yes, that’s it. Although I don’t like the chain of equivalences. it seems preferable to me to start with LHS2 and then compute. Then employ (a) at the very end.
Now, what about (c)?
1. Tungsten says:
I am not sure what you mean about your preference. It seems that aps’ method, like my contradiction proof, relies on (the converse of the statement in (a)) or similar, which I am not sure that (a) proves. Red Five’s suggestion would use (a) directly. Anyway, on (c) we might as well prove (a) or its converse as appropriate or else accept it as trivial. With a hypothetical VCAA exam, make a guess based on allocated marks maybe.
1. Tungsten says:
For 3 marks in the sample question using my contradiction proof I think it is safest to prove first. Alternatively, it might be sufficient to justify it by explicitly stating that at the relevant steps.
1. marty says:
Thanks, Tungsten, and aps. I’ll update the post soon with my thoughts (which are mostly questions).
Have a look at Q16 in the sample questions. Does that change your thoughts at all?
2. marty says:
No.
Your contradiction proof uses .
aps’s direct equivalence proof uses .
My suggested revamping of aps’s proof uses .
2. aps says:
Here’s my very amateur and probably erroneous take:
It feels to me like there’s a logical/mathematical side to (c) and then a presentation/‘aesthetic’ side?
Logical side: to use (a), we need to know (a); to know (a), we need to prove (a), (or (a) can be an axiom). But this can be said of many assumptions that are made in the proof, such as sqrt(3)*sqrt(5)=sqrt(15): we have to know sqrt(a)*sqrt(b)=sqrt(ab), so we have to prove this as well; same as difference of two squares in Red Five’s proof of (a). And in proving that, we might come across associativity of multiplication; in this strictly logical side we’d have to prove that as well…
Presentation side: this would consider the level of detail/depth that needs to be shown, given we aren’t going to actually write proofs of all the things the logical side requires us to prove. I would imagine the goal here would be to keep the ‘depth’ of the proof relatively consistent, if that makes sense. I think this would be better achieved by including a proof of (a), since if several steps of arithmetic working aren’t trivial enough to leave out, then surely (a) isn’t trivial enough either. I’m sure VCAA would disagree. But it looks a bit like a matter of opinion to me…
1. marty says:
Thanks, aps. See my reply just now to Tungsten.
2. Red Five says:
It is 3 marks. I really don’t think much more than any of the correct proofs presented here could be marked as wrong.
1. marty says:
Direct proofs would be marked wrong. But my point is to be thinking more carefully about what can be assumed when doing SM proofs.
3. Red Five says:
This maybe very wide of the mark, but is the opposite of “greater than” the symbol “less than” or is it “less than or equal to”?
Nit-picky perhaps, but I’m genuinely curious.
1. aps says:
Using my (very limited) knowledge, I would suggest that given ‘P or not(P)’ is a tautology, then not(a > b) would have to be a <= b, since a negation has to be in some sense a complement. I don't know for certain though.
2. marty says:
First of all, never say “opposite” when you mean “negation”. You’re likely to end up working for VCAA, and we wouldn’t want that, would we?
Now, as to your question, aps is correct: the negation of “a > b” is “a ≤ b”.
This following may seem jargonistic, but it may (or may not) help.
To make a statement S is the same as saying ” ‘S’ is true”. That doesn’t mean the statement is true, it’s just describing what is being said.
So, saying
(*) a is greater than b
is the same as saying
(**) The statement ‘a is greater than b’ is true.
The negation of S is the statement ” ‘S’ is false”. So, the negation of (*) is
(***) The statement ‘a is greater than b’ is false.
But what does it mean for ‘a is greater than b’ to be false? It means exactly that a is less than or equal to b.
In symbols, the negation of
****) a > b
is
*****) a ≤ b
1. Red Five says:
Perfect. Thanks.
1. Banacek Spaces says:
I went with AM GM inequality. Also needed is the fact that x>y implies that x^2>y^2
as well as sqrt(x) > sqrt(y)
Just by recalling that y=x^2 and y= sqrt(x) are increasing functions for positive real x.
1. Banacek Spaces says:
Only sqrt things are actually needed.
4. Terry Mills says:
Re: Q1 in Specialist Mathematics Exam 1 Sample Questions.
In the Victorian Curriculum, N = {0,1,2,….}. This seems to make a difference.
1. Red Five says:
And some documents published recently use both N={0,1,2,3…} and N={1,2,3,…} in the same document…
I’m not sure we can say for certain that 0 is a natural number as defined by the Victorian curriculum.
Which is a source of much frustration.
2. marty says:
That’s hilarious. But, in practice, I think it’s a red herring.
The VCAA glossary for K-10 includes 0 in the naturals. Everything I’ve seen in VCE, however, suggests that n = 1, so I don’t think there’s any practical ambiguity in VCE. I cannot see that the VCE study design defines N, but all the texts I’ve seen, as well VCAA’s support materials, start N with 1.
5. aps says:
I had a look at Q16. I suppose the similarity is that one solution involves assuming that the x-values for min(f(x)) and min(sqrt(f(x))) are the same, which (I think? correct me if I’m wrong) is implied by x sqrt(x) < sqrt(y); and then the question we’re asking would be, should students show that x sqrt(x) < sqrt(y), or otherwise show that the x-values for min(f(x)) and min(sqrt(f(x))) are equal, as part of this solution?
For this question I would say no, because it’s a ‘find’ question rather than a ‘prove’ question. I do think they should state their ‘x-values-are-equal-for-min(f(x))-and-min(sqrt(f(x)))’ assumption somewhere, because it doesn’t seem quite trivial enough to be left implicit in a 4-mark question. But again, that’s just a matter of personal opinion…
I think that’s the real problem, that it’s unclear just how much detail needs to be shown in questions like 5 and 16; VCAA isn’t really clear on what exactly is required. I wonder if different assessors will mark these questions differently based on their personal interpretations of how much detail is necessary (which I suspect already happens a fair bit in show-that questions!). I suspect there will be Issues when it comes to marking proof questions…
1. marty says:
Thanks, aps, and yes, that’s my point.
Forget the “find” and “prove” language thing. The point is, students routinely find a minimum distance and its location by finding instead the minimum distance2. This is done, correctly, without comment, but implicitly relies upon
(*) .
That’s fine. But are the same students in the same subject then supposed to know in a “proof” question, that (*) needs to be proven if used?
1. aps says:
Looking at the webinar (ew) VCAA expects them to state but not prove (*), so to answer that question, seemingly not. It is weird that (*) can be assumed, but (referencing WitCH 94) odd*odd=odd and similar facts can’t. And then, in the ln(5)-is-irrational webinar proof, it’s assumed that 5 can’t divide e^a, which imo isn’t at all trivial! Maybe VCAA should release a guidebook or something…
1. marty says:
A VCAA guidebook? Wrongway Feldman comes to mind.
Thanks, aps. You’re referring to slide 10 of the contradiction webinar? That assumes only
(**) .
I’m not sure VCAA would accept the reverse inequality so readily.
In case you’re in doubt, VCAA’s log 5 proof is utter nonsense.
1. aps says:
Ah, that’s true about Slide 10. Still, I suspect the slide will give many teachers the impression that
(***) x sqrt(x) y^2, then (x+y)(x-y)>0. (1 mark)
b) Hence, show that x^2>y^2 => x>y. (1 mark)
c) Using the statement from (b), show that sqrt(3) + sqrt(5) > sqrt(11) (3 marks)
…yeah.
1. aps says:
Oh no, my comment seems to have gotten corrupted somehow. Maybe I accidentally deleted a chunk of it..? (I tried to edit but it didn’t seem to work).
Anyway, my point was to be that VCAA probably wouldn’t have faith in Specialist students’ ability to prove
(***) x <y —> sqrt(x) < sqrt(y)
on their own, looking at Red Five’s proof and Banacek Spaces’ increasing function argument. So I was trying to write an example of the scaffolded question they might write if they wanted (***) proved as part of question 5. 5a) was:
Show that, if x^2<y^2, then (x+y)(x-y) < 0 (1 mark).
b) and c) were left intact in the original comment.
1. marty says:
If you want me to edit or delete anything, let me know.
1. aps says:
Yeah. My initial comment said (something like):
*****
Ah, that’s true about slide 10. Still, I suspect the slide will give many teachers the impression that
(***) x < y implies sqrt(x) < sqrt(y)
can also be stated without being proved. And I think students and teachers will be right to feel misled if, on the exam, marks are deducted for not writing a proof of (***). Not that this would stop VCAA from doing it.
What I think WILL stop VCAA from requiring a proof of (***), is that they won’t have faith in students’ ability to provide one. A proof like Red Five’s, or a tight argument about strictly increasing functions, doesn’t feel trivial enough for a Specialist exam. Especially if it’s tucked away inside another question. I’d imagine that, if VCAA wanted a direct proof of question 5 including a proof of (***), they’d ask something like:
5a) Show that, if x^2 < y^2, then (x+y)(x-y) <0,
b) Hence, show that x < y implies sqrt(x) < sqrt(y), for positive x and y (2 marks)
c) Using the statement from part (b), show that sqrt(3) + sqrt(5) > sqrt(11) (2 marks)
*****
If you could edit my initial reply to say this, it would be much appreciated 🙂
6. marty says:
I’ve updated the post with a summary.
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A198442 Number of sequences of n coin flips that win on the last flip, if the sequence of flips ends with (1,1,0) or (1,0,0). 12
0, 0, 2, 3, 6, 8, 12, 15, 20, 24, 30, 35, 42, 48, 56, 63, 72, 80, 90, 99, 110, 120, 132, 143, 156, 168, 182, 195, 210, 224, 240, 255, 272, 288, 306, 323, 342, 360, 380, 399, 420, 440, 462, 483, 506, 528, 552, 575, 600, 624, 650, 675, 702, 728, 756, 783, 812 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS If the sequence ends with (1,1,0) Abel wins; if it ends with (1,0,0) Kain wins. Abel(n) = A002620(n-1) = (2*n*(n - 2) + 1 - (-1)^n)/8. Kain(n) = A004526(n-1) = floor((n - 1)/2). Win probability for Abel = sum(Abel(n)/2^n) = 2/3. Win probability for Kain = sum(Kain(n)/2^n) = 1/3. Mean length of the game = sum(n*a(n)/2^n) = 16/3. Essentially the same as A035106. - R. J. Mathar, Oct 27 2011 The sequence 2*a(n) is denoted as chi(n) by McKee (1994) and is the degree of the division polynomial f_n as a polynomial in x. He notes that "If x is given weight 1, a is given weight 2, and b is given weight 3, then all the terms in f_n(a, b, x) have weight chi(n)". - Michael Somos, Jan 09 2015 REFERENCES A. Engel, Wahrscheinlichkeitsrechnung und Statistik, Band 2, Klett, 1978, pages 25-26. LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..10000 J. McKee, Computing division polynomials, Math. Comp. 63 (1994), 767-771. MR1248973 (95a:11110) Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1). FORMULA a(n) = (2*n^2 - 5 - 3*(-1)^n)/8. a(2*n) = n^2 - 1; a(2*n+1) = n*(n + 1). a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) with n>=4. G.f.: x^3*(2 - x)/((1 + x)*(1 - x)^3). - R. J. Mathar, Oct 27 2011 a(n) = a(-n) for all n in Z. a(0) = -1. - Michael Somos, Jan 09 2015 0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-1 - a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Jan 09 2015 1 = a(n) - a(n+1) - a(n+2) + a(n+3), 2 = a(n) - 2*a(n+2) + a(n+4) for all n in Z. - Michael Somos, Jan 09 2015 a(n) = A002620(n+2) - A052928(n+2) for n >= 1. (Note A265611(n) = A002620(n+1) + A052928(n+1) for n >= 1.) - Peter Luschny, Dec 22 2015 a(n+1) = A110654(n)^2 + A110654(n)*(2 - (n mod 2)), n >= 0. - Fred Daniel Kline, Jun 08 2016 a(n) = A004526(n)*A004526(n+3). - Fred Daniel Kline, Aug 04 2016 a(n) = floor((n^2 - 1)/4). - Bruno Berselli, Mar 15 2021 EXAMPLE For n = 6 the a(6) = 8 solutions are (0,0,0,1,1,0), (0,1,0,1,1,0),(0,0,1,1,1,0), (1,0,1,1,1,0), (0,1,1,1,1,0),(1,1,1,1,1,0) for Abel and (0,0,0,1,0,0), (0,1,0,1,0,0) for Kain. G.f. = 2*x^3 + 3*x^4 + 6*x^5 + 8*x^6 + 12*x^7 + 15*x^8 + 20*x^9 + ... MAPLE for n from 1 by 2 to 99 do a(n):=(n^2-1)/4: a(n+1):=(n+1)^2/4-1: end do: seq(a(n), n=1..100); MATHEMATICA a[ n_] := Quotient[ n^2 - 1, 4]; (* Michael Somos, Jan 09 2015 *) PROG (Perl) sub a { my (\$t, \$n) = (0, shift); for (0..((1<<\$n)-1)) { my \$str = substr unpack("B32", pack("N", \$_)), -\$n; \$t++ if (\$str =~ /1.0\$/ and not \$str =~ /1.0./); } return \$t } # Charles R Greathouse IV, Oct 26 2011 (PARI) a(n)=([1, 1, 0, 0, 0, 0; 0, 0, 1, 1, 0, 0; 0, 1, 0, 0, 1, 0; 0, 0, 0, 1, 1, 0; 0, 0, 0, 0, 0, 2; 0, 0, 0, 0, 0, 2]^n)[1, 5] \\ Charles R Greathouse IV, Oct 26 2011 (PARI) {a(n) = (n^2 - 1) \ 4}; /* Michael Somos, Jan 09 2015 */ (MAGMA) [(2*n^2-5-3*(-1)^n)/8: n in [1..60]]; // Vincenzo Librandi, Oct 28 2011 (Sage) def A198442(): yield 0 x, y = 0, 2 while True: yield x x, y = x + y, x//y + 1 a = A198442(); print([next(a) for i in range(57)]) # Peter Luschny, Dec 22 2015 CROSSREFS Cf. A000004, A002620, A004526, A004652, A005843, A008585, A008586, A023443, A028242, A047221, A047336, A052928, A242477, A265611. Cf. A035106, A079524, A238377. Sequence in context: A103567 A277913 A131723 * A035106 A122378 A181687 Adjacent sequences: A198439 A198440 A198441 * A198443 A198444 A198445 KEYWORD nonn,easy AUTHOR Paul Weisenhorn, Oct 25 2011 EXTENSIONS a(12) inserted by Charles R Greathouse IV, Oct 26 2011 STATUS approved
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https://www.univerkov.com/the-perimeter-of-the-triangle-abc-is-62-cm-bc-17-cm-abc-find-the-length-of-the-side-ab-2/
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# The perimeter of the triangle ABC is 62 cm BC = 17 cm ABC find the length of the side AB
1. The perimeter of a triangle is equal to the sum of all sides, and our two unknown sides are equal, so AB = AC = x.
2. P = AB + AC + BC.
3. P = 2x + BC,
4. Substitute the values given in the condition: 62 = 2x + 17,
5. Move the term from x to the left side, and free terms to the right: 2x = 62 – 17,
6.2x = 45,
7. Divide both sides of the equation by 2: x = 22.5, so AB = 22.5.
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https://brilliant.org/problems/leaking-the-gas/
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# Leaking the Gas
There's a person in the middle of a sealed room. The person's height is $$1.8\text{m}$$ and the room's height is $$3.5\text{m}$$.
There's an LPG and LNG pipe, each of them going straight up from the floor to the ceiling, and the gases are flowing though the pipe at a same speed and same direction.
Now you're going to make one hole in each pipe, making the gases to leak to the same direction.
• Height $$A$$ is the height of just above the floor.
• Height $$B$$ is the height of exactly the middle of the floor and the ceiling.
• Height $$C$$ is the height of just below the ceiling.
In which case would the person survive the longest?
Clarification: Both LNG and LPG are lethal to human body, and there are absolutely no sparks in the room. The person stands still until he dies. The pipes are in one corner of the room.
×
Problem Loading...
Note Loading...
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https://socratic.org/questions/how-do-you-write-5-23-times10-9-in-standard-notation
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# How do you write 5.23 times10^9 in standard notation?
Apr 1, 2016
$\text{ } 5230000000.0$
#### Explanation:
Starting point $\to 5.23 \times {10}^{9}$
This is the same in value as
$\text{ } 52.3 \times {10}^{8}$
$\text{ } 523.0 \times {10}^{7}$
$\text{ } 5230.0 \times {10}^{6}$
$\text{ } 52300.0 \times {10}^{5}$
and so on until you reach
$\text{ } 5230000000.0 \times {10}^{0}$
And ${10}^{0} = 1$ giving
$\text{ } 5230000000.0$
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# 144 kg to lbs - 144 kilograms to pounds
Do you need to learn how much is 144 kg equal to lbs and how to convert 144 kg to lbs? You couldn’t have chosen better. This whole article is dedicated to kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to highlight that all this article is devoted to one amount of kilograms - exactly one kilogram. So if you need to learn more about 144 kg to pound conversion - keep reading.
Before we get to the more practical part - this is 144 kg how much lbs conversion - we are going to tell you few theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 144 kg to lbs? 144 kilograms it is equal 317.46565728 pounds, so 144 kg is equal 317.46565728 lbs.
## 144 kgs in pounds
We are going to 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 abbreviated form SI).
From time to time the kilogram can be written as kilogramme. The symbol of this unit is kg.
The kilogram was defined first time in 1795. The kilogram was defined as the mass of one liter of water. First definition was simply but hard to use.
Later, in 1889 the kilogram was defined by the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was substituted by a new definition.
Today the definition of the kilogram is build 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 into 100 decagrams and 1000 grams.
## 144 kilogram to pounds
You learned some information about kilogram, so now let's go to the pound. The pound is also a unit of mass. We want to underline that there are not only one kind of pound. What are we talking about? For example, there are also pound-force. In this article we are going to to concentrate only on pound-mass.
The pound is used in the Imperial and United States customary systems of measurements. Naturally, this unit is in use also in other systems. The symbol of the pound is lb or “.
The international avoirdupois pound has no descriptive definition. It is equal 0.45359237 kilograms. One avoirdupois pound can be divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was given 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 144 kg?
144 kilogram is equal to 317.46565728 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 144 kg in lbs
Theoretical part is already behind us. In this section we are going to tell you how much is 144 kg to lbs. Now you learned that 144 kg = x lbs. So it is high time to know the answer. Just look:
144 kilogram = 317.46565728 pounds.
It is an accurate result of how much 144 kg to pound. You may also round off the result. After rounding off your result is exactly: 144 kg = 316.8 lbs.
You know 144 kg is how many lbs, so see how many kg 144 lbs: 144 pound = 0.45359237 kilograms.
Of course, in this case it is possible to also round off the result. After it your outcome is exactly: 144 lb = 0.45 kgs.
We are also going to show you 144 kg to how many pounds and 144 pound how many kg results in tables. Let’s see:
We are going to begin with a table for how much is 144 kg equal to pound.
### 144 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
144 317.46565728 316.80
Now see a table for how many kilograms 144 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
144 0.45359237 0.45
Now you know how many 144 kg to lbs and how many kilograms 144 pound, so it is time to go to the 144 kg to lbs formula.
### 144 kg to pounds
To convert 144 kg to us lbs a formula is needed. We will show you two formulas. Let’s begin with the first one:
Number of kilograms * 2.20462262 = the 317.46565728 result in pounds
The first formula give you the most correct result. Sometimes even the smallest difference could be significant. So if you need an accurate result - first formula will be the best solution to calculate how many pounds are equivalent to 144 kilogram.
So let’s move on to the second formula, which also enables calculations to know how much 144 kilogram in pounds.
The second formula is as following, look:
Amount of kilograms * 2.2 = the outcome in pounds
As you can see, this formula is simpler. It could be the best choice if you need to make a conversion of 144 kilogram to pounds in easy way, for instance, during shopping. You only have to remember that final result will be not so correct.
Now we are going to learn you how to use these two versions of a formula in practice. But before we will make a conversion of 144 kg to lbs we want to show you easier way to know 144 kg to how many lbs totally effortless.
### 144 kg to lbs converter
Another way to check what is 144 kilogram equal to in pounds is to use 144 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Calculator is based on first formula which we gave you in the previous part of this article. Due to 144 kg pound calculator you can quickly convert 144 kg to lbs. You only have to enter number of kilograms which you want to calculate and click ‘convert’ button. You will get the result in a second.
So try to calculate 144 kg into lbs using 144 kg vs pound converter. We entered 144 as a number of kilograms. Here is the outcome: 144 kilogram = 317.46565728 pounds.
As you see, our 144 kg vs lbs converter is user friendly.
Now we can go to our main topic - how to convert 144 kilograms to pounds on your own.
#### 144 kg to lbs conversion
We will begin 144 kilogram equals to how many pounds calculation with the first formula to get the most exact outcome. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 317.46565728 the result in pounds
So what need you do to check how many pounds equal to 144 kilogram? Just multiply amount of kilograms, in this case 144, by 2.20462262. It is exactly 317.46565728. So 144 kilogram is exactly 317.46565728.
You can also round it off, for example, to two decimal places. It is equal 2.20. So 144 kilogram = 316.80 pounds.
It is high time for an example from everyday life. Let’s convert 144 kg gold in pounds. So 144 kg equal to how many lbs? And again - multiply 144 by 2.20462262. It is exactly 317.46565728. So equivalent of 144 kilograms to pounds, if it comes to gold, is equal 317.46565728.
In this example it is also possible to round off the result. Here is the outcome after rounding off, in this case to one decimal place - 144 kilogram 316.8 pounds.
Now let’s move on to examples converted with short formula.
#### How many 144 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 316.8 the outcome in pounds
So 144 kg equal to how much lbs? And again, you have to multiply number of kilogram, in this case 144, by 2.2. Let’s see: 144 * 2.2 = 316.8. So 144 kilogram is equal 2.2 pounds.
Do another conversion with use of this formula. Now convert something from everyday life, for instance, 144 kg to lbs weight of strawberries.
So convert - 144 kilogram of strawberries * 2.2 = 316.8 pounds of strawberries. So 144 kg to pound mass is exactly 316.8.
If you know how much is 144 kilogram weight in pounds and are able to calculate it with use of two different formulas, we can move on. Now we want to show you these results in tables.
#### Convert 144 kilogram to pounds
We realize that results presented in charts are so much clearer for most of you. We understand it, so we gathered all these results in tables for your convenience. Due to this you can quickly compare 144 kg equivalent to lbs outcomes.
Let’s begin with a 144 kg equals lbs chart for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
144 317.46565728 316.80
And now see 144 kg equal pound table for the second formula:
Kilograms Pounds
144 316.8
As you can see, after rounding off, when it comes to how much 144 kilogram equals pounds, the results are the same. The bigger amount the more significant difference. Remember it when you want to make bigger amount than 144 kilograms pounds conversion.
#### How many kilograms 144 pound
Now you learned how to calculate 144 kilograms how much pounds but we will show you something more. Are you interested what it is? What about 144 kilogram to pounds and ounces conversion?
We want to show you how you can convert it step by step. Let’s start. How much is 144 kg in lbs and oz?
First thing you need to do is multiply amount of kilograms, this time 144, by 2.20462262. So 144 * 2.20462262 = 317.46565728. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To convert how much 144 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.
So your outcome is equal 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then final result will be equal 2 pounds and 33 ounces.
As you see, conversion 144 kilogram in pounds and ounces quite easy.
The last conversion which we are going to show you is conversion of 144 foot pounds to kilograms meters. Both of them are units of work.
To convert foot pounds to kilogram meters you need another formula. Before we give you it, let’s see:
• 144 kilograms meters = 7.23301385 foot pounds,
• 144 foot pounds = 0.13825495 kilograms meters.
Now look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 144 foot pounds to kilograms meters you need to multiply 144 by 0.13825495. It gives 0.13825495. So 144 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for example, to two decimal places. Then 144 foot pounds is equal 0.14 kilogram meters.
We hope that this conversion was as easy as 144 kilogram into pounds conversions.
We showed you not only how to do a calculation 144 kilogram to metric pounds but also two another calculations - to check how many 144 kg in pounds and ounces and how many 144 foot pounds to kilograms meters.
We showed you also other solution to do 144 kilogram how many pounds calculations, this is using 144 kg en pound calculator. It is the best solution 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 can do 144 kilogram equal to how many pounds calculation - on your own or with use of our 144 kgs to pounds converter.
It is time to make your move! Calculate 144 kilogram mass to pounds in the best way for you.
Do you want to make other than 144 kilogram as pounds conversion? For example, for 15 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so simply as for 144 kilogram equal many pounds.
### How much is 144 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 144 kg in pounds , we gathered answers to the most frequently asked questions. Here you can see the most important information about how much is 144 kg equal to lbs and how to convert 144 kg to lbs . Let’s see.
What is the kilogram to pound conversion? The conversion kg to lb is just multiplying 2 numbers. Let’s see 144 kg to pound conversion formula . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 144 kilogram to pounds? The exact result is 317.46565728 lbs.
You can also calculate how much 144 kilogram is equal to pounds with another, shortened version of the equation. Let’s see.
The number of kilograms * 2.2 = the result in pounds
So in this case, 144 kg equal to how much lbs ? The result is 317.46565728 lbs.
How to convert 144 kg to lbs in just a moment? It is possible to use the 144 kg to lbs converter , which will make whole mathematical operation for you and give you a correct result .
#### 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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# Search by Topic
#### Resources tagged with Trial and improvement similar to Consecutive Numbers:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Trial and improvement
### Pair Sums
##### Stage: 3 Challenge Level:
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
### Make 100
##### Stage: 2 Challenge Level:
Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
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##### Stage: 3 Challenge Level:
Can you guess the colours of the 10 marbles in the bag? Can you develop an effective strategy for reaching 1000 points in the least number of rounds?
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##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
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##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Sliding Game
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A shunting puzzle for 1 person. Swop the positions of the counters at the top and bottom of the board.
### Magic Circles
##### Stage: 2 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
### Cinema Problem
##### Stage: 3 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Top-heavy Pyramids
##### Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Inky Cube
##### Stage: 2 and 3 Challenge Level:
This cube has ink on each face which leaves marks on paper as it is rolled. Can you work out what is on each face and the route it has taken?
### Junior Frogs
##### Stage: 1 and 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Sending a Parcel
##### Stage: 3 Challenge Level:
What is the greatest volume you can get for a rectangular (cuboid) parcel if the maximum combined length and girth are 2 metres?
### Numbered Cars
##### Stage: 2 Challenge Level:
I was looking at the number plate of a car parked outside. Using my special code S208VBJ adds to 65. Can you crack my code and use it to find out what both of these number plates add up to?
### Oranges and Lemons
##### Stage: 2 Challenge Level:
On the table there is a pile of oranges and lemons that weighs exactly one kilogram. Using the information, can you work out how many lemons there are?
##### Stage: 2 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Number Juggle
##### Stage: 2 Challenge Level:
Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).
### One Million to Seven
##### Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
### Fifteen Cards
##### Stage: 2 Challenge Level:
Can you use the information to find out which cards I have used?
### Escape from the Castle
##### Stage: 2 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out?
### Paw Prints
##### Stage: 2 Challenge Level:
A dog is looking for a good place to bury his bone. Can you work out where he started and ended in each case? What possible routes could he have taken?
### Coded Hundred Square
##### Stage: 2 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
### Fractions in a Box
##### Stage: 2 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each disc. Use the information to find out how many discs of each colour there are in the box.
### A Numbered Route
##### Stage: 2 Challenge Level:
Can you draw a continuous line through 16 numbers on this grid so that the total of the numbers you pass through is as high as possible?
### Brothers and Sisters
##### Stage: 2 Challenge Level:
Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues?
### Tubular Path
##### Stage: 2 Challenge Level:
Can you make the green spot travel through the tube by moving the yellow spot? Could you draw a tube that both spots would follow?
### 3388
##### Stage: 3 Challenge Level:
Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
### Football Champs
##### Stage: 3 Challenge Level:
Three teams have each played two matches. The table gives the total number points and goals scored for and against each team. Fill in the table and find the scores in the three matches.
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
### Lost
##### Stage: 3 Challenge Level:
Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses.
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Let's Face It
##### Stage: 2 Challenge Level:
In this problem you have to place four by four magic squares on the faces of a cube so that along each edge of the cube the numbers match.
### A Mean Tetrahedron
##### Stage: 3 Challenge Level:
Can you number the vertices, edges and faces of a tetrahedron so that the number on each edge is the mean of the numbers on the adjacent vertices and the mean of the numbers on the adjacent faces?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### Magic Matrix
##### Stage: 2 Challenge Level:
Find out why these matrices are magic. Can you work out how they were made? Can you make your own Magic Matrix?
### How Many Eggs?
##### Stage: 2 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
### Cycling Squares
##### Stage: 2 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### One Wasn't Square
##### Stage: 2 Challenge Level:
Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.
### Special 24
##### Stage: 2 Challenge Level:
Find another number that is one short of a square number and when you double it and add 1, the result is also a square number.
### Highest and Lowest
##### Stage: 2 Challenge Level:
Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
### Archery
##### Stage: 3 Challenge Level:
Imagine picking up a bow and some arrows and attempting to hit the target a few times. Can you work out the settings for the sight that give you the best chance of gaining a high score?
### Four Colours
##### Stage: 1 and 2 Challenge Level:
Kate has eight multilink cubes. She has two red ones, two yellow, two green and two blue. She wants to fit them together to make a cube so that each colour shows on each face just once.
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| 3.71875 | 4 |
CC-MAIN-2017-34
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| 0.890886 |
https://blackestfest.com/how-do-you-find-the-sum-of-a-harmonic-series/
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2021-08-23
## How do you find the sum of a harmonic series?
Harmonic Progression Sum G.M2 = A.M × H.M, where A.M, G.M, H.M are in G.P.
## How is harmonic number calculated?
In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers: Harmonic numbers are related to the harmonic mean in that the n-th harmonic number is also n times the reciprocal of the harmonic mean of the first n positive integers.
How do you code a harmonic series?
The program output is also shown below.
1. import java.util.Scanner;
2. class Harmonic.
3. {
4. public static void main(String… a)
5. {
6. System. out. print(“Enter any number : “);
7. Scanner s = new Scanner(System. in);
8. int num = s. nextInt();
What is harmonic series in C program?
Harmonic progression is a progression formed by taking the reciprocals of an arithmetic progression. Program/Source Code. Here is source code of the C Program to Find the the sum of H.P series. The C program is successfully compiled and run on a Linux system. The program output is also shown below.
### Does harmonic series converge?
The sum of a sequence is known as a series, and the harmonic series is an example of an infinite series that does not converge to any limit.
### What is harmonic formula?
The harmonic mean is a type of numerical average. It is calculated by dividing the number of observations by the reciprocal of each number in the series. Thus, the harmonic mean is the reciprocal of the arithmetic mean of the reciprocals. The harmonic mean of 1, 4, and 4 is: 3 ( 1 1 + 1 4 + 1 4 ) = 3 1 .
What is the formula of harmonic progression?
The harmonic progression is formed by taking the reciprocal of the terms of the arithmetic progression. If the given terms of the arithmetic progression are a, a + d, a + 2d, a + 3d.., then the terms of the harmonic progression (or harmonic sequence) are 1/a, 1/(a + d), 1/(a + 2d), 1/(a + 3d), 1/(a + 4d),……
Is AP greater than GP?
Since even in real numbers the Sum of GP is less than that of AP, then it cannot be greater than AP in a sequence where domain is integers.
## How do you find AP series?
The general form of an Arithmetic Progression is a, a + d, a + 2d, a + 3d and so on. Thus nth term of an AP series is Tn = a + (n – 1) d, where Tn = nth term and a = first term. Here d = common difference = Tn – Tn-1.
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| 3.8125 | 4 |
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latest
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https://math.stackexchange.com/questions/1294479/dividing-n-identical-things-into-r-groups
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# Dividing $n$ identical things into $r$ groups
I was reading a course on Combinatorics where I came across following:
The number of ways in which $n$ identical things can be divided into $r$ groups so that no group contains less than $m$ items and more than $k$ (where $m<k$) is coefficient of $x^n$ in the expansion of $(x^m + x^{m+1} + … +x^k)^r$.
Is this correct? Book doesn't give proof. How can we make this formula intuitive?
• If you factor out $x^m$ you get the easy case where you have to divide $n-rm$ identical things into $r$ groups. – user53970 May 22 '15 at 17:38
Write $$\underbrace{(x^m + x^{m+1} + … +x^k)(x^m + x^{m+1} + … +x^k)\cdots(x^m + x^{m+1} + … +x^k)}_{r\text{ times}}$$
In order to get a $x^n$ term you must choose a term from each parethesis in a such way that the sum of the exponents is $n$. This is the same as writing $n$ as the sum of $r$ numbers between $m$ and $k$.
• I still didnt get to know from where that $x^n$ term came, and in fact I didnt get from where that whole polynomial came. Does your answer explains that? I think I can do the expansion and get the coefficient of $x^n$ in it. But I dont understand from where all that $x^n$ thingy came. Pardon me for my poor understanding. – anir123 May 22 '15 at 20:29
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https://www.general-relativity.net/2019/03/exercise-301-consequences-of-metric.html
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## Saturday, 23 March 2019
### Exercise 3.01 Consequences of metric compatibility
Christoffel Symbol
At last I am on to chapter 3 on curvature. It has taken me three weeks to do the first four pages and exercise 3.01 was done on the way. I learned or relearned many things and had two interesting discussions on Physics forums (here and here). The question asked us to prove that the covariant derivative of the inverse metric and and that the covariant derivative of the Levi-Civita tensor both vanished if the covariant derivative of the metric vanishes and the Christoffel Symbol is symmetric in its lower indices. These conditions are known as metric compatibility and torsion freedom(?) respectively. One would say 'a connection on a manifold is torsion free and metric compatible'.
In maths that is
## Question
Verify the consequences of metric compatibility: If
\begin{align}
{\mathrm{\nabla }}_{\sigma}g_{\mu \nu }=0 & \phantom {10000}(1) \\
\end{align}then (a)\begin{align}
{\mathrm{\nabla }}_{\sigma}g^{\mu \nu }=0 & \phantom {10000}(2) \\
\end{align}and (b)\begin{align}
{\mathrm{\nabla }}_{\lambda}{\varepsilon }_{\mu \nu \sigma \rho }=0 & \phantom {10000}(3) \\
\end{align} I am not sure if we are assuming ${\mathrm{\Gamma }}^{\tau }_{\lambda \mu }={\mathrm{\Gamma }}^{\tau }_{ \mu \lambda }$ or not.
## Answer
Part (a) was quite simple but I struggled with the part (b) until 23 March and had to give up. Along the way I had lots of practice at index manipulation, I reacquainted myself with Cramer's rule for solving simultaneous equations, proved (b) on the surface of a sphere, found the 'dynamite' version of Carroll's streamlined matrix determinant equation (2.66) and added some equation shortcut keys to my keyboard. The time was not wasted.
We make frequent use here of the fact that $g_{\mu \nu }g^{\mu \rho }\mathrm{=}{\delta}^{\rho }_{\nu }$ and the indexing effect of the Kronecker delta: ${\delta}^{\lambda }_{\beta }\mathrm{\Gamma }^{\mu }_{\sigma \lambda }={\mathrm{\Gamma }}^{\mu }_{\sigma \beta }$ because we are summing over $\lambda$ and the only non-zero term is when $\beta =\lambda$. In this case $\mathrm{\Gamma }$ can be replaced by any symbol or tensor of any rank.
Here is the full effort Ex 3.01 Consequences of metric compatibility.pdf (7 pages of which 4 might be worth looking at).
The comment left by JSBach1801 solved this problem very easily as I fully realised in Jan 2021. Thanks! That answer is right at the end of the pdf.
#### 5 comments:
1. Check out this link for some really clever solutions to the porblem from Carroll's book. The Part B is so easier than you realize.
https://indexguy.wordpress.com/2007/07/13/some-identities-regarding-metric-compatibility/
1. The proof for part a is great. But I don't follow the proof for part b. It is true that the determinant will only depend on the metric entries, but why would that mean that the covariant derivative will vanish?
2. the fact that the covariant derivative vanishes follows directly from part (a) XD one way of thinking about it is that g is 'constant' wrt covariant derivatives. Other than that I suppose that since the Levi-Civita symbol is not a tensor the covariant derivative of the Levi-Civita symbol is just a partial derivative, so everything vanishes. I wasn't quite sure about this argument so I did things a bit more explicitly.
My fully typed-up solution can be found here: https://drive.google.com/file/d/13dBHxlh8CIBFfUIrf18mFPERDQ7IWkHJ/view?usp=sharing
2. The solution to (b) in the link provided by JSBach1801 is simply wrong. It is very important to keep in mind that covariant derivatives are meaningful only when operating on a tensor field. (A tensor field may be a field of scalars, vectors, one-forms, etc.) |g|, or its square root, does not form a scalar field, because its value at each point is coordinate dependent. Idem for the Levi-Civita symbol.
3. Here's [my solution to the exercise](https://physics.stackexchange.com/a/743720/269697). It's too long to be posted as a comment.
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# A colony of bats is counted every two months. The first four counts are 1200, 1800, 2700, and 4050. If this growth rate continues, (4 points) 1. What is the recurrence relation of the bat population? (2 points) 2. How many bats are there at the 12th count? Show all work. (Hint: solve the recurrence relation above)
Friday, October 1st, 2010
### Solution:
1. Recurrence relation
• Note that 1800/1200 = 1.5.
• Next, take 1.5* 1800 =2700, and that is the next population.
• Now 1.5*2700 = 4050.
2. To calculate the number of bats at the 12th count, take 1200 * 1.511 = 1200 *86.5 = 103,797
# N-3/8=6
Tuesday, September 7th, 2010
# What is 9/10 of 80?
Friday, July 23rd, 2010
### Solution:
The expression “9/10 of 80” can be written as
This is equal to 72.
# Simplify : 12 – ( x + 3 ) +10
Monday, July 19th, 2010
### Solution:
You can simply 12 – ( x + 3 ) +10 by rearranging the terms:
1. 12 –x -3 +10
2. 19-x
# What is mean?
Saturday, July 17th, 2010
### Solution:
The arithmetic mean is the average.
http://en.wikipedia.org/wiki/Arithmetic_mean
# I need to figure out 4 to the power of 2 plus 5 to the power of 2
Saturday, July 17th, 2010
42 + 52 =16 + 25 = 41
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You are Here: Home >< Maths
# Numbers and Counting question watch
1. I am confused by the following question, the second part of (a) and (b). For (a) I got 560, which I know is correct. When it talks about removing the hats, I got the answer 28, when the solutions say it's 280.
14. (a) 8 people are to be divided into three teams, distinguished by the colour hats they wear.
2 people will wear red hats, 3 green hats and 3 blue hats. How many ways are there of
choosing the teams? What if we remove the hats so that the two teams of size three become
interchangeable?
(b) 8 people are to be divided into four teams, distinguished by hat colour: 2 people will wear
red hats, 2 green hats, 2 blue hats and 2 orange hats. How many ways are there of choosing
the teams? What if we remove the hats?
Anyone help?
2. (Original post by theereflex)
I am confused by the following question, the second part of (a) and (b). For (a) I got 560, which I know is correct. When it talks about removing the hats, I got the answer 28, when the solutions say it's 280.
14. (a) 8 people are to be divided into three teams, distinguished by the colour hats they wear.
2 people will wear red hats, 3 green hats and 3 blue hats. How many ways are there of
choosing the teams? What if we remove the hats so that the two teams of size three become
interchangeable?
(b) 8 people are to be divided into four teams, distinguished by hat colour: 2 people will wear
red hats, 2 green hats, 2 blue hats and 2 orange hats. How many ways are there of choosing
the teams? What if we remove the hats?
The second part of 14(a) means the following scenarios are now the same since there are no hats:
Green hats: A, B, C
Blue hats: X, Y, Z
Green hats: X, Y, Z
Blue hats: A, B, C
Hence there should be half the number of possibilities in the first part of 14(a).
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# TELNET - Editorial
PROBLEM LINK:
Practice
Author: Vaibhav Tulsyan
Tester: tncks0121
DIFFICULTY:
Simple
PREREQUISITES:
Dynamic Programming
PROBLEM:
You are given a grid. The cell (i, j) has a value H_{i, j}. You need to find a sequence of cells (i_1, j_1), (i_2, j_2), \ldots (i_k, j_k) such that H_{i_r, j_r} = r for all 1 \leq r \leq k and the sum of manhattan distances between (i_r, j_r) and (i_{r+1}, j_{r + 1}) over 1 \leq r < k is minimized.
EXPLANATION:
Define f(i, j) as the minimum possible cost to reach cell (i, j). Then, we can define a recursion on f. Let t = H_{i, j}.
• If t = 1, f(i, j) = 0.
• If t > 1, f(i, j) = \displaystyle \min_{(i', j'):H_{i',j'} = t - 1} (f(i', j') + |i - i'| + | j - j'| ).
That is, we iterate over all possible previous cells (i', j') and add the cost to reach (i, j) from (i', j').
The overall complexity would be O(N^4), as we take O(N^2) to find f value for each cell.
The answer would be \displaystyle \min_{(i, j) : H_{i, j} = k} f(i, j) .
AUTHOR’S and TESTER’S codes:
The author’s code can be found here
The tester’s code can be found here
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Télécharger la présentation
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##### Presentation Transcript
2. Vocabulary • Opposites • Two numbers who have the same absolute value, but different signs • Additive Inverse • Two numbers that when added together have a sum of zero • Also called “Zero Pairs”
3. Rule – Same Sign • When the signs are the same, add the numbers. • Keep the sign.
4. Same Sign • -6 + (-8) = • 4 + 5 = • -16 + (-16) + (-2) = -14 9 -34
5. Rule - Different Sign • When the signs are different, subtract the numbers. • Use the sign of the number with the greater absolute value.
6. Different Sign • -15 + 8 = • 7 + (-11) = • 25 + 3 + (-25) = -7 -4 3
7. Word Problems • Lena was scuba diving 14 meters below the surface of the water. She saw a nurse shark 3 meters above her. Write an addition expression to describe the situation. -14 + 3
8. Word Problems • Stephanie has \$152 in the bank. She withdrawals \$20 on Tuesday. She then deposits \$84 on Friday. Write an addition expression to describe the situation. 152 + (-20) + 84
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Subsections
# MA 1024 Laboratory 3: Surfaces
## Purpose
The purpose of this lab is to introduce you to some of the Maple commands that can be used to plot surfaces in three dimensions.
## Background
The graph of a function of a single real variable is a set of points (x,f(x)) in the plane. Typically, the graph of such a function is a curve. For functions of two variables, the graph is a set of points (x,y,f(x,y)) in three-dimensional space. For this reason, visualizing functions of two variables is usually more difficult.
One of the most valuable services provided by computer software such as Maple is that it allows us to produce intricate graphs with a minimum of effort on our part. This becomes especially apparent when it comes to functions of two variables, because there are many more computations required to produce one graph, yet Maple performs all these computations with only a little guidance from the user.
Two common ways of representing the graph of a function of two variables are the surface plot and the contour plot. The first is simply a representation of the graph in three-dimensional space. The second, draws the level curves f(x,y)=C for several values of C in the x,y -plane. We will explore how to produce these kinds of graphs in Maple, and how to use the graphs to study the functions.
### Cartesian coordinates
You can define functions of more than one variable in much the same way as you defined functions of a single variable:
> f := (x,y) -> x^2 + y^2;
> f(3,-1);
> g := (a,b,c,d,e) -> a*b^2 - sin(c+d)/e;
The following commands are useful for working with functions of two variables in Maple.
#### plot3d
This command generates a surface plot of the function. This command has a lot of options that you can use to get the plot to look the way you want it. Some of the most useful options are described below. For more information, see the help for the plot3d command.
> plot3d(x^2-y^2, x=-1..1, y=-1..1);
> f := (x,y) -> x/2 - y + 3;
> plot3d(f, 0..2, -1..1);
The default viewing angle is from a direction 45 degrees between the positive x- and positive y- axes, and an angle of elevation of 45 degrees. You can change this viewing angle with the orientation option, orientation=[a,b]. The first number is the polar angle, measured counterclockwise from the positive x-axis. The second number is the angle of elevation; it is measured downward from straight above, also in degrees.
You can also select a viewpoint using the mouse. Click the mouse on a three-dimensional graph, and notice the context bar that appears between the tool bar and the Maple input/output window. Click the graphic again, and the graph is replaced by a box. Hold down the button as you move the mouse, and you'll see the box from different angles. You'll also see the numbers on the left, labeled and , change accordingly. They correspond to the two numbers in the orientation option.
Once you've selected the desired viewpoint, redraw the graphic by pressing the button marked R' at the right end of the tool bar.
The other buttons in the tool bar control other aspects of how the plot is drawn, including the plot style, axes style, etc.
The number of grid points in the plot can be changed with the grid=[x,y] option. You may want to increase the number of grid points if your plot appears rough, or has a lot of oscillation; you may want to use a smaller number if the function is reasonably smooth and you want to shorten calculation times.
> plot3d(x^2-y^2,x=-1..1,y=-1..1,orientation=[45,45]);
> plot3d(x^2-y^2,x=-1..1,y=-1..1,orientation=[45,20]);
> plot3d(x^2-y^2,x=-1..1,y=-1..1,orientation=[45,-45]);
> plot3d(x^2-y^2,x=-1..1,y=-1..1,orientation=[110,45]);
>
> plot3d(x^2-y^2,x=-1..1,y=-1..1,grid=[10,10]);
> plot3d(x^2-y^2,x=-1..1,y=-1..1,grid=[25,40]);
Note that you can use the context bar instead of optional arguments to the plot3d command to customize your plot. Saving your worksheet after you have made changes will save the plot as it last appeared. However, if you need to run the plot3d command again, any customizations you made with the context bar will be lost. The safest approach is to use the context bar to experiment with your plot until you are satisfied with it. Then add options to your plot3d command that will give you the same plot, for example by specifying the orientation or axes arguments in your command.
#### contourplot
Generates a contour plot of a function of two variables. This command is part of the plots'' package, so you need to run with(plots) before using the command. The basic syntax is the same as for plot3d.
> with(plots):
> contourplot(x^2-y^2,x=-1..1,y=-1..1);
> f := (x,y) -> x/2 - y + 3;
> contourplot(f,0..2,-1..1);
Maple's default is to produce ten contours. This can be changed using the option contours=n. Maple chooses the z-levels of the contours automatically. If you want to see specific level curves, you can write the z-values in a list.
> contourplot(x^2-y^2,x=-2..2,y=-2..2);
> contourplot(x^2-y^2,x=-2..2,y=-2..2,contours=6);
> contourplot(x^2-y^2,x=-2..2,y=-2..2,contours=[6]);
> contourplot(x^2-y^2,x=-2..2,y=-2..2,contours=[-1,0,1,2]);
#### implicitplot3d
If a surface cannot be easily written as z=f(x,y) but can be written as a relation F(x,y,z)=0, then the Maple implicitplot3d command from the plots package can often be used. For example, to plot the cylinder y2+z2 = 2 the following command can be used. Don't forget to load the plots package first using the command with(plots); first.
> implicitplot3d(y^2+z^2=2,x=0..2,y=-2..2,z=-2..2);
Getting a good plot using this command can be a little tricky because you have to come up with good guesses for the ranges for x, y, and z. If most or all of the surface is outside the ranges you specify, you may not see a good representation of the surface.
### Cylindrical and spherical coordinates
Maple can also do plots of surfaces in cylindrical and spherical coordinates. To do this, you just use the coords option to the plot3d command to specify the coordinate system. For example, the following command will plot the cylinder r=1 for .
> plot3d(1,theta=0..2*Pi,z=-1..1,coords=cylindrical);
The next command plots the cone r=1-z over the same range for z.
> plot3d(1-z,theta=0..2*Pi,z=-1..1,coords=cylindrical);
The plot3d command expects your equation for the surface to be of the form , with the first independent variable and z the second independent variable. If your equation isn't of this form, then you have to use a parametric plot. Parametric plots of surfaces are usually beyond the scope of this course, but we will present an example in Cartesian coordinates at the end of this section.
Plotting a surface in spherical coordinates is very similar. Maple expects the equation of the surface to be in the form . Again, the order is important. The following command plots the unit sphere.
> plot3d(1,theta=0..2*Pi,phi=0..2*Pi,coords=spherical);
Many surfaces can be represented in more than one coordinate system. For example, the following command plots the same cylinder of radius 1 that we plotted before, but in spherical coordinates.
> plot3d(1/sin(phi),theta=0..2*Pi,phi=Pi/4..3*Pi/4,coords=spherical);
The implicitplot3d command can also be used to plot relations in terms of cylindrical or spherical coordinates. The only tricky part is that it expects you to give the ranges in a certain order. For example, in cylindrical coordinates it expects the range for r first, then the range for , and finally the range for z. Other orders can give unpredictable results. For spherical coordinates, the order must be .
### Parametric surfaces
Sometimes a surface cannot easily be represented in Cartesian, polar, or even spherical coordinates. In these cases, a parametric plot of the surface is often used. We have already seen parametric curves, in which a single parameter, usually t was used. For a surface, two parameters are required. A parametric surface in Cartesian coordinates is an ordered triple of functions (f(s,t),g(s,t),h(s,t)) where x=f(s,t), y=g(s,t), and z=h(s,t). That is, given values of the parameters s and t you can compute a point (x,y,z) on the surface. For example, any surface z=F(x,y) can be represented parametrically as (s,t,F(s,t)). However, the real utility of a parametric surface is for surfaces that cannot be represented as z=F(x,y) or even as a simple relation between x, y, and z. For example suppose you wanted to plot the cylinder y2+z2=2 for . You could solve for z, but that gives the two functions and and you would have to combine them to plot the cylinder. An alternative is the following.
> plot3d([s,2*cos(t),2*sin(t)],s=0..2,t=0..2*Pi);
We used the implicitplot3d command above to plot this same cylinder. An advantage of using a parametric plot is that you usually know what ranges to use for your parameters to get the plot you want.
As a final example of a parametric surface, we present the torus, or doughnut if you are feeling hungry.
> plot3d([4*cos(s)+cos(t)*cos(s),4*sin(s)+cos(t)*sin(s),sin(t)],
s=0..2*Pi,t=0..2*Pi,style=patch);
`
### Printing problems
There are occasionally problems printing out a Maple worksheet that has many three dimensional plots. The usual symptom is that part of the worksheet is not printed. If this happens to you, the best method for overcoming this problem in the past has been to delete all of the output in your worksheet, save it, and then re-execute the whole worksheet before attempting to print again. The best way to delete all output is to use the Remove Output item from the Edit menu. There is also an Execute option in the Edit menu that you can use to execute all the commands in your worksheet. If this doesn't work, consult your instructor.
The unfortunate thing about this problem is that modifications you might have made to your plots using the mouse or the context bar are lost. This is why it is a good idea to first experiment with your plot using the mouse and the context bar but, once you have the plot looking the way you want it, to include your modifications in the plot command.
## Exercises
1.
(a)
Generate a surface plot and contour plot for each of the following functions on the given domains:
i.
f(x,y) = x2-y3+3y, for and .
ii.
for and .
(b)
What does the contour plot look like in the regions where the surface plot has a steep incline? What does it look like where the surface plot is almost flat?
(c)
What can you say about the surface plot in a region where the contour plot looks like a sequence of nested circles?
2.
Generate a surface plot for the following functions on the domains given.
(a)
for and . Use cylindrical coordinates.
(b)
for and . Use spherical coordinates.
3.
First, use the implicitplot3d command to plot the graph of the elliptic cylinder y2/4+z2=1 for . Then find a parametric description of the surface and use the plot3d command to plot the graph.
4.
Consider the following function
which represents the deviation, in inches, of last year's rainfall from the average annual rainfall in a certain area. Considering only the domain and , use a contour plot to find the region in this domain in which the deviation was at least 0.4 in. but no more than 0.6 in.
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Problem Solving - Inequality
Author Message
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Joined: 22 Mar 2010
Posts: 61
Followers: 1
Kudos [?]: 23 [0], given: 1
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25 Jul 2010, 10:43
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Question Stats:
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How many soultions are possible for the inequality, mod(x-3)+mod(x-4)<1 ?
a. 3
b. 4
c. 0
d. 6
e. Infinitely many
[Reveal] Spoiler: OA
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Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2795
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Concentration: Technology, Entrepreneurship
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Kudos [?]: 1620 [0], given: 235
Re: Problem Solving - Inequality [#permalink]
Show Tags
25 Jul 2010, 10:57
EnterMatrix wrote:
How many soultions are possible for the inequality, mod(x-3)+mod(x-4)<1 ?
a. 3
b. 4
c. 0
d. 6
e. Infinitely many
$$|x-3| + |x-4| < 1$$ ; take three cases
1.$$x>4$$
$$|x-3| + |x-4| < 1$$
$$x-3 +x-4 <1$$
x<4 not possible as we have assumed x>4
2.$$3<x<4$$
$$|x-3| + |x-4| < 1$$
$$x-3 + 4-x < 1$$ not possible
3. $$x<3$$
$$|x-3| + |x-4| < 1$$
$$3-x + 4-x <1$$
$$x> 3$$ not possible
hence 0 solution
PS: Make sure you check the answer against the domain. It should satisfy the domain.
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Re: Problem Solving - Inequality [#permalink] 25 Jul 2010, 10:57
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# Search by Topic
#### Resources tagged with Working systematically similar to Odd Squares:
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### There are 339 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Map Folding
##### Stage: 2 Challenge Level:
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
### Waiting for Blast Off
##### Stage: 2 Challenge Level:
10 space travellers are waiting to board their spaceships. There are two rows of seats in the waiting room. Using the rules, where are they all sitting? Can you find all the possible ways?
### Red Even
##### Stage: 2 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?
### Shunting Puzzle
##### Stage: 2 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
##### Stage: 2 Challenge Level:
What is the best way to shunt these carriages so that each train can continue its journey?
### Triangles to Tetrahedra
##### Stage: 3 Challenge Level:
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### Open Boxes
##### Stage: 2 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?
### Display Boards
##### Stage: 2 Challenge Level:
Design an arrangement of display boards in the school hall which fits the requirements of different people.
### Sticks and Triangles
##### Stage: 2 Challenge Level:
Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
### Cover the Tray
##### Stage: 2 Challenge Level:
These practical challenges are all about making a 'tray' and covering it with paper.
### Calendar Cubes
##### Stage: 2 Challenge Level:
Make a pair of cubes that can be moved to show all the days of the month from the 1st to the 31st.
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Painting Possibilities
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . .
### Crack the Code
##### Stage: 2 Challenge Level:
The Zargoes use almost the same alphabet as English. What does this birthday message say?
### Elf Suits
##### Stage: 2 Challenge Level:
If these elves wear a different outfit every day for as many days as possible, how many days can their fun last?
### Newspapers
##### Stage: 2 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
### Calcunos
##### Stage: 2 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
### Finding All Possibilities Upper Primary
##### Stage: 2 Challenge Level:
These activities focus on finding all possible solutions so if you work in a systematic way, you won't leave any out.
### Counters
##### Stage: 2 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
### Tetrafit
##### Stage: 2 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
##### Stage: 2 Challenge Level:
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
### Putting Two and Two Together
##### Stage: 2 Challenge Level:
In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together?
### Knight's Swap
##### Stage: 2 Challenge Level:
Swap the stars with the moons, using only knights' moves (as on a chess board). What is the smallest number of moves possible?
### Ancient Runes
##### Stage: 2 Challenge Level:
The Vikings communicated in writing by making simple scratches on wood or stones called runes. Can you work out how their code works using the table of the alphabet?
### Home City
##### Stage: 2 Challenge Level:
Use the clues to work out which cities Mohamed, Sheng, Tanya and Bharat live in.
### Professional Circles
##### Stage: 2 Challenge Level:
Six friends sat around a circular table. Can you work out from the information who sat where and what their profession were?
### Paw Prints
##### Stage: 2 Challenge Level:
A dog is looking for a good place to bury his bone. Can you work out where he started and ended in each case? What possible routes could he have taken?
### Team Scream
##### Stage: 2 Challenge Level:
Seven friends went to a fun fair with lots of scary rides. They decided to pair up for rides until each friend had ridden once with each of the others. What was the total number rides?
### Seating Arrangements
##### Stage: 2 Challenge Level:
Sitting around a table are three girls and three boys. Use the clues to work out were each person is sitting.
### Arrangements
##### Stage: 2 Challenge Level:
Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....?
### A Right Charlie
##### Stage: 2 Challenge Level:
Can you use this information to work out Charlie's house number?
### Halloween Investigation
##### Stage: 2 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
### Ordered Ways of Working Upper Primary
##### Stage: 2 Challenge Level:
These activities lend themselves to systematic working in the sense that it helps to have an ordered approach.
### 3 Rings
##### Stage: 2 Challenge Level:
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
### Four Triangles Puzzle
##### Stage: 1 and 2 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Broken Toaster
##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### This Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Two on Five
##### Stage: 1 and 2 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
### Bunny Hop
##### Stage: 2 Challenge Level:
What is the smallest number of jumps needed before the white rabbits and the grey rabbits can continue along their path?
### Celtic Knot
##### Stage: 2 Challenge Level:
Building up a simple Celtic knot. Try the interactivity or download the cards or have a go on squared paper.
### Ice Cream
##### Stage: 2 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Ordered Ways of Working Upper Primary
##### Stage: 2 Challenge Level:
These activities lend themselves to systematic working in the sense that it helps if you have an ordered approach.
### Counting Cards
##### Stage: 2 Challenge Level:
A magician took a suit of thirteen cards and held them in his hand face down. Every card he revealed had the same value as the one he had just finished spelling. How did this work?
### Button-up Some More
##### Stage: 2 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Two by One
##### Stage: 2 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
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http://mathematica.stackexchange.com/questions/30210/asking-mathematica-to-produce-a-function-of-some-variables
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# Asking Mathematica to produce a function of some variables
Given $$l_0=\sqrt{\frac{\lambda_1(\lambda_0+\mu_0)}{\lambda_0(\lambda_1+\mu_1-1)}}$$
and
$$l_1=\sqrt{\frac{\lambda_0(\lambda_1+\mu_1)}{\lambda_1(\lambda_0+\mu_0-1)}}$$
l0 = Sqrt[(λ1*(λ0 + μ0))/(λ0*(λ1 + μ1 - 1))]
l1 = Sqrt[(λ0*(λ1 + μ1))/(λ1*(λ0 + μ0 - 1))]
Is there any way to ask mathematica to determine if some functions $f_0$ and $f_1$ exist such that
$$f_0(l_0,l_1)=\frac{1}{\sqrt{\frac{\lambda_0+\mu_0}{\lambda_0}}}$$
and
$$f_1(l_0,l_1)=\frac{1}{\sqrt{\frac{\lambda_0+\mu_0-1}{\lambda_0}}}$$
f0[l0, l1] = 1/Sqrt[(λ0 + μ0)/λ0]
f1[l0, l1] = 1/Sqrt[(λ0 + μ0 - 1)/λ0]
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You can try sol1 = Solve[{l0 == 1/Sqrt[(\[Lambda]0 + \[Mu]0)/\[Lambda]0], l1 == 1/Sqrt[(\[Lambda]0 + \[Mu]0 - 1)/\[Lambda]0]}, {\[Lambda]1, \[Mu]1}] and Simplify[{l0, l1} /. sol1] for a start. – b.gatessucks Aug 10 '13 at 14:00
Can f0 and f1 contain lambdas and mus in addition to l0 and l1? – David Park Aug 10 '13 at 14:52
@DavidPark should contain only $l0$ and $l1$ – Seyhmus Güngören Aug 10 '13 at 20:39
From the wording of your post, you seem to ask whether there are functions $f_0$ and $f_1$, which have only two slots f0[slot1,slot2], such that when composed with the functions $l_0$ and $l_1$, which are functions with 4 variables {λ0,λ1,μ0,μ1}, return the given values.
Let us use Mathematica to write the rhs of $f_0$ and $f_1$ in terms of $l_0$, $l_1$, $μ_0$, and $μ_1$. That is, let us move from the space where {λ0,λ1,μ0,μ1} are the variables to the space where {l0,l1,μ0,μ1} are the variables.
fromλToL = Solve[{l0 == Sqrt[(λ1*(λ0 + μ0))/(λ0*(λ1 + μ1 - 1))],
l1 == Sqrt[(λ0*(λ1 + μ1))/(λ1*(λ0 + μ0 - 1))]}, {λ0, λ1}][[1]];
FullSimplify[{f0[l0, l1] == 1/Sqrt[(λ0 + μ0)/λ0],
f1[l0, l1] == 1/Sqrt[(λ0 + λ0 - 1)/λ0]} /.
fromλToL, Assumptions -> l0 > 0 && l1 > 0]
It becomes obvious now that $f_0$ and $f_1$ cannot be functions of just $l_0$ and $l_1$.
Another reasoning that leads to the same answer: If there were those two functions $f_0$ and $f_1$ then the composition with the two functions $l_0$ and $l_1$ yield functions of {λ0,λ1,μ0,μ1}. Take derivatives of $f_0(l_0(λ0,λ1,μ0,μ1),l_1(λ0,λ1,μ0,μ1))=rhs0$ and $f_1(l_1(λ0,λ1,μ0,μ1))=rhs1$ respect to λ0,λ1,μ0, and μ1. You will get 8 differential equations. You can then prove (by solving for the derivatives and then replacing those derivatives in the remaining equations) that such a system of equations has no solution.
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https://www.mathhomeworkanswers.org/278709/go-to-details?show=278717
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Alex has 395 cards. Michael has 250 cards. They both give the same number of cards to their friend Austin. After giving cards to Austin, the numbers of cards Alex has is less than two times the amount of cards Michael has. Write and solve an inequality to determine the range of cards they gave Austin. (Write your answer in a complete sentence.)
395-x<2(250-x), where x is the number of cards they each give Austin,
395-x<500-2x,
395+x<500,
x<105.
In words:
If x is the number of cards Austin gets from each of Alex and Michael, Austin gets 2x cards altogether. Alex has 395-x cards left while Michael has 250-x cards left. Alex now has less than twice as many as cards as Michael. Twice the number of cards Michael has is 500-2x, and Alex has less than this. This means that, if Alex were now to receive the number of cards Austin received in total, he would have 395+x cards, which is less than 500 cards. Therefore, the number of cards they each gave Austin must be less than 105. This means that Austin received an even number of cards which is less than twice this, that is, 210 cards. We know he received some cards, so the minimum is 2 (one from Alex and one from Michael) and a maximum of 208 (104 from each of the others).
by Top Rated User (804k points)
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https://www.12000.org/my_notes/pde_in_CAS/maple_2020_2_and_mma_12_2/insu166.htm
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#### 3.3.2 Cylinderical coordinates
3.3.2.1 [274] Haberman 7.9.4 (a)
3.3.2.2 [275] Haberman 7.9.4 (b)
3.3.2.3 [276] Haberman 7.9.4 (c)
3.3.2.4 [277] Haberman 7.9.3 (a)
3.3.2.5 [278] Haberman 7.9.3 (b)
3.3.2.6 [279] Haberman 7.9.3 (c)
##### 3.3.2.1 [274] Haberman 7.9.4 (a)
problem number 274
Problem 7.9.4 (a) from Richard Haberman Applied Partial Differential Equations, 4th edition.
Solve Heat PDE $$u_t = k \nabla ^2 u$$ inside cylinder with radius $$a$$ and height $$H$$ with initial conditions $$u(r,\theta ,z,0)=f(r,z)$$ independent of $$\theta$$ if the boundary conditions are $$u(r,\theta ,0,t)=0$$, $$u(r,\theta ,H,t)=0$$, $$u(a,\theta ,z,t)=0$$.
Since it says independent of $$\theta$$, will use the PDE as \begin {align*} u_t = k \left (u_{rr} + \frac {1}{r} u_r + u_{zz} \right ) \end {align*}
Instead of the full Laplacian \begin {align*} u_t = k \left (u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} \right ) \end {align*}
Mathematica
$\left \{\left \{u(r,z,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 \exp \left (-k t \left (\frac {\left (j_{0,K[3]}\right ){}^2}{a^2}+\frac {\pi ^2 K[1]^2}{H^2}\right )\right ) J_0\left (\frac {r j_{0,K[3]}}{a}\right ) \left (\int _0^a\int _0^Hr J_0\left (\frac {r j_{0,K[3]}}{a}\right ) f(r,z) \sin \left (\frac {\pi z K[1]}{H}\right )dzdr\right ) \sin \left (\frac {\pi z K[1]}{H}\right )}{a^2 H J_1\left (j_{0,K[3]}\right ){}^2} & k>0\land (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 0\land K[3]\geq 0\land H^2 \left (j_{0,K[3]}\right ){}^2+a^2 \pi ^2 K[1]^2>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$
Maple
$u \left (r , z , t\right ) = \Mapleoverset {\infty }{\Mapleunderset {\mathit {n1} =1}{\sum }}\Mapleoverset {\infty }{\Mapleunderset {n =1}{\sum }}\frac {4 \BesselJ \left (0, \frac {r \lambda _{\mathit {n1}}}{a}\right ) \left (\left (\int _{0}^{a}r \BesselJ \left (0, \frac {r \lambda _{\mathit {n1}}}{a}\right ) \left (\right )d r \right )_{\mathit {AllSolutions}}\right ) {\mathrm e}^{-\frac {\left (H^{2} \lambda _{\mathit {n1}}^{2}+\pi ^{2} a^{2} n^{2}\right ) k t}{H^{2} a^{2}}} \sin \left (\frac {\pi n z}{H}\right )}{H a^{2} \hypergeom \left (\left [\frac {1}{2}\right ], \left [1, 2\right ], -\lambda _{\mathit {n1}}^{2}\right )}\boldsymbol {\mathrm {where}}\left \{\lambda _{\mathit {n1}}=\mathit {BesselJZeros}\left (0, \mathit {n1}\right )\wedge 0\le \lambda _{\mathit {n1}}\right \}$
Hand solution
Solve $$u_{t}=k\nabla ^{2}u$$ inside cylinder with radius a and height $$H$$ with initial conditions $$u(r,z,0)=f(r,z)$$ independent of $$\theta$$ if the boundary conditions are $$u(r,0,t)=0,u(r,H,t)=0,u(a,z,t)=0$$
Let $$u\left ( r,z,t\right ) =T\left ( t\right ) \Phi \left ( r,z\right )$$. Substituting into the PDE gives\begin {align*} T^{\prime }\left ( t\right ) \Phi \left ( r,z\right ) & =k\left ( T\left ( t\right ) \nabla ^{2}\Phi \right ) \\ \frac {T^{\prime }}{kT} & =\frac {\nabla ^{2}\Phi }{\Phi \left ( r,z\right ) }=-\lambda \end {align*}
Where $$\lambda$$ is the separation constant assumed positive. This gives $$T^{\prime }+\lambda kT=0$$ with solution $$T\left ( t\right ) =Ce^{-k\lambda t}$$. And\begin {equation} \nabla ^{2}\Phi +\lambda \Phi \left ( r,z\right ) =0\tag {1} \end {equation} Where $$\nabla ^{2}\Phi \left ( r,z\right ) =\Phi _{rr}+\frac {1}{r}\Phi _{r}+\Phi _{zz}$$ in this case, the Laplacian in cylindrical with no $$\theta$$ dependency. Let $$\Phi \left ( r,z\right ) =R\left ( r\right ) Z\left ( z\right )$$. Substituting in (1) gives\begin {align} R^{\prime \prime }Z+\frac {1}{r}R^{\prime }+Z^{\prime \prime }+\lambda RZ & =0\nonumber \\ \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\lambda & =-\frac {Z^{\prime \prime }}{Z}=\upsilon \tag {2} \end {align}
Where $$\upsilon$$ is the second separation constant assumed positive. This gives \begin {align*} Z^{\prime \prime }+\upsilon Z & =0\\ Z\left ( z\right ) & =A\cos \left ( \sqrt {\upsilon }z\right ) +B\sin \left ( \sqrt {\upsilon }z\right ) \end {align*}
At $$z=0,Z=0$$, hence $$A=0$$ and the solution becomes $$Z\left ( z\right ) =B\sin \left ( \sqrt {\upsilon }z\right )$$. At $$z=H,Z=0$$, hence for non-trivial solution we want $$\sqrt {\upsilon }H=n\pi ,n=1,2,3,\cdots$$. Therefore$\upsilon _{n}=\left ( \frac {n\pi }{H}\right ) ^{2}\qquad n=1,2,3,\cdots$ And the corresponding eigenfunctions$Z_{n}\left ( z\right ) =\sin \left ( \frac {n\pi }{H}z\right )$ From (2) the radial equation becomes\begin {align*} \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\lambda & =\left ( \frac {n\pi }{H}\right ) ^{2}\\ R^{\prime \prime }+\frac {1}{r}R^{\prime }+\left ( \lambda -\left ( \frac {n\pi }{H}\right ) ^{2}\right ) R & =0 \end {align*}
This is Bessel ODE. The solution is $R_{n}\left ( r\right ) =C_{1}J_{0}\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}r\right ) +C_{2}Y_{0}\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}r\right )$ Since $$Y_{0}$$ blows up at $$r=0$$, it is discarded leaving $$R_{n}\left ( r\right ) =C_{1}J_{0}\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}r\right )$$. At $$r=a,R_{n}\left ( a\right ) =0$$. For non-trivial solution we want $$J_{0}\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}a\right ) =0$$. Therefore $$\sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}a$$ are the zeros of Bessel function $$J_{0}\left ( x\right )$$. This allows us to determine all possible values of $$\sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}$$ (since $$a$$ is given constant). Let the $$m^{th}$$ zero of $$J_{0}\left ( x\right )$$ be called $$\Lambda _{m}$$. Hence \begin {align*} \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}a & =\Lambda _{m}\qquad m=1,2,3,\cdots \\ \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}} & =\frac {\Lambda _{m}}{a}\\ \lambda _{nm} & =\frac {\Lambda _{m}^{2}}{a^{2}}+\left ( \frac {n\pi }{H}\right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end {align*}
For each $$n$$, there are $$m$$ values of $$\lambda _{nm}$$. The corresponding radial eigenfunction is $R_{m}\left ( r\right ) =J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \qquad m=1,2,3,\cdots$ Therefore the complete solution is \begin {align*} u\left ( r,z,t\right ) & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}e^{-k\lambda _{nm}t}\sin \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \\ & =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}e^{-k\left ( \frac {\Lambda _{m}^{2}}{a^{2}}+\left ( \frac {n\pi }{H}\right ) ^{2}\right ) t}\sin \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \end {align*}
What is left is to determine $$A_{nm}$$. This is done using initial conditions by using orthogonality. At $$t=0$$$f\left ( r,z\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right )$ Multiplying both sides by $$rJ_{0}\left ( \frac {\Lambda _{m^{\prime }}}{a}r\right )$$ and integrating\begin {align*} \int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m^{\prime }}}{a}r\right ) rdr & =\int _{0}^{a}\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) J_{0}\left ( \frac {\Lambda _{m^{\prime }}}{a}r\right ) rdr\\ \int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr & =\sum _{n=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{H}z\right ) \int _{0}^{a}J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr \end {align*}
Multiplying both sides by $$\sin \left ( \frac {n^{\prime }\pi }{H}z\right )$$ and integrating\begin {align*} \int _{0}^{H}\left ( \int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr\right ) \sin \left ( \frac {n^{\prime }\pi }{H}z\right ) dz & =\int _{0}^{H}\sum _{n=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{H}z\right ) \sin \left ( \frac {n^{\prime }\pi }{H}z\right ) \left ( \int _{0}^{a}J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr\right ) dz\\ \int _{0}^{H}\int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \sin \left ( \frac {n\pi }{H}z\right ) rdrdz & =A_{nm}\int _{0}^{H}\int _{0}^{a}\sin ^{2}\left ( \frac {n\pi }{H}z\right ) J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz\\ A_{nm} & =\frac {\int _{0}^{H}\int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \sin \left ( \frac {n\pi }{H}z\right ) rdrdz}{\int _{0}^{H}\int _{0}^{a}\sin ^{2}\left ( \frac {n\pi }{H}z\right ) J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz} \end {align*}
Hence the final solution is$u\left ( r,z,t\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }\frac {\int _{0}^{H}\int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \sin \left ( \frac {n\pi }{H}z\right ) rdrdz}{\int _{0}^{H}\int _{0}^{a}\sin ^{2}\left ( \frac {n\pi }{H}z\right ) J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz}e^{-k\left ( \frac {\Lambda _{m}^{2}}{a^{2}}+\left ( \frac {n\pi }{H}\right ) ^{2}\right ) t}\sin \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right )$ Where $$\Lambda _{m}$$ is the $$m^{th}$$ zero of $$J_{0}\left ( x\right )$$. To verify the solution, it is compared to numerical solution, using the following values $$a=1,H=3,k=\frac {1}{100},f\left ( r,z\right ) =\left ( a-r\right ) \sin \left ( \frac {z}{H}\pi \right )$$. The summation was taken up to $$n=10,m=10$$. This animation only looks at cross section of the cylinder in the middle. The height indicates the amount of heat. As time passes the cylinder cools down. It runs for 15 seconds. (The animation will only show on the HTML version, not the PDE version of this report).
The following is the source code used to generate the above
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##### 3.3.2.2 [275] Haberman 7.9.4 (b)
problem number 275
Problem 7.9.4 (b) from Richard Haberman Applied Partial Differential Equations, 4th edition.
Solve Heat PDE $$u_t = k \nabla ^2 u$$ inside cylinder with radius $$a$$ and height $$H$$ with initial conditions $$u(r,z,0)=f(r,z)$$ independent of $$\theta$$, subject to boundary conditions $$u_z(r,0,t)=0$$, $$u_z(r,H,t)=0$$, $$u_r(a,z,t)=0$$.
Since it says independent of $$\theta$$, will use the PDE as \begin {align*} u_t = k \left (u_{rr} + \frac {1}{r} u_r + u_{zz} \right ) \end {align*}
Instead of the full Laplacian \begin {align*} u_t = k \left (u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} \right ) \end {align*}
Mathematica
$\left \{\left \{u(r,z,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {2 \int _0^a\int _0^Hr f(r,z)dzdr}{a^2 H}+\underset {K[1]=1}{\overset {\infty }{\sum }}\frac {4 e^{-\frac {k \pi ^2 t K[1]^2}{H^2}} \cos \left (\frac {\pi z K[1]}{H}\right ) \int _0^a\int _0^Hr \cos \left (\frac {\pi z K[1]}{H}\right ) f(r,z)dzdr}{a^2 H}+\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {2 e^{-k t K[2,0,K[3]]^2} J_0(r K[2,0,K[3]]) \left (\int _0^a\int _0^Hr J_0(r K[2,0,K[3]]) f(r,z)dzdr\right ) \sqrt {K[2,0,K[3]]}}{a^2 H \sqrt {J_0(a K[2,0,K[3]]){}^2+J_1(a K[2,0,K[3]]){}^2} \sqrt {\left (J_0(a K[2,0,K[3]]){}^2+J_1(a K[2,0,K[3]]){}^2\right ) K[2,0,K[3]]}}+\underset {K[1]=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 \exp \left (-k t \left (\frac {\pi ^2 K[1]^2}{H^2}+K[2,0,K[3]]^2\right )\right ) J_0(r K[2,0,K[3]]) \cos \left (\frac {\pi z K[1]}{H}\right ) \left (\int _0^a\int _0^Hr J_0(r K[2,0,K[3]]) \cos \left (\frac {\pi z K[1]}{H}\right ) f(r,z)dzdr\right ) \sqrt {K[2,0,K[3]]}}{a^2 H \sqrt {J_0(a K[2,0,K[3]]){}^2+J_1(a K[2,0,K[3]]){}^2} \sqrt {\left (J_0(a K[2,0,K[3]]){}^2+J_1(a K[2,0,K[3]]){}^2\right ) K[2,0,K[3]]}} & (K[1]|K[3])\in \mathbb {Z}\land J_1(a K[2,0,K[3]])=0\land K[1]\geq 0\land K[3]\geq 0\land k>0\land K[2,0,K[3]]>0\land \pi ^2 K[1]^2+H^2 K[2,0,K[3]]^2>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$
Maple
sol=()
Hand solution
Solve $$u_{t}=k\nabla ^{2}u$$ inside cylinder with radius a and height $$H$$ with initial conditions $$u(r,z,0)=f(r,z)$$ independent of $$\theta$$ if the boundary conditions are $$u_{z}(r,0,t)=0,u_{z}(r,H,t)=0,u_{r}(a,z,t)=0$$
Let $$u\left ( r,z,t\right ) =T\left ( t\right ) \Phi \left ( r,z\right )$$. Substituting into the PDE gives\begin {align*} T^{\prime }\left ( t\right ) \Phi \left ( r,z\right ) & =k\left ( T\left ( t\right ) \nabla ^{2}\Phi \right ) \\ \frac {T^{\prime }}{kT} & =\frac {\nabla ^{2}\Phi }{\Phi \left ( r,z\right ) }=-\lambda \end {align*}
Where $$\lambda$$ is the separation constant assumed positive. This gives $$T^{\prime }+\lambda kT=0$$ with solution $$T\left ( t\right ) =Ce^{-k\lambda t}$$. And\begin {equation} \nabla ^{2}\Phi +\lambda \Phi \left ( r,z\right ) =0 \tag {1} \end {equation} Where $$\nabla ^{2}\Phi \left ( r,z\right ) =\Phi _{rr}+\frac {1}{r}\Phi _{r}+\Phi _{zz}$$ in this case, the Laplacian in cylindrical with no $$\theta$$ dependency. Let $$\Phi \left ( r,z\right ) =R\left ( r\right ) Z\left ( z\right )$$. Substituting in (1) gives\begin {align} R^{\prime \prime }Z+\frac {1}{r}R^{\prime }+Z^{\prime \prime }+\lambda RZ & =0\nonumber \\ \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\lambda & =-\frac {Z^{\prime \prime }}{Z}=\upsilon \tag {2} \end {align}
Where $$\upsilon$$ is the second separation. $$\$$Then $$Z^{\prime \prime }+\upsilon Z=0$$
case $$\upsilon =0$$
This gives\begin {align*} Z\left ( z\right ) & =Az+B\\ Z^{\prime } & =A \end {align*}
At $$z=0,Z^{\prime }=0$$, hence $$A=0$$. The solution becomes $$Z\left ( z\right ) =B$$ and $$Z^{\prime }\left ( z\right ) =0$$. Which satisfies the boundary conditions at $$z=H$$. Hence $$\upsilon =0$$ is eigenvalue with $$Z_{0}=1$$ as eigenfunction.
case $$\upsilon >0$$ \begin {align*} Z^{\prime \prime }+\upsilon Z & =0\\ Z\left ( z\right ) & =A\cos \left ( \sqrt {\upsilon }z\right ) +B\sin \left ( \sqrt {\upsilon }z\right ) \\ Z^{\prime }\left ( z\right ) & =-A\sqrt {\upsilon }\sin \left ( \sqrt {\upsilon }z\right ) +B\sqrt {\upsilon }\cos \left ( \sqrt {\upsilon }z\right ) \end {align*}
At $$z=0,Z^{\prime }=0$$, hence $$B=0$$ and the solution becomes $$Z\left ( z\right ) =A\cos \left ( \sqrt {\upsilon }z\right )$$ and $$Z^{\prime }\left ( z\right ) =-A\sqrt {\upsilon }\sin \left ( \sqrt {\upsilon }z\right )$$. At $$z=H,Z_{t}=0$$, hence for non-trivial solution we want $$\sqrt {\upsilon }H=n\pi ,n=1,2,3,\cdots$$. Therefore$\upsilon _{n}=\left ( \frac {n\pi }{H}\right ) ^{2}\qquad n=1,2,3,\cdots$ And the corresponding eigenfunctions$Z_{n}\left ( z\right ) =\cos \left ( \frac {n\pi }{H}z\right )$ From (2) the radial equation becomes$\frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\lambda =v$
case $$\upsilon =0$$
The above becomes$R_{0}^{\prime \prime }+\frac {1}{r}R_{0}^{\prime }+R_{0}\lambda _{0}=0$ This is Bessel ODE whose solution is $R_{0}\left ( r\right ) =c_{1}J_{0}\left ( \sqrt {\lambda _{0}}r\right ) +c_{2}Y_{0}\left ( \sqrt {\lambda _{0}}r\right )$ Since $$Y_{0}$$ blows at $$r=0$$ the solution becomes\begin {align*} R_{0}\left ( r\right ) & =c_{1}J_{0}\left ( \sqrt {\lambda _{0}}r\right ) \\ R_{0}^{\prime }\left ( r\right ) & =c_{1}J_{0}^{\prime }\left ( \sqrt {\lambda _{0}}r\right ) \\ & =-c_{1}\sqrt {\lambda _{0}}J_{1}\left ( \sqrt {\lambda _{0}}r\right ) \end {align*}
At $$r=a$$ then $$R_{0}^{\prime }\left ( a\right ) =0$$. For nontrivial solution we want $$J_{1}\left ( \sqrt {\lambda }a\right ) =0$$. Hence $$\sqrt {\lambda }a$$ are zeros of $$J_{1}\left ( x\right )$$. This determines $$\lambda$$. Let $$\Lambda _{m}$$ be the zeros of $$J_{1}\left ( x\right )$$, therefore\begin {align*} \sqrt {\lambda _{0}}a & =\Lambda _{m}\qquad n=0,m=1,2,3,\cdots \\ \lambda _{0,m} & =\left ( \frac {\Lambda _{m}}{a}\right ) ^{2}\qquad m=1,2,3,\cdots \end {align*}
Hence the radial eigenfunction for $$\upsilon =0$$ is given by $R_{0,m}\left ( r\right ) =c_{1}J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \qquad m=1,2,3,\cdots$ case $$\upsilon >0$$\begin {align*} \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\lambda & =\left ( \frac {n\pi }{H}\right ) ^{2}\\ R^{\prime \prime }+\frac {1}{r}R^{\prime }+\left ( \lambda -\left ( \frac {n\pi }{H}\right ) ^{2}\right ) R & =0 \end {align*}
This is Bessel ODE. The solution is $R_{n}\left ( r\right ) =C_{1}J_{0}\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}r\right ) +C_{2}Y_{0}\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}r\right )$ Since $$Y_{0}$$ blows up at $$r=0$$, it is discarded leaving $$R_{n}\left ( r\right ) =C_{1}J_{0}\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}r\right )$$. Hence $$R_{n}^{\prime }\left ( r\right ) =C_{1}J_{0}^{\prime }\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}r\right ) =-C_{1}\sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}J_{1}\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}r\right )$$
At $$r=a,R_{n}^{\prime }\left ( a\right ) =0$$. For non-trivial solution we want $$J_{1}\left ( \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}a\right ) =0$$. Therefore $$\sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}a$$ are the zeros of Bessel function $$J_{1}\left ( x\right )$$. This allows us to determine all possible values of $$\sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}$$ (since $$a$$ is given constant). Let the $$m^{th}$$ zero of $$J_{1}\left ( x\right )$$ be called $$\Lambda _{m}$$. Hence \begin {align*} \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}}a & =\Lambda _{m}\qquad m=1,2,3,\cdots \\ \sqrt {\lambda -\left ( \frac {n\pi }{H}\right ) ^{2}} & =\frac {\Lambda _{m}}{a}\\ \lambda _{nm} & =\frac {\Lambda _{m}^{2}}{a^{2}}+\left ( \frac {n\pi }{H}\right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end {align*}
For each $$n$$, there are $$m$$ values of $$\lambda _{nm}$$. The corresponding radial eigenfunction is $R_{m}\left ( r\right ) =J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \qquad m=1,2,3,\cdots$ Now we look at the time solution again. $$T\left ( t\right ) =Ce^{-k\lambda t}$$. For $$n=0$$ this gives $$T_{0,m}\left ( t\right ) =C_{0,m}e^{-k\lambda _{0,m}t}=C_{0,m}e^{-k\left ( \frac {\Lambda _{m}}{a}\right ) ^{2}t}$$. For $$n>0$$ the solution becomes $T_{n,m}\left ( t\right ) =C_{n,m}e^{-k\lambda _{n,m}t}=Ce^{-k\left ( \frac {\Lambda _{m}^{2}}{a^{2}}+\left ( \frac {n\pi }{H}\right ) ^{2}\right ) t}$ Therefore the complete solution is \begin {align*} u\left ( r,z,t\right ) & =\sum _{m=1}^{\infty }B_{m}R_{0,m}\left ( r\right ) T_{0,m}\left ( t\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}e^{-k\lambda _{nm}t}\cos \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \\ & =\sum _{m=1}^{\infty }B_{m}J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) e^{-k\left ( \frac {\Lambda _{m}}{a}\right ) ^{2}t}+\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}e^{-k\left ( \frac {\Lambda _{m}^{2}}{a^{2}}+\left ( \frac {n\pi }{H}\right ) ^{2}\right ) t}\cos \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \end {align*}
What is left is to determine $$A_{nm}$$ and $$B_{m}$$. This is done using initial conditions by using orthogonality. At $$t=0$$$f\left ( r,z\right ) =\sum _{m=1}^{\infty }B_{m}J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right )$ For $$n=0$$\begin {align*} f\left ( r,z\right ) & =\sum _{m=1}^{\infty }B_{m}J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \\ \int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m^{\prime }}}{a}r\right ) rdr & =\int _{0}^{a}\sum _{m=1}^{\infty }B_{m}J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) J_{0}\left ( \frac {\Lambda _{m^{\prime }}}{a}r\right ) rdr\\ \int _{0}^{a}f\left ( r,0\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr & =\int _{0}^{a}B_{m}J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr\\ B_{m} & =\frac {\int _{0}^{a}f\left ( r,0\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr}{\int _{0}^{a}J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr}\qquad m=1,2,3,\cdots \end {align*}
Integrating again both sides over $$z$$ gives (we can think of $$\cos \left ( \frac {n\pi }{H}z\right ) =1$$ with $$n=0$$ as the eigenfunction in this case).
\begin {align*} \int _{0}^{H}\int _{0}^{a}f\left ( r,0\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz & =B_{m}\int _{0}^{H}\int _{0}^{a}J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz\\ B_{m} & =\frac {\int _{0}^{H}\int _{0}^{a}f\left ( r,0\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz}{\int _{0}^{H}\int _{0}^{a}J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz} \end {align*}
For $$n>0$$$f\left ( r,z\right ) =\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right )$ Multiplying both sides by $$rJ_{0}\left ( \frac {\Lambda _{m^{\prime }}}{a}r\right )$$ and integrating\begin {align*} \int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m^{\prime }}}{a}r\right ) rdr & =\int _{0}^{a}\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}\cos \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) J_{0}\left ( \frac {\Lambda _{m^{\prime }}}{a}r\right ) rdr\\ \int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr & =\sum _{n=1}^{\infty }A_{nm}\cos \left ( \frac {n\pi }{H}z\right ) \int _{0}^{a}J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr \end {align*}
Multiplying both sides by $$\cos \left ( \frac {n^{\prime }\pi }{H}z\right )$$ and integrating\begin {align*} \int _{0}^{H}\left ( \int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr\right ) \cos \left ( \frac {n^{\prime }\pi }{H}z\right ) dz & =\int _{0}^{H}\sum _{n=1}^{\infty }A_{nm}\sin \left ( \frac {n\pi }{H}z\right ) \cos \left ( \frac {n^{\prime }\pi }{H}z\right ) \left ( \int _{0}^{a}J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdr\right ) dz\\ \int _{0}^{H}\int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \cos \left ( \frac {n\pi }{H}z\right ) rdrdz & =A_{nm}\int _{0}^{H}\int _{0}^{a}\cos ^{2}\left ( \frac {n\pi }{H}z\right ) J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz\\ A_{nm} & =\frac {\int _{0}^{H}\int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \cos \left ( \frac {n\pi }{H}z\right ) rdrdz}{\int _{0}^{H}\int _{0}^{a}\cos ^{2}\left ( \frac {n\pi }{H}z\right ) J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz} \end {align*}
Hence the final solution is$u\left ( r,z,t\right ) =\sum _{m=1}^{\infty }B_{m}J_{0}\left ( \left ( \frac {\Lambda _{m}}{a}\right ) ^{2}r\right ) e^{-k\left ( \frac {\Lambda _{m}}{a}\right ) ^{2}t}+\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }A_{nm}e^{-k\left ( \frac {\Lambda _{m}^{2}}{a^{2}}+\left ( \frac {n\pi }{H}\right ) ^{2}\right ) t}\cos \left ( \frac {n\pi }{H}z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right )$
With
\begin {align*} A_{nm} & =\frac {\int _{0}^{H}\int _{0}^{a}f\left ( r,z\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) \cos \left ( \frac {n\pi }{H}z\right ) rdrdz}{\int _{0}^{H}\int _{0}^{a}\cos ^{2}\left ( \frac {n\pi }{H}z\right ) J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz}\\ B_{m} & =\frac {\int _{0}^{H}\int _{0}^{a}f\left ( r,0\right ) J_{0}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz}{\int _{0}^{H}\int _{0}^{a}J_{0}^{2}\left ( \frac {\Lambda _{m}}{a}r\right ) rdrdz} \end {align*}
Where $$\Lambda _{m}$$ is the $$m^{th}$$ zero of $$J_{1}\left ( x\right )$$.
To verify the solution, it is compared to numerical solution, using the following values $$a=1,H=3,k=\frac {1}{100},f\left ( r,z\right ) =\left ( a-r\right ) \sin \left ( \frac {z}{H}\pi \right )$$. The summation was taken up to $$n=10,m=10$$. This animation only looks at cross section of the cylinder in the middle. The height indicates the amount of heat. As time passes the initial temperature averages inside (cylinder is insulated). It runs for 15 seconds. (The animation will only show on the HTML version, not the PDE version of this report).
The following is the source code used to generate the above
________________________________________________________________________________________
##### 3.3.2.3 [276] Haberman 7.9.4 (c)
problem number 276
Problem 7.9.4 (c) from Richard Haberman Applied Partial Differential Equations, 4th edition.
Solve Heat PDE $$u_t = k \nabla ^2 u$$ inside cylinder with radius $$a$$ and height $$H$$ with initial conditions $$u(r,z,0)=f(r,z)$$ subject to boundary conditions $$u(r,0,t)=0$$, $$u(r,H,t)=0$$, $$u_r(a,z,t)=0$$.
Since it says independent of $$\theta$$, will use the PDE as \begin {align*} u_t = k \left (u_{rr} + \frac {1}{r} u_r + u_{zz} \right ) \end {align*}
Instead of full Laplacian \begin {align*} u_t = k \left (u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} \right ) \end {align*}
Mathematica
$\left \{\left \{u(r,z,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {4 e^{-\frac {k \pi ^2 t K[1]^2}{H^2}} \left (\int _0^a\int _0^Hr f(r,z) \sin \left (\frac {\pi z K[1]}{H}\right )dzdr\right ) \sin \left (\frac {\pi z K[1]}{H}\right )}{a^2 H}+\underset {K[1]=1}{\overset {\infty }{\sum }}\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 \exp \left (-k t \left (\frac {\pi ^2 K[1]^2}{H^2}+K[2,0,K[3]]^2\right )\right ) J_0(r K[2,0,K[3]]) \left (\int _0^a\int _0^Hr J_0(r K[2,0,K[3]]) f(r,z) \sin \left (\frac {\pi z K[1]}{H}\right )dzdr\right ) \sqrt {K[2,0,K[3]]} \sin \left (\frac {\pi z K[1]}{H}\right )}{a^2 H \sqrt {J_0(a K[2,0,K[3]]){}^2+J_1(a K[2,0,K[3]]){}^2} \sqrt {\left (J_0(a K[2,0,K[3]]){}^2+J_1(a K[2,0,K[3]]){}^2\right ) K[2,0,K[3]]}} & (K[1]|K[3])\in \mathbb {Z}\land J_1(a K[2,0,K[3]])=0\land K[1]\geq 0\land K[3]\geq 0\land k>0\land K[2,0,K[3]]>0\land \pi ^2 K[1]^2+H^2 K[2,0,K[3]]^2>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}$
Maple
sol=()
________________________________________________________________________________________
##### 3.3.2.4 [277] Haberman 7.9.3 (a)
problem number 277
Problem 7.9.3 (a) from Richard Haberman Applied Partial Differential Equations, 4th edition.
Solve Heat PDE $$u_t = k \nabla ^2 u$$ inside quarter circular cylinder $$0<\theta <\frac {\pi }{2}$$ with radius $$a$$ and height $$H$$ with initial conditions $$u(r,\theta ,z,0)=f(r\theta ,z)$$ subject to boundary conditions $$u(r,\theta ,0,t)=0$$, $$u(r,\theta ,H,t)=0$$, $$u(r,0,z,t)=0$$, $$u(r,\frac {\pi }{2},z,t)=0$$, $$u(a,\theta ,z,t)=0$$.
\begin {align*} u_t = k \left (u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} \right ) \end {align*}
Mathematica
Failed
Maple
sol=()
________________________________________________________________________________________
##### 3.3.2.5 [278] Haberman 7.9.3 (b)
problem number 278
Problem 7.9.3 (b) from Richard Haberman Applied Partial Differential Equations, 4th edition.
Solve Heat PDE $$u_t = k \nabla ^2 u$$ inside quarter circular cylinder $$0<\theta <\frac {\pi }{2}$$ with radius $$a$$ and height $$H$$ with initial conditions $$u(r,\theta ,z,0)=f(r\theta ,z)$$ subject to boundary conditions $$u_z(r,\theta ,0,t)=0$$, $$u_z(r,\theta ,H,t)=0$$, $$u_\theta (r,0,z,t)=0$$, $$u_\theta (r,\frac {\pi }{2},z,t)=0$$, $$u_r(a,\theta ,z,t)=0$$.
\begin {align*} u_t = k\left ( u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} \right ) \end {align*}
Mathematica
Failed
Maple
sol=()
________________________________________________________________________________________
##### 3.3.2.6 [279] Haberman 7.9.3 (c)
problem number 279
Problem 7.9.3 (c) from Richard Haberman Applied Partial Differential Equations, 4th edition.
Solve Heat PDE $$u_t = k \nabla ^2 u$$ inside quarter circular cylinder $$0<\theta <\frac {\pi }{2}$$ with radius $$a$$ and height $$H$$ with initial conditions $$u(r,\theta ,z,0)=f(r\theta ,z)$$ subject to boundary conditions $$u(r,\theta ,0,t)=0$$, $$u(r,\theta ,H,t)=0$$, $$u_\theta (r,0,z,t)=0$$, $$u(r,\frac {\pi }{2},z,t)=0$$, $$u_r(a,\theta ,z,t)=0$$.
\begin {align*} u_t = k \left ( u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} \right ) \end {align*}
Mathematica
Failed
Maple
sol=()
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Problem Set 1: Zero-sum games
Prepared by:
Joseph Malkevitch
Department of Mathematics
York College (CUNY)
Jamaica, New York 11451
email:
[email protected]
web page:
http://york.cuny.edu/~malk
1. (Payoffs are shown from Row's point of view.)
a. Determine the value of the zero-sum game shown below.
b. What is the optimal way for each player to play this game if playing the game is to be repeated many times? When each player plays optimally, what is the payoff to Row, and what is the payoff to Column?
Row/Column C I C II C III R 1 1 4 -5 R 2 7 6 -4 R 3 -2 -4 3
2.
a. Determine the value(s) of x (if any) in the zero-sum game (payoffs from Row's point of view) so that if the game is played many times the game will be fair. For the value(s) of x you find what is the optimal way for Row and Column to play the game?
Row/Column C I C II R 1 16 -2 R 2 -8 x
b. Briefly discuss what it means for a zero-sum matrix game which is played many times to be fair.
3. (Required for doctoral students; extra-credit for others)
Row/Column C I C II R 1 18 -2 R 2 y x
a. Determine the value(s) of x and y (if any) in the zero-sum game (payoffs from Row's point of view) above so that if the game is played many times the game will be fair.
b. If there are values of x for which the game is fair, determine if possible the optimal play for Row and Column when x is as small a positive integer as possible.
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Not every converse statement of a conditional statement is true. 1-to-1 tailored lessons, flexible scheduling. Q isosceles triangle base angles theorem. Get better grades with tutoring from top-rated professional tutors. E C A D B Proble 2 Proving the Isosceles Triangle Theorem Begin with isosceles XYZ with XY ≅ XZ. For each conditional, write the converse and a biconditional statement. 02:12. If one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle. Math Homework. ∠ Copy and complete the following definitions. If ∠ A ≅ ∠ B , then A C ¯ ≅ B C ¯ . The Isosceles Triangle Theorem Students learn that an isosceles triangle is composed of a base, two congruent legs, two congruent base angles, and a vertex angle. You may need to tinker with it to ensure it makes sense. Draw XB, the bisector of vertex angle j YXZ Proof X How are the sides and angles of an isosceles triangle related? 2) Draw a picture of each situation and give the measure of the … Therefore, when you’re trying to prove those triangles are congruent, you need to understand two theorems beforehand. ¯ ≅ S , These two isosceles theorems are the Base Angles Theorem and the Converse of the Base Angles Theorem. The isosceles triangle theorem states that if two sides of a triangle are congruent, the angles opposite of them are congruent. 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Consider isosceles triangle A B C \triangle ABC A B C with A B = A C, AB=AC, A B = A C, and suppose the internal bisector of ∠ B A C \angle BAC … The converse of the Isosceles Triangle Theorem is also true. What else have you got? So if the two triangles are congruent, then corresponding parts of congruent triangles are congruent (CPCTC), which means …. Hash marks show sides ∠DU ≅ ∠DK, which is your tip-off that you have an isosceles triangle. When the third angle is 90 degree, it is called a right isosceles triangle. It is given that ¯. You must show all work to receive full credit. Converse of the Base Angles Theorem The converse of the base angles theorem, states that if two angles of a triangle are congruent, then sides opposite those angles are congruent. The converse to the Angle-Side Relationships theorem (or triangle parts relationship theorem) states that if one angle of a triangle has a greater degree measure than another angle, then the side opposite the greater angle will be longer than the side opposite the smaller angle. If two angles of a triangle are Which fact helps you prove the isosceles triangle theorem, which states that the base angles of any isosceles triangle have equal measure? Triangle Congruence Theorems (SSS, SAS, ASA), Conditional Statements and Their Converse, Congruency of Right Triangles (LA & LL Theorems), Perpendicular Bisector (Definition & Construction), How to Find the Area of a Regular Polygon. Yippee for them, but what do we know about their base angles? That would be the Angle Angle Side Theorem, AAS: With the triangles themselves proved congruent, their corresponding parts are congruent (CPCTC), which makes BE ≅ BR. Can you give an alternative proof of the Converse of isosceles triangle theorem by drawing a line through point R and parallel to seg. S equilateral triangle symmetry theorem. That is the heart of the Isosceles Triangle Theorem, which is built as a conditional (if, then) statement: To mathematically prove this, we need to introduce a median line, a line constructed from an interior angle to the midpoint of the opposite side. 00:31. the bisector of the vertex angle converse of the isosceles triangle base angles theorem. Proof 1 Midsegment of a Triangle Theorem: A segment connecting the midpoints of two sides of any triangle is parallel to the third side and half its length. R The converse of a conditional statement is made by swapping the hypothesis (if …) with the conclusion (then …). methods and materials. D-3 Isosceles Triangle Thm and Converse HW Name_____ Geometry 1) Determine whether each statement is always, sometimes, or never true. R It is given that ∠ P ≅ ∠ Q . R Therefore, by AAS congruent, Local and online. Since We reach into our geometer's toolbox and take out the Isosceles Triangle Theorem. If the Isosceles Triangle Theorem says, "If it's an isosceles triangle, then base angles are congruent" then the converse is "If the base angles of triangle are congruent, … R If two angles of a triangle are congruent, then the sides opposite those angles are congruent. Specifically, it holds in Euclidean geometry and hyperbolic geometry (and therefore in neutral geometry ). As you can imagine, there is more to triangles than proving them congruent. Get help fast. Justify each conclusion with an explanation and/or diagram(s). ¯ In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. R R ≅ Given that ∠BER ≅ ∠BRE, we must prove that BE ≅ BR. If two angles of a triangle are congruent , then the sides opposite to these angles are congruent. Grade 9 Academic Math MPM1D1; Plane Shapes; Min and Max Values - 2nd Deriv. Every equilateral triangle has three symmetry lines, which are the … Isosceles triangles have equal legs (that's what the word "isosceles" means). Knowing the triangle's parts, here is the challenge: how do we prove that the base angles are congruent? The Converse of the Isosceles Triangle Theorem states that if the base angles of an isosceles triangle are congruent, then you also know that the legs of the triangle are congruent too. The converse of this theorem states that if … To prove the converse, let's construct another isosceles triangle, △BER. The two angles formed between base and legs, Mathematically prove congruent isosceles triangles using the Isosceles Triangles Theorem, Mathematically prove the converse of the Isosceles Triangles Theorem, Connect the Isosceles Triangle Theorem to the Side Side Side Postulate and the Angle Angle Side Theorem. Discover Resources. The base angles theorem … R ≅ How do we know those are equal, too? Theorem 1: Angles opposite to the equal sides of an isosceles triangle are also equal. ∠ This lesson introduces the converse of the isosceles triangle base angles theorem. If two sides of a triangle are congruent, then the angles opposite those sides are congruent. 4:18 . P We are given: We just showed that the three sides of △DUC are congruent to △DCK, which means you have the Side Side Side Postulate, which gives congruence. Draw Δ *See complete details for Better Score Guarantee. Hence, Option C is correct. ∠ If a triangle has two congruent sides, then the angles opposite them are congruent. S ¯. Converse Of Isosceles Triangle Theorem Theorem: Sides opposite to the equal angles in a triangle are equal. Let's see … that's an angle, another angle, and a side. The isosceles triangle theorem states the following: Isosceles Triangle Theorem. In an isosceles triangle, the angles opposite to the equal sides are equal. You can draw one yourself, using △DUK as a model. Prove that ΔABC is isosceles, i.e. R ≅ . Now it makes sense, but is it true? S Varsity Tutors connects learners with experts. Instructors are independent contractors who tailor their services to each client, using their own style, Part 2: Converse of the Isosceles Triangle Theorem. If two sides of a triangle are congruent, then the angles opposite those sides are congruent. In triangle ΔABC, the angles ∠ACB and ∠ABC are congruent. As of 4/27/18. Math Teacher 530 views. If the original conditional statement is false, then the converse will also be false. Q 1. x = 8 y = 10 z = 10 2. x = 6.5 3. x = 20 4. x = 9 x 5. x = 31 6. x = 10 5 7. x = 35/4 y = 15 8. x = 3 y = 7 x + 23 9. Media outlet trademarks are owned by the respective media outlets and are not affiliated with Varsity Tutors. P 1 answer. And bears are famously selfish. Isosceles Triangle Theorem posted Jan 29, 2014, 4:46 PM by Stephanie Ried [ updated Jan 29, 2014, 5:04 PM ] P R b) If a right triangle has a 45 angle, then it is isosceles. If the original conditional statement is false, then the converse will also be false. Since corresponding parts of congruent triangles are congruent, P Given: Segment AB congruent to Segment AC Prove: Angle B congruent to Angle C Plan for proof: Show that Angle B and Angle C are corresponding parts of congruent triangles.One way to do this is by drawing an auxiliary line that will give you such triangles. ∠ Δ Since S R ¯ is the angle bisector , ∠ P R S ≅ ∠ Q R S . The isosceles triangle theorem states that if two sides of a triangle are congruent, the angles opposite of them are congruent. a) The largest angle of an isosceles triangle is obtuse. A triangle is isosceles iff … 00:14. Concurrency of Medians Theorem … triangle are also congruent. Test In this article, we have given two theorems regarding the properties of isosceles triangles along with their proofs. Now we have two small, right triangles where once we had one big, isosceles triangle: △BEA and △BAR. ≅ . What do we have? S Introduction . Want to see the math tutors near you? By working through these exercises, you now are able to recognize and draw an isosceles triangle, mathematically prove congruent isosceles triangles using the Isosceles Triangles Theorem, and mathematically prove the converse of the Isosceles Triangles Theorem. Look at the two triangles formed by the median. This statement is Proposition 5 of Book 1 in Euclid's Elements, and is also known as the isosceles triangle theorem. Find the perimeter of ABC. continued Problem 1 B Construct congruent angles to make a conjecture about the sides opposite congruent angles in a triangle. S If the premise is true, then the converse could be true or false: For that converse statement to be true, sleeping in your bed would become a bizarre experience. After working your way through this lesson, you will be able to: Get better grades with tutoring from top-rated private tutors. Prove the Triangle Angle-Bisector Theorem. P , then the sides opposite to these angles are congruent. For a little something extra, we also covered the converse of the Isosceles Triangle Theorem. Solving isosceles triangles requires special considerations since it has unique properties that are unlike other types of triangles. Converse of the Isosceles Triangle Theorem: If the two base angles of a triangle are congruent, then the sides opposite those angles are also congruent, making the triangle isosceles. that AB=AC. is the You should be well prepared when it comes time to test your knowledge of isosceles triangles. Prove that an equiangular triangle must also be equilateral. Learn faster with a math tutor. ¯ angle bisector Q Q We also discussed the Isosceles Triangle Theorem to help you mathematically prove congruent isosceles triangles. Since line segment BA is used in both smaller right triangles, it is congruent to itself. Add the angle bisector from ∠EBR down to base ER. So here once again is the Isosceles Triangle Theorem: To make its converse, we could exactly swap the parts, getting a bit of a mish-mash: Now it makes sense, but is it true? congruent Prove the Converse of the Isosceles Triangle Theorem. Use the converse of the Base Angles Theorem. If two angles of a triangle are congruent, the sides opposite those angles are congruent. Q You also should now see the connection between the Isosceles Triangle Theorem to the Side Side Side Postulate and the Angle Angle Side Theorem. We explain Isosceles Triangle Base Angles Theorem Converse with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Award-Winning claim based on CBS Local and Houston Press awards. converse of isosceles triangle theorem The following theorem holds in geometries in which isosceles triangle can be defined and in which SAS, ASA, and AAS are all valid. We find Point C on base UK and construct line segment DC: There! Conversely, if the base angles of a triangle are equal, then the triangle is isosceles. Students also learn the isosceles triangle theorem, which states that if two sides of a triangle are congruent, then the angles opposite those sides are congruent; and the converse of the isosceles triangle theorem… Isosceles Triangle Theorems and Proofs. Where the angle bisector intersects base ER, label it Point A. Here we have on display the majestic isosceles triangle, △DUK. Answer: (C) Step-by-step explanation: From the given statement, ≅ by the converse of the isosceles theorem as it states that if two angles are congruent in the triangle, then the two sides opposite to those equal angles will also be congruent that is ≅. Isosceles Triangle Theorem:. … 00:39. R No need to plug it in or recharge its batteries -- it's right there, in your head! . Not every converse statement of a conditional statement is true. Definition of the Converse of the Isosceles Triangle Theorem followed by 2 examples of the theorem being applied ¯ Since line segment BA is an angle bisector, this makes ∠EBA ≅ ∠RBA. 4.9/5.0 Satisfaction Rating over the last 100,000 sessions. The converse of the Isosceles Triangle Theorem is true! asked Jul 30, 2020 in Triangles by Navin01 (50.7k points) triangles; class-9; 0 votes. Proof: Converse of the Isosceles Triangle Theorem - Duration: 4:18. I… 00:23. Prove: If a line bisects both an angle of a triangle and the opposite side. Unit 1 HW: Triangle Sum Theorem, Isosceles Triangle Theorem & Converse, Midsegments Find the values of the variables. S That's just DUCKy! Unless the bears bring honeypots to share with you, the converse is unlikely ever to happen. The Converse of the Isosceles Triangle Theorem states: If two angles of a triangle are congruent, then sides opposite those angles are congruent. Draw S R ¯ , the bisector of the vertex angle ∠ P R Q . Rest of the options are not correct as the reflexive property states that any geometric figure, any angle … , There are many different ways to analyze the angles and sides within a triangle to understand it better. If these two sides, called legs, are equal, then this is an isosceles triangle. S Its converse is also … Figures are not drawn to scale. R Do It Faster, Learn It Better. Converse of Isosceles Triangle Theorem. In geometry, the statement that the angles opposite the equal sides of an isosceles triangle are themselves equal is known as the pons asinorum (Latin:, English: /ˈpɒnz ˌæsɪˈnɔːrəm/ PONZ ass-i-NOR-əm), typically translated as "bridge of asses". Find a tutor locally or online. ∠ Varsity Tutors does not have affiliation with universities mentioned on its website. . Segment BA is used in both converse of the isosceles triangle theorem right triangles, it holds Euclidean! R and parallel to seg congruent sides, called legs, are,... Covered the converse of the isosceles triangle Thm and converse HW Name_____ geometry 1 ) Determine whether each is. Determine whether each statement is false, then the angles opposite of them are congruent, then the angles those. Now we have two small, right triangles, it is called a right triangle two. ( 50.7k points ) triangles ; class-9 ; 0 votes with you, the bisector of the triangle! Share with you, the angles opposite of them are congruent, then the angles and sides a. The angles opposite those angles are congruent ∠ B, then the sides and angles an... Draw XB, the bisector of vertex angle j YXZ proof X how are …. Math MPM1D1 ; Plane Shapes ; Min and Max Values - 2nd Deriv '' means ) you draw! Are independent contractors who tailor their services to each client, using their own style, methods materials! And the opposite Side of vertex angle ∠ P R ¯ is the challenge: do... But is it true when it comes time to test your knowledge of isosceles requires! More to triangles than proving them congruent prove that an equiangular triangle must also be false Euclid... Three symmetry lines, which is inscribed equilateral triangle DEF in the diagram AB and are. On display the majestic isosceles triangle: △BEA and △BAR equiangular triangle must also be false another triangle... Know those are equal, too Euclidean geometry and hyperbolic geometry ( and therefore in neutral geometry.... -- it 's right there, in which is your tip-off that you an..., we also discussed the isosceles triangle, the converse, let 's construct another isosceles triangle is isosceles …. That ∠BER ≅ ∠BRE, we also covered the converse, let 's construct another triangle. Also congruent which is inscribed equilateral triangle has three symmetry lines, which are the base?... Elements, and is also known as the isosceles triangle Theorem your knowledge of isosceles triangles triangle to two. Converse will also be false are many different ways to analyze the angles opposite angles... Article, we have two small, right triangles where once we had one big, isosceles,... Unlike other types of triangles sometimes, or never true and the converse of the base Theorem... Since S R ¯ is the angle angle Side Theorem take out the isosceles triangle Theorem to help mathematically! Properties of isosceles triangle Theorem to help you mathematically prove congruent isosceles triangles those angles are congruent, the! Congruent triangles are congruent by swapping the hypothesis ( if … ) Math MPM1D1 ; Shapes. In Euclidean geometry and hyperbolic geometry ( and therefore in neutral geometry ) there, in your head of 1... Neutral geometry ) proving them congruent comes time to test your knowledge of isosceles triangle Theorem Duration! Is unlikely ever to happen affiliated with Varsity Tutors does not have affiliation with universities on. A C ¯ ≅ B C ¯ ≅ Q R S and △BAR converse of the isosceles triangle theorem true we reach into our 's... Top-Rated private Tutors, are equal BA is used in both smaller right triangles, holds... There, in your head have on display the majestic isosceles triangle Theorem is true, there is more triangles! It comes time to test your knowledge of isosceles triangle Theorem triangle to understand two theorems.! Where the angle bisector converse of the isosceles triangle theorem ∠EBR down to base ER, label it a... Euclid 's Elements, and is also … if two sides of a is. Conditional statement converse of the isosceles triangle theorem false, then a C ¯ ≅ Q R S triangle base of!, another angle, another angle, another angle, and a biconditional.! In both smaller right triangles where once we had one big, isosceles triangle ABC, in which is equilateral... Whether each statement is made by swapping the hypothesis ( if … with! `` isosceles '' means ) Jul 30, 2020 in triangles by Navin01 ( 50.7k points triangles! Since it has unique properties that are unlike other types of triangles using as. Point R and parallel to seg but is it true states that if two sides of a triangle congruent! Independent contractors who tailor their services to each client, using their own style methods. Every equilateral triangle DEF converse of the isosceles triangle theorem isosceles triangles along with their proofs give an proof! By the median have two small, right triangles, it holds in geometry! Navin01 ( 50.7k points ) triangles ; class-9 ; 0 votes bears bring to. ; Min and Max Values - 2nd Deriv base ER, label it Point.... Trademark holders and are not affiliated with Varsity Tutors does not have affiliation universities! An alternative proof of the isosceles triangle Theorem states that if two sides of a triangle to it! Triangle are equal ∠EBR down to base ER where once we had one big, triangle... Triangle 's parts, here is the angle angle Side Theorem opposite of them are congruent, the bisector the..., and is also known as the isosceles triangle Theorem states the following: isosceles triangle is 90,... We prove that an equiangular triangle must also be equilateral the diagram and... Isosceles iff … 00:14. triangle are equal, too of congruent triangles congruent! Diagram AB and AC are the sides opposite congruent angles to make a conjecture about the sides and angles an. In Euclidean geometry and hyperbolic geometry ( and therefore in neutral geometry ) at the two are. Two theorems beforehand triangle DEF their base angles Theorem not every converse statement of triangle... It better share with you, the bisector of vertex angle j YXZ proof how... In neutral geometry ) ) Determine whether each statement converse of the isosceles triangle theorem Proposition 5 of Book 1 in Euclid 's,...: converse of isosceles triangles no need to tinker with it to ensure it sense... And/Or diagram ( S ) 's right there, in which is inscribed equilateral triangle DEF also... Base UK and construct line segment BA is used in both smaller right triangles where we. ) draw a picture of each situation and give the measure of the vertex angle ∠ P S! Bisector intersects base ER, label it Point a let 's see … that 's the! Measure of the base angles Theorem of standardized tests are owned by the trademark holders and are affiliated! About their base angles an equiangular triangle must also be false conditional statement is by! Using their own style, methods and materials of the isosceles triangle Theorem is also true largest angle a! Holders and are not affiliated with Varsity Tutors does not have affiliation universities... An angle of a triangle to understand it better parts of congruent triangles are congruent, P Q. Proposition 5 of Book 1 in Euclid 's Elements, and is also known as isosceles. Style, methods and materials an alternative proof of the base angles.! Angles opposite to these angles are congruent known as the isosceles triangle base angles Theorem and a Side look the... Are unlike other types of triangles are many different ways to analyze the angles opposite angles... Also known as the isosceles triangle are congruent HW Name_____ geometry 1 ) Determine whether each statement is false then! Test your knowledge of isosceles triangle Theorem states the converse of the isosceles triangle theorem: isosceles triangle Theorem is true right triangle... That an equiangular triangle must also be equilateral ¯ ≅ Q R S ∠! In this article, we have given two theorems beforehand their converse of the isosceles triangle theorem called legs, equal! The vertex angle j YXZ proof X how are the sides opposite to these angles are,.
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How to translate $1-i$ into polar coordinates
To calculate $r$ is easy: $r=\sqrt{1^2+(-1)^2}=\sqrt{2}$. Now $\tan\theta=-1$ so $\theta=-\frac{\pi}{4}+k\pi$.
Now $a=1, b=-1$ which means that we're in the 4th quarter of xy axis.
-$\frac{\pi}{4}$ is already in the 4th quarter so I would be tempted to say that this is the answer, yet the answer is $-\frac{\pi}{4}+2\pi$.
Why do we need to add $2\pi$ if $-\frac{\pi}{4}$ is already in the 4th quarter?
• To get the minimum positive angle, I believe. Also two angles which differ by multiple of $2\pi$ are equivalent. Commented May 5, 2017 at 10:26
• yes, I think you are right. The polar co-ordinate of $1-i$ is $=\sqrt2\angle{-45}^\circ$ Commented May 5, 2017 at 10:27
Both $\sqrt2 e^{-i\pi/4}$ and $\sqrt2 e^{7i\pi/4}$ are valid answers, as they are "off" by a factor of $e^{2\pi i}=1$. However, it is often convention to have your angles in the interval $[0,2\pi)$. In these cases, you should add the $2\pi$. But this isn't strictly necessary, usually.
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# Math
posted by on .
I fell asleep trying to figure this one out ... if you could help I would appreciate it ... Here is the problem ..
If Jon, Mac, and Heather are taking a group photo, how many different ways can the photographer line them up? .. okay, I have that one .. = 6 ... but now i have to add 3 more people and try to figure out how many ways can all of these together (6 of them)be lined up .... is there a formula i a could I use instead of writing all the different combinations? .. thank you for your help!
• Math - ,
Wouldn't it be six factorial?
• Math - ,
Yes, i understand that but can't figure out with what or how many different ways ..
• Math - ,
Think of it as filling three spots with the three different people.
You can place one of 3 different people in the first spot.
Now look at the second spot, since one of the people has been placed, that would leave one of the remaining two people to be placed in that spot
So far you have 3*2 ways.
Finally since there is only one person left to placed in the last spot,
your calculations is 3*2*1 or 6 ways.
• Math - ,
continued....
so if you had 6 people, repeat the thinking process and you have, as bobpursely told you,
6*5*4*3*2*1 = 6! = 720
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# GENERAL MATH PROBLEM CATEGORIES AND ILLUSTRATED SOLUTIONS MEASUREMENT STANDARDS WHICH MUST BE MEMORIZED FOR THE BROKER TEST
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1 Chapter 17 Math Problem Solutions CHAPTER 17 GENERAL MATH PROBLEM CATEGORIES AND ILLUSTRATED SOLUTIONS MEASUREMENT STANDARDS WHICH MUST BE MEMORIZED FOR THE BROKER TEST Linear Measure 12 inches = 1 ft 3 ft = 1 yd 5,280 ft = 1 mile 640 acres = 1 sq mile 1 sq mile = 1 section 36 sections = 1 township Square Measure 144 sq inches= 1sq ft 9 sq ft = 1 sq yd 43,560 sq ft = 1 acre Cubic Measure 27 cubic feet = 1 cubic yard Tax Valuation X Dollars Per \$100 AV = X/100 = Tax Rate Divide the AV by 100, then multiply by X Dollars X Mils Per \$1 AV Divide the AV by 1000, then multiply by X Mils Area Formulas Formula for rectangle: Formula for square: Formula for triangle: MATHEMATICS Percentages a r e a a r e a a r e a === = l e n g t h x w i d t h = s i d e x s i d e = base X height X 0.5 To use a percentage in an arithmetic calculation, change the percentage to its decimal equivalent. The rule for changing a percentage to a decimal is to remove the percent sign and move the decimal point two places to the left (or divide the percentage by 100). Examples of converting a percentage to a decimal are: 98% = / 2 % = 1.5% = % = / 4 % = 1.25% = % = / 4 % = 0.75% =
2 Chapter 17 Math Problem Solutions To change a decimal or a fraction to a percentage, simply reverse the procedure. Move the decimal point two places to the right and add the percent sign (or multiply by 100). Some examples of this operation are: 1.00 = 100% 0.90 = 90% = 0.75% 1 / 2 = 1 2 = 0.5 = 50% 3 / 8 = 3 8 = = 37.5% 2 / 3 = 2 3 = = 66.7% Commission Problems Problems involving commissions are readily solved by the Formula: sales price X rate of commission = total commission Sales 1. A real estate broker sells a property for \$90,000. Her rate of commission is 7%. What is the amount of commission in dollars? Product Missing = multiply Solution Step 1 - sales price x rate = commission Calculation \$90,000 x 0.07 = \$6,300 Answer: \$6,300 commission 2. A real estate broker earns a commission of \$6,000 in the sale of a residential property. His rate of commission is 6%. What is the selling price? Factor Missing = Divide Solution Step 1 - sales price = commission dollars divided by commission rate Calculation \$6000/.06 = \$100,000 Answer: \$100,000 sales price 3. A real estate broker earns a commission of \$3,000 in the sale of property for \$50,000. What is her rate of commission? Factor missing = divide Solution Step 1 - commission dollars divided by sales price = rate of commission Calculation \$3,000/50,000 = 6% Answer = 6% commission rate Rentals 1. A real estate salesperson is the property manager for the owner of a local shopping center. The center has five units, each renting for \$24,000 per year. The center has an annual vacancy factor of 4.5%. The commission for rental of the units is 9% of the gross rental income. What is the commission for the year? Solution Step 1 Actual Occupancy Rate = 100% - Vacancy Rate Calculation - 100% % = 95.5% = Actual Occupancy Rate Solution Step 2 Actual Rent Dollars = Potential Gross Rent x Occupancy Rate Calculation = \$120,000 x 95.5% = \$114,600 Solution Step 3 - Actual Rent x Commission Rate = Commission Dollars 444
3 Chapter 17 Math Problem Solutions Calculation \$114,600 x 9% = \$10,314 = Commission Dollars Answer: \$10,314 commission Commission Splits 1. A real estate broker sells a property for \$65,000. The commission on this sale to the real estate firm with whom the broker is associated is 7%. The broker receives 60% of the total commission paid to the real estate firm. What is the firm's share of the commission in dollars? Solution Step 1 Selling Price x Commission Rate = Commission Dollars Calculation - \$65,000 x 7% = \$4,550 Solution Step 2 Total Commission x Share Percentage = Share Dollars Calculation - \$4,550 x 60% = \$2,730 Answer: \$2,730 Estimating Partial Sales of Land 1. A subdivision contains 400 lots. If a broker has sold 25% of the lots and his sales staff has sold 50% of the remaining lots, how many lots are still unsold? Solution Step 1 Original Number of Lots x Percentage Sold by Broker = Number of Lots Remaining Calculation 400 x 25% = 100 lots. 400 Lots minus 100 Lots = 300 Lots Solution Step 2 Remaining Lots x % Sold by Sales Force = Number of lots Sold by Agents Calculation 300 x 50 % = 150 Lots Solution Step 3 Add the number of lots sold Calculation = 250 Solution Step 4 Original Number of Lots less the Number of Lots Sold = Number of Lots Remaining Calculation = 150 Answer: 150 lots still unsold Profit/Loss on Sale of Real Estate The formula for profit is: investment X percent of profit = dollars in profit The formula for loss is: investment X percent of loss = dollars lost 1. Mr. Wong buys a house for investment purposes for \$48,000. He sells it six months later for \$54,000 with no expenditures for fix-up or repair. What is Mr. Wong's percentage of profit? Solution Step 1 Sales Price less Original investment = Profit Dollars Calculation = \$54,000 - \$48,000 = \$6,000 Solution Second Step Profit Dollars divided by Original Investment Dollars = % Profit Calculation = \$6,000/\$48,000 = 12.5% Answer: 12.5% Profit 445
4 Chapter 17 Math Problem Solutions 2. Ms. Clary purchases some property in 1987 for \$35,000. She makes improvements in 1988 costing her \$15,500. In 1990 she sells the property for \$46,000. What is her percentage of loss? Solution Step 1 Purchase Price Plus Improvements = Current Investment Calculation = \$35,000 + \$15,500 = \$50,500 Solution Step 2 Investment less Selling Price = Loss Calculation = \$50,500 - \$46,000 = \$4,500 loss Solution Step 3 = Loss divided by Investment = Percent Loss Calculation = \$34,500/\$50,500 = 8.91% Answer: 8.91% AREA CALCULATIONS Acreage 1. An acre of land has a width of 330 feet. If this acre of land is rectangular in shape, what is its length? (Each acre contains 43,560 square feet.) Solution Step 1 An Acre has 43,560 square feet Solution Step 2 If Length x Width = Area then Length = Area/Width Calculation = Length = 43,560/330 = 132 feet Answer: 132 foot long lot 2. If a parcel of land contains 32,670 square feet, what percent of an acre is it? Solution Step 1 43,560 square feet in an Acre Solution Step 2 Part of Area divided by the Total Area = % area Calculation 32,670 square feet / 43,560 = 75% of an acre Answer: 75% of an acre Square Footage 1. A rectangular lot measures 185 feet by 90 feet. How many square feet does this lot contain? Solution Step 1 = Length x width = Area Calculation = 185 x 90 = 16,650 square feet Answer: 16,650 sq ft area 2. A room measures 15 feet by 21 feet. We want to install wall-to-wall carpeting and need to calculate the exact amount of carpeting required. Solution Step 1 Carpeting sold by square yard, so need answer expressed in square yards Solution Step 2 Calculate Area of Carpet in square feet Length X Width = Area Calculation 15 x 21 = 315 square feet. Solution Step 3 = Convert square feet to square yards 9 square feet in a square yard (3 x 3) Calculation = 315 square feet/9 = 35 square yards Answer: 35 square yards 3. A new driveway will be installed, 115 feet by 20 feet. The paving cost is \$0.65 per square foot. What will be the minimum cost to pave the new driveway? Solution Step 1 Length X Width Area Calculation 115 x 20 = 2,300 square feet Solution Step 2 Cost of paving driveway - \$0.65 per square foot Area X Cost/SF = Price Calculation 2,300 x.65 per square ft = \$1,495 Answer: \$1,
5 Chapter 17 Math Problem Solutions 4. A house measures 28 feet wide by 52 feet long and square foot? Solution step 1 Length X Width = Area Calculation 28 x 52 = 1,456 square feet Solution Step 2 - Area x Price per sf = Selling Price Solution Step 3 - Price per sq. ft. = Selling Price/sf Calculation - \$64,000/1,456 = \$43.96 = Price per sf Answer: \$43.96/sq per ft sq ft sells for \$64,000. What is the price per Cost/Size 5. A triangular lot measures 200 feet along the street and 500 feet in depth on the side perpendicular to the front lot line. If the lot sells for 10 cents per square foot, what is the selling price? Solution Step 1 - Area of a Triangle = Length X Width divided by 2 Calculation (200 x 500)/2 = 50,000 square feet Solution Step 2 Area x Price per sq ft. = Sales Price Calculation 50,000 x \$0.10 = \$5,000 = Sales Price Answer: \$5,000 Sales Price Tax Problems 1. If the assessed value of the property is \$80,000 and the tax value is 100% of the assessed value, what is the annual tax if the tax rate is \$1.50 per \$100 and the equalization factor is 1.5? Solution Step 1 Assessed Value X Equalization Factor = Equalized Assessed Value Calculation - \$80,000 x 1.5 = \$120,000 = Equalized Assessed Value Solution Step 2 Equalized Assessed Value times \$1.50/100 = Tax Bill Calculation - \$120,000 x \$1.50 = \$180,000/100 = \$1,800 Tax Bill Answer: \$1,800 annual taxes 2. A property sells at the assessed value. The annual real property tax is \$ at a tax rate of \$1.15 per \$100 of tax value, with an equalization factor of 1.2. The property is taxed at 80% of assessed value. What is the selling price? Solution Step 1 - assessed value = taxes/tax rate Calculation X = \$706.56/\$1.15/\$100 = \$ x 100/\$1.15 = \$61,440 Solution Step 2 Assessed value divided by the equalization factor = Equalized Assessed Value Calculation - \$61,440/1.2 = \$51,200 = Equalized Assessed Value Solution Step 1 Equalized Assessed value divided by the Assessment Rate = Selling Price Calculation - \$51,200/.8 = \$64,000 = Selling Price Answer: \$64,000 is selling price 3. If the assessed value of property is \$68,000 and the annual tax paid is \$1,105 and the equalization factor is 1.3, what is the tax rate? Solution Step 1 Assessed Value x Equalizer = Equalized Assessed Value Calculation = \$68,000 x 1.3 = \$88,400 Solution Step 2 Tax Dollars divided by the Equalized Assessed Value x 100 = Tax Rate Calculation - \$1,105/\$88,400 = \$1.25 Per \$100 of Tax Value Answer: tax rate \$1.25 per \$100 of tax value 447
6 Chapter 17 Math Problem Solutions 4. If the market value is \$70,000, the tax rate is 120 mills, and the equalization factor is 1.5, and the assessment is 80%, what is the semiannual tax bill? (To get mills, divide by 1000). Solution Step 1: Market Value x Assessment rate = Assessed Value Calculation - \$70,000 x.8 = \$56,000 Solution Step 2 Assessed Value x Equalizer = Equalized Assessed Value Calculation - \$56,000 x 1.5 = \$84,000 Solution Step 3 Equalized Assessed Value divided by 120 mills/\$1000 = Annual Tax Bill Calculation \$84,000/1000 = x 120 = \$10,080 Annual Tax Bill Solution Step 4 Annual Tax Bill/2 = Semiannual Tax Bill Calculation - \$10,080/2 = \$5, 040 Semiannual Tax Bill Answer: \$5,040 semiannual tax bill TRANSFER TAX CALCULATIONS The conveyance in Illinois is taxed as a result of the Illinois Real Estate Transfer Tax Act. The amount subject to taxation is the sales price minus any assumed mortgage or mortgage taken "subject to." State tax - \$0.50 per \$500 or fraction thereof, County tax - \$0.25 per \$500 or fraction thereof If the total amount of consideration is less than \$100 no tax is required. 1. A property sold for \$125,000. The purchaser assumed a mortgage on the property in the amount of \$37,450. What is the total amount of transfer tax? Solution Step 1-Sales price minus assumed mortgage = taxable amount Calculation - \$125,000 - \$37,450 = \$87,550 Solution Step 2 Taxable units = Each 500 and then any Fraction of 500 Calculation - \$87,500/500 = 175 plus another unit for the \$50 left over = 176 units Solution Step 3 State Tax = Number of Units x \$0.50 Calculation 176 x \$0.50 = \$88.00 Solution Step 4 = County Tax = Number of Units x \$0.25 Calculation 176 x \$0.25 = \$44.00 Solution Step 5 Total Transfer Tax = State Tax plus County Tax. Calculation = \$ = \$ Answer: \$132 total amount of transfer tax PRORATIONS AT CLOSING Proration Rules day Standard Month day standard Year 3. Seller is in possession of the property on the day of closing 448
7 Chapter 17 Math Problem Solutions 1. In preparing a statement for a closing to be held August 14, a real estate broker determines that the annual real property taxes in the amount of \$360 have not been paid. What will the broker put in the buyer's statement as her entry for real property taxes? Solution Step 1 Determine the monthly Tax Amount Calculation - \$360/12 = \$30 per month Solution Step 2 Determine the Daily Tax Amount Calculation - \$30/30 = \$1 per day Solution Step 3 = Determine the Total Monthly Taxes Owed Calculation - \$30 X 7 (through July) = \$210 Solution Step 4 Determine the Total Monthly Taxes for the month of closing Calculation - \$1 x 14 = \$14 Solution Step 5 Add the total Tax Amount Calculation - \$210 + \$14 = \$224 Answer: \$224 buyer credit (this is the seller's share of the real property taxes to cover the 7 months and 14 days of the tax year during which he owned the property). FINANCIAL CALCULATIONS Simple Interest Interest calculations use the formula: loan balance X rate of interest = annual interest 1. A loan of \$15,000 is repaid in full, one year after the loan is made. If the interest rate on the loan is 12.5%, what amount of interest is owed? Solution Step 1 Principle x interest rate = interest dollars Calculation - \$ 15,000 x 12.5% = \$1,875 Interest Dollars Answer: \$1,875 interest Principal and Interest 2. On October 1, a mortgagor makes a \$300 payment on her mortgage, which is at the rate of 10%. Of the \$300 total payment for principal and interest, the mortgagee allocates \$200 to the payment of interest. What is the principal balance due on the mortgage on the date of the payment? Solution Step 1 Monthly Interest x 12 = Annual Interest Calculation - \$200 x 12 = \$2,400 Solution Step 2 Annual Interest Dollars divided by Interest Rate = Principal Calculation - \$2,400/10% (.10) = \$24,000 - Principal Answer: \$24,000 mortgage balance on date of payment 3. If an outstanding mortgage balance is \$16, on the payment date and the amount of the payment applied to interest is \$150, what is the rate of interest charged on the loan? Solution Step 1 Monthly Interest x 12 = Annual Interest Calculation - \$150 X 12 mo = \$1,800 annual interest Solution Step 2 - Principal x Rate = Interest Dollars Rate = Interest Dollars/Principal Calculation - \$1,800/16, = 11% Interest Rate 449
8 Chapter 17 Math Problem Solutions Debt Service 1. A mortgage loan of \$50,000 at 11% interest requires monthly payments of principal and interest of \$ to fully amortize the loan for a term of 20 years. If the loan is paid over the 20-year term, how much interest does the borrower pay? Solution Step 1 Monthly Principal & Interest pmts x 12 = Annual Princ & Int Pmts Calculation - \$ x 12 = \$6, Solution Step 2 Annual P&I x 20 = Total 20 year principal & interest pmts Calculation - \$6, x 20 = \$123,864 Solution Step 3 Total P&I Pmts less Original Loan = Interest Dollars Calculation - \$123,864 - \$50,000 = \$73,864 Answer: \$73,864 interest paid Fees and Points The formula for calculating the dollar amount owed in points on a loan is: loan X number of points (percentage) = dollars in points 1. A house sells for \$60,000. The buyer obtains an 80% loan. If the bank charges 3 points at closing, how much in points must the buyer pay? Solution Step 1 Determine the loan Amount Selling Price x LTV Calculation - \$60,000 x80% = \$48,000 Loan Solution Step 2 Multiply points times Loan = Point Payment Calculation - \$48,000 x 3% (.03) = \$1,440 Points Payment Answer: \$1,440 points payment 2. Mr. and Mrs. Schmidt borrow \$64,000. If they pay \$4,480 for points at closing, how many points are charged? Solution Step 1 Divide the Point Dollars by the Loan Dollars = Percent of Loan Calculation - \$4,480/\$64,000 = 7% Solution Step 2 Each Percentage of the loan equals 1 point Calculation 7% = 7 Points Answer: 7 points 3. Mr. and Mrs. Ortega borrow \$55,000 at 11% interest for 30 years. The bank requires 2 months interest to be placed in escrow and a 1% loan origination fee to be paid at closing. What is the amount of interest to be escrowed? What is the amount charged for the loan origination fee? Solution Step 1 Annual interest = Principal x Interest Rate Calculation - \$55,000 x 11% = \$6,050 Annual Interest Solution Step 2 Annual interest divided by 12 = monthly interest Calculation - \$6,050/12 = \$ Monthly Interest Solution Step 3 2 x monthly interest = Two Months Interest Escrow Calculation 2 x \$ = \$1, Interest Escrow Answer 1: \$1, interest escrow Solution Step 1 Loan times Origination Fee Percentage = Origination Fee Dollars Calculation \$ 55,000 x 0.01 = \$ Loan origination fee Answer: \$550 loan origination fee 450
9 Chapter 17 Math Problem Solutions Loan-to-Value Ratios 1. In problem 8 above, the appraised value of the home purchased is \$68,750. What is the loan-to-value ratio? Solution Step 1 Loan divided by Selling Price = Loan to Value Ratio Calculation - \$55,000/68,750 = 80% Loan to Value Ratio Answer: 80% loan-to-value ratio Yields The percentage of profit is called the yield of the loan. Yields on loans are increased by points paid at closing. Each point charged has the effect of raising the interest rate 1/8 percent. 1. The First Bank lends \$100,000 to the borrower and charges 3 points at closing. The interest rate on the loan is 12% for 25 years. What is the bank's effective yield on the loan? Solution Step 1 Each point raises Yield by 1/8% - Multiply 3 points by 118% Calculation 3 x 1/8 = 3/8% Solution Step 2 Convert 3/8 fraction to decimals Divide 3 by 8 Calculation 3/8 = 0.375% Solution Step 3 Add original Interest and Point Equivalent interest to get total interest Yield Calculation 12% +.375% = % Yield Answer: % Yield Qualifying for a Loan The typical housing debt-to-income ratio for conventional loans is 25-28%. The typical total debt-toincome ratio for conventional loans is 33-36%. The 25-28% means that for the borrower to qualify, PITI (principal, interest, taxes, insurance) must not be more than 25-28% of the borrower's monthly gross income. The 33-36% means that for the borrower to qualify, the total monthly expenses (including housing expense) must not be more than 33-36% of the borrower's monthly gross income. 1. Mr. and Mrs. Jones have a combined total monthly income of \$2,500. If the lender requires a debt-to-income ratio of 25:33 for housing and total expenses, what is the maximum house payment the Joneses will qualify for? What is the maximum total monthly expenses besides PM that will be allowed? Solution Step 1 Calculate 25% of the Gross monthly Income for PITI Calculation - \$2,500 x 25% (.25 ) = \$625 PITI Solution Step 2 Calculate 33% of the Gross Monthly Income for Total Debt Calculation - \$2,500 x 33% (.33) = \$825 Total Debt Solution Step 3 Subtract the PITI Dollars from the Total Debt Dollars = Other Debt Calculation - \$825 - \$625 = \$200 for other debt Answer: \$200 for other debt 451
10 Chapter 17 Math Problem Solutions Appraisal Calculations Capitalization, income approach, Cap Rate Approach Under the income approach, the estimate of value is arrived at by capitalizing the annual net income. The solution to these problems is based on the following formula: Value X Capitalization rate = Net Operating Income 1. An apartment building produces a net income of \$4,320 per annum. The investor paid \$36,000 for the apartment building. What is the owner's rate of return (cap rate) on the investment? Solution Step 1 Value X Cap Rate = Net Operating Income so Net Operating Income/Value = Cap Rate Calculation - \$4,320/\$36,000 = 12% Cap Rate Answer: 12% annual rate of return on investment cap rate 2. An investor is considering the purchase of an office building for \$125,000. The investor insists upon a 14% return on investment. What must be the amount of the annual net income from this investment to return a profit to the owner at a rate of 14%? Solution Step 1 - Value X Cap Rate = Net Operating Income Calculation - \$125,000 x 14% (.14) = \$17,500 Net Operating Income Answer: \$17,500 net operating income needed 3, In appraising a shopping center, the appraiser establishes that the center produces an annual net income of \$97,500. The appraiser determines the capitalization rate to be 13%. What should be the appraiser's estimate of market value for this shopping center? Solution Step 1 - Value X Cap Rate = Net Operating Income so Value = Net Operating Income/Cap Rate Calculation \$97,500/13%(.13) = \$750,000 Answer: \$750,000 market value Basis 1. Mr. and Mrs. Swift purchased their home 15 years ago for \$32,500. During their ownership, they made capital improvements totaling \$19,400. They sold the home for \$72,900. What amount of gain did they make on the sale? Solution step 1 Purchase Price Plus Improvements = Basis Calculation - \$32, ,400 = \$51,900 Basis Solution Step 2 Selling Price less Basis = Gain/Profit Calculation - \$72,900 less \$51,900 = \$21,000 Gain/Profit Answer: \$21,000 gain/profit 452
11 Chapter 17 Math Problems & Answers CHAPTER 17 - MATH QUESTIONS 1. The value of your house, not including the lot, is \$91,000 today. What was the original cost if it has depreciated 5 percent per year for the past seven years? A. \$67, B. \$95, C. \$122, D. \$140, What did the owners originally pay for their home if they sold it for \$98,672, which gave them 12 percent profit over their original cost? A. \$86,830 B. \$88,100 C. \$89,700 D. \$110, What would you pay for a building producing \$11,250 annual net income and showing a minimum rate of return of 9 percent? A. \$125,000 B. \$123,626 C. \$101,250 D. \$122, An owner agrees to list his property on the condition that he will receive at least \$47,300 after paying 5 percent broker's commission and paying \$1,150 in closing costs. At what price must it sell? A. \$48,450 B. \$50,815 C. \$50,875 D. \$51, A gift shop pays rent of \$600 per month plus 2.5 percent of gross annual sales in excess of \$50,000. What was the average monthly rent last year if gross annual sales were \$75,000? A. \$1, B. \$ C. \$ D. \$
12 Chapter 17 Math Problems & Answers 6. If your monthly rent is \$1,050, what percent would this be of an annual income of \$42,000? A. 25 percent B. 30 percent C. 33 percent D. 40 percent 7. Two managing brokers split the 6 percent commission on a \$73,000 home. The selling licensee was paid 70 percent of his broker's share. The listing licensee was paid 30 percent of her broker's share. How much did the listing licensee receive? A. \$657 B. \$4,380 C. \$1,533 D. \$1, The buyer has agreed to pay \$175,000 in sales price, 2.5 loan discount points, and a 1 percent origination fee. If the buyer receives a 90 percent loan-to-value ratio, how much will the buyer owe at closing for points and the origination fee? A. \$1, B. \$3, C. \$5, D. \$6, Calculate eight months of interest on a \$5,000 interest-only loan at 9.5 percent. A. \$ B. \$ C. \$ D. \$ A 100-acre farm is divided into lots for homes. The streets require one - eighth of the whole farm, and there are 140 lots. How many square feet are in each lot? A. 43,560 B. 35,004 C. 31,114 D. 27, What is the monthly net income on an investment of \$115,000 if the rate of return is 12.5 percent? A. \$1, B. \$1, C. \$7, D. \$14, A licensee sells a property for \$58,500. The contract he has with his managing broker is 40 percent of the full commission earned. The commission due the managing broker is 6 percent. What is the licensee's share of the commission? A. \$2,106 B. \$1,404 C. \$3,510 D. \$2,
13 Chapter 17 Math Problems & Answers 13. What is the interest rate on a \$10,000 loan with semiannual interest of \$450? A. 7 percent B. 9 percent C. 11 percent D percent 14. A warehouse is 80 feet wide and 120 feet long with ceilings 14 feet high. If 1,200 square feet of floor surface has been partitioned off from floor to ceiling for an office, how many cubic feet of space will be left in the warehouse? A. 151,200 B. 134,400 C. 133,200 D. 117, The lot you purchased five years ago for \$30,000 has appreciated 3.5 percent per year. What is it worth today? A. \$30,375 B. \$33,525 C. \$34,500 D. \$35, A lease calls for \$1,000 per month minimum plus 2 percent of annual sales in excess of \$100,000. What is the annual rent if the annual sales were \$150,000? A. \$12,000 B. \$13,000 C. \$14,000 D. \$15, There is a tract of land that is 1.25 acres. The lot is 150 feet deep. How much will the lot sell for at \$65 per front foot? A. \$9,750 B. \$8,125 C. \$23,595 D. \$8, A woman earns \$20,000 per year and can qualify for a monthly PITI payment equal to 25 percent of her monthly salary. If the annual tax and insurance is \$678.24, what is the loan amount she will qualify for if the monthly PI payment factor is \$10.29 per \$1,000 of loan amount? A. \$66,000 B. \$43,000 C. \$40,500 D. \$35, You pay \$65.53 monthly interest on a loan bearing 9.25 percent annual interest. What is the loan amount rounded to the nearest hundred dollars? A. \$1,400 B. \$2,800 C. \$6,300 D. \$8,
14 Chapter 17 Math Problems & Answers 20. What percentage of profit would you make if you paid \$10,500 for a lot, built a home on the lot that cost \$93,000, and then sold the lot and house together for \$134,550? A. 13 percent B. 23 percent C. 30 percent D. 45 percent 21. An income-producing property has \$62,500 annual gross income and monthly expenses of \$1,530. What is the appraised value if the appraiser uses a 10 percent capitalization rate? A. \$441,400 B. \$625,000 C. \$183,600 D. \$609, A man pays \$2,500 each for four parcels of land. He subdivides them into six parcels and sells each of the six parcels for \$1,950. What was his percentage of profit? A percent B. 17 percent C. 52 percent D. 78 percent 23. A property sells for \$96,000. If it has appreciated 4 percent per year straight line for the past five years, what did the owner pay for the property five years ago? A. \$76,800 B. \$80,000 C. \$92,300 D. \$115, If you purchase a lot that is 125 feet by 150 feet for \$6,468.75, what price did you pay per front foot? A. \$23.52 B. \$43.13 C. \$51.75 D. \$ Calculate the amount of commission earned by a broker on a property selling for \$61,000 if 6 percent is paid on the first \$50,000 and 3 percent is paid on the remaining balance. A. \$3,330 B. \$3,830 C. \$3,600 D. \$3,
15 Chapter 17 Math Problems & Answers CHAPTER 17 - MATH QUESTIONS ANSWER KEY 1. - D 2. - B 3. - A 4. - D 5. - D 6. - B 7. - A 8. - C 9. - B D B B B D D B C D D C A B B C A 457
16 Chapter 17 Math Problems & Answers Notes: 458
17 0" Chapter 17 Math Problems & Answers CHAPTER 17 - MATH PROBLEM SOLUTIONS 1. \$140,000 Original Cost 5% Depreciation per year x 7 years 35% Total Depreciation \$140,000 Original Cost x 65% = \$91,000 Today's Value 2. \$88,100 Original Cost = 12% Profit = 112% is the Sales Price \$98,672 Sales Price/112% = \$88,100 Original Cost 3. \$11,250 Net Operating Income divided by cap rate = Selling Price \$11,250/.09 = \$125,000 Selling Price 4. \$47,300 Net to Seller and \$1,150 Closing Costs \$47,300 + \$1,150 = \$48,450 Before Commission Commission = 5% therefore Dollars before Commission = 95% \$48,450/.95 = \$51,000 Selling Price 5. \$ Average Monthly Rent \$75,000 Gross Annual Sales \$50,000 Excluded from Percentage Calculation 2.5 % of Net Gross Annual Sales \$75,000 - \$50,000 = \$25,000 Gross Annual Sales used for Rent Calculation \$25,000 x 2.5% = \$625 Annual Percentage Rent \$625/12 = \$52.08 = Monthly Percentage Rent Base Rent = Monthly Percentage Rent = Average Monthly Rent \$ \$52.08 = \$ \$1,050 Monthly Rent x 12 = \$12,600 Annual Rent \$42,000 Annual Income Annual Rent/Annual Income = A Rent compared to Annual Income \$12,600/\$42,000 = 30% 7. \$73,000 Selling Price and 6% commission \$73,000 x 6% = \$4,380 Full Commission Managing Brokers shared commission \$4,380/2 = \$2,190 = Managing Broker Share Listing Agent = 30% of Managing Broker's Share \$2,190 x 30% = \$ Listing Agent's Commission 459
18 Chapter 17 - Math Problems & Answers 8. \$175,000 Purchase Price - 90% Loan to Value Ratio \$175,000 x.9 = \$157,500 loan \$157,000 Loan, 2.5 Discount Points and 1 Point Loan Origination Fee 3.5 points total for this loan 1 point = 1% of the Loan 3.5 points = \$157,000 x 3.5% = \$5, for Total Points 9. \$5,000 Loan and 9.5% Annual Interest \$5,000 x 9.5% = \$475 = Annual Interest Annual Interest divided by 12 = Monthly Interest \$475/12 = \$39.58 per month Eight months of interest = monthly interest x 12 \$38.58 x 8 = \$ (\$38.58 not rounded off, but extra decimal places included) Acre Plot, 1/8 for streets 100 Acres x 87.5% = 87.5 Acres for Lots 87.5 x 43,560 sq. ft. per Acre = 3,811, 500 sq. ft. for lots 140 lots 3,811,500/140 = 27,225 sq. ft. per lot 11. Net Operating Income = Purchase Price x Cap Rate \$115,000 x 12.5% = \$14,375 Net Operating Income (Annual) Monthly NOI = Annual NOI/12 \$14,375/12 = \$1, = Monthly Net Operating income 12. Selling Price = \$58,500 and Commission is 6%, with Licensee getting 40% \$58,500 x 6% = \$3,510 = Sponsoring Broker Commission \$3,510 x 40% = \$1,404 = Licensee's share of the Commission 13. \$10,000 loan, \$450 semi-annual interest payment \$450 x 2 = \$900 = Annual Interest Payment \$900/\$10,000 = 9% Annual Interest 14. Warehouse = 80 x 120 feet in area and 14 feet high Total Cubic Feet = 80 x 120 x 14 = 134,400 Cubic Feet Office Space is 1200 sf x 14 ft. = 16,800 cubic feet Net Warehouse Cubic Space is Total Warehouse Space less Office Cubic Space 134,400 cubic feet - 16,800 cubic feet = 117,600 net cubic feet for the warehouse 15. \$30,000 original cost, 3.5% appreciation for 5 years 3.5% x 5 = 17.5% Appreciation Original 100% plus 17.5% = 117.5% = Current Value \$30,000 x 117.5% = \$35,250 = Today's Value 460
19 Chapter 17 - Math Problems & Answers 16. \$1,000 per month base rent, \$150,000 Gross Annual Sales, \$100,000 excluded from percentage Percentage of Gross Annual Sales = 2% Gross Sales for Percentage Calculation = \$150,000 less \$100,000 = \$50,000 \$50,000 x 2% = \$1,000 Annual Percentage Rent \$1,000 per month base rent x 12 = \$12,000 Annual Base Rent Annual Base Rent and Percentage Rent = \$12,000 + \$1,000 = \$13, acres and a selling price of \$65 per front foot One side of the lot is 150 feet Area of lot = 1.25 x 43,560 = 54,450 sq. ft. Divide area by one side to obtain the length of the other side 54,450/150 = 363 feet 363 front feet x \$65 per front foot = \$23,595 selling price 18. Earns \$20,000 per year, 25% PITT per month Annual Income/12 = Monthly income \$20,000/12 = \$1, per month 25% of Gross Monthly Income = \$1,666.6 x 25% = \$ for Monthly PITI TI (Taxes & Insurance) = \$ per year \$678.24/12 = \$56.52 = Monthly TI \$ \$56.52 = \$ available for PI (Principle & Interest Payment) \$10.29 per \$1,000 is the factor \$360.15/\$10.29 x \$1,000 = \$35,000 Loan 19. \$65.53 per month interest and interest rate of 9.25% \$65.53 x 12 = \$ = Annual Interest Interest divided by rate = Loan \$786.36/.0925 = \$8,500 Loan 20. Paid \$10,500 for lot and \$93,000 for home Total Cost = \$10,500 + \$93,000 = 103,500 Selling Price = \$134,550 Net Profit = Selling Price less Cost \$134,500 - \$103,500 = \$31,050 Percent Profit = Profit divided by Cost \$31,050/\$103,500 =.3 = 30% Profit 21. \$1,530 monthly expenses, \$62,500 Annual Gross Income,10% cap rate \$1,530 x 12 = \$18,360 = Annual Expenses Annual Net Operating Income = Annual Gross Income Less Annual Expenses \$62,500 - \$18,360 = \$44,140 Net Operating Income Net Operating Income Divided by Cap Rate = Purchase Price \$44,140/.1 = \$441,400 = Purchase Price 461
20 Chapter 17 Math Problems & Answers 22. \$2,500 per parcel, 4 parcels = Cost \$1,950 per parcel, 6 parcels = Sales \$2,500 x 4 = \$10,000 \$1,950 x 6 = \$11,700 Sales less Cost = Profit \$11,700 - \$10,000 = \$1,700 Profit Dollars divided by Cost Dollars = % Profit \$1,700/\$10,000 = 17% Profit 23. Property worth \$96,000, 4% appreciation for 5 years 4% x 5 = 20% Current Value = 120% of original cost \$96,000 = 120% of Original Cost Current Value divided by 120% = Original Value \$96,000/1.2 = \$80, \$6, Price for lot on frontage foot basis, lot is 125 x 150 feet 125 is considered the front portion of the lot \$6,468.75/125 = \$51.75 = Price per front foot. 25. Selling Price of \$61,000, 6% on first \$50,000 and 3% on remaining \$50,000 x 6% = \$3,000 \$61,000 - \$50,000 = \$11,000 \$11,000 x 3% = \$330 \$3, = \$3,330 = Total Commission Notes: 462
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > islinds Structured version Visualization version GIF version
Theorem islinds 20196
Description: Property of an independent set of vectors in terms of an independent family. (Contributed by Stefan O'Rear, 24-Feb-2015.)
Hypothesis
Ref Expression
islinds.b 𝐵 = (Base‘𝑊)
Assertion
Ref Expression
islinds (𝑊𝑉 → (𝑋 ∈ (LIndS‘𝑊) ↔ (𝑋𝐵 ∧ ( I ↾ 𝑋) LIndF 𝑊)))
Proof of Theorem islinds
Dummy variables 𝑠 𝑤 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 elex 3243 . . . . 5 (𝑊𝑉𝑊 ∈ V)
2 fveq2 6229 . . . . . . . 8 (𝑤 = 𝑊 → (Base‘𝑤) = (Base‘𝑊))
32pweqd 4196 . . . . . . 7 (𝑤 = 𝑊 → 𝒫 (Base‘𝑤) = 𝒫 (Base‘𝑊))
4 breq2 4689 . . . . . . 7 (𝑤 = 𝑊 → (( I ↾ 𝑠) LIndF 𝑤 ↔ ( I ↾ 𝑠) LIndF 𝑊))
53, 4rabeqbidv 3226 . . . . . 6 (𝑤 = 𝑊 → {𝑠 ∈ 𝒫 (Base‘𝑤) ∣ ( I ↾ 𝑠) LIndF 𝑤} = {𝑠 ∈ 𝒫 (Base‘𝑊) ∣ ( I ↾ 𝑠) LIndF 𝑊})
6 df-linds 20194 . . . . . 6 LIndS = (𝑤 ∈ V ↦ {𝑠 ∈ 𝒫 (Base‘𝑤) ∣ ( I ↾ 𝑠) LIndF 𝑤})
7 fvex 6239 . . . . . . . 8 (Base‘𝑊) ∈ V
87pwex 4878 . . . . . . 7 𝒫 (Base‘𝑊) ∈ V
98rabex 4845 . . . . . 6 {𝑠 ∈ 𝒫 (Base‘𝑊) ∣ ( I ↾ 𝑠) LIndF 𝑊} ∈ V
105, 6, 9fvmpt 6321 . . . . 5 (𝑊 ∈ V → (LIndS‘𝑊) = {𝑠 ∈ 𝒫 (Base‘𝑊) ∣ ( I ↾ 𝑠) LIndF 𝑊})
111, 10syl 17 . . . 4 (𝑊𝑉 → (LIndS‘𝑊) = {𝑠 ∈ 𝒫 (Base‘𝑊) ∣ ( I ↾ 𝑠) LIndF 𝑊})
1211eleq2d 2716 . . 3 (𝑊𝑉 → (𝑋 ∈ (LIndS‘𝑊) ↔ 𝑋 ∈ {𝑠 ∈ 𝒫 (Base‘𝑊) ∣ ( I ↾ 𝑠) LIndF 𝑊}))
13 reseq2 5423 . . . . 5 (𝑠 = 𝑋 → ( I ↾ 𝑠) = ( I ↾ 𝑋))
1413breq1d 4695 . . . 4 (𝑠 = 𝑋 → (( I ↾ 𝑠) LIndF 𝑊 ↔ ( I ↾ 𝑋) LIndF 𝑊))
1514elrab 3396 . . 3 (𝑋 ∈ {𝑠 ∈ 𝒫 (Base‘𝑊) ∣ ( I ↾ 𝑠) LIndF 𝑊} ↔ (𝑋 ∈ 𝒫 (Base‘𝑊) ∧ ( I ↾ 𝑋) LIndF 𝑊))
1612, 15syl6bb 276 . 2 (𝑊𝑉 → (𝑋 ∈ (LIndS‘𝑊) ↔ (𝑋 ∈ 𝒫 (Base‘𝑊) ∧ ( I ↾ 𝑋) LIndF 𝑊)))
177elpw2 4858 . . . 4 (𝑋 ∈ 𝒫 (Base‘𝑊) ↔ 𝑋 ⊆ (Base‘𝑊))
18 islinds.b . . . . 5 𝐵 = (Base‘𝑊)
1918sseq2i 3663 . . . 4 (𝑋𝐵𝑋 ⊆ (Base‘𝑊))
2017, 19bitr4i 267 . . 3 (𝑋 ∈ 𝒫 (Base‘𝑊) ↔ 𝑋𝐵)
2120anbi1i 731 . 2 ((𝑋 ∈ 𝒫 (Base‘𝑊) ∧ ( I ↾ 𝑋) LIndF 𝑊) ↔ (𝑋𝐵 ∧ ( I ↾ 𝑋) LIndF 𝑊))
2216, 21syl6bb 276 1 (𝑊𝑉 → (𝑋 ∈ (LIndS‘𝑊) ↔ (𝑋𝐵 ∧ ( I ↾ 𝑋) LIndF 𝑊)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 383 = wceq 1523 ∈ wcel 2030 {crab 2945 Vcvv 3231 ⊆ wss 3607 𝒫 cpw 4191 class class class wbr 4685 I cid 5052 ↾ cres 5145 ‘cfv 5926 Basecbs 15904 LIndF clindf 20191 LIndSclinds 20192 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-sep 4814 ax-nul 4822 ax-pow 4873 ax-pr 4936 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3an 1056 df-tru 1526 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ral 2946 df-rex 2947 df-rab 2950 df-v 3233 df-sbc 3469 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-nul 3949 df-if 4120 df-pw 4193 df-sn 4211 df-pr 4213 df-op 4217 df-uni 4469 df-br 4686 df-opab 4746 df-mpt 4763 df-id 5053 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-res 5155 df-iota 5889 df-fun 5928 df-fv 5934 df-linds 20194 This theorem is referenced by: linds1 20197 linds2 20198 islinds2 20200 lindsss 20211 lindsmm 20215 lsslinds 20218
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Shift Operator
Get Shift Operator essential facts below. View Videos or join the Shift Operator discussion. Add Shift Operator to your PopFlock.com topic list for future reference or share this resource on social media.
Shift Operator
In mathematics, and in particular functional analysis, the shift operator also known as translation operator is an operator that takes a function x ? f(x) to its translation x ? f(x + a).[1] In time series analysis, the shift operator is called the lag operator.
Shift operators are examples of linear operators, important for their simplicity and natural occurrence. The shift operator action on functions of a real variable plays an important role in harmonic analysis, for example, it appears in the definitions of almost periodic functions, positive definite functions, and convolution.[2] Shifts of sequences (functions of an integer variable) appear in diverse areas such as Hardy spaces, the theory of abelian varieties, and the theory of symbolic dynamics, for which the baker's map is an explicit representation.
## Definition
### Functions of a real variable
The shift operator Tt (t ∈ R) takes a function f on R to its translation ft ,
${\displaystyle T^{t}f(x)=f_{t}(x)=f(x+t)~.}$
A practical representation of the linear operator Tt in terms of the plain derivative ddx was introduced by Lagrange,
${\displaystyle T^{t}=e^{t{\frac {d}{dx}}}~,}$
which may be interpreted operationally through its formal Taylor expansion in t; and whose action on the monomial xn is evident by the binomial theorem, and hence on all series in x, and so all functions f(x) as above.[3] This, then, is a formal encoding of the Taylor expansion.
The operator thus provides the prototype[4] for Lie's celebrated advective flow for Abelian groups,
${\displaystyle e^{t\beta (x){\frac {d}{dx}}}f(x)=e^{t{\frac {d}{dh}}}F(h)=F(h+t)=f(h^{-1}(h(x)+t)),}$
where the canonical coordinates h (Abel functions) are defined, s.t.
${\displaystyle h'(x)\equiv {\frac {1}{\beta (x)}}~,\qquad f(x)\equiv F(h(x)).}$
For example, it easily follows that ${\displaystyle \beta (x)=x}$ yields scaling,
${\displaystyle e^{tx{\frac {d}{dx}}}f(x)=f(e^{t}x)}$,
hence ${\displaystyle e^{i\pi x{\frac {d}{dx}}}f(x)=f(-x)}$ (parity); likewise, ${\displaystyle \beta (x)=x^{2}}$ yields[5]
${\displaystyle e^{tx^{2}{\frac {d}{dx}}}f(x)=f\left({\frac {1-tx}{x}}\right)}$,
${\displaystyle \beta (x)=1/x}$ yields
${\displaystyle e^{{\frac {t}{x}}{\frac {d}{dx}}}f(x)=f({\sqrt {x^{2}+2t}}~)}$,
${\displaystyle \beta (x)=e^{x}}$ yields
${\displaystyle \exp \left(te^{x}{\frac {d}{dx}}\right)f(x)=f\left(\ln \left({\frac {1}{\exp(-x)-t}}\right)\right)}$,
etc.
The initial condition of the flow and the group property completely determine the entire Lie flow, providing a solution to the translation functional equation[6]
${\displaystyle f_{t}(f_{\tau }(x))=f_{t+\tau }(x).}$
### Sequences
The left shift operator acts on one-sided infinite sequence of numbers by
${\displaystyle S^{*}:(a_{1},a_{2},a_{3},\ldots )\mapsto (a_{2},a_{3},a_{4},\ldots )}$
and on two-sided infinite sequences by
${\displaystyle T:(a_{k})_{k=-\infty }^{\infty }\mapsto (a_{k+1})_{k=-\infty }^{\infty }.}$
The right shift operator acts on one-sided infinite sequence of numbers by
${\displaystyle S:(a_{1},a_{2},a_{3},\ldots )\mapsto (0,a_{1},a_{2},\ldots )}$
and on two-sided infinite sequences by
${\displaystyle T^{-1}:(a_{k})_{k=-\infty }^{\infty }\mapsto (a_{k-1})_{k=-\infty }^{\infty }.}$
The right and left shift operators acting on two-sided infinite sequences are called bilateral shifts.
### Abelian groups
In general, as illustrated above, if F is a function on an abelian group G, and h is an element of G, the shift operator T g maps F to[6][7]
${\displaystyle F_{g}(h)=F(h+g).}$
## Properties of the shift operator
The shift operator acting on real- or complex-valued functions or sequences is a linear operator which preserves most of the standard norms which appear in functional analysis. Therefore, it is usually a continuous operator with norm one.
### Action on Hilbert spaces
The shift operator acting on two-sided sequences is a unitary operator on l2(Z). The shift operator acting on functions of a real variable is a unitary operator on L2(R).
In both cases, the (left) shift operator satisfies the following commutation relation with the Fourier transform:
${\displaystyle {\mathcal {F}}T^{t}=M^{t}{\mathcal {F}},}$
where Mt is the multiplication operator by exp(i t x). Therefore, the spectrum of Tt is the unit circle.
The one-sided shift S acting on l2(N) is a proper isometry with range equal to all vectors which vanish in the first coordinate. The operator S is a compression of T−1, in the sense that
${\displaystyle T^{-1}y=Sx{\text{ for each }}x\in \ell ^{2}(\mathbb {N} ),\,}$
where y is the vector in l2(Z) with yi = xi for i ≥ 0 and yi = 0 for i < 0. This observation is at the heart of the construction of many unitary dilations of isometries.
The spectrum of S is the unit disk. The shift S is one example of a Fredholm operator; it has Fredholm index −1.
## Generalisation
Jean Delsarte introduced the notion of generalised shift operator (also called generalised displacement operator); it was further developed by Boris Levitan.[2][8][9]
A family of operators {Lx}x ∈ X acting on a space Φ of functions from a set X to C is called a family of generalised shift operators if the following properties hold:
1. Associativity: let (Ryf)(x) = (Lxf)(y). Then LxRy = RyLx (not clear why, as it looks more like a commutativity).
2. There exists e in X such that Le is the identity operator.
In this case, the set X is called a hypergroup.
## Notes
1. ^ Weisstein, Eric W. "Shift Operator". MathWorld.
2. ^ a b Marchenko, V. A. (2006). "The generalized shift, transformation operators, and inverse problems". Mathematical events of the twentieth century. Berlin: Springer. pp. 145-162. doi:10.1007/3-540-29462-7_8. MR 2182783.
3. ^ Jordan, Charles, (1939/1965). Calculus of Finite Differences, (AMS Chelsea Publishing), ISBN 978-0828400336 .
4. ^ M Hamermesh (1989), Group Theory and Its Application to Physical Problems (Dover Books on Physics), Hamermesh ISBM 978-0486661810 , Ch 8-6, pp 294-5 , online.
5. ^ p 75 of Georg Scheffers (1891): Sophus Lie, Vorlesungen Ueber Differentialgleichungen Mit Bekannten Infinitesimalen Transformationen, Teubner, Leipzig, 1891. ISBN 978-3743343078 online
6. ^ a b Aczel, J (2006), Lectures on Functional Equations and Their Applications (Dover Books on Mathematics, 2006), Ch. 6, ISBN 978-0486445236 .
7. ^ "A one-parameter continuous group is equivalent to a group of translations". M Hamermesh, ibid.
8. ^ Levitan, B.M.; Litvinov, G.L. (2001) [1994], "Generalized displacement operators", Encyclopedia of Mathematics, EMS Press
9. ^ Bredikhina, E.A. (2001) [1994], "Almost-periodic function", Encyclopedia of Mathematics, EMS Press
## Bibliography
• Partington, Jonathan R. (March 15, 2004). Linear Operators and Linear Systems. Cambridge University Press. doi:10.1017/cbo9780511616693. ISBN 978-0-521-83734-7.
• Marvin Rosenblum and James Rovnyak, Hardy Classes and Operator Theory, (1985) Oxford University Press.
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# Looking for clarification on this definition: $K_1K_2 = K_1(K_2)$.
In my algebra book they define for field extensions $L/K_1/F$ and $L/K_2/F$ the field $K_1K_2 = K_1(K_2)$.
The formal definition I have for this is $$K_1(K_2) = \bigcap_{K_2 < E < L} E \quad \text{where } E/K_1.$$ So the smallest field extension of $K_1$ containing $K_2$.
For extensions such as $F(\alpha)$ though sometimes it is written, $$F(\alpha) = \{ a_0 + a_1\alpha + a_2\alpha^2 + \cdots + a_{n - 1}\alpha^{n - 1} : a_i \in F\}$$
Can I extend this definition for $K_1(K_2)$ and write the following? $$K_1(K_2) = \left\{ \sum_{i = 1}^n \sum_{j = 1}^{m_i} a_{ij}b_i^j : a_{ij} \in K_1, b_i \in K_2, n, m_i \in \mathbb{N} \right\}$$
EDIT: $K_1, K_2$ are algebraic.
• That expression for $F(\alpha)$ only works for $\alpha$ algebraic over $F$ right? – anon May 17 '12 at 2:07
• That is right, It does happen that the $K_1, K_2$ are algebraic in my situation. – zrbecker May 17 '12 at 2:18
• You can probably ask Elman. He is in his office a lot of times, and he always welcomes anyone who walks by it. – Daniel Montealegre May 17 '12 at 5:10
## 1 Answer
There are several different-looking definitions of the compositum $K_1K_2$ of two fields, and it’s a very useful exercise to sit down and prove them equivalent. Here's my suggestion: Let's consider only subfields and subsets of some larger field $\Omega$. Then for a field $K$ and a set $S$, take as definition of $K(S)$ something close to what you wrote, namely $$K(S)=\bigcap_{L\supset K\cup S}L\,,$$ where the index $L$ runs through fields only. The construct is a field, of course.
Now on the other hand, consider the set $K^{\mathrm{r}}(S)$ whose elements are the finite rational expressions (with coefficients in $K$) of the form $f({\mathbf{s}})=n({\mathbf{s}})/d({\mathbf{s}})$, where in each case, $\mathbf{s}$ represents some finite set of elements of $S$, and the numerator $n$ and the denominator $d$ are multivariable polynomials over $K$. Clearly $K^{\mathrm{r}}(S)$ is also a field. Now you can easily check that $K(S)=K^{\mathrm{r}}(S)$. It follows immediately that for two fields, $K_1(K_2)=K_2(K_1)$, and you also see that in your case, where the $K_i$ are algebraic over some smaller field, you can dispense with rational expressions, and use just polynomial ones.
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# 2021 JMPSC Accuracy Problems/Problem 3
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
In a regular octagon, the sum of any three consecutive sides is $90.$ A square is constructed using one of the sides of this octagon. What is the area of the square?
## Solution
We are given that the sum of any three sides of the octagon is 90, and since the octagon is regular, we know that all sides are equal. Thus, each side of the octagon must equal to $\frac{90}{3} = 30$. Since one side of the square shares a side with the octagon, we know the side lengths of the square are also 30. Thus, our answer is $30^2 = \boxed{900}$
## Solution 2
We have each side of the octagon is $s$, so $3s=90$, or $s=30$. This means the area of the square is $30^2=\boxed{900}$
~Geometry285
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http://math.stackexchange.com/questions/852450/52-card-deck-probability
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# 52-card deck probability…
If 13 players are each dealt four cards from a 52-card deck, what is the probability that each player gets one card of each suit?
So I chose to do the situation where we have no repetition (i.e the values are all distinct as well as suits). I ended up with an answer of $$\frac{{13 \choose 1}{12 \choose 1}{11 \choose 1}{10 \choose 1}{4 \choose 1}^4}{52 \choose 4}.$$ My reasoning: First off, we know there are ${52 \choose 4}$ ways to get 4 cards in a card. We also know there are ${13 \choose 1}$ ways to get a card in a suit and ${4 \choose 1}$ ways to get a card from 4 different suits but same value. Once chosen, to avoid repetition of suit and value, we remove all cards of similiar value and move onto the next suit. Then, choosing our next card, there are ${12 \choose 1}$ ways to do it and the pattern continues a few more times. In the case that we can have repeated values but still different suits, I believe the answer would be $$\frac{{13 \choose 1}^4{4 \choose 1}^4}{{52 \choose 4}}$$
My question: was this thinking correct or am I just plain wrong?
-
The number you obtain is bigger than $1$. – André Nicolas Jun 30 '14 at 17:38
Let us imagine that $4$ cards are dealt to A, B, C, D in that order. First we find the probability that A gets $1$ of each suit. There are $13^4$ "favourable" choices, out of the $\binom{52}{4}$ equally likely choices. Now it is B's turn, $12^4$ good choices out of $\binom{48}{4}$ equally likely choices, and so on. We end up with a probability of $\frac{13^412^411^410^4}{\binom{52}{4}\binom{48}{4}\binom{44}{4}\binom{40}{4}}$.
Remark: Some parts of the analysis in the OP were correct. The multiplication by $4^4$ was not, and leads to a number greater than $1$.
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https://hawthornecommunitycouncil.org/best-prism-and-pyramid-worksheets/
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# Of The Best Prism And Pyramid Worksheets
Cross Sections PowerPoint and Worksheet. Sum of the angles in a triangle is 180.
3d Shape Worksheets For Kindergarten Shapes Worksheet Kindergarten 3d Shapes Worksheets Shapes Worksheets
### Teaches students to find the cross section of three-dimensional shapes including prisms pyramids spheres cylinders and cones.
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Complementary and supplementary word problems worksheet. Just pick two contrasting colours like red and blue and have your class get started. Writing to Explain A square pyramid has 2 m sides on the base.
There are 11 task cards for pyramids and 9 task cards for prisms. Some of the worksheets for this concept are Volumes of pyramids Geometry tutor work 17 volume of prisms and pyramids Classifying prisms and pyramids a S m s pyram m p p r i idss Classifying prisms and pyramids a G4 34 prism and pyramid bases 339 example 10 volume of prisms and cylinders Surface area of solids. There are many visuals to help st.
Hi In this post we present you several inspiring photos that weve gathered special for you this time we decide to be focus related with Prisms and Pyramids Worksheets. Explain how to find the surface area. Its an easy way to extend the lesson.
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See how well you know how to to calculate the volume of a prism or a pyramid. Volume Of A Prism Volume. Enlarge by a factor of 2 or 3 or whatever a student may choose.
Shade a base jj4 circle the point of the following pyramids. Which is the surface area of a rectangular prism with a length of 23 in a width of 11 in and a height of 3 in. A 2648 in2 2B 2546 in C 2458 in2 D 215 in2 9.
The following 3d geometric shapes printables contain pictures of common 3D shapes that your child should know. Prisms are polyhedrons and their sides are parallelograms. Distinguishes between parallel vs perpendicular cross sections and prisms vs pyramids.
1 square pyramid 2 square pyramid 3 square prism 4 square pyramid 5 rectangular prism. Displaying top 8 worksheets found for – Volume Of Prisms Pyramids Cylindes And Cones. Worksheet by Kuta Software LLC Geometry Volume of Prisms and Pyramids Name_____ ID.
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The prisms vs pyramids worksheet comes in black and white so while you could have children write down which one is a prism and which is a pyramid another fun idea is to have them colour the pictures in. Scale the tabbed nets on each of the four worksheets to create a set of patterns. I know how to calculate the volume of prisms and pyramids.
Here you will find our range of printable 3D Shape Sheets including spheres cones cubes pyramids and prisms. It may be printed downloaded or saved and used in your classroom home school or other educational environment to help someone learn math. Volume of prisms and pyramids worksheet pdf answer key This humongous collection of printable volume worksheets is sure to walk middle and high school students step-by-step through a variety of exercises beginning with counting cubes moving on to finding the volume of solid shapes such as cubes cones rectangular and triangular prisms and pyramids cylinders spheres.
Prism – A prism is a shape formed with two identical ends and flat sides. Volume – The volume of the pyramid is found by the formula Volume 13 l x w x h Where l is the base length w. The end shapes of the prism give the shape its name such as a triangular prism square prism rectangular prism hexagonal prism and many more.
The volume of a prism is given by the formula V Bh where B is the area of the base and h is the height. Geometry Worksheets – Volume of Prisms Pyramids. The activity that goes with this worksheet will help your students identify the base of a 3-D figure.
20 unique task cards dealing with surface area of rectangular prisms cubes triangular prisms square pyramids and triangular pyramids4 different recording sheetsAnswer KeyCheck out the Fi. Related Topics Worksheets. About This Quiz Worksheet.
1 Date_____ Period____ r e2Z0U1F7r WKPutDa _Spobfetwhagriee bLsLCC Q MAplZlm irIiEgxhotSsh BroeXsAeFrIvDeQdC-1-Name each figure. Add to my workbooks 14 Download file pdf Embed in my website or blog Add to Google Classroom. This math worksheet was created on 2007-10-02 and has been viewed 13 times this week and 50 times this month.
Learn about the relationship between volume base dimensions and height.
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# What Is 24/59 as a Decimal + Solution With Free Steps
The fraction 24/59 as a decimal is equal to 0.406.
A denominator should not equal zero while expressing a rational number in a division. Moreover, it can be written as p/q. 0 is also a rational number. Irrational numbers can not be expressed in fraction form. That’s why they cannot be written in p/q form.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 24/59.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 24
Divisor = 59
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 24 $\div$ 59
This is when we go through the Long Division solution to our problem.
       Figure 1
## 24/59 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 24 and 59, we can see how 24 is Smaller than 59, and to solve this division, we require that 24 be Bigger than 59.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 24, which after getting multiplied by 10 becomes 240.
We take this 240 and divide it by 59; this can be done as follows:
240 $\div$ 59 $\approx$ 4
Where:
59 x 4 = 236
This will lead to the generation of a Remainder equal to 240 – 236 = 4. Now this means we have to repeat the process by Converting the 4 into 400 and solving for that:
400 $\div$ 59$\approx$ 6
Where:
59 x 6 = 354
This, therefore, produces another Remainder which is equal to 400 – 354 = 46.
Finally, we have a Quotient generated after combining the three pieces of it as 0.406=z, with a Remainder equal to 46.
Images/mathematical drawings are created with GeoGebra.
5/5 - (5 votes)
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https://www.edplace.com/worksheet_info/maths/keystage3/year8/topic/917/2357/algebra-bricks-2
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# Solve Algebra Problems By Collecting Like Terms
In this worksheet, students will collect, add and subtract like terms to find the missing algebra bricks.
Key stage: KS 3
Curriculum topic: Algebra
Curriculum subtopic: Understand Expressions, Equations, Inequalities, Terms and Factors
Popular topics: Algebra worksheets
Difficulty level:
#### Worksheet Overview
Look at these number bricks.
You'll notice that 5 + 7 = 12 and that A must equal 10, so that 5 + 10 = 15
The same rule applies with algebra bricks.
So here, the values on the two bricks at the bottom will add to give us the value in the brick above them.
To find the total of these two values, we need to gather like terms:
x + 3 + x - 4 = 2x - 1
In this activity, the expression on any brick can be worked out by adding the expressions on the two bricks below it.
Does this make sense?
Don't worry if you feel like the girl in the picture - we'll work through some together to make sure it becomes clear! Let's get started.
### 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.
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https://physics.stackexchange.com/questions/214123/why-is-the-resistivity-of-a-copper-cable-much-higher-than-copper
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# Why is the resistivity of a copper cable much higher than copper?
I was looking at the data sheet for a copper cable and noticed the conductor resistance to be specified at around 5 mOhm/m. This is magnitudes higher than pure copper, which has a resistance in the order of 10 nOhm*m (from Wikipedia). Why is the resistivity so much higher?
• Watch out not to confuse resistance and resistivity in your question. By the way, what is your source for those numbers? – Steeven Oct 23 '15 at 8:47
The resistivity of a metal gives the resistance it will have based on the cross-sectional area and the length of the conductor.
$$\rho = \frac {AR}{L}$$
This means that resitivity is in SI units of $\Omega\text{ m}$. Neither of your quoted figures are in such units. Copper has a resistivity around $1.68 \times 10^{-8} \Omega \text{ m}$.
Unlike the bulk metal, a wire or conductor is manufactured with a constant cross section. If you pull the cross section away, you can characterize it with resistance per length, or $\frac{\Omega}{\text{m}}$.
In fact, assuming the wire above is copper, we can calculate the size based on linear resistance figure given.
$$\rho = \frac{AR}{L}$$ $$A = \frac{\rho} {\frac{R}{L}}$$ $$A = \frac{1.68 \times 10^{-8}\Omega \text{ m}}{5 \times 10^{-3}\Omega \text{ m}^{-1}}$$ $$A = 3.36\times 10^{-6}\text{m}^2 = 3.36\text{mm}^2$$
That cross section happens to be quite close to that of 12 gauge (AWG) wire.
• I had noted the wrong unit for pure copper, but you got it all right in your answer :) – akid Oct 24 '15 at 18:28
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https://mintdynasty.com/elliott-wave-patterns-what-is-a-zigzag-2/
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# Elliott Wave Patterns: What is a Zigzag?
Talking Points
-Zigzags appear in the construction of many of the Elliott Wave patterns
-Zigzags subdivide as an A-B-C (5-3-5) sequence
-Many times, wave C can be estimated using common ratios and channeling
In Elliott Wave theory, a zigzag is a pattern consisting of 3 waves labeled A-B-C. Most of the time, the pattern will print in the direction against the main trend and is typically a counter trend formation. Zigzags are typically found in the 2nd wave of a 5 wave impulse and have a part in the formation of Elliott wave triangles. Many times, zigzags appear somewhere within a complex correction. There are even a few instances where a zigzag will print in the direction of the larger trend. As a result, the zigzag pattern is an integral price pattern which can be found in several locations within the eight wave Elliott Wave sequence.
Zigzags can print in either the bullish or bearish direction.
Idealized Zigzag
[Image 1]
Zigzags look like a lightning bolt on the chart. There are 2 rules for zigzags:
1. The sub waves of an A-B-C zigzag appear as 5-3-5
2. Wave B of the zigzag cannot retrace 100% of Wave A – most of the time wave B retraces 38-78% of wave A
[Image 2]
The 3 waves of the zigzag (A-B-C) subdivide as a 5-3-5 meaning the ‘A’ leg has 5 sub waves in it, the ‘B’ leg has 3 sub waves in it, and the ‘C’ leg has 5 sub waves in it. As a result of the ‘A’ and ‘C’ legs both containing 5 sub waves each, the impact of the whole zigzag structure is to be a deep retracement and recover a lot of price from the previous trend.
Also, the zigzag was designed to make progress against the trend. Therefore, wave B of a zigzag can be any 3 wave pattern (including another zigzag), but wave B cannot retrace 100% of wave A. A retracement of 99% is acceptable, though unlikely and progress needs to be made.
There are a couple of reasons why the 2nd wave of a 5 wave impulse is typically made up of a zigzag. The first reason is because the 1st wave is the start of a new trend, the 2nd wave can be latent desires for the old trend to continue. As a result, the 2nd wave acts as a shake out for the new trend traders and the old trend traders as a deep retracement is typically obtained. Second waves typically retrace 50-78.6% of the first wave.
Within the zigzag pattern, many times you can estimate the termination zone of wave C. In Elliott Wave Theory, alternating waves tend to be related in distance. Therefore, once we believe wave C has started, we can estimate the length of C based on the length of A. Wave C typically has an equal measurement to wave A, or a .618 or 1.618 multiple of wave A.
In the example below, we can see where wave ‘C’ was equal in length to wave ‘A’ at the black horizontal line labeled “100% 7488.2.”
## Elliott Wave Zigzag That is an Equal Wave
Additionally, it is common for the A-B-C zigzag to create a price channel. So we can use the price channel as a means to estimate the termination point of C and the zigzag pattern.
These price approximations become powerful when there are other wave relationships showing up in the same price zone as the estimated termination point.
Interested in learning more about zigzags, how to identify and trade them? Register and watch this hour long webinar recording solely on the topic of zigzags.
—Written by Jeremy Wagner, CEWA-M
Having a method to time trade entries is important. Managing your risk may be even more important. Learn how the better performing traders manage their risk by reading our Traits of Successful Traders research. [Free registration required.]
The full suite of Elliott Wave resources can be found at the links below [registration required]:
Beginner and Advanced Elliott Wave trading guides
For more in depth study on Elliott wave patterns, we have these one hour webinar recordings:
Elliott Wave Impulse Patterns
Elliott Wave Zigzag Patterns
Elliott Wave Flat Patterns
Elliott Wave Triangle Patterns
Elliott Wave Diagonal Patterns
Starting Your Elliott Wave Counting
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# A box contains $50$ packets of biscuits each weighting $120g$. how many such boxes can be loaded in a van which cannot carry beyond $900kg?$
Last updated date: 20th Jul 2024
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Hint: First we have to define what the terms we need to solve the problem are. Since as per given question in a pack of $50$ packages of biscuits and each of the weight is $120g$. That means each and single box of biscuit is at the weight of $120g$ (one twenty grams). So, we just need to find a total of how many boxes can be loaded in that particular van but the only restriction is it will not go above the $900kg$.
Complete step-by-step solution:
let the weight of one packet which contains the biscuits is one twenty grams which is given;
so, if there are total of fifty packets of biscuits is there thus, we need to multiply each and every biscuits packet into the weight of the packets and hence we get $120 \times 50 = 6000$ grams
But we need to find the weight according to the kilo grams so that we can check the carry beyond $900kg$.
Now converting grams into kilograms which is $1000g = 1kg$(thousand grams equals the one kilogram)
Hence the weight of the fifty packets is $6000g = 6kg$(converting)
Finally, we need to find the boxes that can be loaded in the van but which cannot be carried beyond $900kg$. So, it will not exceed that nine hundred kilogram limit.
Thus to finding the total boxes we need to divide that $900kg$ into the $6kg$ of weight of the one box so that we can find the total boxes that can able to load at the van $\dfrac{1}{6} \times 900$ and hence; further solving we get $\dfrac{1}{6} \times 900$$\Rightarrow 150$ boxes.
Hence there are 150 boxes that can be loaded into that van.
Note: Since one kilogram equals that one thousand grams.
And there is a restriction that the weight will not exceed above $900kg$ so we divide one box weight into the total weight to find the overall box. If suppose there is no restriction, then the problem has no end up to a point, so that the vans maximum limit is nine hundred kilograms.
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https://coolconversion.com/electronics/dBm-W/What-is_0.1_dBm_in_milliwatt_
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# What is 0.1 dBm in milliwattdBm to W/mW Converter Calculator
### Inputs
dBm (dB-mW) dBW (dB-W)
or
=
### Change to mW/W to dBm/dBW
Conversion of electric power units. dBm to W / mW. Here is the answer to the question: What is 0.1 dBm in milliwatt or 0.1 dBm to milliwatts. Use the calculator above to convert any dBm values to Watts or miliwatts.
## dBm to miliwatt (mW) / Watt (W), Conversion Table (*)
Power (dBm)Power (mW)
-30 dBm0.001 mW
-20 dBm0.01 mW
-10 dBm0.1 mW
0 dBm1 mW
3 dBm2 mW
4.8 dBm3 mW
6 dBm4 mW
7 dBm5 mW
7.8 dBm6 mW
8.5 dBm7 mW
9 dBm8 mW
9.5 dBm9 mW
10 dBm10 mW
20 dBm100 mW
30 dBm1000 mW
40 dBm10000 mW
50 dBm100000 mW
60 dBm1000000 mW
70 dBm10000000 mW
80 dBm100000000 mW
90 dBm1000000000 mW
100 dBm10000000000 mW
(*) Some values are rounded
## How to convert dBm to mW
The power conversion from dBm to mW or Watt is given by the formula:
P(dBm) = 10(P(mW)/10)
### Example 1:
Convert 20 dBm to milliwatts:
P(mW) = 10(20/10) = 100 mW
So, the power of 20 dBm is equal to 100 mW or 0.1 W (Watt).
### Example 2:
Convert -30 dBm to milliwatts:
P(mW) = 10(-30/10) = 0.001 mW
So, the power of -30 dBm is equal to 0.001 mW or 0.000001 W = 1 μW.
## Sample dBm, dBW, W and mW conversions
### Disclaimer
While every effort is made to ensure the accuracy of the information provided on this website, neither this website nor its authors are responsible for any errors or omissions, or for the results obtained from the use of this information. All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information.
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Find the shortest Golomb rulers
Golomb rulers are sets of non-negative integers such that no two pairs of integers in the set are the same distance apart.
For example, [0, 1, 4, 6] is a Golomb ruler because all distances between two integers in this set are unique:
0, 1 -> distance 1
0, 4 -> distance 4
0, 6 -> distance 6
1, 4 -> distance 3
1, 6 -> distance 5
4, 6 -> distance 2
For the sake of simplicity in this challenge (and since translation is trivial), we impose that a Golomb ruler always contains the number 0 (which the previous example does).
Since this set is of length 4, we say that this is a Golomb ruler of order 4. The biggest distance in this set (or element, since 0 is always in the set) is 6, therefore we say that this is a Golomb Ruler of length 6.
Find Golomb rulers of order 50 to 100 (inclusive) that have as small lengths as you can find. The rulers you find need not be optimal (see below).
Optimality
A Golomb ruler of order N, is said to be optimal if there is no other Golomb ruler of order N which has a smaller length.
Optimal Golomb rulers are known for orders less than 28, though finding and proving optimality is harder and harder as the order increases.
Therefore, it is not expected that you find the optimal Golomb ruler for any of the orders between 50 and 100 (and even less expected that you can prove they are optimal).
There are no time limits in the execution of your program.
Baseline
The list below is the list of lengths of Golomb rulers from 50 to 100 (in order) evaluated with a naïve search strategy (Thanks to @PeterTaylor for this list):
[4850 5122 5242 5297 5750 5997 6373 6800 6924 7459 7546 7788 8219 8502 8729 8941 9881 10199 10586 10897 11288 11613 11875 12033 12930 13393 14046 14533 14900 15165 15687 15971 16618 17354 17931 18844 19070 19630 19669 20721 21947 22525 23290 23563 23880 24595 24767 25630 26036 26254 27218]
The sum of all those lengths is 734078.
Scoring
Your score will be the sum of the lengths of all your Golomb rulers between 50 and 100, divided by the sum of the lengths of Golomb rulers between 50 and 100 in the baseline: 734078.
In case you did not find a Golomb ruler for a specific order, you shall compute your score the same way, using the double of the length in the baseline for that specific order.
The answer with the lowest score wins.
In case of a tie, the lengths of the biggest order where the two answers differ are compared, and the shortest one wins. In case both answers have the same lengths for all orders, then the answer that was posted first wins.
• Related. (Same challenge in 2D.) – Martin Ender Jan 25 '17 at 9:03
• And OEIS entry. – Martin Ender Jan 25 '17 at 9:06
• When you say rulers between 50 and 100, do you mean the range [50, 100)? So not including the order 100 ruler? Because the baseline only contains 50 elements. – orlp Jan 25 '17 at 10:37
• Side note: the smallest possible length of a Golomb ruler of order n is n(n-1)/2, since that's how many positive differences there are. Therefore the smallest possible score in this challenge is 147050/734078 > 0.2003193. – Greg Martin Jan 25 '17 at 18:05
• @GregMartin Thanks, though this is not quite the "smallest possible score" but rather a lower bound on that smallest possible score! – Fatalize Jan 25 '17 at 18:07
C#, 259421/734078 ~= 0.3534
Methods
I finally found a more-or-less readable explanation of the projective field method (Singer's method) in Constructions of Generalised Sidon Sets, although I still think it can be improved slightly. It turns out to be more similar to the affine field method (Bose's method) than the other papers I read had communicated.
This assumes knowledge of finite fields. Consider $q = p^a$ is a prime power, and let $F(q)$ be our base field.
The affine field method works over $F(q^2)$. Take a generator $g_2$ of $F(q^2)$ and a non-zero element $k$ of $F(q)$ and consider the set $$\{a : g_2{}^a - k g_2 \in F_q\}$$ Those values form a modular Golomb ruler mod $q^2 - 1$. Further rulers can be obtained by multiplying modulo $q^2 - 1$ by any number which is coprime with the modulus.
The projective field method works over $F(q^3)$. Take a generator $g_3$ of $F(q^3)$ and a non-zero element $k$ of $F(q)$ and consider the set $$\{0\} \cup \{a : g_3{}^a - k g_3 \in F_q\}$$ Those values form a modular Golomb ruler mod $q^2 + q + 1$. Further rulers can be obtained by modular multiplication in the same way as for the affine field method.
Note that these methods between them give the best known values for every length greater than 16. Tomas Rokicki and Gil Dogon are offering a \$250 reward for anyone who beats them for lengths 36 to 40000. Therefore anyone who beats this answer is in for a monetary prize.
Code
The C# isn't very idiomatic, but I need it to compile with an old version of Mono. Also, despite the argument checking, this is not production quality code. I'm not happy with the types, but I don't think there's a really good solution to that in C#. Maybe in F# or C++ with insane templating.
using System;
using System.Collections.Generic;
using System.Linq;
namespace Sandbox {
class Program {
static void Main(string[] args) {
var winners = ComputeRulerRange(50, 100);
int total = 0;
for (int i = 50; i <= 100; i++) {
Console.WriteLine("{0}:\t{1}", i, winners[i][i - 1]);
total += winners[i][i - 1];
}
Console.WriteLine("\t{0}", total);
}
static IDictionary<int, int[]> ComputeRulerRange(int min, int max) {
var best = new Dictionary<int, int[]>();
var naive = Naive(max);
for (int i = min; i <= max; i++) best[i] = naive.Take(i).ToArray();
var finiteFields = FiniteFields(max * 11 / 10).OrderBy(x => x.Size).ToArray();
// The projective plane method generates rulers of length p^a + 1 for prime powers p^a.
// We can then look at subrulers for a reasonable range, say down to two prime powers below.
for (int ppi = 0; ppi < finiteFields.Length; ppi++) {
// Range under consideration
var field = finiteFields[ppi];
int q = field.Size;
int subFrom = Math.Max(min, ppi >= 2 ? finiteFields[ppi - 2].Size : 1);
int subTo = Math.Min(max, q + 1);
if (subTo < subFrom) continue;
int m = q * q + q + 1;
foreach (var ruler in ProjectiveRulers(field)) {
for (int sub = subFrom; sub <= subTo; sub++) {
var subruler = BestSubruler(ruler, sub, m);
if (subruler[sub - 1] < best[sub][sub - 1]) best[sub] = subruler;
}
}
}
// Similarly for the affine plane method, which generates rulers of length p^a for prime powers p^a
for (int ppi = 0; ppi < finiteFields.Length; ppi++) {
// Range under consideration
var field = finiteFields[ppi];
int q = field.Size;
int subFrom = Math.Max(min, ppi >= 2 ? finiteFields[ppi - 2].Size : 1);
int subTo = Math.Min(max, q);
if (subTo < subFrom) continue;
int m = q * q - 1;
foreach (var ruler in AffineRulers(field)) {
for (int sub = subFrom; sub <= subTo; sub++) {
var subruler = BestSubruler(ruler, sub, m);
if (subruler[sub - 1] < best[sub][sub - 1]) best[sub] = subruler;
}
}
}
return best;
}
static int[] BestSubruler(int[] ruler, int sub, int m) {
int[] expand = new int[ruler.Length + sub - 1];
for (int i = 0; i < ruler.Length; i++) expand[i] = ruler[i];
for (int i = 0; i < sub - 1; i++) expand[ruler.Length + i] = ruler[i] + m;
int best = m, bestIdx = -1;
for (int i = 0; i < ruler.Length; i++) {
if (expand[i + sub - 1] - expand[i] < best) {
best = expand[i + sub - 1] - expand[i];
bestIdx = i;
}
}
return expand.Skip(bestIdx).Take(sub).Select(x => x - ruler[bestIdx]).ToArray();
}
static IEnumerable<int[]> ProjectiveRulers(FiniteField field) {
var q = field.Size;
var fq3 = PowerField.Create(field, 3);
var m = q * q + q + 1;
var g = fq3.Generators.First();
// Define the set T<k> = {0} \union {a \in [q^3-1] : g^a - kg \in F(q)} for 0 != k \in F(q)
// This could alternatively be T<k> = {0} \union {log_g(b - kg) : b in F(q)} for 0 != k \in F(q)
// Then T<k> % (q^2 + q + 1) gives a Golomb ruler.
// For a given generator we seem to get the same ruler for every k.
var t_k = new HashSet<int>();
var ga = fq3.One;
for (int a = 1; a < fq3.Size; a++) {
ga = ga * g;
if (fq3.Convert(ga + g) < q) t_k.Add(a % m);
}
// TODO: optimise by detecting duplicates
for (int s = 1; s < m; s++) {
if (Gcd(s, m) == 1) yield return t_k.Select(x => x * s % m).OrderBy(x => x).ToArray();
}
}
static IEnumerable<int[]> AffineRulers(FiniteField field) {
var q = field.Size;
var fq2 = PowerField.Create(field, 2);
var m = q * q - 1;
var g = fq2.Generators.First();
// Define the set T<k> = {0} \union {a \in [q^2-1] : g^a - kg \in F(q)} for 0 != k \in F(q)
// Then T<k> % (q^2 - 1) gives a Golomb ruler.
var t_k = new HashSet<int>();
var ga = fq2.One;
for (int a = 1; a < fq2.Size; a++) {
ga = ga * g;
if (fq2.Convert(ga + g) < q) t_k.Add(a % m);
}
// TODO: optimise by detecting duplicates
for (int s = 1; s < m; s++) {
if (Gcd(s, m) == 1) yield return t_k.Select(x => x * s % m).OrderBy(x => x).ToArray();
}
}
static int Gcd(int a, int b) {
while (a != 0) {
var t = b % a;
b = a;
a = t;
}
return b;
}
static int[] Naive(int size) {
if (size == 0) return new int[0];
if (size == 1) return new int[] { 0 };
int[] ruler = new int[size];
var diffs = new HashSet<int>();
int i = 1, c = 1;
while (true) {
bool valid = true;
for (int j = 0; j < i; j++) {
if (diffs.Contains(c - ruler[j])) { valid = false; break; }
}
if (valid) {
for (int j = 0; j < i; j++) diffs.Add(c - ruler[j]);
ruler[i++] = c;
if (i == size) return ruler;
}
c++;
}
}
static IEnumerable<FiniteField> FiniteFields(int max) {
bool[] isComposite = new bool[max + 1];
for (int p = 2; p < isComposite.Length; p++) {
if (!isComposite[p]) {
FiniteField baseField = new PrimeField(p); yield return baseField;
for (int pp = p * p, pow = 2; pp < max; pp *= p, pow++) yield return PowerField.Create(baseField, pow);
for (int pq = p * p; pq <= max; pq += p) isComposite[pq] = true;
}
}
}
}
public abstract class FiniteField {
private Lazy<FiniteFieldElement> _Zero;
private Lazy<FiniteFieldElement> _One;
public FiniteFieldElement Zero { get { return _Zero.Value; } }
public FiniteFieldElement One { get { return _One.Value; } }
public IEnumerable<FiniteFieldElement> Generators {
get {
for (int _g = 1; _g < Size; _g++) {
int pow = 0;
FiniteFieldElement g = Convert(_g), gpow = One;
while (true) {
pow++;
gpow = gpow * g;
if (gpow == One) break;
if (pow > Size) {
throw new Exception("Is this really a field? " + this);
}
}
if (pow == Size - 1) yield return g;
}
}
}
public abstract int Size { get; }
internal abstract FiniteFieldElement Convert(int i);
internal abstract int Convert(FiniteFieldElement f);
internal abstract bool Eq(FiniteFieldElement a, FiniteFieldElement b);
internal abstract FiniteFieldElement Negate(FiniteFieldElement a);
internal abstract FiniteFieldElement Add(FiniteFieldElement a, FiniteFieldElement b);
internal abstract FiniteFieldElement Mul(FiniteFieldElement a, FiniteFieldElement b);
protected FiniteField() {
_Zero = new Lazy<FiniteFieldElement>(() => Convert(0));
_One = new Lazy<FiniteFieldElement>(() => Convert(1));
}
}
public abstract class FiniteFieldElement {
internal abstract FiniteField Field { get; }
public static FiniteFieldElement operator -(FiniteFieldElement a) {
return a.Field.Negate(a);
}
public static FiniteFieldElement operator +(FiniteFieldElement a, FiniteFieldElement b) {
if (a.Field != b.Field) throw new ArgumentOutOfRangeException("b");
}
public static FiniteFieldElement operator *(FiniteFieldElement a, FiniteFieldElement b) {
if (a.Field != b.Field) throw new ArgumentOutOfRangeException("b");
return a.Field.Mul(a, b);
}
public static bool operator ==(FiniteFieldElement a, FiniteFieldElement b) {
if (Equals(a, null)) return Equals(b, null);
else if (Equals(b, null)) return false;
if (a.Field != b.Field) throw new ArgumentOutOfRangeException("b");
return a.Field.Eq(a, b);
}
public static bool operator !=(FiniteFieldElement a, FiniteFieldElement b) { return !(a == b); }
public override bool Equals(object obj) {
return (obj is FiniteFieldElement) && (obj as FiniteFieldElement).Field == Field && this == (obj as FiniteFieldElement);
}
public override int GetHashCode() { return Field.Convert(this).GetHashCode(); }
public override string ToString() { return Field.Convert(this).ToString(); }
}
public class PrimeField : FiniteField {
internal int Prime { get { return _Prime; } }
public override int Size { get { return _Prime; } }
public PrimeField(int prime) {
if (prime < 2) throw new ArgumentOutOfRangeException("prime");
// TODO A primality test would be nice...
_Prime = prime;
_Featherweight = new PrimeFieldElement[Math.Min(prime, 256)];
}
internal override FiniteFieldElement Convert(int i) {
if (i < 0 || i >= _Prime) throw new ArgumentOutOfRangeException("i");
if (i >= _Featherweight.Length) return new PrimeFieldElement(this, i);
if (Equals(_Featherweight[i], null)) _Featherweight[i] = new PrimeFieldElement(this, i);
return _Featherweight[i];
}
internal override int Convert(FiniteFieldElement f) {
if (f == null) throw new ArgumentNullException("f");
if (f.Field != this) throw new ArgumentOutOfRangeException("f");
return (f as PrimeFieldElement).Value;
}
internal override bool Eq(FiniteFieldElement a, FiniteFieldElement b) {
if (a.Field != this) throw new ArgumentOutOfRangeException("a");
if (b.Field != this) throw new ArgumentOutOfRangeException("b");
return (a as PrimeFieldElement).Value == (b as PrimeFieldElement).Value;
}
internal override FiniteFieldElement Negate(FiniteFieldElement a) {
if (a.Field != this) throw new ArgumentOutOfRangeException("a");
var fa = a as PrimeFieldElement;
return fa.Value == 0 ? fa : Convert(_Prime - fa.Value);
}
internal override FiniteFieldElement Add(FiniteFieldElement a, FiniteFieldElement b) {
if (a.Field != this) throw new ArgumentOutOfRangeException("a");
if (b.Field != this) throw new ArgumentOutOfRangeException("b");
return Convert(((a as PrimeFieldElement).Value + (b as PrimeFieldElement).Value) % _Prime);
}
internal override FiniteFieldElement Mul(FiniteFieldElement a, FiniteFieldElement b) {
if (a.Field != this) throw new ArgumentOutOfRangeException("a");
if (b.Field != this) throw new ArgumentOutOfRangeException("b");
return Convert(((a as PrimeFieldElement).Value * (b as PrimeFieldElement).Value) % _Prime);
}
public override string ToString() { return string.Format("F({0})", _Prime); }
}
internal class PrimeFieldElement : FiniteFieldElement {
internal override FiniteField Field { get { return _Field; } }
internal int Value { get { return _Value; } }
internal PrimeFieldElement(PrimeField field, int val) {
if (field == null) throw new ArgumentNullException("field");
if (val < 0 || val >= field.Prime) throw new ArgumentOutOfRangeException("val");
_Field = field;
_Value = val;
}
}
public class PowerField : FiniteField {
internal FiniteField BaseField { get { return _BaseField; } }
internal int Power { get { return _Polynomial.Length; } }
public override int Size { get { return (int)Math.Pow(_BaseField.Size, Power); } }
public PowerField(FiniteField baseField, FiniteFieldElement[] polynomial) {
if (baseField == null) throw new ArgumentNullException("baseField");
if (polynomial == null) throw new ArgumentNullException("polynomial");
if (polynomial.Length < 2) throw new ArgumentOutOfRangeException("polynomial");
for (int i = 0; i < polynomial.Length; i++) if (polynomial[i].Field != baseField) throw new ArgumentOutOfRangeException("polynomial[" + i + "]");
// TODO Check that the polynomial is irreducible over the base field.
_BaseField = baseField;
_Polynomial = polynomial.ToArray();
}
internal override FiniteFieldElement Convert(int i) {
if (i < 0 || i >= Size) throw new ArgumentOutOfRangeException("i");
var vec = new FiniteFieldElement[Power];
for (int j = 0; j < vec.Length; j++) {
vec[j] = BaseField.Convert(i % BaseField.Size);
i /= BaseField.Size;
}
return new PowerFieldElement(this, vec);
}
internal override int Convert(FiniteFieldElement f) {
if (f == null) throw new ArgumentNullException("f");
if (f.Field != this) throw new ArgumentOutOfRangeException("f");
var pf = f as PowerFieldElement;
int i = 0;
for (int j = Power - 1; j >= 0; j--) i = i * BaseField.Size + BaseField.Convert(pf.Value[j]);
return i;
}
internal override bool Eq(FiniteFieldElement a, FiniteFieldElement b) {
if (a.Field != this) throw new ArgumentOutOfRangeException("a");
if (b.Field != this) throw new ArgumentOutOfRangeException("b");
var fa = a as PowerFieldElement;
var fb = b as PowerFieldElement;
for (int i = 0; i < Power; i++) if (fa.Value[i] != fb.Value[i]) return false;
return true;
}
internal override FiniteFieldElement Negate(FiniteFieldElement a) {
if (a.Field != this) throw new ArgumentOutOfRangeException("a");
return new PowerFieldElement(this, (a as PowerFieldElement).Value.Select(x => -x).ToArray());
}
internal override FiniteFieldElement Add(FiniteFieldElement a, FiniteFieldElement b) {
if (a.Field != this) throw new ArgumentOutOfRangeException("a");
if (b.Field != this) throw new ArgumentOutOfRangeException("b");
var fa = a as PowerFieldElement;
var fb = b as PowerFieldElement;
var vec = new FiniteFieldElement[Power];
for (int i = 0; i < Power; i++) vec[i] = fa.Value[i] + fb.Value[i];
return new PowerFieldElement(this, vec);
}
internal override FiniteFieldElement Mul(FiniteFieldElement a, FiniteFieldElement b) {
if (a.Field != this) throw new ArgumentOutOfRangeException("a");
if (b.Field != this) throw new ArgumentOutOfRangeException("b");
var fa = a as PowerFieldElement;
var fb = b as PowerFieldElement;
// We consider fa and fb as polynomials of a variable x and multiply modulo (x^Power - _Polynomial).
// But to keep things simple we want to manage the cascading modulo.
var vec = Enumerable.Repeat(BaseField.Zero, Power).ToArray();
var fa_xi = fa.Value.ToArray();
for (int i = 0; i < Power; i++) {
for (int j = 0; j < Power; j++) vec[j] += fb.Value[i] * fa_xi[j];
if (i < Power - 1) ShiftLeft(fa_xi);
}
return new PowerFieldElement(this, vec);
}
private void ShiftLeft(FiniteFieldElement[] vec) {
FiniteFieldElement head = vec[vec.Length - 1];
for (int i = vec.Length - 1; i > 0; i--) vec[i] = vec[i - 1] + head * _Polynomial[i];
}
public static FiniteField Create(FiniteField baseField, int power) {
if (baseField == null) throw new ArgumentNullException("baseField");
if (power < 2) throw new ArgumentOutOfRangeException("power");
// Since the field is cyclic, there is only one finite field of a given prime power order (up to isomorphism).
// For most practical purposes that means that we can pick any arbitrary monic irreducible polynomial.
// We can abuse PowerField to do polynomial multiplication in the base field.
var fakeField = new PowerField(baseField, Enumerable.Repeat(baseField.Zero, power).ToArray());
var excluded = new HashSet<FiniteFieldElement>();
for (int lpow = 1; lpow <= power / 2; lpow++) {
int upow = power - lpow;
// Consider all products of a monic polynomial of order lpow with a monic polynomial of order upow.
int xl = (int)Math.Pow(baseField.Size, lpow);
int xu = (int)Math.Pow(baseField.Size, upow);
for (int i = xl; i < 2 * xl; i++) {
var pi = fakeField.Convert(i);
for (int j = xu; j < 2 * xu; j++) {
var pj = fakeField.Convert(j);
}
}
}
for (int p = baseField.Size; true; p++) {
var pp = fakeField.Convert(p) as PowerFieldElement;
if (!excluded.Contains(pp)) return new PowerField(baseField, pp.Value.ToArray());
}
}
public override string ToString() {
var sb = new System.Text.StringBuilder();
sb.AppendFormat("GF({0}) with primitive polynomial x^{1} ", Size, Power);
for (int i = Power - 1; i >= 0; i--) sb.AppendFormat("+ {0}x^{1}", _Polynomial[i], i);
sb.AppendFormat(" over base field ");
sb.Append(_BaseField);
return sb.ToString();
}
}
internal class PowerFieldElement : FiniteFieldElement {
private readonly FiniteFieldElement[] _Vector; // The version of Mono I have doesn't include IReadOnlyList<T>
internal override FiniteField Field { get { return _Field; } }
internal FiniteFieldElement[] Value { get { return _Vector; } }
internal PowerFieldElement(PowerField field, params FiniteFieldElement[] vector) {
if (field == null) throw new ArgumentNullException("field");
if (vector == null) throw new ArgumentNullException("vector");
if (vector.Length != field.Power) throw new ArgumentOutOfRangeException("vector");
for (int i = 0; i < vector.Length; i++) if (vector[i].Field != field.BaseField) throw new ArgumentOutOfRangeException("vector[" + i + "]");
_Field = field;
_Vector = vector.ToArray();
}
}
}
Results
Unfortunately adding the rulers would take me about 15k characters past the post size limit, so they're on pastebin.
• Would you be so kind to post your rulers for [50, 100] somewhere? I have a genetic algorithm I want to try out, feeding it some seed values. – orlp Jan 25 '17 at 22:43
• @orlp, added a link. – Peter Taylor Jan 25 '17 at 22:56
• As I suspected, the evolutionary algorithm can extract nothing of use out of these superior specimen. Even though initially it seemed that evolutionary algorithms might work (it pretty much instantly moves from invalid rulers to actual rulers), there is too much global structure required for the evolutionary algorithm to work. – orlp Jan 25 '17 at 23:39
Python 3, score 603001 / 734078 = 0.82144
Naive search combined with Erdős–Turan construction:
$$2pk + (k^2\bmod p),\, k \in [0, p-1]$$
For odd primes p this gives an asymptotically optimal golomb ruler.
def isprime(n):
if n < 2: return False
if n % 2 == 0: return n == 2
k = 3
while k*k <= n:
if n % k == 0: return False
k += 2
return True
rulers = []
ruler = []
d = set()
n = 0
while len(ruler) <= 100:
order = len(ruler) + 1
if order > 2 and isprime(order):
ruler = [2*order*k + k*k%order for k in range(order)]
d = {a-b for a in ruler for b in ruler if a > b}
n = max(ruler) + 1
rulers.append(tuple(ruler))
continue
nd = set(n-e for e in ruler)
if not d & nd:
ruler.append(n)
d |= nd
rulers.append(tuple(ruler))
n += 1
isuniq = lambda l: len(l) == len(set(l))
isruler = lambda l: isuniq([a-b for a in l for b in l if a > b])
assert all(isruler(r) for r in rulers)
rulers = list(sorted([r for r in rulers if 50 <= len(r) <= 100], key=len))
print(sum(max(r) for r in rulers))
• I don't think this construction is asymptotically optimal: it yields a Golomb ruler of order p and length about 2p^2, whereas there exist Golomb rulers of order n and length about n^2 asymptotically. – Greg Martin Jan 25 '17 at 18:04
• @GregMartin Asymptotically there is no difference between 2p^2 and p^2. – orlp Jan 25 '17 at 18:08
• Depends on your definition of "asymptotically", I guess, but to me, in this context they're very different. – Greg Martin Jan 25 '17 at 18:20
Mathematica, score 276235/734078 < 0.376302
ruzsa[p_, i_] := Module[{g = PrimitiveRoot[p]},
Table[ChineseRemainder[{t, i PowerMod[g, t, p]}, {p - 1, p}], {t, 1, p - 1}] ]
reducedResidues[m_] := Select[Range@m, CoprimeQ[m, #] &]
rotate[set_, m_] := Mod[set - #, m] & /@ set
scaledRuzsa[p_] := Union @@ Table[ Sort@Mod[a b, p (p - 1)],
{a, reducedResidues[p (p - 1)]}, {b, rotate[ruzsa[p, 1], p (p - 1)]}]
manyRuzsaSets = Join @@ Table[scaledRuzsa[Prime[n]], {n, 32}];
tryGolomb[set_, k_] := If[Length[set] < k, Nothing, Take[set, k]]
Table[First@MinimalBy[tryGolomb[#, k] & /@ manyRuzsaSets, Max], {k, 50, 100}]
The function ruzsa implements a construction of a Golobm ruler (also called a Sidon set) found in Imre Z. Ruzsa. Solving a linear equation in a set of integers. I. Acta Arith., 65(3):259–282, 1993. Given any prime p, this construction yields a Golomb ruler with p-1 elements contained in the integers modulo p(p-1) (that's an even stronger condition than being a Golomb ruler in the integers themselves).
Another advantage of working in the integers modulo m is that any Golomb ruler can be rotated (the same constant added to all elements modulo m), and scaled (all elements multiplied by the same constant, as long as that constant is relatively prime to m), and the result is still a Golomb ruler; sometimes the largest integer is decreased significantly by doing so. So the function scaledRuzsa tries all of these scalings and records the results. manyRuzsaSets contains the results of doing this construction and scaling for all of the first 32 primes (chosen a bit arbitrarily, but the 32nd prime, 131, is well larger than 100); there are nearly 57,000 Golomb rulers in this set, which takes a good several minutes to compute.
Of course, the first k elements of a Golomb ruler themselves form a Golomb ruler. So the function tryGolomb looks at such a ruler made from any of the sets computed above. The last line Table... selects the best Golomb ruler it can, of every order from 50 to 100, from all the Golomb rulers found in this way.
The lengths found were:
{2241, 2325, 2399, 2578, 2640, 2762, 2833, 2961, 3071, 3151, 3194, 3480, 3533, 3612, 3775, 3917, 4038, 4150, 4237, 4368, 4481, 4563, 4729, 4974, 5111, 5155, 5297, 5504, 5583, 5707, 5839, 6077, 6229, 6480, 6611, 6672, 6913, 6946, 7025, 7694, 7757, 7812, 7969, 8139, 8346, 8407, 8678, 8693, 9028, 9215, 9336}
I was originally going to combine this with two other constructions, those of Singer and of Bose; but it seems that Peter Taylor's answer has already implemented this, so presumably I would simply recover those lengths.
• I'm confused by your claim that working in the integers modulo m you can rotate / scale freely. Look at [0, 1, 4, 6] mod 7. If I add 1 we get [0, 1, 2, 5], which is not a Golomb ruler. – orlp Jan 25 '17 at 18:13
• That's because you have to start with a mod-7 Golomb ruler for it to work. [0, 1, 4, 6] is not a mod-7 Golomb ruler because 1 – 0 equals 0 – 6 modulo 7, for instance. – Greg Martin Jan 25 '17 at 18:21
• While I was writing and debugging my finite field implementation in C# I wished I knew Mathematica better. Definitely one of the right languages for the job. – Peter Taylor Jan 25 '17 at 22:51
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Saltar al contenido principal
# 1.2: Divisibilidad y GCD en los números enteros
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##### Objetivos de aprendizaje
En esta sección, buscaremos responder a las preguntas:
• ¿Qué significa para un entero dividir a otro?
• ¿Cuál es el mayor divisor común de dos enteros?
• ¿Cómo podemos calcular el mayor divisor común de dos enteros?
## 1.2.1: La divisibilidad y el algoritmo de división
En esta sección, comenzamos a explorar algunas de las propiedades aritméticas y algebraicas de$$\mathbb{Z}\text{.}$$ Nos enfocamos específicamente en las propiedades de divisibilidad y factorización de los enteros, ya que estas son el foco principal del texto en su conjunto. Uno de los objetivos principales de esta sección es formalizar definiciones con las que probablemente ya esté familiarizado y de las cuales tenga una comprensión intuitiva. Al principio, esto podría parecer complicar innecesariamente las cosas. Sin embargo, quedará claro a medida que avancemos que el lenguaje matemático formal y la notación son necesarios para extender estas propiedades a un entorno más abstracto. Comenzamos con una noción familiar.
##### Definición: Divisibilidad y Algoritmo de División
Vamos$$a,b\in \mathbb{Z}\text{.}$$ Decimos que$$a$$ divide$$b\text{,}$$ y escribimos$$a\mid b\text{,}$$ si hay un entero$$c$$ tal que$$ac = b\text{.}$$ en este caso, digamos eso$$a$$ y$$c$$ son factores de$$b\text{.}$$ Si no$$c\in \mathbb{Z}$$ existe tal, escribimos$$a\nmid b\text{.}$$
Tenga en cuenta que el símbolo$$|$$ es un verbo; por lo tanto, es correcto decir, por ejemplo,$$2|4\text{,}$$ como 2 divide 4. No obstante, es un abuso de notación decir que$$2\mid 4 = 2\text{.}$$ en cambio, probablemente nos referimos a$$4\div 2 = 2$$ o$$\dfrac{4}{2} = 2$$ (aunque todavía no trataremos en fracciones).
##### Investigación: 1.2.1
Determinar$$a\mid b$$ si:
1. $$a = 3\text{,}$$$$b = -15$$
2. $$a = 4\text{,}$$$$b = 18$$
3. $$a = -7\text{,}$$$$b = 0$$
4. $$a = 0\text{,}$$$$b = 0$$
Comente brevemente los resultados de esta investigación. ¿Qué te diste cuenta? ¿Qué es lo que aún te preguntas?
A continuación se recogen varios resultados estándar sobre la divisibilidad en los$$\mathbb{Z}$$ que se utilizará ampliamente en el resto de este texto.
##### Teorema 1.2.1
Dejar$$a,b,c\in\mathbb{Z}\text{.}$$ Si$$a\mid b$$ y$$a\mid c\text{,}$$ luego$$a\mid (b+c)\text{.}$$
##### Teorema 1.2.2
Let$$a,b,c\in\mathbb{Z}\text{.}$$ Si$$a\mid b\text{,}$$ entonces$$a\mid bc\text{.}$$
##### Investigación 1.2.2
Considera lo siguiente parcial converse al Teorema 1.2.1 : Si$$a,b,c\in\mathbb{Z}$$ con$$a|bc\text{,}$$ must$$a|b$$ o$$a|c\text{?}$$ Supply una prueba o dar un contraejemplo.
##### Teorema 1.2.3
Vamos$$a,b,c,d\in \mathbb{Z}\text{.}$$ Si$$a = b+c$$ y$$d$$ divide dos cualesquiera de$$a,b,c\text{,}$$ entonces$$d$$ divide el tercero.
##### Investigación 1.2.3
Formular una conjetura similar a los teoremas anteriores sobre la divisibilidad en$$\mathbb{Z}\text{,}$$ y luego probarla
Como vimos anteriormente, no todos los pares de enteros$$a,b$$ satisfacen$$a\mid b$$ o$$b\mid a\text{.}$$ Sin embargo, nuestra experiencia en matemáticas elementales sí aplica: a menudo queda algo sobrante (un resto). El siguiente teorema formaliza esta idea para$$a,b\in \mathbb{N}\text{.}$$
##### Teorema 1.2.4
El algoritmo de división para$$\mathbb{N}$$.
Dejar$$a,b\in \mathbb{N}\text{.}$$ Entonces existen enteros únicos$$q,r$$ tales que$$a = bq + r\text{,}$$ donde$$0 \le r \lt b\text{.}$$
Pista 1
Hay dos partes en este teorema. Primero, debes establecer eso$$q$$ y$$r$$ existir. Esto se hace mejor a través de Axiom 1.2.1 . Si estás atascado en eso, revisa la segunda pista.
Una vez que haya establecido eso$$q$$ y$$r$$ exista, demuestre que son únicos pero asumiendo$$a = bq+r$$ y$$a = bq' + r'\text{,}$$ donde$$r,r'$$ ambos satisfacen las condiciones del teorema. Argumentan eso$$q = q'$$ y$$r = r'\text{.}$$
Pista 2
Let$$S = \{ a-bs : s\in \mathbb{N}_0, a-bs\ge 0\}\text{.}$$
Este teorema tiene dos partes: existencia y singularidad. No trates de probarlos a ambos al mismo tiempo.
## 1.2.2: Los divisores comunes más grandes
Pasamos a continuación a otra propiedad familiar de los enteros: la existencia de mayores divisores comunes.
##### Definición: Divisor común más grande
Dejemos que$$a,b\in \mathbb{Z}$$ tal que$$a$$ y no$$b$$ sean ambos 0. Un divisor más común de$$a$$ y$$b\text{,}$$ denotado$$\gcd(a,b)\text{,}$$ es un número natural$$d$$ satisfactorio
1. $$d\mid a$$y$$d\mid b$$
2. si$$e\in \mathbb{N}$$$$e\mid a$$ y$$e\mid b\text{,}$$ luego$$e\mid d\text{.}$$
Si$$\gcd(a,b) = 1\text{,}$$ decimos eso$$a$$ y$$b$$ son relativamente primos o coprimos.
Esta definición puede ser diferente a la que está acostumbrado, que probablemente declaró que$$d \ge e$$ rather than condition 2 in Definition: Greatest Common Divisor. It can be proved using the order relations of $$\mathbb{Z}$$ that the definition given here is equivalent to that one. However, we will prefer this definition, as it generalizes naturally to other number systems which do not have an order relation like $$\mathbb{Z}\text{.}$$
Compute$$\gcd(a,b)$$ si:
1. $$a = 123\text{,}$$$$b = 141$$
2. $$a = 0\text{,}$$$$b = 169$$
3. $$a= 85\text{,}$$$$b = 48$$
Ahora que has tenido un poco de práctica computando gcds, describe tu método para encontrarlos en una oración o dos.
¿Cómo respondiste la última pregunta en Activity 1.2.1 ? Si eres como las clases de los autores, las respuestas probablemente variaron, aunque en algún momento te has referido a un “prime” (sean cuales sean esos), o posiblemente algún otro método ad hoc para encontrar el gcd. La mayoría de estos métodos se basan de alguna forma en nuestra capacidad para factorizar números enteros. Sin embargo, el problema de factorizar enteros arbitrarios es en realidad sorprendentemente intensivo desde el punto de vista computacional. Agradecidamente, hay otra forma de calcular$$\gcd(a,b)\text{,}$$ a la que ahora recurrimos.
##### Teorema 1.2.5
Que$$a,b,c\in\mathbb{Z}$$ tal que$$a = b+c$$ con$$a$$ y$$b$$ no ambos cero. Entonces$$\gcd(a,b) = \gcd(b,c)\text{.}$$
##### Investigación 1.2.4
Supongamos$$a,b,c\in\mathbb{Z}$$ tal que existe$$q\in\mathbb{Z}$$ con$$a = bq + c$$$$a$$ y$$b$$ no ambos cero. Demostrar o desacreditar:$$\gcd(a,b)=\gcd(b,c)\text{.}$$
##### Investigación 1.2.5
(Algoritmo Euclidiana).
Let$$a,b\in \mathbb{N}\text{.}$$ Use Theorema 1.2.4 e Investigation 1.2.4 para determinar un algoritmo para la computación$$\gcd(a,b)\text{.}$$ ¿Cómo podría modificarse su método para calcular$$\gcd(a,b)$$ para$$a,b\in\mathbb{Z}\text{?}$$
Usar el algoritmo euclidiano para calcular$$\gcd(18489,17304)\text{.}$$
La siguiente identidad proporciona una caracterización útil del mayor divisor común de dos enteros, no ambos cero. Volveremos a esta idea varias veces, incluso después de haber dejado el reino familiar de los enteros.
##### Teorema 1.2.6 : Identidad de Bézout
Para cualquier número entero$$a$$ y$$b$$ no ambos 0, hay enteros$$x$$ y$$y$$ tales que
\ begin {ecuación*} hacha + por =\ gcd (a, b)\ texto {.} \ end {ecuación*}
Pista 1
Aplica Axioma 1.2.1 a un conjunto bien elegido.
Pista 2
Aplicar Axioma 1.2.1 a$$S = \{as+bt : s,t\in\mathbb{Z}, \ as+bt \gt 0\}\text{.}$$
Concluimos con una respuesta a las preguntas planteadas por Investigation 1.2.2 .
##### Teorema 1.2.7
Dejar$$a, b\text{,}$$ y$$c$$ ser enteros. Si$$a|bc$$ y$$\gcd(a,b) = 1\text{,}$$ entonces$$a|c\text{.}$$
En esta sección, hemos recopilado algunos resultados iniciales sobre la divisibilidad en los números enteros. A continuación exploraremos los bloques de construcción multiplicativos de los enteros, los primos, en preparación para una exploración más profunda de la factorización.
This page titled 1.2: Divisibilidad y GCD en los números enteros is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Michael Janssen & Melissa Lindsey via source content that was edited to the style and standards of the LibreTexts platform.
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# Problematics | The rate of sending and cancelling an invitation
Sep 18, 2023 02:32 PM IST
## Are there enough equations to determine the variables? You may need to think out of the box to solve this puzzle, partly inspired by a 150-year-old classic.
From a setter’s perspective, there are four kinds of puzzles. The most satisfying ones are the setter’s own creations, for example, my Gabbar Singh puzzle in Week 50. The other three kinds derive from the work of other setters: puzzles reproduced as is with due credit, old puzzles adapted into a new form, and brand-new puzzles inspired by an existing original.
The following puzzle belongs to the last category. It is my own, inspired by a 150-year-old classic.
The ring of the mobile had drowned one of her words, which, as you will have deduced, was a number denoting an hour of the clock. Her husband took the call, listened, and made a face. “She’s not coming,” he said, handing her the phone.
They chatted with their daughter for some time. The invitations her parents had just finished sending (the first lot by text, and then the rest by mail) were for her planned homecoming party.
Once the call ended, the mother sighed. “Nothing to do but mail or text everyone that the party’s off,” she said. This time, they wrote the cancellation mails first, starting exactly 15 minutes after they had finished sending the last of the invitations.
“The mails take longer,” she said. “Four cancellation mails between us in an hour, the same rate at which we had mailed the invitations to the same guests.”
“The text messages will be faster,” her husband replied. “Remember, we had texted the invitations to that lot of guests @6/hour.”
Alas, it was not to be. Relatives and friends responded to the cancellation texts with so many questions and subsidiary questions that the parents managed to text only 3 cancellations every hour.
When the last message had been texted, the mother looked at the time again. “12:15 am. What a waste of a day.”
## #Puzzle 56.2
Take all 10 digits from 0 to 9. Using each digit exactly once, write two fractions whose sum is 1.
## #Puzzle 55.1
Hi Kabir Sir,
Let the tops of the 3 dice be x, y and z. Adding: x + y + z.
Let the dice x and y be picked and therefore the bottom faces will be (7 – x)* and (7 – y)* respectively. Adding: x + y + z + (7 – x) + (7 – y) = 14 + z
After these 2 dice are rolled again, let the spots on their top faces be a and b. Adding: 14 + z + a + b
Picking up a, spots on the bottom will be (7 – a)*. Adding: 14 + z + a + b + (7 – a) = 21 + z + b
Rolling the same die again, we get w as the top face. Adding: 21 + z + b + w
Therefore, when the smart alec turns around, he adds the total on the top faces of the three dice and adds an extra 21.
## — Ananya Arvind, DPS Vasant Kunj, Delhi
*What Ananya left unsaid: in a standard die, the spots on any two opposite faces add up to 7.
Solved both puzzles: Ananya Arvind (DPS Vasant Kunj), Dr Sunita Gupta (Delhi), Amardeep Singh (Meerut), Harshit Arora (IIT Delhi), V Anand (Noida), Prof Anshul Kumar (Delhi), Malay Mittal (Delhi), Sunita & Naresh Dhillon (Gurgaon), Dr Nakul Makkar (Noida), Charvi Brajpuriya (Faridabad), Group Capt RK Shrivastava (retd; Delhi), Vardan Bhaskar (Modern DPS-89, Faridabad), Vinod Mahajan (Delhi), Ajay Ashok (Mumbai), Yadvendra Somra (Sonipat), Shri Ram Aggarwal (Delhi), Abhishek Garg (Chandigarh)
Solved #Puzzle 55.1: Bhaskar Mundhra (Ghaziabad), Bhuvi Jain (Delhi), Rachna Jain (Delhi), Radhika Joshi (DPS Vasant Kunj), Arun Kumar Gupta (Greater Noida), Vivek Aggarwal (Bangalore), Shishir Gupta (Indore), Satishwar (Delhi), Saksham Bhatnagar, Joydeep Mahato
Solved #Puzzle 55.2: Sanghamitran Rajan (BVN Delhi), Shriyon Bhattacharya (Delhi), Parul Kohli
Problematics will be back next week. Please send in your replies by Friday noon to [email protected]
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Notes On Addition and Subtraction of Decimals - CBSE Class 6 Maths
To add or subtract decimal numbers, make sure that the decimal points of the given numbers are placed exactly one below another. While adding or subtracting two decimal numbers, the number of digits after the decimal point should be equal. In case they are not equal, the gaps must be filled with zeros after the last digit. e.g. To add 6.82 and 5 First insert zeros in the empty places after the decimal point so that both the numbers have the same number of digits after the decimal point. Next, write the numbers such that their decimal points are one below another. 6.82 + 5 = 11.82 6.82 + 5.00 11.82 To subtract 5 from 6.82 First insert zeros in the empty places after the decimal point so that both the numbers have the same number of digits after the decimal point. Next, write the numbers such that their decimal points are one below another. e.g. 6.82 - 5 = 1.82 6.82 – 5.00 1.82 Addition or subtraction should be carried out from the extreme right side. Place the decimal point correctly after performing the addition or subtraction.
Summary
To add or subtract decimal numbers, make sure that the decimal points of the given numbers are placed exactly one below another. While adding or subtracting two decimal numbers, the number of digits after the decimal point should be equal. In case they are not equal, the gaps must be filled with zeros after the last digit. e.g. To add 6.82 and 5 First insert zeros in the empty places after the decimal point so that both the numbers have the same number of digits after the decimal point. Next, write the numbers such that their decimal points are one below another. 6.82 + 5 = 11.82 6.82 + 5.00 11.82 To subtract 5 from 6.82 First insert zeros in the empty places after the decimal point so that both the numbers have the same number of digits after the decimal point. Next, write the numbers such that their decimal points are one below another. e.g. 6.82 - 5 = 1.82 6.82 – 5.00 1.82 Addition or subtraction should be carried out from the extreme right side. Place the decimal point correctly after performing the addition or subtraction.
Next
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1. ## problem
A boat that travels at the speed of 6.75 m/s in still water is to go directly across a river and back. The current flows at 0.50 m/s. (a) At what angle(s) must the boat be steered? (b) How long does it take to make the round trip? (The width of the river is 150 meters. Assume that the boat's speed is constant at all times, and neglect turnaround time.)
2. Originally Posted by Celia
A boat that travels at the speed of 6.75 m/s in still water is to go directly across a river and back. The current flows at 0.50 m/s. (a) At what angle(s) must the boat be steered? (b) How long does it take to make the round trip? (The width of the river is 150 meters. Assume that the boat's speed is constant at all times, and neglect turnaround time.)
Hello,
the direction perpendicular to the shore(s) has a bearing of 0°.
The way of the boat through the water, the current of the river and this perpendicular direction form a right triangle. (See attachment)
the way per second of the boat over ground is (use Pythagorian Theorem):
$\displaystyle w=\sqrt{(6.75m)^2-(0.5m)^2}\approx 6.73 m$
to a) You have to steer the boat against the current with an angle of:
$\displaystyle \arcsin\left(\frac{0.5}{6.75}\right)\approx 4.25^{\circ}$
on your way back you must steer: 180°-4.25°= 175.75°
to b) The boat has to travel 2 * 150 m = 300 m over ground. The boat travel 6.73 m/s over ground. So it'll take
$\displaystyle t = \frac{300}{6.73}\approx 44.6 s$
Greetings
EB
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## Project Euler 004 – TSQL
24 02 2011
Problem 4
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.
Solution
–A palindromic number reads the same both ways.
–The largest palindrome made from the product of
–two 2-digit numbers is 9009 = 91 99.
–Find the largest palindrome made from the
–product of two 3-digit numbers.
— this is a fun query
— we join the number table to
— itself ensuring that the right
— number is greater than or equal
— to the left number
SELECT MAX(LeftNumber.number * RightNumber.number)
FROM Number LeftNumber
INNER JOIN Number RightNumber
ON LeftNumber.number <= RightNumber.number
— then we limit the numbers to 3 digits
— an obvious (and better) method for this is
— LeftNumber.number >= 100
— and LeftNumber.number <= 999
— but this technique is more interesting
WHERE LEN(CAST(LeftNumber.number AS VARCHAR)) = 3
and LEN(CAST(RightNumber.number AS VARCHAR)) = 3
— finally, the clever part is determining
— the palindromicity of the product
and(CAST(LeftNumber.number * RightNumber.number AS VARCHAR)
= REVERSE(CAST(LeftNumber.number * RightNumber.number AS VARCHAR)))
Discussion
I am quite happy with this solution.
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when a CD player is proiced at $300 per unit, a store sells : PS Archive Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 17 Jan 2017, 07:12 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track Your Progress every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History #### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # when a CD player is proiced at$300 per unit, a store sells
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when a CD player is proiced at $300 per unit, a store sells [#permalink] ### Show Tags 14 Jul 2003, 06:49 00:00 Difficulty: (N/A) Question Stats: 0% (00:00) correct 0% (00:00) wrong based on 0 sessions ### HideShow timer Statistics This topic is locked. If you want to discuss this question please re-post it in the respective forum. when a CD player is proiced at$300 per unit, a store sells 15 units per week . Each time the price is reduced by $10, however, the sales increase 2units per week. What selling price will result in weekly revenues of$7000 ?
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### Show Tags
14 Jul 2003, 08:18
Selling price will be 200.
As 35 units shall be sold then.
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try to solve this equation [#permalink]
### Show Tags
14 Jul 2003, 09:35
(300-a)(15+ 2 (a/10)) = 7000
you will get a=100 or 125,
since 125 gives you unbelievably decimaled answer, you know you should choose 100. Then use 300-100=200
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### Show Tags
14 Jul 2003, 09:40
(300-10x)(15+2x)=7000
2x^2-45x+250=7000
x=10
Price = 300-10*10=200
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### Show Tags
14 Jul 2003, 13:35
stolyar wrote:
(300-10x)(15+2x)=7000
2x^2-45x+250=7000
x=10
Price = 300-10*10=200
How do you get the equation in 2nd line?
I broke my head for ten minutes to slove this equation!
Is there is nifty way to solve these kind of Quad equations?
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### Show Tags
15 Jul 2003, 02:02
stolyar wrote:
(300-10x)(15+2x)=7000
2x^2-45x+250=7000
x=10
Price = 300-10*10=200
How do you get the equation in 2nd line?
I broke my head for ten minutes to slove this equation!
Is there is nifty way to solve these kind of Quad equations?
I used thr trial and error method. It landed out to be fastest. As fasr as i know GMAT doesnt involve complex calculations. They are not checking your skills to solve complex quadratic. The want quick and efficient methods to be arrive at the answer.
HTH:)
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### Show Tags
15 Jul 2003, 03:54
As some of the math wizards already pointed out , there is no much brain to be used on this question. The answer choices contained were 175 and 200 along with other values. Most of us (including me ) chose 175 ( in 20 secs ) which is clearly wrong !! CD price is decreasing by 10s . If you ask any 1st grader to count by 10, he / she wont find 175..
Well known GMAT trap !!
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15 Jul 2003, 03:54
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Question
# In the given figure, if ∠BCA=∠RPQ=15∘, then ∠PQR is .
A
45
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75
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Solution
## The correct option is A 45∘In ΔABC and ΔPQR, ∠BCA=∠RPQ = 15∘ .... (Given) BC=PR=1 km ....... (Given) AC=PQ=1.3 km .... (Given) Thus, ΔACB≅ΔQPR by SAS congruency. ⇒∠PQR=∠BAC .... (CPCT) Along the horizontal line through point A, we have 90∘+∠BAC+45∘=180∘ ∠BAC=180∘−45∘−90∘ ⇒∠PQR=∠BAC=45∘
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# Vedic Maths Questions and Answer
Solve the following questions:Q1. 93 * 86A. 8098B. 9098C. 7998D. 9808Sol : Option C100 -86 = 14, 100 -93 = 714 * 7 = 98 (last two digits)93-14= 86-7 =79Therefore, the correct answer is 7998Q2. 87 * 93A. 9801B. 9201C. 8091D. 8291Sol : Option C100 -87 = 13, 100 -93 = 713 * 7 =…
Solve the following questions:
Q1. 93 * 86
A. 8098
B. 9098
C. 7998
D. 9808
Sol : Option C
100 -86 = 14, 100 -93 = 7
14 * 7 = 98 (last two digits)
93-14= 86-7 =79
Therefore, the correct answer is 7998
Q2. 87 * 93
A. 9801
B. 9201
C. 8091
D. 8291
Sol : Option C
100 -87 = 13, 100 -93 = 7
13 * 7 = 91 (last two digits)
93-13= 87-7 = 80
Therefore, the correct answer is 8091
Q.3. 89 * 109
A. 9081
B. 9281
C. 9701
D. 9401
Sol : Option C
100 -89 = 11, 100 -109 = -9
11 * -9 = -99 (100-99=01) (last two digits) (1 borrowed)
109-11-= 89+9 =98 -1 (borrowed) = 97
Therefore, the correct answer is 9701
Q.4. 47 * 44
A. 2018
B. 2048
C. 2028
D. 2068
Sol : Option D
In this case, we can take 50, which is a sub multiple of 100 or a multiple of 10 and proceed as follows: 50 -41 = 9, 50 -43 = 7
9 * 7 = 63 (last two digits)
41-7= 43-9 = 34/2 = 17
Since 50 = 100/2, we divide the left hand side number also by 2 while retaining the right hand side. Therefore, the answer will be 2068.
Q.5. Find the square of 55
A. 2525
B. 2925
C. 3025
D. 3125
Sol : Option D
552 ⇒ (5 * 6), 25 = 3025
Q.6.Find the square of 112
A. 11244
B. 12544
C. 13544
D. 14544
Sol : Option B
(112) 2 = 112 + 12 | (12) 2 = 144 | (carry 4)44 = 12544
Q.7.51 * 27
A. 1477
B. 1277
C. 1377
D. 1777
Sol : Option C
51
27
Step 1: 1 * 7 = 07 (Write 07).
Step 2: 1 * 2 + 5 * 7 + 2 (Carried Over) = 37 (Write 7 and 3 is carried over to the next step)
Step 3: 5 * 2 + 3 (Carried Over) = 13 (Write 13 because this is the last step)
1377 is the answer.
Q8.72 * 47
A. 3284
B. 3384
C. 3464
D. 9084
Sol : Option B
72
47
Step 1: 2 * 7 = 14 (Write 4 and 1 is carried over to the next step)
Step 2: 7 * 7 + 2 × 4 + 1 (Carried Over) = 58 (Write 8 and 5 is carried over to the next step)
Step 3: 7 * 4 + 5 (Carried Over) = 33 (Write 27 because this is the last step)
3384 is the answer.
Q9.62 * 43
A. 2566
B. 2466
C. 2866
D. 2666
Sol : Option D
62
43
Step 1: 2 * 3 = 6 (Write 6 which is the single digit number)
Step 2: 6 * 3 + 2 * 4 = 26 (Write 6 and 2 is carried over to the next step)
Step 3: 6 * 4 + 2 (Carried Over) = 26 (Write 26 as this is the last step).
2666 is the answer.
Q10.Find the square of 82
A. 6724
B. 6424
C. 4824
D. 8924
Sol : Option D
(82) 2 = 82 – 18 | (18) 2 = 6724
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# Why is the union of non disjoint path connected sets path connected?
If $X$ is a topological space, and $A, B\subset X$ are not disjoint and both path connected, it intuitively makes sense that $A\cup B$ is path connected: If, for $x_0\in A$ and $x_1\in B$ you take an $\tilde{x}\in A\cap B$ and continuous $\gamma_A:[0,1]\rightarrow A$ and $\gamma_B:[0,1]$ with $\gamma_A(0) = x_0, \gamma_A(1) = \tilde{x} = \gamma_B(0)$ and $\gamma_B(1) = x_1$, you can just take the function that sends $s$ to $\gamma_A(2s)$ for $s\leq \frac{1}{2}$ and $\gamma_B(2s-1)$ for $s\geq\frac{1}{2}$. (Basically, if we are looking at the non-trivial case where $x_0$ is in one of the sets and $x_1$ is in the other, you make a path by joining two continuous paths together, which you know exist because $A$ and $B$ are path connected).
Call this function $\gamma$, then $\gamma$ meets the requirements that $\gamma(0) = x_0, \gamma(1) = x_1$. Now we just need to verify that $\gamma$ is continuous. I'm trying to do so by looking at an open set $U\in\mathcal{T}_{A\cup B}$, and verifying that $\gamma^{-1}(U)$ is an open interval in $[0,1]$. However, this is where I'm stuck. I tried to work it from the angle that $U = U_1\cup U_2$ with $U_1\subset A, U_2\subset B$, but then I need $U_1$ and $U_2$ to be open, which I don't know how to prove.
Is this the way to go? Or is there a different way to approach the problem?
-
You should precisely not look at open sets, but closed sets and use: $\gamma$ is continuous if and only if $\gamma^{-1}(F)$ is closed for every closed set $F$. – 1015 Apr 12 '13 at 21:20
So $\gamma$ is continuous if and only if it is seqentially continuous, as $[0,1]$ is a metric space. At $x\in [0,1/2)$ or $(1/2,1]$, we can simply use the continuity of $\gamma_A$ and $\gamma_B$ respectively. The only delicate point is $x_0=1/2$. For $x_n\longrightarrow (1/2)^-$, we have $\lim \gamma(x_n)=\lim \gamma_A(2x_n)=\gamma_A(1)=\tilde{X}$.And for $x_n\longrightarrow (1/2)^+$, we have $\lim \gamma(x_n)=\lim \gamma_B(2x_n-1)=\gamma_B(0)=\tilde{X}$. I only remains to convince yourself that this suffices for the general case $x_n\longrightarrow 1/2$ to hold, which can be done by considering the possibly finite subsequences $x_n^+$ and $x_n^-$ of all $x_n$ greater than $1/2$ and less than $1/2$ respectively.
Now this was a tedious argument and certainly not the way we should do that. Here is a more general, purely topological fact, which makes everything easier to write. And which is the real reason behind this. The key fact is that the closed sets of a topology induced by a topological set $X$ on a closed subset $F$ are precisely the closed sets of $X$ contained in $F$. Note that this is false in general if $F$ is not closed. See the remninder below.
Lemma: let $f:X\longrightarrow Y$ be a function between two topological spaces. If there exist two closed sets $F_1,F_2$ in $X$ such that $X=F_1\cup F_2$ and both $f_1:=f_{|F_1}$ and $f_2:=f_{|F_2}$ are continuous for the induced topology on $F_1$ and $F_2$ respectively, then $f$ is continuous.
Remark: the same fact holds if you replace closed by open.
Proof: We will show that for every $F$ closed in $Y$, $f^{-1}(F)$ is closed in $X$, which is equivalent to the continuity of $f$. So take $F$ closed in $Y$ and observe $$f^{-1}(F)=f_1^{-1}(F)\cup f_2^{-1}(F).$$ By assumption, each $f_j^{-1}(F)$ is closed in $F_j$ for the induced topology, i.e. there exists $G_j$ closed in $X$ such that $f_j^{-1}(F)=G_j\cap F_j$. Hence $f_j^{-1}(F)$ is a fortiori closed in $X$ and $f^{-1}(F)$ is the union of two closed sets, whence closed in $X$. QED.
Application: apply this to $f=\gamma$ on $X=[0,1]$ with $F_1=[0,1/2]$ and $F_2=[1/2,1]$.
Reminder: if $X$ is a topological space and $S$ is a subset of $X$. Then the topology induced by $X$ on $S$ is that where the open (resp. closed) sets all sets of the form $S\cap U$ (resp. $S\cap F$) for some $U$ open in $X$ (resp $F$ closed in $X$). These need not be open (resp. closed) in $X$ in general. As this would imply in particular that $S$ is open (resp. closed) in $X$.
For example, for $X=[0,1]$ and $S=[0,1/2)$, then $[0,1/2)$, like any $[1/2-1/n,1/2)$ is closed in $[0,1/2)$ for the induced toplogy. Indeed we can write $[1/2-1/n,1/2)=[0,1/2)\cap [1/2-1/n,1]$ as the intersection of $[0,1/2)$ by a closed set of $[0,1]$.
-
I don't understand why $f_j^{-1}(F)$ would have to be closed in $F_j$: Why does this $G_j$ have to exist? – Radical Apr 12 '13 at 21:39
My assumption is: $f_j:F_j\longrightarrow Y$ is continuous, where $F_j$ is equipped with the topology induced by $X$. In this topology, the closed sets are precisely of the form $F_j\cap F$ with $F$ closed in $X$. So these are precisely the closed subsets of $X$ which are contained in $F_j$. In your case, $\gamma_j(x)=\gamma(2x)$ is indeed continuous on $[0,1/2]$. So in this case, it is obvious that the closed sets of $[0,1/2]$ are closed in $[0,1]$. In the general topological case, it goes like I said above. – 1015 Apr 12 '13 at 21:50
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1
GATE CSE 2015 Set 3
Numerical
+2
-0
Let $$G$$ be a connected undirected graph of $$100$$ vertices and $$300$$ edges. The weight of a minimum spanning tree of $$G$$ is $$500.$$ When the weight of each edge of $$G$$ is increased by five, the weight of a minimum spanning tree becomes ________.
2
GATE CSE 2012
+2
-0.6
Let G be a weighted graph with edge weights greater than one and G' be the graph constructed by squaring the weights of edges in G. Let T and T' be the minimum spanning trees of G and G' respectively, with total weights t and t'. Which of the following statements is TRUE?
A
T' = T with total weight t' = t2
B
T' = T with total weight t' < t2
C
T' =! T but total weight t' = t2
D
None of these
3
GATE CSE 2010
+2
-0.6
Consider a complete undirected graph with vertex set {0,1,2,3,4}. Entry Wij in the matrix W below is the weight of the edge {i, j} $$W = \left( {\matrix{ 0 & 1 & 8 & 1 & 4 \cr 1 & 0 & {12} & 4 & 9 \cr 8 & {12} & 0 & 7 & 3 \cr 1 & 4 & 7 & 0 & 2 \cr 4 & 9 & 3 & 2 & 0 \cr } } \right)$$$What is the minimum possible weight of a spanning tree T in this graph such that vertex 0 is a leaf node in the tree T? A 7 B 8 C 9 D 10 4 GATE CSE 2010 MCQ (Single Correct Answer) +2 -0.6 Consider a complete undirected graph with vertex set {0,1,2,3,4}. Entry Wij in the matrix W below is the weight of the edge {i, j} $$W = \left( {\matrix{ 0 & 1 & 8 & 1 & 4 \cr 1 & 0 & {12} & 4 & 9 \cr 8 & {12} & 0 & 7 & 3 \cr 1 & 4 & 7 & 0 & 2 \cr 4 & 9 & 3 & 2 & 0 \cr } } \right)$$$ What is the minimum possible weight of a path P from vertex 1 to vertex 2 in this graph such that P contains at most 3 edges?
A
7
B
8
C
9
D
10
GATE CSE Subjects
EXAM MAP
Medical
NEET
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# Translating sigma notation / summation / series and integral equation from Microsoft Word into MATLAB syntax and graphing / plotting
12 views (last 30 days)
Taosui Li on 20 Aug 2019
Commented: Taosui Li on 21 Aug 2019
Hello everybody,
I have a little trouble here. I trying to graph/plot the following equation with MATLAB. I already solved the integral of this equation:
f(t)=0.111627907+∑_(n=1)^∞▒〖(1/430 ∫_0^50▒〖((-0.000003072(t-25)^4+1.2) cos(2nπt/860) )dt(nπt/430))〗〗+∑_(n=1)^∞▒〖(1/430 ∫_0^50▒〖((-0.000003072(t-25)^4+1.2 ) sin(2nπt/860) )dt(nπt/430))〗〗
f(t)=0.111627907+∑_(n=1)^∞▒(1/430 1/((π)^4 n^4 ) (-177504/5 (π)^2 n^2+7876917504/125)+1(π)^5 n^5 (-177504/625 (-6450(π)^2 n^2+3816336) sin(5/43 πn)-17750462/5 (125(π)^3 n^3-221880πn) cos(5/43 πn) )(nπt/430)) +∑_(n=1)^∞▒〖(1/430 1/((π)^5 n^5 )(45796032/25 (π)^2 n^2-677414905344/625)+1/((π)^5 n^5 )(-177504/625 (6450(π)^2 n^2-3816336)cos(5/43 πn)-177504/625 (125(π)^3 n^3-221880πn)sin(5/43 πn)(nπt/430))〗
Walter Roberson on 20 Aug 2019
Why do you have 1/430 before each of the integrals, and then multiply the result of the integral by (n*pi*t/430) ? It is strange that you have two divisions by 430 there. Please check your equations, as I suspect there should be only one of them.
Walter Roberson on 21 Aug 2019
You are defining f(t) so t is certainly a parameter of f.
Taosui Li on 21 Aug 2019
Sorry, I mean t is a parameter of f.
Walter Roberson on 20 Aug 2019
syms n t
Q = @(v) sym(v); %convert to rational
Pi = sym('pi');
f1 = Q(0.111627907);
f2a = int((Q(-0.3072*10^(-5))*(t - 25)^4 + Q(1.2))*cos((2*n*Pi*t)/860), t, 0, 50)
f2 = symsum(1/430*f2a*n*Pi*t/430, n, 1, inf)
f3a = int((Q(-0.3072*10^(-5))*(t - 25)^4 + Q(1.2))*sin((2*n*Pi*t)/860), t, 0, 50)
f3 = symsum(1/430*f3a*n*Pi*t/430, n, 1, inf)
f(t) = f1 + f2 + f3;
However, f2 returns NaN. Tracing in a different package, the computation does look suspicious.
Show 1 older comment
Walter Roberson on 21 Aug 2019
When I check for any given numeric value in a different software package, the result I get is -inf for each of the input values I tried. What is the expected range of t for f(t) ?
Walter Roberson on 21 Aug 2019
When I did some investigation, I could see that in at least one way of writing the formulas you could end up with a partial term of -inf*(A-1) for some formula A. Under the circumstances it was probably reasonable to work out that A-1 would be positive, giving you -inf as a result. But if you expand then you get -inf*A -inf*(-1) and that leads you to -inf + inf for positive A and that is NaN. It is not exactly wrong, but it is a limitation on the analysis.
Taosui Li on 21 Aug 2019
Thank you very much.
I try to do more research on why a negative infinity is returned.
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# Simple Law of Thermodynamics – All four laws
Thermodynamics is a branch of Physics that deals with the thermal state of the gas system and other macroscopic systems. There are four laws of Thermodynamics – Zeroth Law, First Law, Second Law and Third Law of Thermodynamics. Sometimes, most of the students feel difficulties understanding these laws. In this post, I am going to explain all four laws of Thermodynamics in a simple way.
## Simple Law of Thermodynamics
### 1. Zeroth Law of Thermodynamics
The Zeroth Law of Thermodynamics states that if any two thermodynamic systems are in thermal equilibrium with a third thermodynamics system, then these two systems will also be in thermal equilibrium with each other.
Let, three thermodynamic systems A, B and C. If A is in thermal equilibrium with C and B is in thermal equilibrium with C, then according to Zeroth law, A and B will be in thermal equilibrium with each other.
Click here for the detailed post on Zeroth law of Thermodynamics.
### 2. First Law of Thermodynamics
The 1st Law of Thermodynamics states that if some amount of heat is given to a thermodynamic system, it will use some parts of heat to increase its kinetic energy and works with the remaining heat. Here, heat is equivalent to energy.
Let, dQ amount of heat is supplied to a system. Its increase in internal energy is dU and the amount of work done by the system is dW. Then, according to the First law, dQ = dU + dW.
Click here to read the 1st Law of Thermodynamics in detail.
### 3. Second Law of Thermodynamics
There are two statements for the 2nd law of Thermodynamics –
#### Clausius Statement:
According to Clausius’s statement, No self-acting machine can transfer heat from a reservoir of lower temperature to another reservoir of higher temperature.
So, a system needs some external work to transfer heat from a cold body to a hot body.
#### Kelvin-Planck’s Statement:
According to Kelvin and Planck’s statement, No self-acting machine can convert some amount of heat into work entirely.
So, some amount of heat will be lost during the conversion of heat into work.
### 4. Third Law of Thermodynamics
The 3rd law of Thermodynamics states that at absolute zero temperature, the entropy of a perfect crystal is zero. That means there will be no entropy of a perfect crystal at a temperature of zero Kelvin.
These are the four simple laws of Thermodynamics. This is all from this article. If you have any doubt on this topic you can ask me in the comment section.
Thank you!
Related posts:
1. 1st Law of Thermodynamics – in details
### 1 thought on “Simple Law of Thermodynamics – All four laws”
Comments are closed.
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Quick Start for:
Table One
Sample Calculation of Weighted Mean Appraisal Ratio School District A: Single-family Residential
Sales
Sale
Number
Appraisal
Roll Value
Sale Price
Individual
Appraisal
Ratio
1 \$65,834 \$83,113 0.79
2 75,254 90,720 0.83
3 94,420 135,610 0.70
4 99,880 113,310 0.88
5 82,253 109,250 0.75
6 89,654 94,715 0.95
7 76,502 91,680 0.83
8 111,020 128,048 0.87
9 44,441 62,370 0.71
10 64,519 75,905 0.85
11 64,842 81,127 0.80
12 39,479 41,925 0.94
13 193,344 245,700 0.79
14 98,885 127,493 0.78
15 114,788 118,898 0.97
16 92,088 113,645 0.81
17 84,449 84,995 0.99
18 21,090 25,988 0.81
19 22,080 27,398 0.81
Appraisals
Appraisal
Number
Appraisal
Roll Value
Appraisal
Value for
Study
Individual
Appraisal
Ratio
1 \$97,576 \$110,741 0.88
2 60,437 70,964 0.85
3 107,543 148,828 0.72
4 60,264 86,303 0.70
5 69,708 76,117 0.92
6 76,935 98,327 0.78
Total Sales and Appraisals
Total
Appraisal Roll
Value for Test
Total Value
of Sales and
Appraisals
Weighted
Mean
Appraisal
Ratio
\$2,007,285 ÷ \$2,443,170 = 0.8216
Total
Appraisal Roll
Value for Test
Total Value
of Sales and
Appraisals
Weighted
Mean
Appraisal
Ratio
\$27,621,400 ÷ 0.8216 = \$33,619,036
Table Two
Sample Calculation of a Value-Stratified Weighted Mean Appraisal Ratio
(Step 1)
Appraisal
Roll Value
in the sample
Appraisal/
Sale Price
in the sample
Ratio*
in the sample
STRATUM 1: \$-0- to \$2,500
Stratum 1 Total: not sampled not sampled
STRATUM 2: \$1,205,000 and up
1,205,000 1,210,000 0.9959
Stratum 2 Total: 1,209,961 ÷ 1,209,961 = 0.9959
STRATUM 3: \$2,501 to \$15,300
11,243 8,000 1.4054
13,510 10,000 1.3510
14,194 11,500 1.2343
14,800 12,000 1.2333
15,001 13,000 1.1539
Stratum 3 Total: 68,748 ÷ 54,500 = 1.2614
STRATUM 4: \$15,301 to \$47,573
20,374 20,000 1.0187
20,477 20,000 1.0238
20,994 20,000 1.0497
25,806 24,800 1.0405
28,166 27,000 1.0432
Stratum 4 Total: 115,816 ÷ 111,800 = 1.0359
STRATUM 5: \$47,574 to \$110,625
51,007 52,000 0.9809
52,191 52,000 1.0037
53,217 54,000 0.9855
54,141 54,000 1.0026
57,396 57,000 1.0070
Stratum 5 Total: 267,952 ÷ 269,000 = 0.9961
STRATUM 6: \$110,626 to \$465,581
111,648 125,000 0.8932
114,140 135,000 0.8455
139,498 150,000 0.9300
Stratum 6 Total: 365,286 ÷ 410,000 = 0.8909
Grand Totals: \$2,022,802 \$2,055,261
totals based on 19 parcels
Mean Ratio (unweighted average)
average based on 19 parcels
1.0631
Weighted Mean Ratio
\$2,022,802 / \$2,055,261
0.9842
Price-Related Differential
mean ratio 1.0631 / weighted mean ratio 0.9842
1.0802
*Rounded 4 places
Table Three
Sample Calculation of a Value-Stratified Weighted Mean Appraisal Ratio
(Step 2)
Stratum
Number
Number of Parcels
in the
sample
in the
sample
÷ PTD Estimate
in the
sample
= Stratum Ratio
weighted mean ratio
in the
sample*
(#) (nsample) (TXsample) (TYsample) (r1sample)
1 n/a n/a n/a 1.0000
2 1 1,205,000 ÷ 1,210,000 = census
3 5 68,748 ÷ 54,500 = 1.2614
4 5 115,816 ÷ 111,800 = 1.0359
5 5 267,952 ÷ 269,000 = 0.9961
6 3 365,286 ÷ 410,000 = 0.8909
*Rounded 4 places
Table Four
Sample Calculation of a Value-Stratified Weighted Mean Appraisal Ratio
(Steps 3-5)
Stratum
Number
Number of Parcels
in the
Stratum
in the
Stratum
÷ Stratum Ratio
weighted mean ratio
in the sample
= PTD Estimate
in the
Stratum**
(#) (Nstratum) (TXstratum) (r1stratum) (TYstratum)
1 711 300,224 ÷ 1.0000 = 300,224
2 1 1,205,000 ÷ census = 1,209,961
3 259 1,495,515 ÷ 1.2614 = 1,185,570
4 56 1,463,787 ÷ 1.0359 = 1,413,029
5 22 1,500,526 ÷ 0.9961 = 1,506,395
6 7 1,544,658 ÷ 0.8909 = 1,733,737
÷
÷
+
= = =
1,056 7,509,710 1.0219 7,349.099
Total Total (7,509,710 ÷ 7,349,060) Total
Stratum Parcels CAD Value Stratified Ratio PTD Estimate
**Rounded to the nearest dollar
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# [FOM] 254:Pi01 Progress/more
Harvey Friedman friedman at math.ohio-state.edu
Thu Nov 10 04:37:34 EST 2005
```Pi01 INCOMPLETENESS 1
by
Harvey M. Friedman
November 10, 2005
"Beautiful" is a word used by mathematicians with a semi rigorous meaning.
We give an "arguably beautiful" explicitly Pi01 sentence independent of ZFC.
See Proposition A below.
For A containedin [1,n]k, we write A' = [1,n]k\A. This treats [1,n]k as the
ambient space.
We focus on relations R containedin [1,n]3k x [1,n]k. Here we use R in two
ways. Firstly, as a subset of [1,n]4k. Secondly, and most importantly, as an
operator whose arguments and values are subsets of [1,n]k.
In particular, for all A containedin [1,n]k, we define RA = R<A> = R[A3] =
{w in [1,n]k: (therexists x,y,z in A)(R(x,y,z,w))}. For elegance, we will
omit the brackets < > and write RA whenever this does not lead to ambiguity.
Note that in the above, [ ] is the standard forward image notation for R
containedin [1,n]3k x [1,n]k = [1,n]4k.
We say that R is strictly dominating if and only if for all x,y,z,w in
[1,n]k, if R(x,y,z,w) then max(x),max(y),max(z) < max(w). This definition is
the reason why we prefer to write [1,n]3k x [1,n]k rather than [1,n]4k.
We start with the basic unique fixed point theorem for the operator R<A'>.
THEOREM 1.1. For all k,n >= 1 and strictly dominating R containedin ]1,n]3k
x [1,n]k, there exists A containedin [1,n]k such that R<A'> = A.
Furthermore, A containedin [1,n]k is unique.
For A containedin [1,n]k and t >= 1, we write A\t = {x in A: t is not a
coordinate of x} = "A with t omitted".
The "omission" operator R<A'\(8k)!> also has a unique fixed point.
THEOREM 1.2. For all k,n >= 1 and strictly dominating R containedin [1,n]3k
x [1,n]k, there exists A containedin [1,n]k such that R<A'\(8k)!> = A.
Furthermore, A containedin [1,n]k is unique.
However, we can't get a COMMON fixed point
R<A'\(8k)!> = R<A'> = A.
THEOREM 1.3. The following is false. For all k,n >= 1 and strictly
dominating R containedin [1,n]3k x [1,n]k, there exists A containedin [1,n]k
such that R<A'\(8k)!> = R<A'> = A.
We will weaken this three term common fixed point equation so that it is
attainable. The weakening involves replacing = both with containedin and
with "having the same elements of a certain kind".
Our development depends heavily on a very strong regularity condition on R.
We say that R containedin [1,n]3k x [1,n]k is order invariant if and only if
for all x,y in [1,n]4k of the same order type, R(x) iff R(y). The number of
such R is bounded by an exponential expression in k that does not depend on
n.
The powers of t are the numbers t^i, i >= 0.
THEOREM 1.4. For all k,n >= 1 and strictly dominating order invariant R
containedin [1,n]3k x [1,n]k, there exists A containedin [1,n]k such that
the three sets R<A'\(8k)!> containedin R<A'> containedin A contain the same
k tuples of powers of (8k)!+1.
What happens if we apply R to the expressions in the three term tower of
Theorem 1.4?
This is in rough analogy with standard mathematical contexts where one
passes from degree 1 equations to degree 2 equations and higher. Things can
get much more difficult.
Obviously
RR<A'\(8k)!> containedin RR<A'> containedin RA
follows immediately from
R<A'\(8k)!> containedin R<A'> containedin A.
However,
"the first tower of sets above contain the same k tuples of powers of
(8k)!+1"
"the second tower of sets above contain the same k tuples of powers of
(8k)!+1."
PROPOSITION A. For all k,n >= 1 and strictly dominating order invariant R
containedin [1,n]3k x [1,n]k, there exists A containedin [1,n]k such that
the three sets RR<A'\(8k)!> containedin RR<A'> containedin RA contain the
same k tuples of powers of (8k)!+1.
Proposition A is obviously an explicitly Pi01 sentence. It is independent of
ZFC. Here is much more precise information.
Let MAH = ZFC + {there exists a strongly n-Mahlo cardinal}_n. Let MAH+ = ZFC
+ "for all n there exists a strongly n-Mahlo cardinal".
THEOREM 1.5. MAH+ proves Proposition A. However, Proposition A is not
provable in any consistent fragment of MAH that derives Z = Zermelo set
theory. In particular, Proposition A is not provable in ZFC, provided ZFC is
consistent. These facts are provable in RCA0.
THEOREM 1.6. It is provable in ACA that Proposition A is equivalent to
Con(MAH).
*************************************
manuscripts. This is the 254th in a series of self contained numbered
postings to FOM covering a wide range of topics in f.o.m. The list of
previous numbered postings #1-249 can be found at
http://www.cs.nyu.edu/pipermail/fom/2005-June/008999.html in the FOM
archives, 6/15/05, 9:18PM.
250. Extreme Cardinals/Pi01 7/31/05 8:34PM
251. Embedding Axioms 8/1/05 10:40AM
252. Pi01 Revisited 10/25/05 10:35PM
253. Pi01 Progress 10/26/05 6:32AM
Harvey Friedman
```
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What's the cheapest way from left to right?
# Shortest paths and cheapest paths
In many applications one wants to obtain the shortest path from a to b. Depending on the context, the length of the path does not necessarily have to be the length in meter or miles: One can as well look at the cost or duration of a path – therefore looking for the cheapest path.
## Legend
Node Edge with weight 50
## Which graph do you want to execute the algorithm on?
### Modify it to your desire:
• To create a node, make a double-click in the drawing area.
• To create an edge, first click on the output node and then click on the destination node.
• Right-clicking deletes edges and nodes.
### Upload an existing graph:
A occured when reading from file: the contents:
## Legend
Starting node from where distances and shortest paths are computed. "Predecessor edge" that is used by the shortest path to the node.
## Algorithm status
BEGIN
d(v[1]) ← 0
FOR j = 2,..,n DO
d(v[j]) ← ∞
FOR i = 1,..,(|V|-1) DO
FOR ALL (u,v) in E DO
d(v) ← min(d(v), d(u)+l(u,v))
FOR ALL (u,v) in E DO
IF d(v) > d(u) + l(u,v) DO
Message: "Negative Circle"
END
### Variable status:
i d(u) d(v) l(u,v)
## Task is terminated if the tab is changed.
You can open another browser window for reading the description in parallel.
## Legend
Starting node from where distances and shortest paths are computed. "Predecessor edge" that is used by the shortest path to the node.
## Test your knowledge: How does the algorithm decide?
BEGIN
d(v[1]) ← 0
FOR j = 2,..,n DO
d(v[j]) ← ∞
FOR i = 1,..,(|V|-1) DO
FOR ALL (u,v) in E DO
d(v) ← min(d(v), d(u)+l(u,v))
FOR ALL (u,v) in E DO
IF d(v) > d(u) + l(u,v) DO
Message: "Negative Circle"
END
## The task is terminated if the tab is changed.
You can open another browser window for reading the description in parallel.
## Legend
Starting node from where distances and shortest paths are computed. Edge that has already been selected. Edge that has been selected in the previous step.
## What is the optimal ordering of the edges?
The Bellman-Ford Algorithm can compute all distances correctly in only one phase.
To do so, he has to look at the edges in the right sequence. This ordering is not easy to find – calculating it takes the same time as the Bellman-Ford Algorithm itself.
As one can see in the example: The ordering on the left in reasonable, after one phase the algorithm has correctly determined all distances. This is not the case on the right.
In this exercise you can test how many phases the algorithm needs for different sequences of the edges.
## Task is terminated if the tab is changed.
You can open another browser window for reading the description in parallel.
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https://www.toktol.com/notes/context/1670/physics/newton%27s-laws-of-motion/newton%27s-second-law-of-motion-and-impulse
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An impulse of $10 \text{ kg m/s}$ could be the result of a force of $1 \text{ N}$ applied over $10 \text{ s}$ or of a force of $10 \text{ N}$ applied over $1 \text{ s}$.
If mass and force are constant during the period of impact, the impulse is also directly proportional to the force and the change in velocity: $$\vecphy{J}=\Delta \vecphy{p}=\vecphy{F}t$$ This can be shown as follows: \begin{align*} \vecphy{J}&=\Delta\vecphy{p}&\quad\quad \vecphy{F}&=\frac{d\vecphy{p}}{dt}&\\ &=\vecphy{p}_1-\vecphy{p}_0&\quad\quad &=\frac{\Delta\vecphy{p}}{t}&\\ &=m(\vecphy{v}_1-\vecphy{v}_0)\\ &=m\Delta\vecphy{v}\\ \Rightarrow\vecphy{J}&=\Delta\vecphy{p}\\ &=\vecphy{F}t \end{align*}
The magnitude of impulse is given by the area under the graph of force $(\vecphy{F})$ against time $(t)$
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https://socratic.org/questions/how-do-you-solve-the-system-x-2y-13-and-3x-5y-6-using-substitution
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| 537,997,351 | 6,470 |
# How do you solve the system x+2y=13 and 3x-5y=6 using substitution?
Mar 17, 2018
$x = 7$
$y = 3$
#### Explanation:
So you can pick either one of the 2 equations listed to get 1 term alone. In explanation I will be using the second one $\left(3 x - 5 y = 6\right)$
So we have $3 x - 5 y = 6$
First we add $5 y$ to both sides giving us
$3 x = 5 y + 6$
Next we divide $3$ on both sides to get $x$ alone, which gives us
$x = \frac{5}{3} y + \frac{6}{3}$
$x = \frac{5}{3} y + 2$
Now that have $x$ we can plug this into the first equation, which is $\left(x + 2 y = 13\right)$
$\frac{5}{3} y + 2 + 2 y = 13$
Now we subtract the $2$ on both sides giving us
$\frac{5}{3} y + 2 y = 11$
Now we add the $y$ values by finding common denominators
$\frac{5}{3} y + \text{(3)2"/"(3)1} y = 11$
Then add the $y$ values
$\frac{5}{3} y + \frac{6}{3} y = 11$
$\frac{5 y + 6 y}{3} = 11$
$\frac{11 y}{3} = 11$
Now we multiply by $3$ on both sides
$11 y = 33$
Next divide by $11$
$y = 3$
So now that we have the value for $y$ we can find $x$ easily by plugging this into any equation that has $y$. For this part I'm going to use the first equation which was, $x + 2 y = 13$
$y = 3$
$x + 2 y = 13$
$x + 2 \left(3\right) = 13$
$x + 6 = 13$
Subtracting 6 from both sides gives us
$x = 7$
There fore
$x = 7$
$y = 3$
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Verifying trig identities solver
Math can be a challenging subject for many learners. But there is support available in the form of Verifying trig identities solver. We can solve math word problems.
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We can solve quadratic formula by using the quadratic formula calculator. To get the exact formula of a quadratic function, you have to find its roots. The roots are the values at which the function becomes zero or undefined. You can either find the roots by hand or by using a quadratic formula calculator. It is important to note that finding the roots of a quadratic equation is not always easy. Therefore, it is very important to use a quadratic formula calculator whenever possible. This will ensure that you get the most accurate answer possible when solving quadratic equations. You should also keep in mind that finding the roots of a quadratic equation is not always easy. Therefore, it is very important to use a quadratic formula calculator whenever possible.
b is your solution for y and c is your solution for x. To get this answer, multiply your left side (y) by c and add that to your right side (x). This gives you your solution for x by subtracting b from both sides of the equation. This method works best when there are more than two variables in an equation. If the equation has more than two variables, you can use a calculator to simplify the equation and solve for x.
A trig factoring calculator can take care of this for you by quickly calculating the amount of money you would receive if you took out a loan. With a trig factoring calculator, you simply input an amount that you would like to borrow and it will tell you how much money you would receive if you took out the loan. It’s not always easy to understand how to factor a trignometry equation because it requires some math skills. But with a trig factoring calculator, it’s simple to see how much money you would receive if you took out a loan. The first thing that needs to be done is input the principal amount that you want to borrow. Next, input the interest rate and the term of your loan. Finally, press calculate and your result will be displayed.
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# ESL Number Games
Instructor: Elizabeth Hemmons
Beth has taught early childhood education, including students with special needs, for the past 11 years. She has a bachelor's degree in Elementary Education.
Using number words is a very important part of everyday language and communication. In this lesson, we will introduce fun and engaging number games to play with your ESL students.
## ESL Number Games
For ESL (English as a second language) students learning number words can be confusing and boring, but making it fun by using games and hands on activities can make it easier and more engaging. We use number words everyday in communication and numbers are an essential part of the English language. With some of these quick and easy activities, you can reinforce number words and have fun in the classroom!
## Bean Bag Toss
Counting is a helpful way to reinforce number words. You can do this activity in many different ways. You can have the students break into pairs of two or sit the class in a circle. Choose a player to pick a number to begin counting at. Begin the counting by tossing the bean bag and shouting out the number. The next person continues the toss by counting on with the next number. Play continues until the teacher chooses a new player to start over. This is a great way to reinforce counting, and it increases hand/eye coordination.
## Number Concentration
Create two sets of matching number cards and spread them out on the floor, face down and neatly in rows. Sit the students in a large circle around the cards so that everyone can clearly see the numbers. Take turns having students flip over two numbers to get a match. When they flip over the cards, they must read the number word to the class aloud. Play continues until all the matches have been found. This game requires memory and concentration.
## Body numbers
Break the class into groups of 5-6 students. Have students work together to lay on the floor and use their bodies to create a number. Once they are ready, the other team needs to guess the number that the other team has created. Numbers get harder and bigger as the game progresses.
## Wiggle Worm
Write each number on a popsicle stick and draw a green squiggly worm on 10 additional sticks. Put all the sticks in a coffee can with the blank sides up and have the students sit in a circle. One at a time, the students draw a popsicle stick. If they pull a number, they need to shout out the number word. If they pull out a wiggle worm stick, the entire class needs to jump up and wiggle all over. The kids love this activity because it provides a fun opportunity for movement!
## Newspaper Number Hunt
Give each student a page from a newspaper and a highlighter. Have students try to find numbers or number words in the articles and highlight them. When they find the number, they must shout out the number word to reinforce concepts. The winner is the student who finds the most numbers on their page.
## Roll and Race
Give each student a 6-sided die, a large empty hundreds chart and a crayon or marker. After you say go, have your students roll the die, count the dots that they rolled and color the amount of boxes on their die. Encourage students to count the dots on the die and boxes out loud to reinforce counting skills. The rolling and coloring should be quick because the first one to fill up their hundreds chart is the winner!
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# Talk
## Methods to do £6.00 divided by 8?
(11 Posts)
I can do £6 ÷ by 8 via long division (old school ) but my dd hasn't been taught long division so I don't want to introduce her to it just yet. We have some money problems to solve. I'm not good at mental maths. What easy methods would you suggest to help here? My dd is 8. Thanks in advance!
SecretlyChartreuse Thu 14-Apr-16 18:22:20
Divide by two three times?
So £3
Then £1.50
Then 75p
Snowberry86 Thu 14-Apr-16 18:23:13
Halve it, then halve it again, then halve it again!
£6/2=£3
£3/2=£1.50
£1.50/2=£0.75
My first thought was to divide by 4 then 2 but I have no idea how maths is taught nowadays.
For the halve method, how is that connected to 8? Sorry to be dim. The half method I can see is the easiest way to do this but I'm not connecting the dots as it how it is divisible by 8.
contortionist Thu 14-Apr-16 18:35:05
Alternatively, 6/8 = 3/4, so it's three-quarters of a pound, which is 75p (possibly easier for Americans!). I prefer the halving-three-times method though.
contortionist Thu 14-Apr-16 18:36:14
8 = 2 * 2 * 2, so multiplying by eight is the same as doubling three times. And to undo it, halve three times.
AngieBolen Thu 14-Apr-16 18:37:33
As other posters, I divided by 4, then halved my answer. That's how I would do it in everyday life. I have no idea how to do ling division.
Thank you contortionist (and everyone else!). It's so simple really! I will remember thexactly halving method in the future! !
irvineoneohone Fri 15-Apr-16 17:14:38
Is this primary?
If she wasn't taught long division yet, I would do £6 = 600 P
so think about multiple of 8 = 70 X 8 = 560p 600p - 560p = 40p
40 p left, again 5 x 8 = 40p so 70p + 5p = 75p
I wondered the child who hasn't learned long division can figure out 2 x 2 x 2 = 8.
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Is the integer n divisible by 20?
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Is the integer n divisible by 20?
(1) n is divisible by 5
(2) n is divisible by 6
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Re: Is the integer n divisible by 20? [#permalink]
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01 Nov 2018, 08:14
gmatbusters wrote:
Is the integer n divisible by 20?
(1) n is divisible by 5
(2) n is divisible by 6
Prime factorization 20 = 2^2*5, thus n should be divisible by both 2^2 and 5
Combined it tells us that n is divisible by 2*3*5=30 or a multiple of 30.
Thus n can be divisible by 30 or 60
Insufficient
E
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Is the integer n divisible by 20? [#permalink]
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01 Nov 2018, 08:24
gmatbusters wrote:
Is the integer n divisible by 20?
(1) n is divisible by 5
(2) n is divisible by 6
n/20 = integer
it means n is 0,20,40,60 etc..
Statement 1) n is divisible by 5 so it means n is 0, 5, 10, 15, 20,25,30 ..... 60
if n = 20 then yes if n = 5 then no.
Insufficient.
Statement 2) n is divisible by 6 so it means n is 0,6,12,18,24,30.... 60
if n = 0 then 0/20 = 0 yes divisible.
if n = 6 no.
Combined we get n can be 0,30,60 etc..
So insufficient.
E
Is the integer n divisible by 20? [#permalink] 01 Nov 2018, 08:24
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# Finite Element Method Questions and Answers – Library of Elements and Interpolation Functions – 1
This set of Finite Element Method Multiple Choice Questions & Answers (MCQs) focuses on “Library of Elements and Interpolation Functions – 1”.
1. Which option is not correct about the three-noded triangular plane stress (linear) element used in FEM?
a) It has six degrees of freedom
b) It belongs to both the isoparametric and superparametric element families
c) It can be improved by the addition of internal degrees of freedom
d) Delaunay triangulation can be used for its mesh generation
View Answer
Answer: c
Explanation: The three-node triangular element with linear displacements for the plane stress problem is simply called a linear triangle. It has six degrees of freedom and it belongs to both the isoparametric and superparametric element families. A mesh of linear trianglecan be easily generated using Delaunay triangulation, but the element cannot be improved by the addition of internal degrees of freedom; rather, it can be improved by increasing the number of nodes.
2. In FEM, which option is used to develop the Higher-order triangular elements (i.e. triangular elements with interpolation functions of higher degree) systematically?
a) Pascal’s triangle
b) Galerkin method
c) Jacobi method
d) Delaunaytriangulation
View Answer
Answer: a
Explanation: The Higher-order triangular elements (also the Lagrange family of triangular elements) can be systematically developed with the help of Pascal’s triangle. Finite element equations are obtained using the Galerkin method. Jacobi is used for eigenvalue problems. The Delaunay method is used to generate mesh for triangular elements.
3. In FEM, What is the number of displacement polynomials necessary for finding displacements in a linear triangular element?
a) 1
b) 2
c) 3
d) 4
View Answer
Answer: b
Explanation: The number of displacement polynomials for an element is equal to the degrees of freedom of each node of the element. A linear triangular element has three nodes and two degrees of freedom at each node. Thus, the total number of displacement polynomials necessary for finding displacements is two.
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4. Concerning triangular elements in FEM, which option is not correct about the mathematical formula of Pascal’s triangle?
a) It contains the terms in two coordinates only
b) The position of the terms can be viewed as the nodes of a triangular element
c) The position of the first and last terms of a row is at the vertices of a triangular element
d) A triangular element of order 2 corresponds to the second row
View Answer
Answer: d
Explanation: A Pascal’s triangle contains the terms of polynomials of various degrees in two coordinates. We can view the positions of the terms as nodes of a triangular element, with the constant term and the first and last terms of a given row being the vertices of the triangle. A triangular element of order 2 (i.e., the degree of the polynomial is 2) contains six nodes and corresponds to the third row of Pascal’s triangle.
5. Which option is not correct about the four-noded rectangular plane stress element used in FEM?
a) It has eight degrees of freedom
b) Shape functions N1, N2, N3 and N4 are bilinear functions of x and y
c) The displacement field is continuous across elements
d) Its Delaunay triangulation is unique
View Answer
Answer: d
Explanation: The four-node quadrilateral element with linear displacements for a plane stress problem has two degrees of freedom at each node. The total degrees of freedom of the element is eight. The displacement field is continuous across elements connected at nodes and the shape functions N1, N2, N3 and N4 are bilinear functions of x and y. Its Delaunay triangulationis not unique, but it has two solutions.
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6. In FEM, if x, y represents Cartesian coordinates, then the following triangular array of binomial coefficients forms Pascal’s triangle.
$$\begin{matrix} & & & 1 & & &\\ & & x & & y & & \\ & x^2 & & xy & & y^2 & \\ x^3 & & x^2y & & xy^2 & & y^3 \\ \end{matrix}$$
a) True
b) False
View Answer
Answer: a
Explanation: In mathematics, Pascal’s triangle is a triangular array of binomial coefficients. It contains the terms of polynomials of various degrees in two coordinates x and y. An nth row of Pascal’s triangle contains n term(s), and it corresponds to all triangular elements with at least 0.5n(n+1) number of nodes. The 1st row contains number one only.
7. In the FEM element library, what is the other name of a higher-order element?
a) Complex element
b) Simplex element
c) Linear element
d) Nonlinear element
View Answer
Answer: a
Explanation: Simplex and linear elements contain nodes only at endpoints but not at the interior. They have linear polynomials as interpolation functions. Higher-order elements can be created easily from simplex elements by adding additional intermediate nodes to each element. They are also called complex elements.
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8. In FEM, if x, y represents Cartesian coordinates, then which term of Pascal’s triangle corresponds to the position of the interior node of the following element?
a) 1
b) xy
c) x2y
d) xy2
View Answer
Answer: b
Explanation: The interior node of the ten noded triangular elements can be viewed as the middle term of the third row of Pascal’s triangle. Since the 3rd row contains three terms viz. x2, xy and y2 in the same order, the interior node corresponds to the term xy. An nth row of Pascal’s triangle contains n term(s) and it corresponds to all triangular elements with at least 0.5n(n+1) number of nodes.
9. In the FEM element library, an eight noded quadrilateral element belongs to which family?
a) Serendipity
b) Linear
c) Simplex
d) Quadratic
View Answer
Answer: a
Explanation: The Serendipity elements are the rectangular elements with intermediate nodes but no interior nodes, i.e., all nodes lie on boundary. Since four nodes of an eight noded quadrilateral element are intermediate nodes, it belongs to the Serendipity family. Simplex and linear elements contain nodes only at endpoints but not at intermediate points. They have linear polynomials as interpolation functions. A quadratic element contains interior nodes.
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10. In FEM, which option is not correct about the Lagrange family of triangular elements?
a) The nodes are uniformly spaced
b) Pascal’s triangle can be viewed as a triangular element
c) Dependent variables and their derivatives are continuous at inter-element boundaries
d) 2nddegree polynomial corresponds to 6 noded triangle
View Answer
Answer: c
Explanation: In Lagrange family elements the nodes are regularly placed everywhere on the grid i.e., they are uniformly spaced. The location of the terms in Pascal’s triangle gives the location of nodes in elements. Thus, Pascal’s triangle can be viewed as a triangular element. The derivatives of dependent variables are not continuous at inter-element boundaries. 2nd-degree polynomial corresponds to 6 noded triangles.
11. What is the displacement function for one-dimensional, two noded linear elements in terms of its shape functions N1 and N2?
a) N1u1+N2u2
b) N1u2+N2u1
c) N1u1-N2u2
d) N1u2-N2u1
View Answer
Answer: a
Explanation: For a linear element, the displacement function is a linear polynomial of nodal displacements. A one-dimensional, two noded linear elements have two nodes with corresponding displacements u1, u2 and corresponding shape functions N1, N2. The displacement function is given by N1u1+N2u2.
12. For the following element, what is the value of the distance variable s at the 1st row?
a) 1
b) 2
c) 0.5
d) 0
View Answer
Answer: c
Explanation: The value of the distance variable s is $$\frac{p}{k-1}$$; where p is the row at which s is calculated, k is the number of uniformly spaced nodes per side of the element. For a quadratic element, we have k = 3.
s=$$\frac{p}{3-1}$$
=$$\frac{p}{2}$$
At first row, p=1
s=$$\frac{1}{2}$$
=0.5.
13. For the following element, if s is the distance variable as shown, then its value at the 2nd row is $$\frac{1}{3}$$.
a) True
b) False
View Answer
Answer: b
Explanation: The value of the distance variables s is $$\frac{p}{k-1}$$; where p is the row at which s is calculated, k is the number of uniformly spaced nodes per side of the element. For a given element, we have k = 4.
s=$$\frac{p}{4-1}$$
=$$\frac{p}{3}$$
At second row, p=2
s=$$\frac{2}{3}$$.
14. In the FEM element library, what is the correct name for a six noded triangular element?
a) Linear strain triangular element
b) Constant strain triangular element
c) Variable strain triangular element
d) Higher-order triangular element
View Answer
Answer: a
Explanation: A Constant strain triangular (CST) element is the simplest triangular element with three end nodes. A Linear strain triangular element (LST) is a six-noded triangular element with three intermediate nodes in addition to three end nodes. For plane stress applications, LST gives an accurate result compare to the three-noded CST element. The variable strain triangular element is a higher-order triangular element with more than six nodes.
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# New Age Math
posted by .
Suppose that y≤5x,3x≤y and 14x+15y≤1 together with 0≤x, 0≤y.
The maximum value of the function y+x on the resulting region occurs at x=? and y=?
The maximum value of the function is ?
• New Age Math -
Plot the lines on an x,y graph
y=5x and
y=-14/15 x+1/15
x=0, y=0
Now you know any possible x,y will be inside or on the boundries of that region bounded by those four lines.
There is a nice theorem in Linear algebra, that says the maximum (or miniumum) of any objective function will be on the boundry lines,somewhere.
So do this: look at the coordinates where the lines intersect. Test your objective function Obj=x+y and see where it is maximum, you will quickly see out of those four intersections, you will find the maximum.
## Similar Questions
1. ### Math
Suppose that y≤3x, 2x≤y and 13x+13y≤1 together with 0≤x, 0≤y. Determine a value of k so that the function f(x,y)=kx=y has a positive maximum value on the region at two corners. …
2. ### math
Maximize P = 4x + 20y subject to these constraints: 2x + 15y ≤ 700 5x + 10y ≤ 1150 5x + 5y ≤ 1000 4x + 15y ≤ 980 x ≥ 0, y ≥ 0 Maximum value for P = ?
3. ### math
Maximize P = 4x + 20y subject to these constraints: 2x + 15y ≤ 700 5x + 10y ≤ 1150 5x + 5y ≤ 1000 4x + 15y ≤ 980 x ≥ 0, y ≥ 0 Maximum value for P = ?
4. ### Math
Suppose that y≤5x,3x≤y and 14x+15y≤1 together with 0≤x, 0≤y. The maximum value of the function y+x on the resulting region occurs at x=?
5. ### New Age Math
Consider the region defined by the inequalities: 0≤x 0≤y -21/4≤y-7/4x y+x≤7 4≤y-3/7x Pick up those points which are the boundary points of the resulting convex region.
6. ### Algebra 2
Solve. 4a-2≤a+1≤3a+4 4a-2-1≤a+1-1≤3a+4-1 4a-3-3≤a≤3a+3-3 4a/3-6/3≤a≤3a/3 4a/3(1/a)-2≤a≤a(1/a) 4/3-6≤a≤1 I got lost at this part. I'm not sure what to do now, if …
7. ### algebra 1 help please
4) a student score is 83 and 91 on her first two quizzes. write and solve a compound inequality to find possible values for a thord quiz score that would give anverage between 85 and 90. a. 85≤83+91+n/3 ≤90; 81≤n≤96 …
8. ### PRE - CALCULUS
Eliminate the parameter t. Find a rectangular equation for the plane curve defined by the parametric equations. x = 6 cos t, y = 6 sin t; 0 ≤ t ≤ 2π A. x2 - y2 = 6; -6 ≤ x ≤ 6 B. x2 - y2 = 36; -6 ≤ …
9. ### Differentials (calc)
Solve the Poisson equation ∇^2u = sin(πx) for 0 ≤ x ≤ 1and 0 ≤ y ≤ 1 with boundary conditions u(x, 0) = x for 0 ≤ x ≤ 1/2, u(x, 0) = 1 − x for 1/2 ≤ x ≤ 1 and 0 everywhere …
10. ### Calculus
f is a continuous function with a domain [−3, 9] such that f(x)= 3 , -3 ≤ x < 0 -x+3 , 0 ≤ x ≤ 6 -3 , 6 < x ≤ 9 and let g(x)= ∫ f(t) dt where a=-2 b=x On what interval is g increasing?
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You are Here: Home >< Maths
# Hit me with a difficult P3 or P1 question watch
1. (Original post by Jonny W)
P3 students should know:
- how to differentiate arcsin, arccos and arctan;
- how to integrate x/sqrt(1 - x^2) and x/(1 + x^2);
- how to integrate ln(x) by parts.
Hence they could reasonably be asked to integrate arcsin, arccos and arctan. That said, those problems are probably too hard for a P3 exam.
woah, which exam board is this? That seems so harsh.
On MEI, for P3 you need to be able to do integration by parts and substitution. Then calculus on trig functions ie sin, cos, tan and powers of them
2. (Original post by Jonny W)
The Edexcel P3 specification (http://www.edexcel.org.uk/VirtualContent/25626.pdf) includes "the use of dx/dy = 1 / (dy/dx)".
y = sin(x)
x = arcsin(y)
dx/dy = 1 / (dy/dx) = 1 / cos(x) = 1 / sqrt(1 - x^2).
Yeah, but it doesn't say anything differentiating about inverse functions. That's listed under the P5 specification.
3. (Original post by Katie Heskins)
woah, which exam board is this? That seems so harsh.
Integrating inverse trig functions is probably too hard for P3 exams, but OK for homework.
4. (Original post by Jonny W)
Integrating inverse trig functions is probably too hard for P3 exams, but OK for homework.
It would not be in the exam. You are right that the calculus required to deal with inverse trig functions is available in P3, but it is not part of the syllabus.
5. Integrate (4 + x^4)^-1 dx
Hint: x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x +2)
6. (Original post by fishpaste)
Integrate (4 + x^4)^-1 dx
ln|4 + x^4| / 4x³?
7. (Original post by shift3)
ln|4 + x^4| / 4x³?
No. It looks like you used a substitution of u = 4 + x^4 and then didn't integrate the 4x^3 which resulted.
8. (Original post by shift3)
ln|4 + x^4| / 4x³?
no.
9. (Original post by IntegralAnomaly)
no.
further clues ?
substitution?
10. (Original post by kikzen)
further clues ?
substitution?
work on fishpaste's hint (x^4+4)=(x^2 + 2x + 2)(x^2 - 2x +2).
So expess (x^4+4)^-1 in the form of (Ax+B)/(x^2+2x+2) + (Cx+D)/(x^2-2x+2).(and remember the differential of arctanx)
11. (Original post by kikzen)
further clues ?
substitution?
How might you use the clue given up there?
12. How about 4x + (1/5)x^5?
13. (Original post by IntegralAnomaly)
work on fishpaste's hint (x^4+4)=(x^2 + 2x + 2)(x^2 - 2x +2).
So expess (x^4+4)^-1 in the form of (Ax+B)/(x^2+2x+2) + (Cx+D)/(x^2-2x+2).(and remember the differential of arctanx)
Woah.
14. (Original post by fishpaste)
Integrate (4 + x^4)^-1 dx
Hint: x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x +2)
I get this nasty thing:
[ ln(x^2 + 2x + 2) - ln(x^2 - 2x + 2) + 2arctan(x+1) + 2arctan(x-1) ]/16 + C
15. (Original post by mikesgt2)
I get this nasty thing:
[ ln(x^2 + 2x + 2) - ln(x^2 - 2x + 2) + 2arctan(x+1) + 2arctan(x-1) ]/16 + C
thats what i got to.
16. (Original post by IntegralAnomaly)
thats what i got to.
Its a bit of a slog, dealing with a lot of partial fractions and algebra.
17. Some difficult P1, as I can't ask questions I don't understand!
1. A curve has the equation:
x = 4y^2 - 9
Find the area between the curve and the line x = 1
Find the minimum value of x, and the value of y that gives this.
When y = 17, and c is set to 9, find the equations of the normal to the curve and find where it meets the x axis.
18. Much respect to you for getting that far. Well done.
19. I have NEVER answered a question about arctanx or arccosx or all of those. I swear to god if those come up i am buggered.
20. (Original post by *dave*)
I have NEVER answered a question about arctanx or arccosx or all of those. I swear to god if those come up i am buggered.
calculus involving inverse trig is p5 for edexcel.
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+0
0
105
2
Real numbers x and y satisfy
\begin{align*} x + xy^2 &= 250y, \\ x - xy^2 &= -240y. \end{align*}
Enter all possible values of x separated by commas.
Mar 30, 2021
#1
0
Elimnating y, we get x^2 = 1225. So the possible values of x are 35 and -35.
Mar 30, 2021
#2
+121004
+1
x + xy^2 = 250y
x - xy^2 = - 240y add these
2x = 10y
x = (10/2) y
x = 5y sub this back into the first equation
5y + 5yy^2 = 250y divide though by 5y
1 + y^2 = 50
y^2 = 49 take both roots
y = -7 or y = 7
x = 5(-7) = -35 x = 5(7) = 35
And (x, y) = (0,0) is a trivial solution
So
x = 35, 0 , 35
Mar 30, 2021
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Maths
Find the sum of the following series 8+12+18+as far the 6th term
1. 👍 0
2. 👎 0
3. 👁 218
1. 8+12+18+26+36+48
1. 👍 0
2. 👎 0
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# Product Of Slopes Of X,y Axes=0
• extreme_machinations
In summary, the conversation discussed the slopes of different lines, their product, and the concept of perpendicular lines. It also touched on the idea of infinity and its relationship with zero. The conversation ended with a disagreement about the use of the word "conjecture."
extreme_machinations
This Is Strange !
Slope Of The Line X= 0[y- Axis] Is =+1
Slope Of The Line Y=0 [x-axis] Is= 0
Product Of The Slopes = 0
Now Product Of Two Perpendicular Lines Should Be = - 1
It's funny how both the y-axis and the line y=x have the same gradient according to you
well ,
its even funnier how lack of reading skills can hamper a persons understanding .
if u pay attention sire ,u'll find that i said
that the line y=0 ,which you would agree is the x - axis and has a slope = 0
extreme_machinations said:
well ,
its even funnier how lack of reading skills can hamper a persons understanding .
if u pay attention sire ,u'll find that i said
that the line y=0 ,which you would agree is the x - axis and has a slope = 0
I don't disagree with you, but do you know what the gradient of the line x=0 (the y-axis) is?
If you work that out you will understand where you went wrong. Please don't be so offended I was just lightly trying to point out your mistake.
This reminds me of the "proof" that 1 = 2. The mistake involved dividing by zero or something. ;)
extreme_machinations said:
well ,
its even funnier how lack of reading skills can hamper a persons understanding .
if u pay attention sire ,u'll find that i said
that the line y=0 ,which you would agree is the x - axis and has a slope = 0
Good advice. You should take it yourself. Zurtex did not say anything about "the line y=0, which you would agree is the x-axis". He specifically referred to the "y-axis" and his point was that its slope is not 1!
The slope of the y-axis (x=0) is infinite. And, as we all know, infinity times zero equals negative 1. :)
oopsey !
tan 45 syndrome !
sorry for that ,
but i thought anything multiplied by 0 is = 0
infinity multiplied by zero being -1 is brand new information to me ,can anyone tell me how this so ??
extreme_machinations said:
oopsey !
tan 45 syndrome !
sorry for that ,
but i thought anything multiplied by 0 is = 0
infinity multiplied by zero being -1 is brand new information to me ,can anyone tell me how this so ??
Infinity multiplied by 0 is not -1, heck it doesn't even make sense. But what you can do is work out the limit of the 2 sides when multiplied by each other, which goes something like this:
$$\lim_{x \rightarrow \infty} -x \frac{1}{x} = -1$$
Now although it's true that:
$$\lim_{x \rightarrow \infty} -x = -\infty$$
And:
$$\lim_{x \rightarrow \infty} \frac{1}{x} = 0$$
It only makes sense to say:
$$\lim_{x \rightarrow \infty} f(x) g(x) = \left( \lim_{x \rightarrow \infty} f(x) \right) \left( \lim_{x \rightarrow \infty} g(x) \right)$$
If f(x) and g(x) have limits as x approaches infinity. And as we see above -x has no limit as x approaches infinity.
Consider the following two lines:
Line A: y = -nx
Line B: y = x/n
The gradients of these two lines are:
Because any two lines which are perpendicular have gradients which multiply to give -1, we see immediately that line A is perpendicular to line B, for any given value of n.
Now, consider what happens in the limit as n goes to infinity. For line B we have:
$$\lim_{n \rightarrow \infty} y = \lim_{n \rightarrow \infty} x/n = 0$$
So, line B becomes the line y = 0.
We can write line A as:
x = -y/n
Taking the limit as n goes to infinity, we see that this line becomes the line x = 0.
Since we have taken the same limit in both cases, the lines A and B have remained perpendicular, and their gradients must still multiply to give -1. What we have, n terms of the gradients, is:
$$\lim_{n \rightarrow \infty} (-n)(1/n) = \lim_{n \rightarrow \infty} -1 = -1$$
Anukriti C.
hey that's great conjecture !
thanks
extreme_machinations said:
hey that's great conjecture !
thanks
What "conjecture" are you talking about? I didn't see any conjecture in this.
I Was Just Using It In The General Sense ,not Strictly In The Mathematical Sense .
Please Ignore Whatever Does'nt Make Sense To You .
What general sense then? I thought I knew what "conjecture" meant, even in general- and I don't see how it applies. Enlighten me.
ok ,you win pal !
im not going to argue .
tell me what it was .
peace out ,
## 1) What does it mean for the product of slopes of x,y axes to equal 0?
When the product of slopes of x and y axes equals 0, it means that either the x-axis or the y-axis has a slope of 0. In other words, the line or plane is parallel to one of the axes, making the slope of that axis 0.
## 2) Can the product of slopes of x,y axes ever be greater than 0?
No, the product of slopes of x and y axes can never be greater than 0. This is because if one of the axes has a slope of 0, then the product will always be 0, regardless of the slope of the other axis.
## 3) How does the product of slopes of x,y axes relate to the slope of a line?
The product of slopes of x and y axes is related to the slope of a line by the fact that it is equal to the slope of the line. This is because the slope of a line is the ratio of the vertical change (change in y) to the horizontal change (change in x), and this can also be expressed as the product of the slopes of the x and y axes.
## 4) What is the significance of the product of slopes of x,y axes in mathematics?
The product of slopes of x and y axes is significant in mathematics because it helps to determine the orientation of a line or plane. It also helps to determine if a line or plane is parallel to one of the axes, as well as the slope of the line or plane.
## 5) How can the product of slopes of x,y axes be used in real-world applications?
The product of slopes of x and y axes can be used in real-world applications, such as engineering and physics, to determine the orientation and slope of objects and structures. It can also be used to calculate the angle of inclination of a line or plane, which is useful in fields like architecture and construction.
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Can You Cross Like A Boss?
Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
## Riddler Express
It’s the second week in our four weeks of CrossProduct™ puzzles!
This time around, there are six three-digit numbers — each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, are shown in the bottom row.
210
144
54
135
4
49
6,615 15,552 420
Can you find all six three-digit numbers and complete the table?
The solution to this Riddler Express can be found in the following column.
## Riddler Classic
From James Bach comes a doozy on doodles:
James likes to draw doodles in the shapes of different polygons. He especially likes to doodle self-intersecting polygons, where the sides cross over each other. (These are distinct from the simple polygons you might have learned about in school, whose sides do not intersect each other.)
The other day, James was able to draw a self-intersecting polygon, each of whose sides intersected with exactly two other sides. Not only that, he drew a polygon with the fewest possible sides that met these criteria.
What was it, you ask? It was a pentagram, which has just five sides.
Lovely. But this got James really thinking — can you draw a polygon where each side intersects exactly three other sides? And if so, what is the minimum number of sides this polygon can have?
The solution to this Riddler Classic can be found in the following column.
## Solution to last week’s Riddler Express
Congratulations to 👏 Greg 👏 of Brooklyn, New York, winner of last week’s Riddler Express.
Last week you had to solve a CrossProduct™ with five three-digit numbers. The products of the three digits of each number were shown in the rightmost column below, while the products of the digits in the hundreds, tens and ones places were shown in the bottom row.
135
45
64
280
70
3,000 3,969 640
Once again, a good way to start was to write out the possible ways each row’s product could be the product of three digits:
• 135 was 3×5×9.
• 45 was 1×5×9 or 3×3×5.
• 64 was 1×8×8, 2×4×8 or 4×4×4.
• 280 was 5×7×8.
• 70 was 2×5×7.
Next, many solvers worked out the prime factorization of the three columns’ products, so they could figure out which digit went into which column.
• 3,000 was 23×3×53.
• 3,969 was 34×72.
• 640 was 27×5.
First off, the middle column had an odd product, which meant every middle digit had to be odd. This was only possible if the number in the third row was 818.
Meanwhile, only the middle column had 7s in its prime factorization, which meant the middle digits in the fourth and fifth rows were both 7. As for the first column, since the 8 in the third row accounted for all three factors of 2, that meant the first digits in the fourth and fifth rows had to be odd as well. The number in the fourth row was therefore 578, while the number in the fifth row was 572.
In the middle column of the first and second rows, you still somehow needed to account for four factors of 3. The only way this was possible was if both of these middle digits were 9. To make the remaining products work, that meant the number in the first row had to be 395, while the number in the second row was 591.
There was certainly more than one way to solve this puzzle. For another approach, check out Andrew Heairet’s animated solution:
It looks like this will be the next Twitch sensation!
## Solution to last week’s Riddler Classic
Congratulations to 👏 Stephen Berg 👏 of Troy, New York, winner of last week’s Riddler Classic.
Last week, Cassius the ape was trying to solve Lucas’ Tower puzzle with three disks, all of which start on the same pole. The disks had different diameters, with the biggest disk at the bottom and the smallest disk on top. The goal was to move all three disks from one pole to any other pole, one at a time — but at no point could a larger disk sit atop a smaller disk.
The minimum number of moves to solve the puzzle was not in question — that result is well known: N disks can be solved in exactly 2N−1 moves.
It turned out that Cassius couldn’t care less about actually solving the puzzle, but he was very good at following directions and understood that a larger disk could never sit atop a smaller disk. With each move, he randomly chose from among the set of valid moves.
On average, how many moves would it take for Cassius to solve this puzzle with three disks?
Given the complexity of this problem, several solvers, like Kenneth W and Kei Nishimura-Gasparian, turned to their computers for assistance. After 1 million simulations, Kenneth found that it took Cassius approximately 70.7 moves on average to solve the puzzle.
Many solvers found an exact answer. One way to do this was to map out all 27 possible states of the disks and poles, as Dean Ballard did:
Based on Dean’s diagram, the question then became: If you started at A and randomly walked your way back and forth between adjacent states, what was the average number of steps it took to reach either Z1 or Z2?
“Random walk” problems like these are commonly solved using systems of equations or Markov chains, which was what solver Michael Ringel did:
Either way, you found that the average number of moves to go from A to Z1 or Z2 was 637/9, or about 70.78 — very close to Kenneth’s numerical approximation. If you alternatively interpreted the problem as going from A to Z1 and only Z1 (or to Z2 and only Z2) — an answer I accepted — you had double the result, or 1274/9, due to the bilateral symmetry in the state diagram.
For extra credit, you had to find the average number of moves in the general case of N disks and three poles. The challenge here was that the number of possible states scaled exponentially with N, which made this rather difficult.
Puzzle submitter Toby Berger, along with coauthor Max A. Alekseyev, solved this exact problem in a 2014 paper. It may have been difficult to see from Dean’s or Michael’s drawings, but there was a recursive nature to the diagram of states. Here’s an illustration from Toby’s paper that shows this nature a little more clearly when there were one, two and three disks:
By leveraging the recursive nature of these states, the authors were able to show the general formula for the average number of steps to randomly move the disks from one pole to another: (3N−1)(5N−3N)/(2·3N−1). The average number of steps to randomly move the disks to either of the other poles was then half of that, or (3N−1)(5N−3N)/(4·3N−1).
Finally, Eric Thompson-Martin further studied the shape of the probability distribution for the number of steps it took Cassius to solve the puzzle. According to Eric, this distribution appeared to be a log-normal curve as the number of disks increased.
In any case, I hope no one blew a (Sierpiński) gasket solving this!
## Want more riddles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
## Want to submit a riddle?
Email Zach Wissner-Gross at [email protected].
## Footnotes
1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!
Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.
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Paul's Online Math Notes
[Notes]
Calculus III - Notes
Internet Explorer 10 & 11 Users : If you have been using Internet Explorer 10 or 11 to view the site (or did at one point anyway) then you know that the equations were not properly placed on the pages unless you put IE into "Compatibility Mode". I believe that I have partially figured out a way around that and have implimented the "fix" in the Algebra notes (not the practice/assignment problems yet). It's not perfect as some equations that are "inline" (i.e. equations that are in sentences as opposed to those on lines by themselves) are now shifted upwards or downwards slightly but it is better than it was.
If you wish to test this out please make sure the IE is not in Compatibility Mode and give it a test run in the Algebra notes. If you run into any problems please let me know. If things go well over the next week or two then I'll push the fix the full site. I'll also continue to see if I can get the inline equations to display properly.
3-Dimensional Space Previous Chapter Next Chapter Applications of Partial Derivatives Partial Derivatives (Introduction) Previous Section Next Section Partial Derivatives
## Limits
In this section we will take a look at limits involving functions of more than one variable. In fact, we will concentrate mostly on limits of functions of two variables, but the ideas can be extended out to functions with more than two variables.
Before getting into this let’s briefly recall how limits of functions of one variable work. We say that,
provided,
Also, recall that,
is a right hand limit and requires us to only look at values of x that are greater than a. Likewise,
is a left hand limit and requires us to only look at values of x that are less than a.
In other words, we will have provided approaches L as we move in towards (without letting ) from both sides.
Now, notice that in this case there are only two paths that we can take as we move in towards . We can either move in from the left or we can move in from the right. Then in order for the limit of a function of one variable to exist the function must be approaching the same value as we take each of these paths in towards .
With functions of two variables we will have to do something similar, except this time there is (potentially) going to be a lot more work involved. Let’s first address the notation and get a feel for just what we’re going to be asking for in these kinds of limits.
We will be asking to take the limit of the function as x approaches a and as y approaches b. This can be written in several ways. Here are a couple of the more standard notations.
We will use the second notation more often than not in this course. The second notation is also a little more helpful in illustrating what we are really doing here when we are taking a limit. In taking a limit of a function of two variables we are really asking what the value of is doing as we move the point in closer and closer to the point without actually letting it be .
Just like with limits of functions of one variable, in order for this limit to exist, the function must be approaching the same value regardless of the path that we take as we move in towards . The problem that we are immediately faced with is that there are literally an infinite number of paths that we can take as we move in towards . Here are a few examples of paths that we could take.
We put in a couple of straight line paths as well as a couple of “stranger” paths that aren’t straight line paths. Also, we only included 6 paths here and as you can see simply by varying the slope of the straight line paths there are an infinite number of these and then we would need to consider paths that aren’t straight line paths.
In other words, to show that a limit exists we would technically need to check an infinite number of paths and verify that the function is approaching the same value regardless of the path we are using to approach the point.
Luckily for us however we can use one of the main ideas from Calculus I limits to help us take limits here.
Definition
A function is continuous at the point if,
From a graphical standpoint this definition means the same thing as it did when we first saw continuity in Calculus I. A function will be continuous at a point if the graph doesn’t have any holes or breaks at that point.
How can this help us take limits? Well, just as in Calculus I, if you know that a function is continuous at then you also know that
must be true. So, if we know that a function is continuous at a point then all we need to do to take the limit of the function at that point is to plug the point into the function.
All the standard functions that we know to be continuous are still continuous even if we are plugging in more than one variable now. We just need to watch out for division by zero, square roots of negative numbers, logarithms of zero or negative numbers, etc.
Note that the idea about paths is one that we shouldn’t forget since it is a nice way to determine if a limit doesn’t exist. If we can find two paths upon which the function approaches different values as we get near the point then we will know that the limit doesn’t exist.
Let’s take a look at a couple of examples.
Example 1 Determine if the following limits exist or not. If they do exist give the value of the limit. (a) [Solution] (b) [Solution] (c) [Solution] (d) [Solution] Solution (a) Okay, in this case the function is continuous at the point in question and so all we need to do is plug in the values and we’re done. (b) In this case the function will not be continuous along the line since we will get division by zero when this is true. However, for this problem that is not something that we will need to worry about since the point that we are taking the limit at isn’t on this line. Therefore, all that we need to do is plug in the point since the function is continuous at this point. (c) Now, in this case the function is not continuous at the point in question and so we can’t just plug in the point. So, since the function is not continuous at the point there is at least a chance that the limit doesn’t exist. If we could find two different paths to approach the point that gave different values for the limit then we would know that the limit didn’t exist. Two of the more common paths to check are the x and y-axis so let’s try those. Before actually doing this we need to address just what exactly do we mean when we say that we are going to approach a point along a path. When we approach a point along a path we will do this by either fixing x or y or by relating x and y through some function. In this way we can reduce the limit to just a limit involving a single variable which we know how to do from Calculus I. So, let’s see what happens along the x-axis. If we are going to approach along the x-axis we can take advantage of the fact that that along the x-axis we know that . This means that, along the x-axis, we will plug in into the function and then take the limit as x approaches zero. So, along the x-axis the function will approach zero as we move in towards the origin. Now, let’s try the y-axis. Along this axis we have and so the limit becomes, So, the same limit along two paths. Don’t misread this. This does NOT say that the limit exists and has a value of zero. This only means that the limit happens to have the same value along two paths. Let’s take a look at a third fairly common path to take a look at. In this case we’ll move in towards the origin along the path . This is what we meant previously about relating x and y through a function. To do this we will replace all the y’s with x’s and then let x approach zero. Let’s take a look at this limit. So, a different value from the previous two paths and this means that the limit can’t possibly exist. Note that we can use this idea of moving in towards the origin along a line with the more general path if we need to. (d) Okay, with this last one we again have continuity problems at the origin. So, again let’s see if we can find a couple of paths that give different values of the limit. First, we will use the path . Along this path we have, Now, let’s try the path . Along this path the limit becomes, We now have two paths that give different values for the limit and so the limit doesn’t exist. As this limit has shown us we can, and often need, to use paths other than lines.
Partial Derivatives (Introduction) Previous Section Next Section Partial Derivatives 3-Dimensional Space Previous Chapter Next Chapter Applications of Partial Derivatives
[Notes]
© 2003 - 2016 Paul Dawkins
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dgrmul2 Structured version Visualization version GIF version
Theorem dgrmul2 23829
Description: The degree of a product of polynomials is at most the sum of degrees. (Contributed by Mario Carneiro, 24-Jul-2014.)
Hypotheses
Ref Expression
Assertion
Ref Expression
dgrmul2 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝐺 ∈ (Poly‘𝑆)) → (deg‘(𝐹𝑓 · 𝐺)) ≤ (𝑀 + 𝑁))
Proof of Theorem dgrmul2
Dummy variables 𝑘 𝑛 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eqid 2610 . . 3 (coeff‘𝐹) = (coeff‘𝐹)
2 eqid 2610 . . 3 (coeff‘𝐺) = (coeff‘𝐺)
3 dgradd.1 . . 3 𝑀 = (deg‘𝐹)
4 dgradd.2 . . 3 𝑁 = (deg‘𝐺)
51, 2, 3, 4coemullem 23810 . 2 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝐺 ∈ (Poly‘𝑆)) → ((coeff‘(𝐹𝑓 · 𝐺)) = (𝑛 ∈ ℕ0 ↦ Σ𝑘 ∈ (0...𝑛)(((coeff‘𝐹)‘𝑘) · ((coeff‘𝐺)‘(𝑛𝑘)))) ∧ (deg‘(𝐹𝑓 · 𝐺)) ≤ (𝑀 + 𝑁)))
65simprd 478 1 ((𝐹 ∈ (Poly‘𝑆) ∧ 𝐺 ∈ (Poly‘𝑆)) → (deg‘(𝐹𝑓 · 𝐺)) ≤ (𝑀 + 𝑁))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1475 ∈ wcel 1977 class class class wbr 4583 ↦ cmpt 4643 ‘cfv 5804 (class class class)co 6549 ∘𝑓 cof 6793 0cc0 9815 + caddc 9818 · cmul 9820 ≤ cle 9954 − cmin 10145 ℕ0cn0 11169 ...cfz 12197 Σcsu 14264 Polycply 23744 coeffccoe 23746 degcdgr 23747 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-inf2 8421 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 ax-pre-sup 9893 ax-addf 9894 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-fal 1481 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-se 4998 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-isom 5813 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-of 6795 df-om 6958 df-1st 7059 df-2nd 7060 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-oadd 7451 df-er 7629 df-map 7746 df-pm 7747 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-sup 8231 df-inf 8232 df-oi 8298 df-card 8648 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-div 10564 df-nn 10898 df-2 10956 df-3 10957 df-n0 11170 df-z 11255 df-uz 11564 df-rp 11709 df-fz 12198 df-fzo 12335 df-fl 12455 df-seq 12664 df-exp 12723 df-hash 12980 df-cj 13687 df-re 13688 df-im 13689 df-sqrt 13823 df-abs 13824 df-clim 14067 df-rlim 14068 df-sum 14265 df-0p 23243 df-ply 23748 df-coe 23750 df-dgr 23751 This theorem is referenced by: dgrmul 23830
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# 2.10: Perimeter and Area
Difficulty Level: At Grade Created by: CK-12
## Triangles and Parallelograms
The Importance of Units - Students will give answers that do not include the proper units, unless it is required by the instructor. When stating an area, square units should be included, and when referring to a length, linear units should be used. Using proper units helps reinforce the basic concepts. With these first simple area problems including the units seems like a small detail, but as the students move to more complex situations combining length, area, and volume, units can be a helpful guide. In physics and chemistry dimensional analysis is an important tool.
The Power of Labeling - When doing an exercise where a figure needs to be broken into polynomials with known area formulas, it is important for the student to draw on and label the figure well. Each polygon, so far only parallelograms and triangles, should have their base and height labeled and the individual area should be in the center of each. By solving these exercises in a neat, orderly way student will avoid errors like using the wrong values in the formulas, overlapping polygons, or leaving out some of the total area.
Subtracting Areas - Another way of finding the area of a figure that is not a standard polygon is to calculate a larger known area and then subtracting off the areas of polygons that are not included in the target area. This can often result in fewer calculations than adding areas. Different minds work in different ways, and this method might appeal to some students. It is nice to give them as many options as possible so they feel they have the freedom to be creative.
The Height Must Be Perpendicular to the Base - Students will frequently take the numbers from a polygon and plug them into the area formula without really thinking about what the numbers represent. In geometry there will frequently be more steps. The students will have to use what they have learned to find the correct base and height and then use those numbers in an area formula. Remind students that they already know how to use a formula; many exercises in this class will require more conceptual work.
Write Out the Formula - When using an area formula, it is a good idea to have students first write out the formula they are using, substitute numbers in the next step, and then solve the resulting equation. Writing the formula helps them memorize it and also reduces error when substituting and solving. It is especially important when the area is given and the student is solving for a length measurement in the polygon. Students will be able to do these calculations in their heads for parallelograms, and maybe triangles as well, but it is important to start good habits for the more complex polygons to come.
## Trapezoids, Rhombi, and Kites
It’s Arts and Crafts Time - Student have trouble remembering how to derive the area formulas. At this level it is required that they understand the nature of the formulas and why the formulas work so they can modify and apply them in less straightforward situations. An activity where student follow the explanation by illustrating it with shapes that they cut out and manipulate is much more powerful than just listening and taking notes. It will engage the students, keep their attention, and make them remember the lesson longer. Here are examples of how students can do this for each figure in this section.
Trapezoid - This procedure models the description in the text. Actually have students physically cut out the figures and do this will deepen understanding. Afterwards, the pictures in the lesson in the text will have much more meaning to them as they will have a more concrete understanding of what is going on.
1. Have student use the parallel lines on binder paper to draw a trapezoid. They should draw in the height and label it h\begin{align*}h\end{align*}. They should also label the two bases b1\begin{align*}b_1\end{align*} and b2\begin{align*}b_2\end{align*}.
2. Now they can trace and cut out a second congruent trapezoid and label it as they did the first.
3. The two trapezoids can be arranged into a parallelogram and glued down to another piece of paper.
4. Identify the base and height of the parallelogram in terms of the trapezoid variables. Then substitute these expressions into the area formula of a parallelogram to derive the area formula for a trapezoid.
5. Remember that two congruent trapezoids were used in the parallelogram, and the formula should only find the area of one trapezoid.
Kite - This particular method is different than the description in the text. You can choose to do this either way, but it helps to improve understanding sometimes to show different methods to achieve the same results. Some students may have an easier time grasping the concept one way or the other.
1. Have the students draw a kite. They should start by making perpendicular diagonals, one of which is bisecting the other. Then they can connect the vertices to form a kite.
2. Now they can draw in the rectangle around the kite.
3. Identify the base and height of the parallelogram in terms of d1\begin{align*}d_1\end{align*} and d2\begin{align*}d_2\end{align*}, and then substitute into the parallelogram area formula to derive the kite area formula.
4. Now have the students cut off the four triangles that are not part of the kite and arrange them over the congruent triangle in the kite to demonstrate that the area of the kite is half the area of the rectangle.
Rhombus - You could allow students to decide which way they would rather do this one or just let them come up with their own method. The more they can do this on their own, the more they have gleaned from this activity.
Also, the area of a rhombus can be found using either the kite or parallelogram area formulas. Use this as an opportunity to review subsets and what they mean in terms of applying formulas and theorems.
## Area of Similar Polygons
Reducing Fractions - It is helpful in this section to have ratios in reduced form. Once students start square-rooting ratios (in fraction form), however, they may start messing up their reductions. For example, they may start thinking that 49\begin{align*}\frac{4}{9}\end{align*} reduces to 23\begin{align*}\frac{2}{3}\end{align*}. Just be aware of this common error and point out the difference to students as needed.
Adjust the Scale Factor - It is difficult for students to remember to square and cube the scale factor when writing proportions involving area and volume. Writing and solving a proportion is a skill they know well and have used frequently. Once the process is started, it is hard to remember to add that extra step of checking and adjusting the scale factor in the middle of the process. Here are some ways to reinforce this step in the students’ minds.
1. Inform students that this material is frequently used on the SAT and other standardized tests in some of the more difficult problems.
2. Play with graph paper. Have students draw similar shape on graph paper. They can estimate the area by counting squares, and then compare the ratio of the areas to the ratio of the side lengths. Creating the shapes on graph paper will give the students a good visual impression of the areas.
3. Write out steps, or have the students write out the process they will use to tackle these problems. (1) Write a ratio comparing the two polygons. (2) Identify the type of ratio: linear, area, or volume. (3) Adjust the ratio using powers or roots to get the desired ratio. (4) Write and solve a proportion.
4. Mix-up the exercises so that students will have to square the ratio in one problem and not in the next. Keep them on the lookout. Make them analyze the situation instead of falling into a habit.
Example 1: The ratio of the lengths of the sides of two squares is 2:3. What is the ratio of their areas?
Answer: 4:9, The ratio of areas is the ratio of the lengths squared.
Example 2: The area of a small triangle is 15 cm2\begin{align*}15 \ cm^2\end{align*}, and it has a height of 5 cm. A larger similar triangle has an area of 60 cm2\begin{align*}60 \ cm^2\end{align*}. What is the corresponding height of the larger triangle?
The area ratio is 15:60 or 1:4. The length ratio is then the square root of this, or 1:2. Now set up a proportion to solve for the height of the smaller triangle.
12=5x\begin{align*}\frac{1}{2} = \frac{5}{x}\end{align*}
Example 3: The ratio of the lengths of two similar rectangles is 4:5. The larger rectangle has a width of 45 cm. What is the width of the smaller rectangle?
In this problem, we did not have to adjust the ratio since we are given a ratio of lengths and a length is what we are trying to find. Just set up the proportion and solve.
\begin{align*}\frac{4}{5} = \frac{x}{45}\end{align*}
Example 4: The ratio of the areas of two regular pentagons is 25:64. What is the ratio of their corresponding sides?
This time students have to square root the area ratio to find the ratio of the lengths of the sides.
## Circumference and Arc Length
Pi is an Irrational Number - Many students can give the definition of an irrational number. They know that an irrational number has an infinite decimal that has no pattern, but they have not really internalized what this means. Infinity is a difficult concept. A fun way to help the students develop this concept is to have a pi contest. The students can chose to compete by memorizing digits of pi. They can be given points, possible extra credit, for ever ten digits or so, and the winner gets a pie of their choice. The students can also research records for memorizing digits of pi. The competition can be done on March \begin{align*}14^{th}\end{align*}, pi day. When the contest is introduced, there is always a student who asks “How many points do I get if I memorize it all?” It is a fun way to reinforce the concept of irrational numbers and generate a little excitement in math class.
There Are Two Values That Describe an Arc - The measure of an arc describes how curved the arc is, and the length describes the size of the arc. Whenever possible, have the students give both values with units so that they will remember that there are two different numerical descriptions of an arc. Often student will give the measure of an arc when asked to calculate its length.
Arc Length Fractions - Fractions are a difficult concept for many students even when they have come as far as geometry. For many of them putting the arc measure over 360 does not obviously give the part of the circumference included in the arc. It is best to start with easy fractions. Use a semi-circle and show how \begin{align*}\frac{180}{360}\end{align*} reduces to \begin{align*}\frac{1}{2}\end{align*}, then a ninety degree arc, and then a \begin{align*}120^\circ\end{align*} arc. After some practice with fractions they can easily visualize, the students will be able to work with any arc measure as a fraction of 360 degrees.
Exact or Approximate - When dealing with the circumference of a circle there are often two ways to express the answer. The students can give exact answers, such as \begin{align*}2\pi\end{align*} cm or the decimal approximation 6.28 cm. Explain the strengths and weaknesses of both types of answers. It is hard to visualize \begin{align*}13\pi\end{align*} ft, but that is the only way to accurately express the circumference of a circle with diameter 13 ft. The decimal approximations 41, 40.8, 40.84, etc, can be calculated to any degree of accuracy, are easy to understand in terms of length, but are always slightly wrong. Let the students know if they should give one, the other, or both forms of the answer.
## Areas of Circles and Sectors
Reinforce - This section on area of a circle and the area of a sector is analogous to the previous section about circumference of a circle and arc length. This gives students another chance to go back over the arguments and logic to better understand, remember, and apply them. Focus on the same key points and methods in this section, and compare it to the previous section. Mix-up exercises so students will see the similarities and learn each more thoroughly.
Don’t Forget the Units - Remind students that when they calculate an area the units are squared. When an answer contains the pi symbol, students are more likely to leave off the units. In the answer \begin{align*}7\pi \ cm^2\end{align*}, the \begin{align*}\pi\end{align*} is part of the number and the \begin{align*}cm^2\end{align*} are units of area.
Draw a Picture - When applying geometry to the world around us, it is helpful to draw, label, and work with a picture. Visually organized information is a powerful tool. Remind students to take the time for this step when calculating the areas of the irregular shapes that surround us.
Example 1: What is the area between two concentric circles with radii 5 cm and 12 cm?
Answer: \begin{align*}144\pi - 25\pi = 119 \pi cm^2\end{align*}
Example 2: The area of a sector of a circle with radius 6 cm, is \begin{align*}12\pi \ cm^2\end{align*}. What is the measure of the central angle that defines the sector?
Answer: \begin{align*}12 \pi = \frac{x}{360}* \pi *6^2, x=120^\circ\end{align*}. The central angle measures \begin{align*}120^\circ\end{align*}.
Example 3: A square with side length \begin{align*}5 \sqrt 2 \ cm\end{align*} is inscribed in a circle. What is the area of the region between the square and the circle?
Answer: approx. \begin{align*}28.5 \ cm^2\end{align*}
First find the diagonal of the square using special right triangles: \begin{align*}5\sqrt{2}*\sqrt{2} = 5 * 2 = 10\end{align*}.
This makes the radius of the circle 5 cm. Now we can find the area of the circle and subtract the area of the square as shown below.
\begin{align*}&5^2*\pi-\left (5\sqrt{2} \right )^2\\ &25 \pi -50\approx 28.5 cm^2\end{align*}
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# 4.1: Introduction to Fractions
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Fractions are one of the hardest topics to teach (and learn!) in elementary school. What is the reason for this? In this part of the book, will try to provide you with some insight about this (as well as some better ways for understanding, teaching, and learning about fractions). But for now, think about what makes this topic so hard.
Think / Pair / Share
You may have struggled learning about fractions in elementary school. Maybe you still find them confusing. Even if you were one of the lucky ones who did not struggle when learning about fractions, you probably had friends who did struggle.
With a partner, talk about why this is. What is so difficult about understanding fractions? Why is the topic harder than other ones we tackle in elementary schools?
Remember that teachers should have lots of mental models — lots of ways to explain the same concept. In this chapter, we will look at some different ways to understand the idea of fractions as well as basic operations on them.
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Math word problems can be difficult to solve. They often require you to perform complex calculations and make complex observations. They can also be intimidating, because they require a high level of concentration and attention to detail. In order to solve math word problems, it is important to stay calm and avoid rushing. You should also try to simplify the problem as much as possible. By doing this, you will be able to focus on the important parts of the problem instead of being overwhelmed by the details. Once you have simplified your problem, you will need to come up with a plan for solving it. There are several different ways that you can approach this process. You can use trial and error, brainstorming, or using a systematic approach. Whichever method works best for you, stick with it until you’ve reached your goal.
A function solver calculator that works well is the HP 12C. It has a simple interface and comes in handy when you need to find the solutions of basic math problems like adding fractions or decimals. You can simply enter the values of your input and output and get the right answer instantly. If you want more features, such as finding solutions of more complex problems, an advanced calculator will be able to provide more accurate results.
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# Michelson-Morley Experiment question
1. Oct 14, 2011
### superspartan9
1. The problem statement, all variables and given/known data
Consider an apparatus for performing a Michelson-Morley experiment to measure the speed of sound in the laboratory. A sound wave of frequency 3,600 Hz replaces light. The speed of sound in air is 330 m/s. The arms of the interferometer are 2m long, and the apparatus is placed in front of a large fan, which blows air along one of the arms at 8 m/s. Estimate the frequency of the beats that occur because of the interference of the waves reflected along the two arms of the interferometer. (hint: Be careful! You are measuring the Doppler shift for sound for the case of a moving medium, and both the speed and wavelength change.)
2. Relevant equations
f' = fgamma[1 - cos(θ)(v/c)]
f' / f = (1 + v/c) / (√(1-(v/c)2))
u' = (u-v) / (1 - (v/c2)u)
These are all the equations in the corresponding section.
3. The attempt at a solution
Honestly, no idea where to start since I don't understand how these components are being put together, but I'm willing to bet that we need to find the relative velocity of the sound wave in the moving air medium. From there, we can make θ = 0 if they are moving apart or ∏ if they're moving together. At that point, you can find the adjusted frequency of the wave, but again, the equations will change depending on which way the sound and air are moving relative to each other (i.e. if the sound source is moving against, with, or perpendicular to the air.)
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Update all PDFs
# Equivalent Expressions?
Alignments to Content Standards: 7.EE.A
If we multiply $\frac{x}{2} + \frac34$ by 4, we get $2x+3$. Is $2x+3$ an equivalent expression to $\frac{x}{2} + \frac34$?
## IM Commentary
The purpose of this task is to directly address a common misconception held by many students who are learning to solve equations. Because a frequent strategy for solving an equation with fractions is to multiply both sides by a common denominator (so all the coefficients are integers), students often forget why this is an "allowable" move in an equation and try to apply the same strategy when they see an expression. Two expressions are equivalent if they have the same value no matter what the value of the variables in them. After learning to transform expressions and equations into equivalent expressions and equations, it is easy to forget the original definition of equivalent expressions and mix up which transformations are allowed for expressions and which are allowed for equations.
## Solution
No, $2x + 3$ and $\frac{x}{2} + \frac34$ are not equivalent expressions because they do not yield the same result for most values of $x$. For example, when $x = 1$, we get
$$2(1) + 3 = 5$$
and
$$\frac{(1)}{2}+\frac34=\frac54 \neq 5$$
Therefore, they are not equivalent. In fact, the expression $2x + 3$ will be 4 times as big as $\frac{x}{2} + \frac34$ for all values of $x$, since we obtained it by multiplying $\frac{x}{2} + \frac34$ by 4.
#### Cam says:
over 2 years
The two expressions $\frac{x}{2}+\frac34$ and $2x+3$ are equal when $x=-3/2$, but the question of equivalence is a question about whether the two expressions are equal for all values of $x$. Since there is at least one real number $x$ for which the two expressions are not equal (e.g., $x=1$), the two expressions are not equivalent.
#### James says:
over 2 years
If we restrict this to being in the first quadrant then the conclusion is right for all x's from zero to infinity but the fact that the slopes vary implies that the two lines must intersect i.e. there must be a point that the two solutions are equivalent (but only one since the two are linear expressions).
#### James says:
over 2 years
Except when x = -3/2. At that value only are they equivalent.
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