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# MSG Engineering Computation II [Pengiraan Kejuruteraan II]
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1 UNIVERSITI SAINS MALAYSIA Second Semester Examination 013/014 Academic Session June 014 MSG Engineering Computation II [Pengiraan Kejuruteraan II] Duration : 3 ours [Masa : 3 jam] Please ceck tat tis examination paper consists of SIX pages of printed materials before you begin te examination. [Sila pastikan baawa kertas peperiksaan ini mengandungi ENAM muka surat yang bercetak sebelum anda memulakan peperiksaan ini.] Instructions: [Araan: Answer all four [4] questions. Jawab semua empat [4] soalan.] In te event of any discrepancies, te Englis version sall be used. [Sekiranya terdapat sebarang percanggaan pada soalan peperiksaan, versi Baasa Inggeris endakla diguna pakai]. /-
2 -- [MSG 389] 1. (a) Consider a large tank olding 1000 liter of pure water into wic a salt solution begins to flow at a constant rate of 6 liter/min. Te solution inside te tank is kept well stirred, and is flowing out of te tank at a rate of 6 liter/min. If te concentration of salt in te solution entering te tank is 0.1 kg/liter, determine wen te concentration of salt will reac 0.05 kg/liter? [40 marks] Consider te initial value problem ' y x + 3xy + y = 0 y 1 = 0.5, = 0.1. Solve te initial value problem to determine y (1.3) using: (i) Euler metod (ii) Second order Taylor series metod (iii) Second order Runge Kutta metod (iv) Fourt order Runge Kutta metod (v) Euler metod as predictor metod and Heun metod as corrector metod (vi) Midpoint metod [60 marks] 1. (a) Pertimbangkan sebua tangki besar mengandungi 1000 liter air tulen di mana larutan garam mula mengalir pada kadar yang tetap 6 liter / min. Campuran di dalam tangki itu dikacau seragam, dan mengalir keluar dari tangki pada kadar 6 liter / min. Jika kepekatan garam dalam campuran yang memasuki tangki itu adala 0.1 kg / liter, tentukan masa yang diperlukan untuk kepekatan garam mencapai 0.05 kg / liter? [40 marka] Pertimbangkan masala nilai awalan ' y x xy y y = 0 1 = 0.5, = 0.1. Selesaikan masala nilai awalan ini bagi menentukan y (1.3) menggunakan: (i) Kaeda Euler (ii) Kaeda siri Taylor tertib kedua (iii) Kaeda Runge Kutta tertib kedua (iv) Kaeda Runge Kutta tertib keempat (v) Kaeda Euler metod sebagai penganggar dan kaeda Heun sebagai pembaiki (vi) Kaeda titik tenga [60 marka] 3/-
3 -3- [MSG 389]. (a) Find te approximate value of y(0.5) for te initial value problem ' y = x+ y, y 0 = 1, using te multistep metod yi+ 1 = yi 1+ ( fi fi + fi 3 wit =0.1. Compute te starting value using Runge-Kutta fourt order metod wit te same step lengt =0.1. [50 marks] Solve te initial value problem ' y = x + y 3, y( = 0, on te interval [1, 1.6] using te predictor-corrector metod, Predictor : yi+ 1 = yi + ( 3fi fi Corrector : yi+ 1 = yi + ( 5 fi fi fi 1 wit te step lengt =0.. Perform tree corrector iterations per step. Compute te starting value using Taylor series order metod wit te same step lengt. [50 marks]. (a) Dapatkan nilai anggaran y(0.5) bagi masala nilai awalan ' y = x+ y, y( 0) = 1, dengan menggunakan kaeda multi langka: yi+ 1 = yi 1+ ( fi fi + fi 3 dengan =0.1. Dapatkan nilai awalan menggunakan kaeda Runge-Kutta tertib keempat dengan saiz langka yang sama =0.1. [50 marka] Selesaikan masala nilai awalan ' y = x + y 3, y( = 0, dalam selang [1, 1.6] menggunakan kaeda pengangar dan pembaiki, Penganggar : yi+ 1 = yi + ( 3fi fi Pembaiki : yi+ 1 = yi + ( 5 fi fi fi 1 dengan saiz langka =0.. Lakukan tiga langka pembaiki bagi setiap lelaran. Dapatkan nilai permulaan dengan mengunakan kaeda siri Taylor dengan saiz langka, yang sama. [50 marka] 4/-
4 -4- [MSG 389] 3. (a) Set up te equations for a value of u tat satisfies te Laplace equation of a regular plate wit 0 x 4 and 0 y. Use te conditions: x= y = 1 T( 0, y) = 5 T( x ) T x,0 = 0 and, = 10 T = 3 at x = 4 [40 marks] Solve te boundary value problem " ' y 4y + 3y = 0, y 0 = 1, y 1 = 0. using second order finite difference metod wit (i) =1/ (ii) =1/3. " 1 ' 1 [Hint: yi = ( yi 1 yi + yi+, yi = ( yi+ 1 yi ] [60 marks] 3. (a) Bangunkan persamaan untuk nilai u yang memenui persamaan Laplace bagi sata biasa dengan 0 x 4 dan 0 y. Gunakan syarat : x= y = 1 T( 0, y) = 5 T x,0 = 0 dan T x, = 10 T = 3 pada x = 4 [40 marka] Selesaikan masala nilai sempadan " ' y 4y + 3y = 0, y 0 = 1, y 1 = 0. mengunakan kaeda beza teringga peringkat kedua dengan (i) =1/ (ii) =1/3. " 1 ' 1 [Petunjuk: y = ( y 1, ( 1 i i y + y y y y i i + = i i + i ] [60 marka] 5/-
5 -5- [MSG 389] 4. (a) Te partial differential equation of te temperature in a long tin rod is given by T T = α. t If α = 0.8 cm / s, te initial temperature of rod is 40 C, and te rod is divided into tree equal segments, find te temperature at node 1 (using t = 0.1s) for t=0. sec by using: (i) Crank-Nicolson metod (ii) Forward Time Centered Space (FTCS) sceme node = T = 80 o C T = 0 o C 9 cm [70 marks] A typical steady-state eat flow problem is te following: A tin steel plate is a unit square in m. If two of its edges are eld at 0 C and te oter two are eld at te temperatures sown below: u u u = + = 0, 0 < x< 1, 0 < y< 1. y u 0, x = 100x u y,0 = 100y u x,1 = 100x u 1, y = 100y Write down te five point formula for te following coordinates: (1,, (,, (3,, (1,), (,), (3,), (1,3), (,3) and (3,3). [30 marks] 6/-
6 -6- [MSG 389] 4. (a) Masala pembezaan separa bagi suu suatu rod nipis yang panjang diberikan ole T T = α. t Sekiranya α = 0.8 cm / s, suu awalan rod adala 40 C, dan rod dibaagikan kepada tiga baagian yang seragam, dapatkan suu di nod 1 (mengunakan t = 0.1s) untuk t=0.s dengan menggunakan: (i) (ii) Kaeda Crank- Nicolson Skema Beza Ke Depan Teradap Masa Beza Pusat Teradap Ruang nod = T = 80 o C T = 0 o C 9 cm [70 marka] Aliran aba kaeda mantap biasa diberikan sebagai : Satu plat besi yang nipis adala satu unit persegi dalam m. Sekiranya dua pengujuna diletakkan pada suu 0 C dan dua pengujung lagi diletakkan dalam suu berikut : u u u = + = 0, 0 < x< 1, 0 < y< 1. y u( 0, x) = 100x u( y,0) = 100y u( x, = 100x u 1, y = 100y Tuliskan formula lima titik untuk koordinat berikut: (1,, (,, (3,, (1,), (,), (3,), (1,3), (,3) dan (3,3). [30 marka] -ooo000ooo-
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### INSTRUCTION: This paper consists of SIX (6) essay questions. Answer any FOUR (4) questions only.
INSTRUCTION: This paper consists of SIX (6) essay questions. Answer any FOUR (4) questions only. ARAHAN: Kertas ini mengandungi ENAM (6) soalan esei. Jawab mana-mana EMPAT (4) soalan sahaja. QUESTION 1 | HuggingFaceTB/finemath | |
# How do we know/prove a slope of a line is constant?
• lamp23
#### lamp23
Before I just accepted that the slope of a line is constant, but I'm wondering if there is an even more fundamental definition of a line.
If one draws two right triangles with a certain Δx and Δy as the legs and wants to prove that the connection of the two hypotenuses is a straight line, then should one prove that the angle between them is 180°?
I have drawn the original picture in purple and then by SAS one can prove the two triangles congruent and then prove that the corresponding angles ∅ are congruent. Once one proves there is a right angle adjacent to it and a (90-∅) adjacent to that, the sum gives 180°.
If $f(x) = ax + b$ then $f'(x) = a$, thus the slope is constant. No need to make it any more complicated than that.
If $f(x) = ax + b$ then $f'(x) = a$, thus the slope is constant. No need to make it any more complicated than that.
Then you are assuming that f(x)=ax+b is the equation of the line. Yet in the derivations for an equation of a line that I have seen (one in Stewart's Calculus), that equation is derived from assuming a =Δy/Δx is constant.
And what about the line whose equation is x = 2? This isn't even a function, let alone one that is differentiable, and yet it is a line.
i have to ask, first, for you: what constitues an acceptable definition of a line?
there is more to the question than meets the eye, and what you will regard as an acceptable proof, depends on what you will allow as "given".
to underscore my point, in euclidean geometry, often lines are NOT defined, but are assumed to have certain properties instead (line is a "primitive concept" and any conceivable object with the properties of a line, is said to be a model for a line).
it is not hard to show, that for any set in the plane satisfying:
L = {(x,y): ax+by = c} (where a,b and c are "constants"). that the equation:
y2 - y1= m(x2 - x1)
has a unique solution m that holds for any pair (x1,y1), (x2,y2) in L; unless b = 0, in which case NO m will work.
but perhaps this is not what you're looking for, without more information, i cannot say.
Before I just accepted that the slope of a line is constant, but I'm wondering if there is an even more fundamental definition of a line.
If one draws two right triangles with a certain Δx and Δy as the legs and wants to prove that the connection of the two hypotenuses is a straight line, then should one prove that the angle between them is 180°?
You don't need that the two triangles are congruent- only similar. That way, you can use different length $\Delta x$ and get a different $\Delta y$. But because the hypotenuses of the two right triangles are the same line, and the two horizontal sides are parallel, by "corresponding angles" from geometry, we get that the two angles you have labeled "$\phi$" are congruent so the triangles are similar. Then the ratios of corresponding sides are the same. Since the slope is the ratio of two sides, it is the same at every point no matter what "rise" and "run" you use. | HuggingFaceTB/finemath | |
+0
# Some Help
0
115
2
A number is divisible by 9 if the sum of its digits is divisible by 9. For example, the number 19,836 is divisible by 9 but 19,825 is not.
If D767E89 is divisible by 9, where D and E each represent a single digit, what is the sum of all possible values of the sum D+E?
Jun 23, 2020
#1
+783
+1
We look at the current sum of the digits we have(not including D and E).
The sum is 7+6+7+8+9=37. So 37+D+E must be a multiple of 9. Now let's look at the multiples of 9 greater than 37:
45, 54, 63....
However, the greatest that D+E can add up to is 18, so the only possible multiples of 9 would be 45. Since there are no other possible multiples of 9, then the sum of D+E is 45-37=8.
Jun 23, 2020
edited by gwenspooner85 Jun 23, 2020
#2
0
Except 45 - 37 = 8 not 18. So, you could have: 1+7 =8, 2+6 =8, 3+5 =8, 4+4 =8, 5+3 =8, 6+2 =8, 7+1 =8, 9+8 =17, 8+9 =17,
Grand Total =[8 x 7] + [17 x 2] =90
Guest Jun 23, 2020 | HuggingFaceTB/finemath | |
$\newcommand{\const}{\mathrm{const}}$
Deadline Monday, October 1, 9 pm
### APM 346 (2012) Home Assignment 2
#### Problem 1
Consider equation with the initial conditions \begin{align} & u_{tt}-4u_{xx}=0,\qquad &&t>0, x>vt, \label{eq-1}\\ &u|_{t=0}= e ^{-x}, \qquad &&x>0, \label{eq-2}\\ &u_t|_{t=0}= e^{-x}, \qquad &&x>0, \label{eq-3} \end{align}
1. Let $v=3$. Find which of these conditions (a)-(c) at $x=vt$, $t>0$ could be added to (\ref{eq-1})-(\ref{eq-3}) so that the resulting problem would have a unique solution:
1. None,
2. $u|_{x=vt}=0$ ($t>0$),
3. $u|_{x=vt}=u_x|_{x=vt}=0$ ($t>0$).
Solve the problem you deemed as a good one.
2. Let $v=1$. Find which of these conditions (a)-(c) at $x=vt$, $t>0$ could be added to (\ref{eq-1})-(\ref{eq-3}) so that the resulting problem would have a unique solution:
1. None
2. $u|_{x=vt}=0$ ($t>0$),
3. $u|_{x=vt}=u_x|_{x=vt}=0$ ($t>0$).
Solve the problem you deemed as a good one.
3. Let $v=-3$. Find which of these conditions (a)-(c) at $x=vt$, $t>0$ could be added to (\ref{eq-1})-(\ref{eq-3}) so that the resulting problem would have a unique solution:
1. None
2. $u|_{x=vt}=0$ ($t>0$),
3. $u|_{x=vt}=u_x|_{x=vt}=0$ ($t>0$).
Solve the problem you deemed as a good one.
#### Problem 2
A spherical wave is a solution of the three-dimensional wave equation of the form $u(r, t)$, where r is the distance to the origin (the spherical coordinate). The wave equation takes the form \begin{equation} u_{tt} = c^2 \bigl(u_{rr}+\frac{2}{r}u_r\bigr) \qquad\text{(“spherical wave equation”).} \label{eq-4} \end{equation}
1. Change variables $v = ru$ to get the equation for $v$: $v_{tt} = c^2 v_{rr}$.
2. Solve for $v$ using \begin{equation} v = f(x+ct)+g(x-ct) \label{eq-5} \end{equation} and thereby solve the spherical wave equation.
3. Use \begin{equation} v(r,t)=\frac{1}{2}\bigl[ \phi (r+ct)+\phi (r-ct)\bigr]+\frac{1}{2c}\int_{r-ct}^{r+ct}\psi (s)\,ds \label{eq-6} \end{equation} with $\phi(r)=v(r,0)$, $\psi(r)=v_t(r,0)$ to solve it with initial conditions $u(r, 0) = \Phi (r)$, $u_t(r, 0) = \Psi(r)$.
4. Find the general form of solution continuous as $r=0$.
### Problem 3
By method of continuation combined with D'Alembert formula solve each of the following four problems (a)--(d).
For a solution $u(x, t)$ of the wave equation with $\rho = T =1 \Longrightarrow c = 1$, the energy density is defined as $e=\frac{1}{2}\bigl(u_t^2+u_x^2\bigr)$ and the momentum density as $p = u_t u_x$.
2. Show that both $e(x, t)$ and $p(x,t)$ also satisfy the wave equation. | HuggingFaceTB/finemath | |
# RD Sharma Solutions Class 11 The Straight Lines
## RD Sharma Solutions Class 11 Chapter 23
A straight line is a curve such that every point on the line segment joining any two points on it lies on it. The theorem states that every first-degree equation in x, y represents a straight line. When we say that a first degree equation in x, y i.e., ax + by + c = 0 represents a line, it means that all points (x,y) satisfying ax + by + c = 0 lie along a line. Thus, a line is also defined as the locus of a point satisfying the condition ax + by + c = 0 where a, b, c are constants. Thus, in order to determine a line, we will need two conditions to determine the two unknown.
The trigonometrical tangent of the angle that a line makes with the positive direction of the x-axis in an anticlockwise sense is called the slope or gradient of the line. Learn about to how to find out the slope of a line in terms of coordinates of any two point on it. Practise the solved examples and learn about how to find angles between lines, various conditions of parallelism of lines and conditions of perpendicularity of two lines. In further exercise, we will be discussing intercepts of a line and different forms of an equation of a straight line. Learn these concepts easily by practicing the questions from the exercises given in RD Sharma solution for the chapter “Straight Lines”. | HuggingFaceTB/finemath | |
# Circle theories class 9
2
by hari90485
2016-03-13T10:57:46+05:30
1. A circle is a collection of all the points in a plane, which are equidistant from a fixed point in the plane.
2. Equal chords of a circle (or of congruent circles) subtend equal angles at the centre.
3. If the angles subtended by two chords of a circle (or of congruent circles) at the centre (corresponding centre) are equal, the chords are equal.
4. The perpendicular from the centre of a circle to a chord bisects the chord.
5. The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
6. There is one and only one circle passing through three non-collinear points.
7. Equal chords of a circle (or of congruent circles) are equidistant from the centre (or corresponding centres).
8. Chords equidistant from the centre (or corresponding centres) of a circle (or of congruent circles) are equal.
9. If two arcs of a circle are congruent, then their corresponding chords are equal and conversely, if two chords of a circle are equal, then their corresponding arcs (minor, major) are congruent.
10. Congruent arcs of a circle subtend equal angles at the centre.
11. The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
12. Angles in the same segment of a circle are equal.
13. Angle in a semicircle is a right angle.
14. If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.
15. The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
16. If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.
2016-03-13T11:01:39+05:30
Theorem: 1)The perpendicular from the centre of a circle to a chord bisects the chord.
2)The line drawn from the centre of a circle to bisect a chord is perpendicular to the chord.
3)Equal chords of a circle subtend equal angles at the centre.
4)
Chords that subtend equal angles at the centre of a circle are equal in length. | HuggingFaceTB/finemath | |
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A000955 A sequence satisfying (a(2n+1) + 1)^3 = Sum_{k=1..2n+1} a(k)^3. (Formerly M4073 N1688) 2
1, 6, 8, 262, 2448, 17997702, 44082372248, 5829766629386380698502, 256989942683351711945337288361248, 198131491921177194311506308094238133848780474484255622782351242502 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 REFERENCES N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS D. A. Klarner, Sequences of k-th powers with k-th power partial sums, Math. Mag., 37 (1964), 165-167. D. A. Klarner, Sequences of k-th powers with k-th power partial sums, Math. Mag., 37 (1964), 165-167. Annotated scanned copy. FORMULA a(1)=1, a(2)=6, a(3)=8, a(2n+2) = 3*a(2n+1)^2 + 8*a(2n+1) + 6, a(2n+3) = 3*a(2n+1)^3 + 12*a(2n+1)^2 + 17*a(2n+1) + 8. [Sean A. Irvine, Sep 16 2011] MATHEMATICA a[1] = 1; a[2] = 6; a[3] = 8; a[n_] := a[n] = If[EvenQ[n], 3*a[n-1]^2 + 8*a[n-1] + 6, 3*a[n-2]^3 + 12*a[n-2]^2 + 17*a[n-2] + 8]; Array[a, 10] (* Jean-François Alcover, Feb 15 2016, after Sean A. Irvine *) CROSSREFS Sequence in context: A169971 A295429 A267165 * A027721 A264518 A259129 Adjacent sequences: A000952 A000953 A000954 * A000956 A000957 A000958 KEYWORD nonn AUTHOR EXTENSIONS One more term from Sean A. Irvine, Sep 15 2011 STATUS approved
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# How do you factor 2x^3 + 2x^2 - 84x?
Aug 13, 2016
$2 x \left(x + 7\right) \left(x - 6\right)$
#### Explanation:
The first step is to take out the $\textcolor{b l u e}{\text{common factor}}$ of 2x
$\Rightarrow 2 x \left({x}^{2} + x - 42\right) \ldots \ldots . . \left(A\right)$
Now factorise the $\textcolor{b l u e}{\text{quadratic}}$ inside the bracket.
Using the a-c method, find the factors of - 42 which sum to +1
These are +7 and - 6
$\Rightarrow {x}^{2} + x - 42 = \left(x + 7\right) \left(x - 6\right)$
substituting these into (A) gives the factorised form.
$\Rightarrow 2 {x}^{3} + 2 {x}^{2} - 84 x = 2 x \left(x + 7\right) \left(x - 6\right)$
Aug 13, 2016
$2 x \left({x}^{2} + x - 42\right) = 2 x \left(x - 6\right) \left(x + 7\right)$
${x}^{2} + x - 42 = \left(x - 6\right) \left(x + 7\right)$ (If this were not immediately obvious, I could have used the quadratic equation and solved for $x$ in the quadratic.
And thus, $2 {x}^{3} + 2 {x}^{2} - 84 x$ $=$ $2 x \left(x - 6\right) \left(x + 7\right)$ | HuggingFaceTB/finemath | |
## 15603
15,603 (fifteen thousand six hundred three) is an odd five-digits composite number following 15602 and preceding 15604. In scientific notation, it is written as 1.5603 × 104. The sum of its digits is 15. It has a total of 3 prime factors and 8 positive divisors. There are 8,904 positive integers (up to 15603) that are relatively prime to 15603.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 15
• Digital Root 6
## Name
Short name 15 thousand 603 fifteen thousand six hundred three
## Notation
Scientific notation 1.5603 × 104 15.603 × 103
## Prime Factorization of 15603
Prime Factorization 3 × 7 × 743
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 15603 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 15,603 is 3 × 7 × 743. Since it has a total of 3 prime factors, 15,603 is a composite number.
## Divisors of 15603
1, 3, 7, 21, 743, 2229, 5201, 15603
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 23808 Sum of all the positive divisors of n s(n) 8205 Sum of the proper positive divisors of n A(n) 2976 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 124.912 Returns the nth root of the product of n divisors H(n) 5.24294 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 15,603 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 15,603) is 23,808, the average is 2,976.
## Other Arithmetic Functions (n = 15603)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 8904 Total number of positive integers not greater than n that are coprime to n λ(n) 2226 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1819 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 8,904 positive integers (less than 15,603) that are coprime with 15,603. And there are approximately 1,819 prime numbers less than or equal to 15,603.
## Divisibility of 15603
m n mod m 2 3 4 5 6 7 8 9 1 0 3 3 3 0 3 6
The number 15,603 is divisible by 3 and 7.
## Classification of 15603
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (15603)
Base System Value
2 Binary 11110011110011
3 Ternary 210101220
4 Quaternary 3303303
5 Quinary 444403
6 Senary 200123
8 Octal 36363
10 Decimal 15603
12 Duodecimal 9043
20 Vigesimal 1j03
36 Base36 c1f
## Basic calculations (n = 15603)
### Multiplication
n×i
n×2 31206 46809 62412 78015
### Division
ni
n⁄2 7801.5 5201 3900.75 3120.6
### Exponentiation
ni
n2 243453609 3798606661227 59269659735124881 924784500847153518243
### Nth Root
i√n
2√n 124.912 24.9883 11.1764 6.8967
## 15603 as geometric shapes
### Circle
Diameter 31206 98036.5 7.64832e+08
### Sphere
Volume 1.59116e+13 3.05933e+09 98036.5
### Square
Length = n
Perimeter 62412 2.43454e+08 22066
### Cube
Length = n
Surface area 1.46072e+09 3.79861e+12 27025.2
### Equilateral Triangle
Length = n
Perimeter 46809 1.05419e+08 13512.6
### Triangular Pyramid
Length = n
Surface area 4.21674e+08 4.4767e+11 12739.8
## Cryptographic Hash Functions
md5 b39f604b83278f4b3dfa16efc9670fbf df35411644acb90df56f509c66e87b165a92765a 236a15b22de78301c5d945ef7a238814cd4910f21bab8381e68dc0905b18f1cc 62e0af1cc13aee6d4eb5a9afb7b03ea3515390635830cf70ea14e2034f129848934cfb47233696fbdac4d6a36a6d6ff918fd77a487e9f60ea1fc7029a71eb34c aa2eef14951c77bbabba2ba45f25dc23a951dbfd | HuggingFaceTB/finemath | |
#### Mean of a Discrete Random Variable
The mean of a discrete random variable $$X$$ is denoted $$\mu_X$$ or, when no confusion will arise, simply $$\mu$$. The terms expected value, $$E(X)$$, and expectation are commonly used in place of the term mean.
$E(X) = \sum_{i=1}^{N}x_iP(X=x_i)$
In a large number of independent observations of a random variable $$X$$, the $$E(X)$$ of those observations - the sample - will approximate the mean, $$\mu$$, of the population. The larger the number of observations, the closer $$E(X)$$ is to $$\mu$$ (Weiss 2010).
Let us recall our experiment from the previous section, when we picked 1,000 individuals and asked for the number of siblings. Let us again take a look at the table, summarizing the experiment
$\begin{array}{c|lcr} \text{Siblings} & \text{Frequency} & \text{Relative}\\ \ x & f & \text{frequency}\\ \hline 0 & 205 & 0.205 \\ 1 & 419 & 0.419 \\ 2 & 280 & 0.28 \\ 3 & 65 & 0.065 \\ 4 & 29 & 0.029 \\ 5 & 2 & 0.002 \\ \hline & 1000 & 1 \end{array}$
Let us calculate the expected value (mean) for that experiment.
\begin{align} \\ & E(X) = \sum_{i=1}^{N}x_iP(X=x_i) \\ & = 0 \cdot P(X=0) + 1 \cdot P(X=1)+ 2 \cdot P(X=2) + 3 \cdot P(X=3) +4 \cdot P(X=4)+ 5 \cdot P(X \ge 5) \\ & = 0 \cdot 0.205 + 1 \cdot 0.419 + 2 \cdot 0.28+ 3 \cdot 0.065 + 4 \cdot 0.029 + 5 \cdot 0.002 \\ & = 1.3 \end{align}
The resulting expected value of 1.3 is close to the mean $$\mu$$, which we calculate by using the population’s probabilities (real probabilities are taken from the lower right figure in the previous section).
$\mu = 1 \cdot 0.2 + 2 \cdot 0.425 + 3 \cdot 0.275 + 4 \cdot 0.07 + 5 \cdot 0.025 = 1.31$
##### Exercise
Let us consider a fair six sided dice. We can easily compute the expected value $$E(X)$$ using R. The term fair means that each random variable $$X=x_i,\; x \in 1,2,3,4,5,6$$ is equally likely to occur. Therefore $$P(X=x_i) = \frac{1}{6}$$.
$E(X) = \sum_{i=1}^{6}x_iP(X=x_i) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} = 3.5$
In R we write the following code:
# Expected value of a fair six sided dice
p.die <- 1/6
die <- c(1,2,3,4,5,6)
sum(die*p.die)
## [1] 3.5
However, what if we are not sure if the dice is really fair? How to know that we are not cheated? Or to put it in other words: How often do we need to roll a dice before we can be more confident?
Let us do a computational experiment: We know from reasoning above that the expected value of a 6-sided fair dice is 3.5. We conduct an experiment by rolling a dice over and over again. We store the result and before we roll the dice again we calculate the average of all dice rolls so far. In order to achieve that little experiment, we write a for loop in R.
### Simulation ###
eyes <- seq(1,6) # possible events
probs <- rep(1/length(eyes),length(eyes)) # probabilities
expected.value <- round(sum(eyes*probs),2) # calculate expected value
n <- 500 # number of maximum rolls
values <- NULL # initialize an empty vector to store the value
averages <- NULL # initialize an empty vector to store the average values so far
#for-loop
for (roll in 1:n){
values <- c(values, sample(x = eyes, size = 1, prob = probs)) #sample method, type help(sample()) for further information
averages <- c(averages, mean(values))
}
### Plot ###
par(xpd = FALSE)
plot(x = seq(1:length(averages)),
y = averages,
type = 'l',
ylim = c(min(eyes), max(eyes)),
lwd = 2,
ylab = 'Expected value',
xlab = 'number of trials',
col = "#3366FF")
abline(h = expected.value, lty = 2, col = 'red')
legend('topright',
legend = paste("Expected value: ", as.character(expected.value)),
col = "red",
lty = 2)
The graph shows that after some initial volatile behavior, the curve finally flattens and approximates the $$E(X)$$ of 3.5.
#### Standard Deviation of a Discrete Random Variable
The standard deviation of a discrete random variable $$X$$ is denoted $$\sigma_X$$ or, when no confusion will arise, simply $$\sigma$$. It is defined as
$\sigma = \sqrt{\sum_{i=1}^{N}(x_i-\mu)^2P(X=x_i)}$
##### Exercise
Let us turn to R and calculate the standard deviation for the dice roll experiment from above. During the experiment we rolled 500 times. The outcome of these rolls are stored in the vector values. The probability for each of these numbers in the vector values approximates $$\frac{1}{6}= 0.167$$. So we just put those numbers in the equation for the standard deviation from above. Remember, that the mean is stored in the expected.value variable.
x <- seq(1,6)
p.x <- prop.table(table(values))
p.x
## values
## 1 2 3 4 5 6
## 0.160 0.184 0.170 0.140 0.180 0.166
sqrt(sum((x-mean(values))^2 * p.x))
## [1] 1.712882
Roughly speaking, our experiment showed that after 500 rolls, on average, the value of a dice number is 1.71 from the experimental mean of 3.494. | HuggingFaceTB/finemath | |
# What is the baseline in softball?
Baseline — The baselines in softball are 60 feet. When measuring baselines, the proper way to do it is from the back of home plate to the back white corner of first base. Then, from the back white corner of first base to the center of second base.
## What does baseline mean in softball?
The baseline is an imaginary area between bases. Base runners, especially when they are passing more than one base during a single play, typically do not run in a straight line from one base to the next.
## Can you run out of the baseline in softball?
Plate umpire Nic Lentz called Bell out for running out of the baseline per rule 5.09 (b) (1) that reads, When determining whether a base runner should be called out under Rule 5.09(b)(1), so long as the umpire determines that a play is being made on the runner and an attempt to tag is occurring, i.e. the fielder is
## What is the baseline rule?
And so if the fielder is if he if the runner crosses past with the pewter is attempting to make a play on another runner or add another base. Then he's allowed to run out of the base but but the this
## What is the baseline in baseball?
A baseline in baseball is a straight line between all of the bases. They are not visible on the field in between first base to second base and second base to third base. From home plate to first base and third base to home plate they coincide with the foul lines.
## Where is the first baseline?
The corner of home plate where the two 11-inch sides meet at a right angle is at one corner of a 90-foot (27.43 m) square. The other three corners of the square, in counterclockwise order from home plate, are called first, second, and third base.
## Where is the first base line?
Baseball’s first-base running lane is a 3-foot-wide (0.9-meter-wide) lane. Marked in chalk, it begins halfway between home plate and first base, stretching along the first-base foul line in foul territory. It ends just behind the first-base bag.
## Can a fielder stand in the baseline?
How do I spot obstruction? – Fielders without the ball often stand on a base or in the base path. Doing so does not make them guilty of obstruction. They’re not obstructing unless a runner’s advance or path is altered.
## What are the rules of base running in softball?
BASE RUNNING
• Runners must touch each base in order.
• Runners may overrun 1st base only, all other bases the runner may be tagged and. …
• Runners can not lead off a base, they must be on base until the ball as left the. …
• After a fly ball has been caught the base runner must tag the occupied base before.
## Can a player block the baseline?
AG: Players in the field can block any base as long as they are holding the ball or are in the process of fielding the ball. To do so without those conditions could result in the blocking player being called for obstruction and the base being awarded to the runner.
## How wide is the base path in softball?
The batter should run in the “lane,” which is a three-foot wide path on the foul side of the first base line; the lane starts halfway from home plate to first base and runs to the orange (outside) bag at first base.
## Where is the base path?
The base path is established when a fielder with the ball attempts to tag a runner. Then, and only then, is there a base path. And the base path is a straight line from the runner’s position to the base to which he is advancing or retreating.
## How far can you go out of the baseline?
three foot
Rule 6.05(j) The batter is out when in running the last half of the distance to first base, while the ball is being fielded to first base, the batter runs outside (to the right of) the three foot line, or inside (to the left of) the foul line, and in the umpire’s judgment in so doing interferes with the fielder taking …
## Can you slide in the middle of the base path?
In a nutshell: On a force play, the runner must slide on a line directly between the bases; it is allowable to slide through the base if the runner’s momentum carries him beyond the bag. If carried through the bag, contact with the defender beyond the bag is allowable so long as the slide is legal in all respects.
## Can you run back to home from first base?
3—On a tag play between home plate and first base, a batter-runner may retreat toward home plate to evade a tag, but shall be declared out after touching or passing home plate, or leaving the base line. The ball remains live. Therefore, a runner can back up to just before he would touch home plate. | HuggingFaceTB/finemath | |
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All dogs are the same color! (Posted on 2003-05-30)
Find the mistake in the following proof that all dogs are the same color (if there's any):
Let's use induction. Consider groups of 1 dog. All dogs of every group are the same color, of course. So we now that it's true for 1. Suppose it's true for groups of k dogs, i.e. every group of k dogs are the same color. Then let's consider any group A of k+1 dogs. Consider a subgroup of A containing k dogs. Let's call x the dog in A but not in the subgroup. Then by induction, all dogs in the subgroup are the same color. Now consider a subroup of A of k dogs, with x in the subgroup. All dogs except for x are the same color. Then, since every group of k dogs are the same color (by induction), all dogs in A are the same color. So x and every dog are the same color.
See The Solution Submitted by Fernando Rating: 3.5000 (10 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Error! Comment 23 of 23 |
When k=1, the proof is wrong. If we take k+1=2 dogs, then the subgroups of 1 dog are disjoint. Therefore, they can be of different colors.
Posted by Math Man on 2011-10-13 20:59:15
Search: Search body:
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### Ecology Lab Report
.. ). In our case we added up 1.1+ .0024 + .99=2.0924. The results for eight, the red larvae deloped was , while the obsereved removed was 19. The expected removed were 75*19/200=7.1. The chi square analysis for the red larvae deployed was (19-7.1)^2/7.1=19.9.
For day 8 the blue larvae developed was 70, while the observed removed was 0. The expected removed were 70*19/200=6.7. THe chi-square analysis for the blue larvae was (0-6.7)^2/6.7=6.7. For day 8 of the lime larvae, deployed was 55, while the observed removed was 0. The expected removed were 55*19/200=5.2. THe chi square analyis for the lime was (0-5.2)^2/5.2=5.2. When all of these are added up, the total chi square analysis was 19.9+6.7+5.2=31.8 For the time period in which the number of prey was removed, you calculated the chi anaylsis for the AM and period PM of they eight.
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For the time perid of AM we would take the number, of larvae deployed, times the number removed divided by the total number deployed for the AM time period. Then we would add them up to get the chi analysis. For the expected removed for red larvae, the calculated was 36*9/100=3.24. For the expected removed number of blue, we calculated 35*9/100=3.15. For expected removed number of lime larvae, we calculated 26*9/100=2.34.
To calculate the chi analysis for red larvae we would take the number removed minus the expected , then square it and divide by the number expected. That is (9-3.24)^2/3.24= 10.24. To find the chi analysis for the blue larvae you would calculate, (0-3.15)^2/3.15=3.15. To calculate the chi square analysis for the lime larvae you would calculate (0-2.43)^2/2.34=2.34. When all of these are added up 10.24+3.15+2.43=15.82 For the time period of PM, wwe would also take the number of larvae deployed, time the number removed removed divived by the total number deployed for the PM time period. Then we would add it up to find the chi analysis for the PM period. For the expected removed for red larvae, the calculated was 36(10)/100=3.6.
For the expected removed for blue 35(10)/100=3.5. For the expected removed red larvae, the calculated was 29(10)/100=2.9. To calculate the chi analysis was the red larvae, you take the number removed minus the expected, then square it and divide by the numberexpected. That is (10-3.6)^2/100=11.37. To find the chi analysis for the blue larvae, you would calculate, (0-3.5)^2/3.5=3.5.
To find the chi analysis for the lime larvae, the calculate was (0-2.9)^2/2.9. When all of these are added up 11.37+3.5+2.9=17.8. Discuss The results for this experiment did not surprise me at all. I did not expect that on the first day, that there would be a color favor, amongst the birds. I realized that although lime might be a more pleasant color than red or blue, the birds might not find this favoritism.
Although we knew that the red was the most palable, the birds would have to test the larvae first before coming to a conclusion as to which one they prefered. According to the data, I failed to reject my hypothesis because the chi sqare analysis that I calculated, 2.09, is smaller than the 5.991 which is the chi analysis given in the book. When the chi analysis is smaller, this shows that , my hypothesis was not refused and that there was no color prefrence chosen by the birds on the first day. Also through the graphs, one can also see frrom the pattern of the percentage line of removed food, vs time, that there was no larvae color preference. By the eighth day I expected that the birds would have realize that only the red was since the most palatable larvae, especially since there was only about twenty five percent lime that was palable. However, through tha graph, One can see that toward the eight day the birds were still feeding on some of the lime larvae.
This made it an oblivious indication that mimicry had occured. Also I calculated chi square value was much higher that given in the book, which is about a comparision of 31.8 vs 5.991. Since it is higher, I failed to accept my null hypothesis. This means that the birds up on till that they could still not really tell the difference, or had’t really noticed that some of the food was more palatable than others. In addition to the eight day, I had also seperated the time periods to see if there was any effect on temperature on the out come of the feed in each the birds removed the food. After graphing the data, I failed to accept my null hypothesis.
I thought that the diffent time periods of the day would affect when the birds would usually remove the larvae. I imagined that birds would come out first thig in the morning and then go bacl to their nest. However through the one could see that the results in both days were very similar. In addition when I calculated the chi square analysis for both periods, they were 17.7 for PM and 15.83 for Am. These numbers were both larger than the given value in the book of 3.84, which is why I failed to acceot my null hypothesis.
The results of this lab were sometimes a little difficult to interpret due to various mistakes. One large and costly mistake was missing data. Several groups did not report their data, which resulted in the graphs and calculations being somwhat unusual and often difficult to understand. There were also often large gaps in the graphs. The problem of negative numbers in the data threw off the graphs even more than it already was.
Also, although not proven, it is possible that othe organisms or enviromental conditions could have interfiered with the data collection. The factor of the wind must be taken into account for emoving the prey. If the prey was removed, it doesn’t mean that the predators had eaten it. The predator also could it in his or her mouth and spit it right back out. This response would be very important to mimicry , because the preadator or prey is realizing what is palatable and what isn’t.
The negative numbers that appeared in this experiment could have been the cause of experimental error, due to the fact that there were two extra larvae in each petri dish. There were quite a few mitakes in this data but at least we successfuly obtained viable data. According to the graphs, the birds preyed upon the red larvae the most. The red larvea was 100% palatable and 0% unpalatable. I expected the red larvae to be preyed upon the most, but what shocked me was the amount of lime that was still being preyed on while the birds approched the eighth day.
I thought that since there was only 25% of the lime larvae that was palable the would have picked it up right away. Even the blue larvae was removed about as much as s the lime was another indication that mimicry had occured. The blue was only 25% unpalatable there fore it should have yeilded much higher results than the lime. Some times as was a little confused as to whether my accepting or rejecting a null hypothesis was correct. This is because the chi square analysis isn’t a procedure thatalways ccurately reject or support the null hypothesis.
The chi square analysis is most likely a test of randomness, rather than a test for supporting or rejecting the null hypothesis. This is also why we say that we fail to rejact it, because, just because we fail to reject it doesn’t mean that it is correct, therefore we cannot accept it. This is also why it is good to have graphs, in an experiment like this one, so that they can back up your results and hypothesis, because it is giving you a visual sense of what is going on. Bibliography Charlene Ngong Science. | HuggingFaceTB/finemath | |
# How To Determine Limiting Reactant Given Grams 2021
Coach Gr » happy » How To Determine Limiting Reactant Given Grams 2021 How To Determine Limiting Reactant Given Grams 2021. Subtract the amount (in moles) of the excess reactant that will react from the amount. The determination of the limiting reactant is typically just a piece of a larger puzzle. How To Determine Limiting Reactant Given Grams 2021 from www.scottishindependencereferendum.info
2h2 + o2 = 2h2o. Divide by the coefficients of. Now use the moles of the limiting reactant to calculate the mass of the product.
### Calculate The Number Of Moles H 2 = Given Volume /Molar Volume.
Divide by the coefficients of the hydrogen. 2h2 + o2 = 2h2o. Now use the moles of the limiting reactant to calculate the mass of the product.
### Given The Grams Of Each Reactant, Convert Each Of.
How to determine limiting reactant given grams 2021 the determination of the limiting reactant is typically just a piece of a larger puzzle. Determine the limiting reactant, theoretical yield, and percent yield for the reaction. The stoichiometric coefficients tell us the molar ratio within the chemical equation.
### Subtract The Amount (In Moles) Of The Excess Reactant That Will React From The Amount.
4.3 limiting reactant, theoretical yield, & percent yield youtube from www.youtube.com. Moles of hcl = 0.25. Calculate the number of moles h 2 = 30/22400.
### Divide By The Coefficients Of.
Mole number of hydrogen= (4g hydrogen/ molar atomic mass of hydrogen) to consume 1.5 mole of oxygen, (2×1.5)=3 moles of hydrogen will be required. Answer if you're given the moles present of each reactant, and asked to find the limiting reactant of a certain reaction, then the simplest way to find. In most limiting reactant stoichiometry problems, the real goal is to determine how much product.
### The Determination Of The Limiting Reactant Is Typically Just A Piece Of A Larger Puzzle.
Remember to use the molar ratio between the limiting reactant and the product. | HuggingFaceTB/finemath | |
# Identifying Frieze Patterns Exploration
Objective: Become familiar with identifying border patterns as well as creating them.
## Materials Printed version of the Identifying Frieze Patterns Exploration: File:IdentifyFrieze.pdf
• Printed copy of the Identifying Frieze Patterns Exploration.
## Exploration
### Crystallographic Notation
The crystallographic notation for the frieze patterns is made up of four letters/numbers. The label always starts with a P, and the rest of the label is determined by the symmetries.
P M or 1 M, a or 1 2 or 1 The first symbol is always a P Vertical mirror line gives and M, otherwise we have 1 If the axis is a mirror line we get M, if there is a glide reflection we get a, otherwise we have a 1 Two fold rotation gives a 2, otherwise we get a 1
### Alternative Notation
The border patterns can be given fairly simple names consisting of 2 symbols. The first symbol is either 'M' or '1', depending on if the border pattern has a vertical line of symmetry. The second symbol is 'M', 'G', '2' or '1', depending on what other symmetries are present. (This notational system is derived from these meanings: M designates a mirror symmetry, G a glide-reflection symmetry, 2 a 2-fold rotation symmetry, and 1 the absence of a symmetry.)
Determining the symmetry group can then be accomplished by following a set of questions (we assume that the border patterns run from the left to the right, so that the terms horizontal and vertical are unambiguous). The order of the questions is important!
M or 1 M, G, 2 or 1 Vertical mirror line gives and M, otherwise we have 1 Is the axis is a mirror line? - M; Is there is a glide reflection? - G; Is there a 2-fold rotation? - 2; otherwise we get a 1
The correspondence between this system and the one in the text (the IUC notation) is as follows:
• 1M = p1m1
• 1G = p1a1
• 12 = p112
• 11 = p111
• MM = pmm2
• MG = pma2
• M1 = pm11
## Questions
1. What is the symmetry group for the following border pattern: ... FFFFFFFFFFFFFFFFFFFFFFF...
2. You can form all 7 border patterns if you start with F. Show the other 6.
3. What is the symmetry group for the following border pattern: ... BBBBBBBBBBBBBBBBBBBB...
4. You can form all 7 border patterns if you start with B. Show the other 6.
5. What is the symmetry group for the following border pattern: ... OOOOOOOOOOOOOOOOOO ...
6. You can form all 7 border patterns if you start with O. Show the other 6.
Handin: A sheet with answers to all questions. | HuggingFaceTB/finemath | |
# 6.7E: Exponential and Logarithmic Models (Exercises)
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For the following exercises, use this scenario: A doctor prescribes 300 milligrams of a therapeutic drug that decays by about $$17 \%$$ each hour.
54. To the nearest minute, what is the half-life of the drug?
55. Write an exponential model representing the amount of the drug remaining in the patient's system after $$t$$ hours. Then use the formula to find the amount of the drug that would remain in the patient's system after 24 hours. Round to the nearest hundredth of a gram.
For the following exercises, use this scenario: A soup with an internal temperature of $$350^{\circ}$$ Fahrenheit was taken off the stove to cool in a $$71^{\circ} \mathrm{F}$$ room. After fifteen minutes, the internal temperature of the soup was $$175^{\circ} \mathrm{F}$$.\
56. Use Newton's Law of Cooling to write a formula that models this situation.
57. How many minutes will it take the soup to cool to $$85^{\circ} \mathrm{F} ?$$
For the following exercises, use this scenario: The equation $$N(t)=\frac{1200}{1+199 e^{-0.625 t}}$$ models the number of people in a school who have heard a rumor after $$t$$ days.
58. How many people started the rumor?
59. To the nearest tenth, how many days will it be before the rumor spreads to half the carrying capacity?
60. What is the carrying capacity?
For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic.
61.
x f(x)
1 3.05
2 4.42
3 6.4
4 9.28
5 13.46
6 19.52
7 28.3
8 41.04
9 59.5
10 86.28
62.
x f(x)
0.5 18.05
1 17
3 15.33
5 14.55
7 14.04
10 13.5
12 13.22
13 13.1
15 12.88
17 12.69
20 12.45
63. Find a formula for an exponential equation that goes through the points (-2,100) and (0,4). Then express the formula as an equivalent equation with base $$e$$.
This page titled 6.7E: Exponential and Logarithmic Models (Exercises) is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. | HuggingFaceTB/finemath | |
Top 10 things to know about circles
1. A circle is the set of points in a plane that are equidistant from a given point called the center.
2. The distance from the center is called the radius, and the point is called the center. Twice the radius is known as the diameter, .
3. The angle a circle subtends from its center is a full angle, equal to or radians.
4. The perimeter of a circle is called the circumference, and is given by .
5. The area of a circle is given by .
6. A central angle is an angle with endpoints and located on a circle's circumference and vertex located at the circle's center.
7. For an inscribed angle and central angle with the same endpoints, 8. A tangent to a circle at a point is perpendicular to the radius .
9. The circumcircle is a triangle's circumscribed circle, i.e., the unique circle that passes through each of the triangle's three vertices. The center of the circumcircle is called the circumcenter, and the circle's radius is called the circumradius. A triangle's three perpendicular bisectors (i.e. the line perpendicular to the side from the midpoint of the side) meet at .
10. The incircle is the inscribed circle of a triangle , i.e., the unique circle that is tangent to each of the triangle's three sides. The center of the incircle is called the incenter, and the radius of the circle is called the inradius. The incenter is the point of concurrence of the triangle's angle bisectors.
This document was generated using the LaTeX2HTML translator Version 2002-2-1 (1.70)
Copyright © 1993, 1994, 1995, 1996, Nikos Drakos, Computer Based Learning Unit, University of Leeds.
Copyright © 1997, 1998, 1999, Ross Moore, Mathematics Department, Macquarie University, Sydney.
The command line arguments were:
latex2html 10_things_circles.tex
The translation was initiated by Steven Dunbar on 2006-02-02 Steven R. Dunbar
Department of Mathematics and Statistics | HuggingFaceTB/finemath | |
##### Roderick's credit card company calculates a finance charge based upon a periodic
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Roderick's credit card company calculates a finance charge based upon a periodic rate of 1.2% on all unpaid balances. If Roderick has an unpaid balance of \$200, determine the finance charge that he will be assessed.
\$301.80 \$298.20 \$2.40 \$54.00
Dec 19th, 2014
This question can be reduced to:
" what is 1.2% of 200\$?" Whenever you see the word "of" it means to multiply. So, this problem becomes 1.2% x \$200, and you should change the 1.2% to .012; thus .012 x \$200 = \$2.40
Dec 19th, 2014
...
Dec 19th, 2014
...
Dec 19th, 2014
Sep 25th, 2017
check_circle | HuggingFaceTB/finemath | |
Rule 18:
To add fractions, the denominators must be equal. Complete the following steps to add two fractions.
1.
Build each fraction so that both denominators are equal.
2.
Add the numerators of the fractions.
3.
The denominators will be the denominator of the built-up fractions.
4.
Problem 2:
Solution:
The denominators are different, so you must build each fraction to a form where they both have the same denominator. Since both 5 and 15 will divide evenly into 15, build both fractions to a denominator of 15. Build the fraction to an equivalent fraction whose denominator is 15.
.
can now be written .
4 + 9 = 13. Add the fractions to get .
Check:
Now prove to yourself with your calculator that your answer is correct. Calculate 4 divided by 15, 3 divided by 5, and add the results. Now divide 13 by 15. Both answers should be the same. If you are correct, the answers are the same (equivalent) and you have successfully added two fractions.
[Previous Problem] [Next Problem] [Menu Back] | HuggingFaceTB/finemath | |
### Math and Linear equations
How do you solve 2x-3x/5=7
Subject
Level
Middle School
6
if 3x+4 a factor of 15x2 + 41x + 28
Subject
Level
Middle School
8
x+y=14 by substitution method
Subject
Level
Homeschool
10
How do you solve x + 2 = 10
Subject
Level
Middle School
7
How to solve x +2 =10
Subject
Level
Junior High
6
5x+3y=9; 4y+7x=13
In matrix method
Subject
Level
Highschool
9
The equation of x- axis
Subject
Level
Junior High
10
(- 3) multiply (- 5) x (- 4)
Subject
Level
Highschool
7
7+3÷3·6
Subject
Level
Middle School
3
How to solve 3t+36=69
Subject
Level
Highschool
7
Solve 11x-5=-5
Subject
Level
Highschool
8
$$x+2534=8129$$
Subject
Level
Primary
4
Sam wants to write an equation to represent a proportional relationship with a constant of proportionality of 6. He writes the equation y=6+x. Is he correct. If not what is the correct equation? Explain the answer. Consider the graph of the equation in the response.
Subject
Level
Middle School
8
How do you solve x - 36 〓 36 + 23
Subject
Level
Highschool
1
how do you solve x-22=10
Subject
Level
Highschool
7
How do you solve 5x+2+3x
Subject
Level
Highschool
9
How do you solve 3a-8a+5a+9a-2a
Subject
Level
Highschool
8
2X+53=3X-6
Subject
Level
Highschool
8
2x_3={8}
Subject
Level
Middle School
8
How do you sovle x+2=?
Subject
Level
Middle School
8
How to solve x+12=-34
Subject
Level
Highschool
12
2p=3(p-3)
Subject
Level
Junior High
8
The sum of three consecutive integer is 18.
Subject
Level
Primary
7
2x=12+12
Subject
Level
Middle School
8
How do you solve x/2-7=9
Subject
Level
Middle School
7
How do you solve y-15=-4
Subject
Level
Highschool
9
Solve the equation 6x+2y=9 for y in terms of x
Subject
Level
Highschool
10
How do you convert to standard form
3x+2y+4=0
Solution:
- \frac{3 x}{2} - 2
Derivative
\frac{\partial}{\partial y} 3 x + 2 y + 4 = 0
Complex Roots
- \frac{3 x}{2} - 2
Subject
Level
Highschool
9
What is the highest power of the variable in the linear expressions in one variable?
Subject
Level
Highschool
8
X + 9 = 24
Subject
Level
Highschool
10
2p=3 (p-3)
Subject
Level
Junior High
8
Subject
Level
Middle School
6
x+y+z=3. 13x+2z=2. -x-5z=-5
Subject
Level
Highschool
10
Subject
Level
None
Perimeter of rectangular
Subject
Level
Highschool
8
what will be the answer of this solution?
Subject
Level
Highschool
10
How to solve x + 10 = 12.
Subject
Level
Middle School
7
Ho do I solve 3x-4=5
Subject
Level
Middle School
4
how to solve a+23=45
Subject
Level
Middle School
7
Resolve 27a^3+8b^3 into two real factors
Subject
Level
Highschool
9
9x²-1÷3x²+2x-1
Subject
Level
Highschool
12
How to solve x+3=10
Subject
Level
Middle School
8
How to solve x+2=10
Subject
Level
Highschool
7
How do you solve 8x+7=4x-17
Subject
Level
Highschool | HuggingFaceTB/finemath | |
# Finding maximum likelihood function for 2 normally distributed samples
1. Apr 2, 2012
### Gullik
1. The problem statement, all variables and given/known data
The question is about how to combine to different samples done with 2 different methods of the same phenomena.
Method 1 gives normally distributed variables $X_1,X_2,...X_{n_1}$, with $\mu$ and $\sigma^2_1$
Method 1 gives normally distributed variables $Y_1,Y_2,...,Y_{n_2}$, with $\mu$ and $\sigma^2_2$
All measurements can be considered independent.
Define $\overline{X}=1/n_1*\Sigma^{n_1}_{i=1}X_i$ and $\overline{Y}={1/n_2}*\Sigma^{n_2}_{j=1}Y_j$
Both the samples are taken from the same population so they have the same $\mu$
The question is set up a likelihood function for the $n_1 + n_2$ measurements, and show that the maximum likelihood estimator is given by
$\mu_{mle}={(\sigma^2_2n_1\overline{X}+\sigma_1^2n_2\overline{Y})/(\sigma_2^2n_1+\sigma^2_1n_1)}$
2. Relevant equations
The likelihood function for 1 sample of a normally distributed function.
$L(\mu)=L(X_1,X_2,...X_n;\mu,\sigma^2)=1/((2\pi\sigma^2)^{n/2})*e^{-1/2*\Sigma^{n}_{i=1}(X_i-\mu)/\sigma}$
3. The attempt at a solution
The main problem is that I don't know how I should handle the 2 samples instead of 1, I'm not sure how those two should combine.
Edit; I think I came up with the solution while I was writing up this which took quite some time since I don't know latex. I'll see when I try it out.
Since the measurement is presumed independent does the likelihood function become something like?
$L(\mu)=L(X_1,X_2,...X_{n_1};\mu,\sigma^2_1)*L(Y_1,Y_2,...,Y_{n_2};\mu,\sigma_2^2)$
Last edited: Apr 2, 2012 | HuggingFaceTB/finemath | |
# Decrease in Potential Energy of a Sliding Tile
• MHB
• Shah 72
In summary, a 1.2 kg tile slides 3m down a roof with an angle of 35 degrees to the horizontal, resulting in a decrease in potential energy of 20.6 J. There was an error in the calculation of potential energy due to using 10 m/s^2 instead of 9.8 m/s^2 as the acceleration due to gravity.
Shah 72
MHB
A tile of mass 1.2 kg slides 3m down a roof that makes an angle of 35 degree to the horizontal. Find the decrease in potential energy.
Iam getting the ans 24.8J
PE = mgh= 1.2× 12 sin 35 ×3
The ans in the textbook is 20.6J
You have the right idea but something went wrong.
Your mistake: PE = mgh = 1.2 x 10 x (3 x sin(35)). Where did you get the 12 from?
PE = mgh = 1.2 x 10 x (3 x sin(35)) = 20.6 J.
(You have to use a Scientific calculator anyway, so I don't understand why you would be told to use g = 10 m/s^2 instead of 9.8 m/s^2. I think it's silly!)
-Dan
topsquark said:
You have the right idea but something went wrong.
Your mistake: PE = mgh = 1.2 x 10 x (3 x sin(35)). Where did you get the 12 from?
PE = mgh = 1.2 x 10 x (3 x sin(35)) = 20.6 J.
(You have to use a Scientific calculator anyway, so I don't understand why you would be told to use g = 10 m/s^2 instead of 9.8 m/s^2. I think it's silly!)
-Dan
Thank you!
## 1. What is potential energy?
Potential energy is the stored energy an object has due to its position or configuration. It is the energy that an object has the potential to convert into other forms of energy, such as kinetic energy.
## 2. How is potential energy related to a sliding tile?
In the context of a sliding tile, potential energy refers to the energy that the tile has due to its position on the surface. As the tile slides down a slope, its potential energy decreases, and this energy is converted into kinetic energy.
## 3. What factors affect the decrease in potential energy of a sliding tile?
The decrease in potential energy of a sliding tile is affected by the height of the slope, the mass of the tile, and the acceleration due to gravity. The higher the slope, the greater the decrease in potential energy. Similarly, a heavier tile will have more potential energy than a lighter tile, and thus a greater decrease in potential energy as it slides down the slope.
## 4. Why does the potential energy decrease as the tile slides?
The decrease in potential energy of a sliding tile is due to the conversion of potential energy into kinetic energy. As the tile slides down the slope, its potential energy decreases, and its kinetic energy increases, resulting in a constant total energy.
## 5. How is the decrease in potential energy of a sliding tile calculated?
The decrease in potential energy of a sliding tile can be calculated using the formula PE = mgh, where m is the mass of the tile, g is the acceleration due to gravity, and h is the height of the slope. This formula represents the potential energy an object has at a certain height above the ground.
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989 | HuggingFaceTB/finemath | |
DofotheroU
2020-12-24
For Exercise,
a. Identify the equation as representing a circle, an ellipse, a hyperbola, or a parabola.
b. Graph the curve. c. Identify key features of the graph. That is. If the equation represents a circle, identify the center and radius.
If the equation represents an ellipse, identify the center, vertices, endpoints of the minor axis, foci, and eccentricity.
If the equation represents a hyperbola, identify the center, vertices, foci, equations of the asymptotes, and eccentricity.
If the equation represents a parabola, identify the vertex, focus, endpoints of the latus rectum, equation of the directrix, and equation of the axis of symmetry.
funblogC
Step 1
a)For the given equation simplify it and compare with standard forms of conic sections to determine which conic section or a figure it represents as.
Which is a hyperbola.
Step 2
Now to find the features of the given hyperbola there are standard forms where these features can be determined. Let us determine centre ‘C’ of this hyperbola by comparing with standard form of such hyperbola as.
here
Step 3
Vertices "V" for this hyperbola can be determined to be as.
${a}^{2}=6$
Step 4
Foci "F" for this hyperbola can be determined to be as
Step 5
Equation of asymptotes for this hyperbola can be determined to be as.
Step 6
Eccentricity "e" is calculated by the formula as.
Do you have a similar question? | HuggingFaceTB/finemath | |
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HW11-Solutions
# HW11-Solutions - Version PREVIEW – HW11 – feng...
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Unformatted text preview: Version PREVIEW – HW11 – feng – (58230) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Displacement Current 03 001 (part 1 of 2) 10.0 points A capacitor of capacitance C has a charge Q at t = 0. At that time, a resistor of resistance R is connected to the plates of the charged capacitor. Find the magnitude of the dis- placement current between the plates of the capacitor as a function of time. 1. Q RC e − t/Q 2. RC Q e − t/ ( RC ) 3. Q RC e t/ ( RC ) 4. RC Q e t/ ( RC ) 5. Q RC e − t/ ( RC ) correct Explanation: Basic Concept RC circuits. Displacement Current. The displacement current is defined to be I d = ǫ d Φ E dt . The electric field inside a capacitor is essen- tially uniform and E = q ǫ A . Since the charge on a capacitor in a discharging RC circuit is given by q ( t ) = Q e − t/RC , the displacement current is found by I d = ǫ d Φ E dt = ǫ d dt parenleftbigg q ǫ A A parenrightbigg = dq dt =- Q RC e − t/ ( RC ) . Note that the displacement current equals the actual current in the wires to the capacitor. Thus, the Ampere-Maxwell law tells us that vector B will be the same regardless of which current we evaluate. 002 (part 2 of 2) 10.0 points Given C = 2 μ F, Q = 28 μ C, R = 391 kΩ, and ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 , at what rate is the electric flux between the plates changing at time t = 0 . 17 s? Correct answer:- 3 . 2538 × 10 6 Vm / s. Explanation: Let : ǫ = 8 . 85419 × 10 − 12 C 2 / N · m 2 , t = 0 . 17 s , C = 2 μ F , Q = 28 μ C = 2 . 8 × 10 − 5 C , and R = 391 kΩ = 3 . 91 × 10 5 Ω . From the discussion in the previous part, we know that d Φ E dt = I d ǫ =- Q ǫ RC e − t/ ( RC ) =- 2 . 8 × 10 − 5 C ǫ (3 . 91 × 10 5 Ω) (2 × 10 − 6 F) × e bracketleftbigg- . 17 s (3 . 91 × 10 5 Ω)(2 × 10 − 6 F) bracketrightbigg =- 3 . 2538 × 10 6 Vm / s . Which Maxwell Equation 003 (part 1 of 6) 10.0 points Which of Maxwell’s equations can be used, along with a symmetry argument, to calculate the electric field of a point charge? 1. contintegraldisplay vector B · d vectors = μ I + μ ǫ ∂ Φ E ∂t 2. contintegraldisplay vector E · d vectors =- ∂ Φ B ∂t 3. contintegraldisplay vector B · d vectors = 0 4. contintegraldisplay vector E · d vectors = Q ǫ Version PREVIEW – HW11 – feng – (58230) 2 5. contintegraldisplay vector B · d vector A = μ I 6. contintegraldisplay vector E · d vector A =- ∂ Φ B ∂t 7. contintegraldisplay vector B · d vector A = Q ǫ 8. contintegraldisplay vector E · d vector A = Q ǫ correct 9....
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# Evaluate the following integrals:
Question:
Evaluate the following integrals:
$\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x$
Solution:
Given I $=\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x$
Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$
Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$
$\Rightarrow p x+q=\lambda(2 a x+b)+\mu$
$\Rightarrow 2 x+3=\lambda(2 x+4)+\mu$
$\therefore \lambda=1 / 2$ and $\mu=-1$
Let $2 x+3=2 x+4-1$ and split,
$\Rightarrow \int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x=\int\left(\frac{2 x+4}{\sqrt{x^{2}+4 x+5}}-\frac{1}{\sqrt{x^{2}+4 x+5}}\right) d x$
$=2 \int \frac{x+2}{\sqrt{x^{2}+4 x+5}} d x-\int \frac{1}{\sqrt{x^{2}+4 x+5}} d x$
Consider $\int \frac{x+2}{\sqrt{x^{2}+4 x+5}} d x$
Let $\mathrm{u}=\mathrm{x}^{2}+4 \mathrm{x}+5 \rightarrow \mathrm{dx}=\frac{1}{2 \mathrm{x}+4} \mathrm{du}$
$\Rightarrow \int \frac{x+2}{\sqrt{x^{2}+4 x+5}} d x=\int \frac{1}{2 \sqrt{u}} d u$
$=\frac{1}{2} \int \frac{1}{\sqrt{u}} d u$
We know that $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$
$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{u}} d u=\frac{1}{2}(2 \sqrt{u})$
$=\sqrt{u}=\sqrt{x^{2}+4 x+5}$
Consider $\int \frac{1}{\sqrt{x^{2}+4 x+5}} d x$
$\Rightarrow \int \frac{1}{\sqrt{x^{2}+4 x+5}} d x=\int \frac{1}{\sqrt{(x+2)^{2}+1}} d x$
Let $u=x+2 \rightarrow d x=d u$
$\Rightarrow \int \frac{1}{\sqrt{(x+2)^{2}+1}} d x=\int \frac{1}{\sqrt{u^{2}+1}} d u$
We know that $\int \frac{1}{\sqrt{x^{2}+1}} d x=\sinh ^{-1} x+c$
$\Rightarrow \int \frac{1}{\sqrt{u^{2}+1}} d u=\sinh ^{-1}(u)$
$=\sinh ^{-1}(x+2)$
Then,
$\Rightarrow \int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x=2 \int \frac{x+2}{\sqrt{x^{2}+4 x+5}} d x-\int \frac{1}{\sqrt{x^{2}+4 x+5}} d x$
$=2 \sqrt{x^{2}+4 x+5}-\sinh ^{-1}(x+2)+c$
$\therefore I=\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x=2 \sqrt{x^{2}+4 x+5}-\sinh ^{-1}(x+2)+c$ | HuggingFaceTB/finemath | |
Home Board Paper Solutions Text Book Solutions Articles NCERT Exemplar Solutions
Question 17
# Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure.R = 0.083 bar L K–1 mol–1.
From the gas equation,
PV = (w/M) RT
Where w is the mass of the gas, M is the molar mass of the gas
For CO2 , M = 44 g/mol
Substituting the given values ,we get
1 x V = (8.8 / 44) x 0.083 x 304.1
= 5.05 L
Hence, the volume occupied is 5.05 L. | HuggingFaceTB/finemath | |
# The Magnitude of a Vector
## Homework Statement
For any vector in 2D space, it can be broken down into its horizontal and vertical components.
## Homework Equations
In one of my engineering classes, we are using the following equation to determine the magnitude of a vector:
$$u=v_1 \cdot cos\theta +u_2 \cdot sin\theta$$
Where $\theta$ is the angle with respect to the horizontal, v1 is the horizontal component and v2 is the vertical component of the vector.
I know this equation works but I don't understand why.
I feel like I am missing a fundamental concept, because to determine the magnitude of a vector, I would use Pythagoras theorem, and I cannot derive the above equation from Pythagoras's equation.
## Answers and Replies
gneill
Mentor
Can you provide some context for where this equation is applied? Perhaps give a specific example.
In general, this equation will not work for a single vector whose x and y components are ##u_1## and ##u_2##. Perhaps they are summing the horizontal components of two different vectors to obtain a net horizontal resultant?
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
That equation doesn't give the magnitude of the vector. It gives you the component of the vector in the direction of ##\hat n = \cos\theta\,\hat i + \sin\theta\,\hat j##.
CWatters
Science Advisor
Homework Helper
Gold Member
In one of my engineering classes, we are using the following equation to determine the magnitude of a vector:
u=v1⋅cosθ + u2⋅sinθ
Where $\theta$ is the angle with respect to the horizontal, v1 is the horizontal component and v2 is the vertical component of the vector.
I know this equation works but I don't understand why.
It comes from geometry... See this diagram... If that's not clear do say and I will explain some more. #### Attachments
• CivilSigma
Chestermiller
Mentor
I think you meant to write the equation as $$u=u_1\cos{\theta}+u_2\sin{\theta}\tag{1}$$where $$u_1=u\cos{\theta}\tag{2}$$and$$u_2=u\sin{\theta}\tag{3}$$If you substitute Eqns. 2 and 3 into Eqn. 1, you get:
$$u=u\cos^2{\theta}+u\sin^2{\theta}=u$$
• CivilSigma and SammyS | HuggingFaceTB/finemath | |
Definitions
# What is the definition of adding and subtracting integers?
###### Answered 2012-09-11 04:38:31
adding and subtracting integers is when you add and minus 2 numbers
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## Related Questions
Integers are whole numbers, both positive and negative. Therefore, adding and subtracting integers would be adding and subtracting whole numbers. Examples: 8+2 -8+2 8-2 -8-2
I would think that the commonality of adding and subtracting integers is that the answer itself will always be an integer. In other words, the answer is always gonna be a "whole number".
No- adding negative numbers is like adding positive numbesr , except the answer is negative.
They aren't. The rules are the same as those for adding/subtracting or multiplying integers. Just be careful of the decimal point's location.
It's pretty much the exact same. Subtracting is the same as adding a negative number. Foe example, five minus three is the same as five plus negative three.
to subtrct integers ,rewrite as adding opposites and use the rules for addtion of integers..
Subtracting two positive fractional numbers, or adding one positive and one negative integer.
When adding negative integers, you subtract. (2+-1=1) When subtracting negative integers, you add. (2--3=5)
Adding IntegersTo add integers, one must consider the following two rules to be a successful.If you want to think of it on the number line you start from 0 and when you add a positive number you go that much to the right, and when you add a negative number you go that much to the left. When adding two positive integers, just add like normal. When adding one positive integer, and one negative integer, it is like subtracting a positive number from a positive number. When adding two negative integers, it is like subtracting a positive number from a negative number.
When you can do addition and subtraction without recourse to it. Then it simply becomes a distraction.
It is the additive identity and so it leaves the number(s) unchanged.
Subtracting an integer is the same as adding the additive inverse. In symbols: a - b = a + (-b), where "-b" is the additive inverse (the opposite) of b.
Adding two numbers with different signs means subtracting the two absolute integers (without sign) and vice versa.
for integers with tow different signs, it will always be negative in multiplying and division. for adding and subtracting, the sign is for the bigger number.
Adding integers, if they have the same sign, add their absolute values and keep the same sign. Subtracting, change the sign of the 2nd number and the add using rules of addition. Multiplying and dividing, Divide the absolute values, if the signs are the same the answer is positive, if the signs are different the answer is negative.
Subtracting a fraction is the same as adding its negative.
###### Business & FinanceMath and ArithmeticNumbers AlgebraSchool SubjectsPercentages, Fractions, and Decimal Values
Copyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. | HuggingFaceTB/finemath | |
Find for which value of the parameter $k$ a function is bijective
I have to draw (by hand obviously) the plot of the following function:
$$f(x)= 13\ln(\frac{x}{|x+1|})-12\ln (x+x^2) +kx,$$
for $k \in \mathbb{R}$. To do so, I have to study the first and second derivative, limits at infinity, and so on. Normally, I do these exercises with functions without parameters. Could you show me how to proceed in this case?
Also, I have to determine for which values of $k$ this function is bijective. However, I don’t have a clue about how to proceed, as I’ve never been shown exercises of this kind.
Solutions Collecting From Web of "Find for which value of the parameter $k$ a function is bijective"
Due to the first term, the function is only defined for $x>0$, because $\frac{x}{|x+1|}>0$ if and only if $x>0$. So we’ll assume $f\colon(0,\infty)\to\mathbb{R}$.
So first of all rewrite it as
$$f(x)=13\ln x-13\ln(x+1)-13\ln x-12\ln(1+x)+kx=\ln x-25\ln(x+1)+kx$$
We easily have
$$\lim_{x\to0}f(x)=-\infty,$$
In order that the function is onto $\mathbb{R}$, we need that $\lim_{x\to\infty}f(x)=\infty$. Now
$$f(x)=\ln\frac{x}{(x+1)^{25}}+kx$$
and the limit at $\infty$ of the first term is $-\infty$. So this forces $k>0$, because for $k\le0$ we have $\lim_{x\to-\infty}f(x)=-\infty$. When $k>0$ we indeed have $\lim_{x\to\infty}f(x)=\infty$ (verify it).
Now we need that the function is monotonic, so we look at the derivative
$$f'(x)=\frac{1}{x}+\frac{25}{1+x}+k$$
Under the hypothesis that $k>0$ this is everywhere positive, so the function is indeed increasing. | HuggingFaceTB/finemath | |
# Inequation
Solve the inequation:
5k - (7k - 1)≤ 2/5 . (5-k)-2
Result
k≥: 0.625
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
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# Simple Harmonic Motion (F=-kx) - Help!
1. Apr 9, 2013
### ProgressNation
Simple Harmonic Motion (F=-kx) -- Help!
Q: The formula for Hooke's LAW is Felastic=-kx , so, the "X" is always negative? And the "F" is always negative? I know the "K" is always should be positive but please explain to me because I'm confused which to put negative and which to put positive; I wanna understand!
I'm waiting for an answer ASAP!
Thank you!
2. Apr 9, 2013
### Averki
Consider a spring with one end fixed to a wall and the other end connected to a block. If you were to pull the block in the positive x-direction, in which direction would the spring force act? If you were instead to push the block in the negative x-direction, in which direction would the spring force act?
3. Apr 9, 2013
### ProgressNation
If I pulled the block in the positive x-direction, the Spring Force will act in the negative Direction so the force will be negative. And if was pulling the block in the negative x-direction, the Spring force will act on the opposite direction (the positive direction) so the force will be positive. Am I right?
But the problem is that the question doesn't say to you that if the block was pulled either right or left direction like this question:
If a mass of 0.55 KG attached to a vertical spring stretches the spring 2.0 cm from its original equilibrium position, what is the spring constant?
If you notice there is no force in this question so it's Fg=mg
And so on...
So this question only says "stretches the spring 2.0 cm from it's original equilibrium position", it doesn't say it's either right,left,north,south.
4. Apr 9, 2013
### Averki
Exactly! The spring will always oppose the applied force, in this case being your pulling / pushing the block.
While it is true that the problem does not explicitly specify the direction of the stretching, it is implied. If this spring is vertical, presumably with one end of the spring fixed to, say, a ceiling and the other end attached to the 0.55kg block, logically, which direction would the applied force, in this case the force of gravity, stretch the spring?
5. Apr 10, 2013
### sophiecentaur
This is a correct way of describing what happens - but it took a lot of words. However, a half dozen symbols plus the all-important minus sign describe it better because you can often solve things once they have been put into a mathematical equation. The minus sign just tells you the force is a 'restoring force' - in a direction which is against the displacement. You need to have the confidence to go into the Maths and believe what comes out. (There are times when you do need to give the mathematical answer a credidbility test, though because it can throw up unreal answers).
The oscillation of a mass on a spring is not affected, in principle, by whether the spring is pre-stressed by the mass hanging down, being pushed up or not under gravity at all as long as the force displacement law is not distorted by extra influences - like the spring coils starting to touch each other. | HuggingFaceTB/finemath | |
# Converting Molarity to PPM
Converting molarity to ppm (parts per million) is simple using the factor label method. Let’s first look at what ppm (or pph, or ppt) stands for. PPM means parts per million, pph is parts per hundred, and ppt is parts per thousand. The following expression is another way to express ppm: $\frac{g solute}{g solvent} * 10^6$
Lets try an example where we have a solution of 25mM Ca in water. How would we convert this solution to ppm? To solve this problem we will set up a factor label problem. You are probably quite familiar with the factor label method for converting 1 unit (or tag) to another tag (i.e. 300milliliters to liters). Our calcium problem is a bit more complicated because now we have tags on the top and on the bottom. On the top we have mmol and on the bottom we have liter. Both of those tags need to be converted to grams for us to determine ppm. $25mM = \frac{25mmol}{liter H_20}$
Step 1: Convert the top tag (25mmol) to grams. Since we want to convert a mmol quantity to grams, we will use the element’s molar mass. For an explanation of molar mass, see this post. Molar mass has the units $\frac{grams}{mol}$ so we must first convert mmol to mol: $\frac{25mmol Ca}{L H_20} * \frac{1mol Ca}{1000mmol Ca}$
Now we can add calcium’s molar mass: $\frac{25 mmol Ca}{L H_20} * \frac{1 mol Ca}{1000 mmol Ca} * \frac{40 g Ca}{1 mol Ca}$
We are now done with the top tag; mmol has been converted to grams. Lets tackle the bottom tag (liter H20).
Step 2: Our next step is to convert liters of water to grams of water. The great thing about the factor label method is that we can just continue with our last equation to convert liters to grams. This step requires the density of water (1g/ml). To use density, we must first convert liters to milliliters, then we can insert the density of water: $\frac{25mmol Ca}{L H_20} * \frac{1mol Ca}{1000mmol Ca} * \frac{40g Ca}{1mol Ca} * \frac{1L H_20}{1000 mL H_20} * \frac{1mL H_20}{1gH_20}$
Notice we flipped the density of water around so that mL is on the top. This way mL will cancel out and we will be left with grams H20 in the bottom.
Since the tags cross out at each step, our only remaining tags are $\frac{g Ca}{gH_20}$. Multiply by 1,000,000 (or 100 if you want pph; 1,000 if you want ppt) and you have converted 25mM Ca into ppm! Can you convert ppm to molarity? This entry was posted in Chemistry, Factor-Label Method, Math. Bookmark the permalink. | HuggingFaceTB/finemath | |
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# 2.E: Limits (Exercises)
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## 2.1: A Preview of Calculus
1) [T] Complete the following table with the appropriate values: y-coordinate of Q, the point $$Q(x,y)$$, and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.
$$x$$ $$y$$ $$Q(x,y)$$ $$m_{sex}$$
1.1 a. e. i.
1.01 b. f. j.
1.001 c. g. k.
1.0001 d. h. l.
Solution: a. 2.2100000; b. 2.0201000; c. 2.0020010; d. 2.0002000; e. (1.1000000, 2.2100000); f. (1.0100000, 2.0201000); g. (1.0010000, 2.0020010); h. (1.0001000, 2.0002000); i. 2.1000000; j. 2.0100000; k. 2.0010000; l. 2.0001000
2) Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the line tangent to f at $$x=1$$.
3) Use the value in the preceding exercise to find the equation of the tangent line at point P. Graph $$f(x)$$ and the tangent line.
Solution: $$y=2x$$
For the following exercises, points $$P(1,1)$$ and $$Q(x,y)$$ are on the graph of the function $$f(x)=x^3$$.
1) [T] Complete the following table with the appropriate values: y-coordinate of Q, the point $$Q(x,y)$$, and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.
$$x$$ $$y$$ $$Q(x,y)$$ $$m_{sec}$$
1.1 a. e. i.
1.01 b. f. j.
1.001 c. g. k.
1.0001 d. h. l.2
2) Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at $$x=1$$.
Solution: 3
3) Use the value in the preceding exercise to find the equation of the tangent line at point P. Graph $$f(x)$$ and the tangent line.
For the following exercises, points $$P(4,2)$$ and $$Q(x,y)$$ are on the graph of the function $$f(x)=\sqrt{x}$$.
1) [T] Complete the following table with the appropriate values: y-coordinate of Q, the point $$Q(x,y)$$, and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.
$$x$$ $$y$$ $$Q(x,y)$$ $$m_{sec}$$
4.1 a. e. i.
4.01 b. f. j.
4.001 c. g. k.
4.0001 d. h. l.
Solution: a. 2.0248457; b. 2.0024984; c. 2.0002500; d. 2.0000250; e. (4.1000000,2.0248457); f. (4.0100000,2.0024984); g. (4.0010000,2.0002500); h. (4.00010000,2.0000250); i. 0.24845673; j. 0.24984395; k. 0.24998438; l. 0.24999844
2) Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at $$x=4$$.
3) Use the value in the preceding exercise to find the equation of the tangent line at point P.
Solution: $$y=\frac{x}{4}+1$$
For the following exercises, points $$P(1.5,0)$$ and $$Q(ϕ,y)$$ are on the graph of the function \9f(ϕ)=cos(πϕ)\).
1) [T] Complete the following table with the appropriate values: y-coordinate of Q, the point $$Q(x,y)$$, and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.
$$x$$ $$y$$ $$Q(x,y)$$ $$m_{sec}$$
1.4 a. e. i.
1.49 b. f. j.
1.499 c. g. k.
1.4999 d. h. l.
2) Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the tangent line to f at $$x=4$$.
Solution: $$π$$
3) Use the value in the preceding exercise to find the equation of the tangent line at point P.
For the following exercises, points $$P(−1,−1)$$ and $$Q(x,y)$$ are on the graph of the function $$f(x)=\frac{1}{x}$$.
[T] Complete the following table with the appropriate values: y-coordinate of Q, the point $$Q(x,y)$$, and the slope of the secant line passing through points P and Q. Round your answer to eight significant digits.
$$x$$ $$y$$ $$Q(x,y)$$ $$m_{sec}$$
-1.05 a. e. i.
-1.01 b. f. j.
-1.005 c. g. k.
-1.001 d. h. l.
Solution: a. −0.95238095; b. −0.99009901; c. −0.99502488; d. −0.99900100; e. (−1;.0500000,−0;.95238095); f. (−1;.0100000,−0;.9909901); g. (−1;.0050000,−0;.99502488); h. (1.0010000,−0;.99900100); i. −0.95238095; j. −0.99009901; k. −0.99502488; l. −0.99900100
2) Use the values in the right column of the table in the preceding exercise to guess the value of the slope of the line tangent to f at $$x=−1$$.
3) Use the value in the preceding exercise to find the equation of the tangent line at point P.
Solution: $$y=−x−2$$
For the following exercises, the position function of a ball dropped from the top of a 200-meter tall building is given by $$s(t)=200−4.9t^2$$, where position s is measured in meters and time t is measured in seconds. Round your answer to eight significant digits.
1) [T] Compute the average velocity of the ball over the given time intervals.
1. a. [4.99,5]
2. b. [5,5.01]
3. c. [4.999,5]
4. d. [5,5.001]
2) Use the preceding exercise to guess the instantaneous velocity of the ball at $$t=5$$ sec.
Solution: −49 m/sec (velocity of the ball is 49 m/sec downward)
For the following exercises, consider a stone tossed into the air from ground level with an initial velocity of 15 m/sec. Its height in meters at time t seconds is $$h(t)=15t−4.9t^2$$.
1) [T] Compute the average velocity of the stone over the given time intervals.
1. a. [1,1.05]
2. b. [1,1.01]
3. c. [1,1.005]
4. d. [1,1.001]
2) Use the preceding exercise to guess the instantaneous velocity of the stone at $$t=1$$ sec.
Solution: 5.2m/sec
For the following exercises, consider a rocket shot into the air that then returns to Earth. The height of the rocket in meters is given by $$h(t)=600+78.4t−4.9t^2$$, where t is measured in seconds.
1) [T] Compute the average velocity of the rocket over the given time intervals.
1. a. [9,9.01]
2. b. [8.99,9]
3. c. [9,9.001]
4. d. [8.999,9]
2) Use the preceding exercise to guess the instantaneous velocity of the rocket at $$t=9$$ sec.
Solution: -9.8m/sec
For the following exercises, consider an athlete running a 40-m dash. The position of the athlete is given by $$d(t)=\frac{t^3}{6}+4t$$, where d is the position in meters and t is the time elapsed, measured in seconds.
1) [T] Compute the average velocity of the runner over the given time intervals.
1. a. [1.95,2.05]
2. b. [1.995,2.005]
3. c. [1.9995,2.0005]
4. d. [2,2.00001]
2) Use the preceding exercise to guess the instantaneous velocity of the runner at $$t=2$$ sec.
Solution:6 m/sec
For the following exercises, consider the function$$f(x)=|x|$$.
1) Sketch the graph of f over the interval [$$−1,2$$] and shade the region above the x-axis.
2) Use the preceding exercise to find the exact value of the area between the x-axis and the graph of f over the interval [$$−1,2$$] using rectangles. For the rectangles, use the square units, and approximate both above and below the lines. Use geometry to find the exact answer.
Solution: Under, 1 $$unit^2$$; over: 4 $$unit^2$$. The exact area of the two triangles is $$\frac{1}{2}(1)(1)+\frac{1}{2}(2)(2)=2.5 units^2$$.
For the following exercises, consider the function $$f(x)=\sqrt{1−x^2}$$. (Hint: This is the upper half of a circle of radius 1 positioned at ($$0,0$$).)
1) Sketch the graph of f over the interval [$$−1,1$$].
2) Use the preceding exercise to find the exact area between the x-axis and the graph of f over the interval [$$−1,1$$] using rectangles. For the rectangles, use squares 0.4 by 0.4 units, and approximate both above and below the lines. Use geometry to find the exact answer.
Solution: Under, 0.96 $$unit^2$$; over, 1.92 $$unit^2$$. The exact area of the semicircle with radius 1 is $$\frac{π(1)^2}{2}=\frac{π}{2} unit^2$$
For the following exercises, consider the function $$f(x)=−x^2+1$$.
1) Sketch the graph of f over the interval [$$−1,1$$].
2) Approximate the area of the region between the x-axis and the graph of f over the interval [$$−1,1$$].
Solution: Approximately 1.3333333 $$unit^2$$
## 2.2: The Limit of a Function
1) [T] Complete the following table for the function. Round your solutions to four decimal places.
$$x$$ $$f(x)$$ $$x$$ $$f(x)$$
0.9 a. 1.1 e.
0.99 b. 1.01 f.
0.999 c. 1.001 g.
0.9999 d. 1.0001 h.
2) What do your results in the preceding exercise indicate about the two-sided limit $$lim_{x→1}f(x)$$? Explain your response.
Solution: $$\lim_{x \to 1}f(x)$$ does not exist because
$$\lim_{x \to 1^−}f(x)=−2≠\lim_{x \to 1^+}f(x)=2$$.
For the following exercises, consider the function $$f(x)=(1+x)^{1/x}$$.
3) [T] Make a table showing the values of f for $$x=−0.01,−0.001,−0.0001,−0.00001$$ and for $$x=0.01,0.001,0.0001,0.00001$$. Round your solutions to five decimal places.
$$x$$ $$f(x)$$ $$x)\ \(f(x)$$
-0.01 a. 0.01 e.
-0.001 b. 0.001 f.
-0.0001 c. 0.0001 g.
-0.00001 d. 0.00001 h.
4) What does the table of values in the preceding exercise indicate about the function $$f(x)=(1+x)^{1/x}$$?
Solution: $$\lim_{x \to 0}(1+x)^{1/x}=2.7183$$
5) To which mathematical constant does the limit in the preceding exercise appear to be getting closer?
In the following exercises, use the given values to set up a table to evaluate the limits. Round your solutions to eight decimal places.
6) [T] $$\lim_{x \to 0}\frac{sin2x}{x};±0.1,±0.01,±0.001,±.0001$$
$$x$$ $$\frac{sin2x}{x}$$ $$x$$ $$\frac{sin2x}{x}$$
-0.1 a. 0.1 e.
-0.01 b. 0.01 f.
-0.001 c. 0.001 g.
-0.0001 d. 0.0001 h.
Solution: a. 1.98669331; b. 1.99986667; c. 1.99999867; d. 1.99999999; e. 1.98669331; f. 1.99986667; g. 1.99999867; h. 1.99999999; $$lim_{x \to 0}\frac{sin2x}{x}=2$$
7) [T] $$\lim_{x \to 0}\frac{sin3x}{x} ±0.1, ±0.01, ±0.001, ±0.0001$$
$$x$$ $$\frac{sin3x}{x}$$ $$x$$ $$\frac{sin3x}{x}$$
-0.1 a. 0.1 e.
-0.01 b. 0.01 f.
-0.001 c. 0.001 g.
-0.0001 d. 0.0001 h.
8) Use the preceding two exercises to conjecture (guess) the value of the following limit: $$\lim_{x \to 0}\frac{sinax}{x}$$ for a, a positive real value.
Solution: $$\lim_{x \to 0}\frac{sinax}{x}=a$$
[T] In the following exercises, set up a table of values to find the indicated limit. Round to eight digits.
9) $$lim_{x \to 2}\frac{x^2−4}{x^2+x−6}$$
$$x$$ $$\frac{x^2−4}{x^2+x−6}$$ $$x$$ $$\frac{x^2−4}{x^2+x−6}$$
1.9 a. 2.1 e.
1.99 b. 2.01 f.
1.999 c. 2.001 g.
1.9999 d. 2.0001 h.
10) $$lim_{x \to 1}(1−2x)$$
$$x$$ $$1−2x$$ $$x$$ $$1−2x$$
0.9 a. 1.1 e.
0.99 b. 1.01 f.
0.999 c. 1.001 g.
0.9999 d. 1.0001 h.
Solution: a. −0.80000000; b. −0.98000000; c. −0.99800000; d. −0.99980000; e. −1.2000000; f. −1.0200000; g. −1.0020000; h. −1.0002000; $$\lim_{x \to 1}(1−2x)=−1$$
11) $$lim_{x \to 0}\frac{5}{1−e^{1/x}}$$
$$x$$ $$\frac{5}{1−e^{1/x}}$$ $$x$$ $$\frac{5}{1−e^{1/x}}$$
-0.1 a. 0.1 e.
-0.01 b. 0.01 f.
-0.001 c. 0.001 g.
-0.0001 d. 0.0001 h.
12) $$lim_{z \to 0}\frac{z−1}{z^2(z+3)}$$
$$z$$ $$\frac{z−1}{z^2(z+3)}$$ $$z$$ $$\frac{z−1}{z^2(z+3)}$$
-0.1 a. 0.1 e.
-0.01 b. 0.01 f.
-0.001 c. 0.001 g.
-0.0001 d. 0.0001 h.
Solution: a. −37.931934; b. −3377.9264; c. −333,777.93; d. −33,337,778; e. −29.032258; f. −3289.0365; g. −332,889.04; h. −33,328,889 $$\lim_{x \to 0}\frac{z−1}{z^2(z+3)}=−∞$$
13) $$lim_{t \to 0^+}\frac{cost}{t}$$
$$t) \(\frac{cost}{t}$$
0.1 a.
0.01 b.
0.001 c.
0.0001 d.
14) $$lim_{x \to 2}\frac{1−\frac{2}{x}}{x^2−4}$$
$$x$$ $$\frac{1−\frac{2}{x}}{x^2−4}$$ $$x$$ $$\frac{1−\frac{2}{x}}{x^2−4}$$
1.9 a. 2.1 e.
1.99 b. 2.01 f.
1.999 c. 2.001 g.
1.9999 d. 2.0001 h.
Solution: a. 0.13495277; b. 0.12594300; c. 0.12509381; d. 0.12500938; e. 0.11614402; f. 0.12406794; g. 0.12490631; h. 0.12499063; $$∴\lim_{x \to 2}\frac{1−\frac{2}{x}}{x^2−4}=0.1250=\frac{1}{8}$$
[T] In the following exercises, set up a table of values and round to eight significant digits. Based on the table of values, make a guess about what the limit is. Then, use a calculator to graph the function and determine the limit. Was the conjecture correct? If not, why does the method of tables fail?
15) $$lim_{θ \to 0}sin(\frac{π}{θ})$$
$$θ$$ $$sin(\frac{π}{θ})$$ $$θ$$ $$sin(\frac{π}{θ})$$
-0.1 a. 0.1 e.
-0.01 b. 0.01 f.
-0.001 c. 0.001 g.
-0.0001 d. 0.0001 h.
16) $$lim_{α \to 0^+} \frac{1}{α}cos(\frac{π}{α})$$
$$a$$ $$\frac{1}{α}cos(\frac{π}{α})$$
0.1 a.
0.01 b.
0.001 c.
0.0001 d.
Solution: a. −10.00000; b. −100.00000; c. −1000.0000; d. −10,000.000; Guess: $$\lim_{α→0^+}\frac{1}{α}cos(\frac{π}{α})=∞$$, actual: DNE In the following exercises, consider the graph of the function$$y=f(x)$$ shown here. Which of the statements about $$y=f(x)$$ are true and which are false? Explain why a statement is false. 1) $$lim_{x→10}f(x)=0$$
2) $$lim_{x→−2^+}f(x)=3$$
Solution: False; $$lim_{x→−2^+}f(x)=+∞$$
3) $$lim_{x→−8}f(x)=f(−8)$$
4) $$lim_{x→6}f(x)=5$$
Solution: False; $$lim_{x→6}f(x)$$ DNE since $$lim_{x→6^−}f(x)=2$$ and $$lim_{x→6^+}f(x)=5$$.
In the following exercises, use the following graph of the function $$y=f(x)$$ to find the values, if possible. Estimate when necessary. 1) $$lim_{x→1^−}f(x)$$
2) $$lim_{x→1^+}f(x)$$
Solution: 2
3) $$lim_{x→1}f(x)$$
4) $$lim_{x→2}f(x)$$
Solution: 1
5) $$f(1)$$
In the following exercises, use the graph of the function $$y=f(x)$$ shown here to find the values, if possible. Estimate when necessary. 1) $$lim_{x→0^−}f(x)$$
Solution: 1
2) $$lim_{x→0^+}f(x)$$
3) $$lim_{x→0}f(x)$$
Solution: DNE
4) $$lim_{x→2}f(x)$$
In the following exercises, use the graph of the function $$y=f(x)$$ shown here to find the values, if possible. Estimate when necessary. 1) $$lim_{x→−2^−}f(x)$$
Solution: 0
2) $$lim_{x→−2^+}f(x)$$
3) $$lim_{x→−2}f(x)$$
Solution: DNE
4) $$lim_{x→2^−}f(x)$$
5) $$lim_{x→2^+}f(x)$$
Solution: 2
6) $$lim_{x→}f(x)$$
In the following exercises, use the graph of the function $$y=g(x)$$ shown here to find the values, if possible. Estimate when necessary. 1) $$lim_{x→0^−}g(x)$$
Solution: 3
2) $$lim_{x→0^+}g(x)$$
3) $$lim_{x→0}g(x)$$
Solution: DNE
In the following exercises, use the graph of the function $$y=h(x)$$ shown here to find the values, if possible. Estimate when necessary. 1) $$lim_{x→0^−}h(x)$$
2) $$lim_{x→0^+}h(x)$$
Solution: 0
3) $$lim_{x→0}h(x)$$
In the following exercises, use the graph of the function $$y=f(x)$$ shown here to find the values, if possible. Estimate when necessary. 1) $$lim_{x→0^−}f(x)$$
Solution: −2
2) $$lim_{x→0^+}f(x)$$
3) $$lim_{x→0}f(x)$$
Solution: DNE
4) $$lim_{x→1}f(x)$$
5) $$lim_{x→2}f(x)$$
Solution: 0
In the following exercises, sketch the graph of a function with the given properties.
1) $$lim_{x→2}f(x)=1,lim_{x→4^−}f(x)=3,lim_{x→4^+}f(x)=6,x=4$$ is not defined.
2) $$lim_{x→−∞}f(x)=0,lim_{x→−1^−}f(x)=−∞, lim_{x→−1^+}f(x)=∞,lim_{x→0}f(x)=f(0), f(0)=1, lim_{x→∞}f(x)=−∞$$ 3) $$lim_{x→−∞}f(x)=2,lim_{x→3^−}f(x)=−∞, lim_{x→3^+}f(x)=∞,lim_{x→∞}f(x)=2,f(0)=\frac{−1}{3}$$
4) $$limx→−∞f(x)=2,limx→−2f(x)=−∞, limx→∞f(x)=2,f(0)=0$$ 5) $$lim_{x→−∞}f(x)=0,lim_{x→−1^−}f(x)=∞,lim_{x→−1^+}f(x)=−∞, f(0)=−1,lim_{x→1^−}f(x)=−∞, lim_{x→1^+}f(x)=∞,lim_{x→∞}f(x)=0$$
6) Shock waves arise in many physical applications, ranging from supernovas to detonation waves. A graph of the density of a shock wave with respect to distance, x, is shown here. We are mainly interested in the location of the front of the shock, labeled xSF in the diagram. a. Evaluate $$lim_{x→x_{SF}^+}ρ(x)$$.
b. Evaluate $$lim_{x→x_{SF}^−}ρ(x)$$.
c. Evaluate $$lim_{x→x_{SF}}ρ(x)$$. Explain the physical meanings behind your answers.
Solution: a. $$ρ_2$$ b. $$ρ_1$$ c. DNE unless $$ρ_1=ρ_2$$. As you approach xSF from the right, you are in the high-density area of the shock. When you approach from the left, you have not experienced the “shock” yet and are at a lower density.
7) A track coach uses a camera with a fast shutter to estimate the position of a runner with respect to time. A table of the values of position of the athlete versus time is given here, where x is the position in meters of the runner and t is time in seconds. What is $$lim_{t→2}x(t)$$? What does it mean physically?
$$t(sec)$$ $$x(m)$$
1.75 4.5
1.95 6.1
1.99 6.42
2.01 6.58
2.05 6.9
2.25 8.5
## 2.3: The Limit Laws
In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).
1) $$lim_{x→0}(4x^2−2x+3)$$
Solution: Use constant multiple law and difference law:
$$lim_{x→0}(4x^2−2x+3)=4lim_{x→0}x^2−2lim_{x→0}x+lim_{x→0}3=3$$
2) $$lim_{x→1}\frac{x^3+3x^2+5}{4−7x}$$
3) $$lim_{x→−2}\sqrt{x^2−6x+3}$$
Solution: Use root law: $$lim_{x→−2}\sqrt{x^2−6x+3}=\sqrt{lim_{x→−2}(x2−6x+3)}=\sqrt{19}$$
4) $$lim_{x→−1}(9x+1)^2$$
In the following exercises, use direct substitution to evaluate each limit.
5) $$lim_{x→7}x^2)$$
Solution: 49
6) $$lim_{x→−2}(4x^2−1)$$
7) $$lim_{x→0}\frac{1}{1+sinx}$$
Solution: 1
8) $$lim_{x→2}e^{2x−x^2}$$
9) $$lim_{x→1}\frac{2−7x}{x+6}$$
Solution: $$−\frac{5}{7}$$
10) $$lim_{x→3}lne^{3x}$$
In the following exercises, use direct substitution to show that each limit leads to the indeterminate form $$0/0$$. Then, evaluate the limit.
11) $$lim_{x→4}\frac{x^2−16}{x−4}$$
Solution:$$lim_{x→4}\frac{x^2−16}{x−4}=\frac{16−16}{4−4}=\frac{0}{0}; then, lim_{x→4}\frac{x^2−16}{x−4}= lim_{x→4}\frac{(x+4)(x−4)}{x−4}=8$$
12) $$lim_{x→2}\frac{x−2}{x^2−2x}$$
13) $$lim_{x→6}\frac{3x−18}{2x−12}$$
Solution: $$lim_{x→6}\frac{3x−18}{2x−12}=\frac{18−18}{12−12}=\frac{0}{0}; then, lim_{x→6}\frac{3x−18}{2x− 12}=lim_{x→6}\frac{3(x−6)}{2(x−6)}=\frac{3}{2}$$
14) $$lim_{h→0}\frac{(1+h)^2−1}{h}$$
15) $$lim_{t→9}\frac{t−9}{\sqrt{t−3}}$$
Solution: $$lim_{x→9}\frac{t−9}{\sqrt{t}−3}=\frac{9−9}{3−3}=\frac{0}{0}; then, lim_{t→9}\frac{t−9}{\sqrt{t}−3} =lim_{t→9}\frac{t−9}{\sqrt{t}−3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=lim_{t→9}(\sqrt{t}+3)=6$$
16) $$lim_{h→0}\frac{\frac{1}{a+h}−\frac{1}{a}}{h}$$, where a is a real-valued constant
17) $$lim_{θ→π}\frac{sinθ}{tanθ}$$
Solution: $$lim_{θ→π}\frac{sinθ}{tanθ}=\frac{sinπ}{tanπ}=\frac{0}{0}; then, lim_{θ→π}\frac{sinθ}{tanθ}=lim_{θ→ π}\frac{sinθ}{\frac{sinθ}{cosθ}}=lim_{θ→π}cosθ=−1$$
18) $$lim_{x→1}\frac{x^3−1}{x^2−1}$$
19) $$lim_{x→1/2}\frac{2x^2+3x−2}{2x−1}$$
Solution: $$lim_{x→1/2}\frac{2x^2+3x−2}{2x−1}=\frac{\frac{1}{2}+\frac{3}{2}−2}{1−1}=\frac{0}{0}; then, lim_{x→ 1/2}\frac{2x^2+3x−2}{2x−1}=lim_{x→1/2}frac{(2x−1)(x+2)}{2x−1}=\frac{5}{2}$$
20) $$lim_{x→−3}\frac{\sqrt{x+4}−1}{x+3}$$
In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of Example to simplify the function to help determine the limit.
21) $$lim_{x→−2^−}\frac{2x^2+7x−4}{x^2+x−2}$$
Solution: −∞
22) $$lim_{x→−2^+}\frac{2x^2+7x−4}{x^2+x−2}$$
23) $$lim_{x→1^−}\frac{2x^2+7x−4}{x^2+x−2}$$
Solution: −∞
24) $$lim_{x→1^+}\frac{2x^2+7x−4}{x^2+x−2}$$
In the following exercises, assume that $$lim_{x→6}f(x)=4,lim_{x→6}g(x)=9$$, and $$lim_{x→6}h(x)=6$$. Use these three facts and the limit laws to evaluate each limit
25) $$lim_{x→6}2f(x)g(x)$$
Solution: $$lim_{x→6}2f(x)g(x)=2lim_{x→6}f(x)lim_{x→6}g(x)=72$$
26) $$lim_{x→6}\frac{g(x)−1}{f(x)}$$
27) $$lim_{x→6}(f(x)+\frac{1}{3}g(x))$$
Solution: $$lim_{x→6}(f(x)+\frac{1}{3}g(x))=lim_{x→6}f(x)+\frac{1}{3}lim_{x→6}g(x)=7$$\
28) $$lim_{x→6}\frac{(h(x))^3}{2}$$
29) $$lim_{x→6}\sqrt{g(x)−f(x)}$$
Solution: $$lim_{x→6}\sqrt{g(x)−f(x)}=\sqrt{lim_{x→6}g(x)−lim_{x→6}f(x)}=\sqrt{5}$$
30) $$lim_{x→6}x⋅h(x)$$
31) $$lim_{x→6}[(x+1)⋅f(x)]$$
Solution: $$lim_{x→6}[(x+1)f(x)]=(lim_{x→6}(x+1))(lim_{x→6}f(x))=28$$
32) $$lim_{x→6}(f(x)⋅g(x)−h(x))$$
[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.
33) $$f(x)=\begin{cases}x2 & x≤3,\\ x+4 & x>3\end{cases}$$
1. a. $$lim_{x→3^−}f(x)$$
2. b. $$lim_{x→3^+}f(x)$$
Solution: a. 9; b. 7
34) $$g(x)=\begin{cases}x^3−1 & x≤0\\1 & x>0\end{cases}$$
1. a. $$lim_{x→0^−}g(x)$$
2. b. $$lim_{x→0^+}g(x)$$
35) $$h(x)=\begin{cases}x^2−2x+1 & x<2x≥2\\3−x & x≥2\end{cases}$$
1. a. $$lim_{x→2^−}h(x)$$
2. b. $$lim_{x→2^+}h(x)$$
In the following exercises, use the following graphs and the limit laws to evaluate each limit. 36) $$lim_{x→−3^+}(f(x)+g(x))$$
37) $$lim_{x→−3^−}(f(x)−3g(x))$$
Solution: $$lim_{x→−3^−}(f(x)−3g(x))=lim_{x→−3^−}f(x)−3lim_{x→−3^−}g(x)=0+6=6$$
38) $$lim_{x→0}\frac{f(x)g(x)}{3}$$
39) $$lim_{x→−5}\frac{2+g(x)}{f(x)}$$
Solution: $$lim_{x→−5}\frac{2+g(x)}{f(x)}=\frac{2+(lim_{x→−5}g(x))}{lim_{x→−5}f(x)}=\frac{2+0}{2}=1$$
40) $$lim_{x→1}(f(x))^2$$
41) $$lim_{x→1}\sqrt{f(x)−g(x)}$$
Solution: $$lim_{x→1}\sqrt{f(x)−g(x)}=\sqrt{lim_{x→1}f(x)−lim_{x→1}g(x)}=\sqrt{2+5}=\sqrt{7}$$
42) $$lim_{x→−7}(x⋅g(x))$$
43) $$lim_{x→−9}[x⋅f(x)+2⋅g(x)]$$
Solution: $$lim_{x→−9}(xf(x)+2g(x))=(lim_{x→−9}x)(lim_{x→−9}f(x))+2lim_{x→−9}(g(x))=(−9)(6)+2(4)=−46$$
For the following problems, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions $$f(x),g(x)$$, and $$h(x)$$ when possible.
44) [T] True or False? If $$2x−1≤g(x)≤x^2−2x+3$$, then $$lim_{x→2}g(x)=0$$.
45) [T] $$lim_{θ→0}θ^2cos(\frac{1}{θ}) Solution: The limit is zero. 46) \(lim_{x→0}f(x)$$, where $$f(x)=\begin{cases}0 & x rational\\ x^2 & x irrrational\end{cases}$$
47) [T] In physics, the magnitude of an electric field generated by a point charge at a distance r in vacuum is governed by Coulomb’s law: $$E(r)=\frac{q}{4πε0_r^2}$$, where E represents the magnitude of the electric field, q is the charge of the particle, r is the distance between the particle and where the strength of the field is measured, and \frac{1}{4πε_0} is Coulomb’s constant: $$8.988×109N⋅m^2/C^2$$.
a. Use a graphing calculator to graph $$E(r)$$ given that the charge of the particle is $$q=10^{−10}$$.
b. Evaluate $$lim_{r→0^+}E(r)$$. What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?
Solution: a b. ∞. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate negative distance.
48) [T] The density of an object is given by its mass divided by its volume: $$ρ=m/V.$$
a. Use a calculator to plot the volume as a function of density $$(V=m/ρ)$$, assuming you are examining something of mass 8 kg ($$m=8$$).
b. Evaluate $$lim_{x→0^+}V(\rho)$$ and explain the physical meaning.
## 2.4: Continuity
For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.
1) $$f(x)=\frac{1}{\sqrt{x}}$$
Solution: The function is defined for all x in the interval $$(0,∞)$$.
2) $$f(x)=\frac{2}{x^2+1}$$
3) $$f(x)=\frac{x}{x^2−x}$$
Solution: Removable discontinuity at $$x=0$$; infinite discontinuity at $$x=1$$
4) $$g(t)=t^{−1}+1$$
5) $$f(x)=\frac{5}{e^x−2}$$
Solution: Infinite discontinuity at $$x=ln2$$
6) $$f(x)=\frac{|x−2|}{x−2}$$
7) $$H(x)=tan2x$$
Solution: Infinite discontinuities at $$x=\frac{(2k+1)π}{4}$$, for $$k=0,±1,±2,±3,…$$
8) $$f(t)=\frac{t+3}{t^2+5t+6}$$
For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?
9) $$\frac{2x^2−5x+3}{x−1}$$ at $$x=1$$
Solution: No. It is a removable discontinuity.
10) $$h(θ)=\frac{sinθ−cosθ}{tanθ}$$ at $$θ=π$$
11) $$g(u)=\begin{cases}\frac{6u^2+u−2}{2u−1} & if u≠12\\ \frac{7}{2} & if u=12\end{cases}$$, at $$u=\frac{1}{2}$$
Solution: Yes. It is continuous.
12) $$f(y)=\frac{sin(πy)}{tan(πy)}$$, at $$y=1$$
13) $$f(x)=\begin{cases}x^2−e^x & if x<0\\x−1 & if x≥0\end{cases}$$, at $$x=0$$
Solution: Yes. It is continuous.
14) $$f(x)=\begin{cases}xsin(x) & if x≤π\\ xtan(x) & if x>π\end{cases}$$, at $$x=π$$
In the following exercises, find the value(s) of k that makes each function continuous over the given interval.
15) $$f(x)=\begin{cases}3x+2 & x<k\\2x−3 & k≤x≤8\end{cases}$$
$$k=−5$$
16) $$f(θ)=\begin{cases}sinθ & 0≤θ<\frac{π}{2}\\cos(θ+k) & \frac{π}{2}≤θ≤π\end{cases}$$
17) $$f(x)=\begin{cases}\frac{x^2+3x+2}{x+2} & x≠−2\\ k & x=−2\end{cases}$$
Solution: k=−1
18) $$f(x)=\begin{cases}e^{kx} & 0≤x<4\\x+3 & 4≤x≤8\end{cases}$$
19) $$f(x)=\begin{cases}\sqrt{kx} & 0≤x≤3\\x+1 & 3<x≤10\end{cases}$$
Solution: $$k=\frac{16}{3}$$
In the following exercises, use the Intermediate Value Theorem (IVT).
20) Let $$h(x)=\begin{cases}3x^2−4 & x≤2\\5+4x & x>2\end{cases}$$ Over the interval $$[0,4]$$, there is no value of x such that $$h(x)=10$$, although $$h(0)<10$$ and $$h(4)>10$$. Explain why this does not contradict the IVT.
21) A particle moving along a line has at each time t a position function s(t), which is continuous. Assume $$s(2)=5$$ and $$s(5)=2$$. Another particle moves such that its position is given by $$h(t)=s(t)−t$$. Explain why there must be a value c for $$2<c<5$$ such that $$h(c)=0$$.
Solution:Since both $$s$$ and $$y=t$$ are continuous everywhere, then $$h(t)=s(t)−t$$ is continuous everywhere and, in particular, it is continuous over the closed interval [$$2,5$$]. Also, $$h(2)=3>0$$ and $$h(5)=−3<0$$. Therefore, by the IVT, there is a value $$x=c$$ such that $$h(c)=0$$.
22) [T] Use the statement “The cosine of t is equal to t cubed."
a. Write a mathematical equation of the statement.
b. Prove that the equation in part a. has at least one real solution.
c. Use a calculator to find an interval of length 0.01 that contains a solution.
23) Apply the IVT to determine whether $$2^x=x^3$$ has a solution in one of the intervals [$$1.25,1.375$$] or [$$1.375,1.5$$]. Briefly explain your response for each interval.
Solution: The function $$f(x)=2^x−x^3$$ is continuous over the interval [$$1.25,1.375$$] and has opposite signs at the endpoints.
24) Consider the graph of the function $$y=f(x)$$ shown in the following graph. a. Find all values for which the function is discontinuous.
b. For each value in part a., state why the formal definition of continuity does not apply.
c. Classify each discontinuity as either jump, removable, or infinite.
25) Let $$f(x)=\begin{cases}3x & x>1\\ x^3 & x<1\end{cases}$$.
a. Sketch the graph of $$f$$.
b. Is it possible to find a value k such that $$f(1)=k$$, which makes $$f(x)$$ continuous for all real numbers? Briefly explain.
Solution:
a. b. It is not possible to redefine $$f(1)$$ since the discontinuity is a jump discontinuity.
26) Let $$f(x)=\frac{x^4−1}{x^2−1}$$ for $$x≠−1,1$$.
a. Sketch the graph of $$f$$.
b. Is it possible to find values $$k_1$$ and $$k_2$$ such that $$f(−1)=k$$ and $$f(1)=k_2$$, and that makes $$f(x)$$ continuous for all real numbers? Briefly explain.
27) Sketch the graph of the function $$y=f(x)$$ with properties i. through vii.
1. i. The domain of f is ($$−∞,+∞$$).
2. ii. f has an infinite discontinuity at $$x=−6$$.
3. iii. $$f(−6)=3$$
4. iv. $$lim_{x→−3^−}f(x)=lim_{x→−3^+}f(x)=2$$
5. v. $$f(−3)=3$$
6. vi. f is left continuous but not right continuous at $$x=3$$.
7. vii. $$lim_{x→−∞}f(x)=−∞$$ and $$lim_{x→+∞}f(x)=+∞$$
Solution: Answers may vary; see the following example: 28) Sketch the graph of the function $$y=f(x)$$ with properties i. through iv.
1. i. The domain of f is [$$0,5$$].
2. ii. $$lim_{x→1^+}f(x)$$ and $$lim_{x→1^−}f(x)$$ exist and are equal.
3. iii. $$f(x)$$ is left continuous but not continuous at $$x=2$$, and right continuous but not continuous at $$x=3$$.
4. iv. $$f(x)$$ has a removable discontinuity at $$x=1$$, a jump discontinuity at $$x=2$$, and the following limits hold: $$lim_{x→3^−}f(x)=−∞$$ and $$lim_{x→3^+}f(x)=2$$.
In the following exercises, suppose $$y=f(x)$$ is defined for all x. For each description, sketch a graph with the indicated property.
29) Discontinuous at $$x=1$$ with $$lim_{x→−1}f(x)=−1$$ and $$lim_{x→2}f(x)=4$$
Solution: Answers may vary; see the following example: 30) Discontinuous at $$x=2$$ but continuous elsewhere with $$lim_{x→0}f(x)=\frac{1}{2}$$
Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.
31) $$f(t)=\frac{2}{e^t−e^{−t}}$$ is continuous everywhere.
Solution: False. It is continuous over ($$−∞,0$$) ∪ ($$0,∞$$).
32) If the left- and right-hand limits of $$f(x)$$ as $$x→a$$ exist and are equal, then f cannot be discontinuous at $$x=a$$.
33) If a function is not continuous at a point, then it is not defined at that point.
Solution: False. Consider $$f(x)=\begin{cases}x & if x≠0\\ 4 & if x=0\end{cases}$$.
34) According to the IVT, $$cosx−sinx−x=2$$ has a solution over the interval [$$−1,1$$].
35) If $$f(x)$$ is continuous such that $$f(a)$$ and $$f(b)$$ have opposite signs, then $$f(x)=0$$ has exactly one solution in [$$a,b$$].
Solution: False. Consider $$f(x)=cos(x)$$ on [$$−π,2π$$].
36) The function $$f(x)=\frac{x^2−4x+3}{x^2−1}$$ is continuous over the interval [$$0,3$$].
37) If $$f(x)$$ is continuous everywhere and $$f(a),f(b)>0$$, then there is no root of $$f(x)$$ in the interval [$$a,b$$].
Solution: False. The IVT does not work in reverse! Consider $$(x−1)^2$$ over the interval [$$−2,2$$].
[T] The following problems consider the scalar form of Coulomb’s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation $$F(r)=k_e\frac{|q_1q_2|}{r^2}$$, where $$k_e$$ is Coulomb’s constant, $$q_i$$ are the magnitudes of the charges of the two particles, and r is the distance between the two particles.
38) To simplify the calculation of a model with many interacting particles, after some threshold value $$r=R$$, we approximate F as zero.
a. Explain the physical reasoning behind this assumption.
b. What is the force equation?
c. Evaluate the force F using both Coulomb’s law and our approximation, assuming two protons with a charge magnitude of $$1.6022×10^{−19}$$ coulombs (C), and the Coulomb constant $$k_e=8.988×10^9Nm^2/C^2$$ are 1 m apart. Also, assume $$R<1m$$. How much inaccuracy does our approximation generate? Is our approximation reasonable?
d. Is there any finite value of R for which this system remains continuous at R?
39) Instead of making the force 0 at R, instead we let the force be 10−20 for $$r≥R$$. Assume two protons, which have a magnitude of charge $$1.6022×10^{−19}C$$, and the Coulomb constant $$k_e=8.988×10^9Nm^2/C^2$$. Is there a value R that can make this system continuous? If so, find it.
Solution: $$R=0.0001519m$$
Recall the discussion on spacecraft from the chapter opener. The following problems consider a rocket launch from Earth’s surface. The force of gravity on the rocket is given by $$F(d)=−mk/d^2$$, where m is the mass of the rocket, d is the distance of the rocket from the center of Earth, and k is a constant.
40) [T] Determine the value and units of k given that the mass of the rocket on Earth is 3 million kg. (Hint: The distance from the center of Earth to its surface is 6378 km.)
41) [T] After a certain distance D has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by $$F(d)=\begin{cases}−\frac{mk}{d^2} & if d<D\\ 10,000 & if d≥D\end{cases}$$. Find the necessary condition D such that the force function remains continuous.
Solution: $$D=63.78km$$
42) As the rocket travels away from Earth’s surface, there is a distance D where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as $$F(d)=\begin{cases} −\frac{m_1k}{d^2} & if d<D \\ −\frac{m_2k}{d^2} & if d≥D\end{cases}$$. Is there a D value such that this function is continuous, assuming $$m_1≠m_2?$$
Prove the following functions are continuous everywhere
43) $$f(θ)=sinθ$$
Solution: For all values of $$a$$, $$f(a)$$ is defined, $$lim_{θ→a}f(θ)$$ exists, and $$lim_{θ→a}f(θ)=f(a)$$. Therefore, $$f(θ)$$ is continuous everywhere.
44) $$g(x)=|x|$$
45) Where is $$f(x)=\begin{cases} 0 & \text{if x is irrational}\\ 1 & \text{if x is rational}\end{cases}$$ continuous?
Solution: Nowhere
## 2.5: The Precise Definition of a Limit
In the following exercises, write the appropriate $$ε−δ$$ definition for each of the given statements.
1) $$lim_{x→a}f(x)=N$$
2) $$lim_{t→b}g(t)=M$$
Solution: For every $$ε>0$$, there exists a $$δ>0$$, so that if $$0<|t−b|<δ$$, then $$|g(t)−M|<ε$$
3) $$lim_{x→c}h(x)=L$$
4) $$lim_{x→a}φ(x)=A$$
Solution: For every $$ε>0$$, there exists a $$δ>0$$, so that if $$0<|x−a|<δ$$, then $$|φ(x)−A|<ε$$
The following graph of the function f satisfies $$lim_{x→2}f(x)=2$$. In the following exercises, determine a value of $$δ>0$$ that satisfies each statement. 5) If $$0<|x−2|<δ$$, then $$|f(x)−2|<1$$.
6) If $$0<|x−2|<δ$$, then $$|f(x)−2|<0.5$$.
Solution: $$δ≤0.25$$
The following graph of the function f satisfies $$lim_{x→3}f(x)=−1$$. In the following exercises, determine a value of $$δ>0$$ that satisfies each statement. 7) If $$0<|x−3|<δ$$, then $$|f(x)+1|<1$$.
8) If $$0<|x−3|<δ$$, then $$|f(x)+1|<2$$.
Solution: $$δ≤2$$
The following graph of the function f satisfies $$lim_{x→3}f(x)=2$$. In the following exercises, for each value of ε, find a value of $$δ>0$$ such that the precise definition of limit holds true. 9) $$ε=1.5$$
10) $$ε=3$$
Solution: $$δ≤1$$
[T] In the following exercises, use a graphing calculator to find a number $$δ$$ such that the statements hold true.
11) $$∣sin(2x)−\frac{1}{2}∣<0.1$$, whenever $$∣x−\frac{π}{12}∣<δ$$
12) $$∣\sqrt{x−4}−2∣<0.1$$, whenever $$|x−8|<δ$$
Solution: $$δ<0.3900$$
In the following exercises, use the precise definition of limit to prove the given limits.
13) $$lim_{x→2}(5x+8)=18$$
14) $$lim_{x→3}\frac{x^2−9}{x−3}=6$$
Solution: Let $$δ=ε$$. If $$0<|x−3|<ε$$, then $$|x+3−6|=|x−3|<ε$$.
15) $$lim_{x→2}\frac{2x^2−3x−2}{x−2}=5$$
16) $$lim_{x→0}x^4=0$$
Solution: Let $$δ=\sqrt{ε}$$ If $$0<|x|<\sqrt{ε}$$, then $$∣x^4∣=x^4<ε$$.
17) $$lim_{x→2}(x^2+2x)=8$$
In the following exercises, use the precise definition of limit to prove the given one-sided limits.
18) $$lim_{x→5^−}\frac{5−x}=0$$
Let $$δ=ε_2$$. If $$5−ε^2<x<5$$, then $$∣\sqrt{5−x}∣=\sqrt{5−x}<ε$$.
19) $$lim_{x→0^+}f(x)=−2$$, where $$f(x)=\begin{cases}8x−3 & if x<0\\4x−2 & if x≥0\end{cases}$$.
20) $$lim_{x→1^−}f(x)=3$$, where $$f(x)=\begin{cases}5x−2 & if x<1\\7x−1 & if x≥1\end{cases}$$.
Solution: Let $$δ=ε/5$$. If $$1−ε/5<x<1$$, then $$|f(x)−3|=5x−5<ε$$.
In the following exercises, use the precise definition of limit to prove the given infinite limits.
21) $$im_{x→0}\frac{1}{x^2}=∞$$
22) $$lim_{x→−1}\frac{3}{(x+1)^2}=∞$$
Solution: Let $$δ=\sqrt{\frac{3}{N}}$$. If $$0<|x+1|<\sqrt{\frac{3}{N}}$$, then $$f(x)=\frac{3}{(x+1)^2}>N$$.
23) $$lim_{x→2}−\frac{1}{(x−2)^2}=−∞$$
24) An engineer is using a machine to cut a flat square of Aerogel of area 144 cm2. If there is a maximum error tolerance in the area of 8 cm2, how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to $$δ$$, ε, a, and L?
Solution: $$0.033 cm, ε=8,δ=0.33,a=12,L=144$$
25) Use the precise definition of limit to prove that the following limit does not exist: $$lim_{x→1}\frac{|x−1|}{x−1}.$$
26) Using precise definitions of limits, prove that $$lim_{x→0}f(x)$$ does not exist, given that $$f(x)$$ is the ceiling function. (Hint: Try any $$δ<1$$.)
27) Using precise definitions of limits, prove that $$lim_{x→0}f(x)$$ does not exist: $$f(x)=\begin{cases}1 & \text{if x is rational}\\0 & \text{if x is irrational}\end{cases}$$. (Hint: Think about how you can always choose a rational number $$0<r<d$$, but $$|f(r)−0|=1$$.)
28) Using precise definitions of limits, determine $$lim_{x→0}f(x)$$ for $$f(x)=\begin{cases}x & \text{if x is rational}\\0 & \text{if x is irrational}\end{cases}$$. (Hint: Break into two cases, x rational and x irrational.)
Solution: 0
29) Using the function from the previous exercise, use the precise definition of limits to show that $$lim_{x→a}f(x)$$ does not exist for $$a≠0$$
For the following exercises, suppose that $$lim_{xa}f(x)=L$$ and $$lim_{xa}g(x)=M$$ both exist. Use the precise definition of limits to prove the following limit laws:
30) $$lim_{x→a}(f(x)−g(x))=L−M$$
Solution: $$f(x)−g(x)=f(x)+(−1)g(x)$$
31) $$lim_{x→a}[cf(x)]=cL$$ for any real constant c (Hint: Consider two cases: $$c=0$$ and $$c≠0$$.)
32) $$lim_{x→a}[f(x)g(x)]=LM$$. (Hint: $$|f(x)g(x)−LM|= |f(x)g(x) −f(x)M +f(x)M −LM| ≤|f(x)||g(x) −M| +|M||f(x)−L|.)$$
## Chapter Review Exercises
True or False. In the following exercises, justify your answer with a proof or a counterexample.
1) A function has to be continuous at $$x=a$$ if the $$lim_{x→a}f(x)$$ exists.
2) You can use the quotient rule to evaluate $$lim_{x→0}\frac{sinx}{x}$$.
Solution: False
3) If there is a vertical asymptote at $$x=a$$ for the function $$f(x)$$, then f is undefined at the point $$x=a$$.
4) If $$lim_{x→a}f(x)$$ does not exist, then f is undefined at the point $$x=a$$.
Solution: False. A removable discontinuity is possible.
5) Using the graph, find each limit or explain why the limit does not exist.
a. $$lim_{x→−1}f(x)$$
b. $$lim_{x→1}f(x)$$
c. $$lim_{x→0^+}f(x)$$
d. $$lim_{x→2}f(x)$$ In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.
6) $$lim_{x→2}\frac{2x^2−3x−2}{x−2}$$
Solution: 5
7) $$lim_{x→0}3x^2−2x+4$$
8) $$lim_{x→3}\frac{x^3−2x^2−1}{3x−2}$$
Solution: 8/7
9) $$lim_{x→π/2}\frac{cotx}{cosx}$$
10) $$lim_{x→−5}\frac{x^2+25}{x+5}$$
Solution:DNE
11) $$lim_{x→2}\frac{3x^2−2x−8}{x^2−4}$$
12) $$lim_{x→1}\frac{x^2−1}{x^3−1}$$
Solution: 2/3
13) $$lim_{x→1}\frac{x^2−1}{\sqrt{x}−1}$$
14) $$lim_{x→4}\frac{4−x}{\sqrt{x}−2}$$
Solution: −4
15) $$lim_{x→4}\frac{1}{\sqrt{x}−2}$$
In the following exercises, use the squeeze theorem to prove the limit.
16) $$lim_{x→0}x^2cos(2πx)=0$$
Solution: Since $$−1≤cos(2πx)≤1$$, then $$−x^2≤x^2cos(2πx)≤x^2$$. Since $$lim_{x→0}x^2=0=lim_{x→0}−x^2$$, it follows that $$lim_{x→0}x^2cos(2πx)=0$$.
17) $$lim_{x→0}x^3sin(\frac{π}{x})=0$$
18) Determine the domain such that the function $$f(x)=\sqrt{x−2}+xe^x$$ is continuous over its domain.
Solution: $$[2,∞]$$
In the following exercises, determine the value of c such that the function remains continuous. Draw your resulting function to ensure it is continuous.
19) $$f(x)=\begin{cases}x^2+1 & x>c\\2^x & x≤c\end{cases}$$
20) $$f(x)=\begin{cases}\sqrt{x+1} & x>−1\\x^2+c & x≤−1\end{cases}$$
In the following exercises, use the precise definition of limit to prove the limit.
21) $$lim_{x→1}(8x+16)=24$$
22) $$lim_{x→0}x^3=0$$
Solution: $$δ=\sqrt[ε]{3}$$
23) A ball is thrown into the air and the vertical position is given by $$x(t)=−4.9t^2+25t+5$$. Use the Intermediate Value Theorem to show that the ball must land on the ground sometime between 5 sec and 6 sec after the throw.
24) A particle moving along a line has a displacement according to the function $$x(t)=t^2−2t+4$$, where x is measured in meters and t is measured in seconds. Find the average velocity over the time period $$t=[0,2]$$.
Solution: $$0$$ m/sec
25) From the previous exercises, estimate the instantaneous velocity at $$t=2$$ by checking the average velocity within $$t=0.01$$ sec. | HuggingFaceTB/finemath | |
# Class 10 Maths Extra Questions for Linear equation
Given below are the Class 10 Maths Extra Questions And Important Questions for Linear equations in two variables
a. Concepts questions
b. True and False
c. Multiple choice questions
## Concept Questions
Question 1..
Which of these equation have (i) Unique solution (ii) Infinite solutions (iii) no solutions
1. $152x-378y=-74$ , $-378x +152y=-604$
2. $x+2y =10$ , $3x+6y=30$
3. $3x+4y=6$ , $12x+16x=30$
4. $7x-11y=53$ ,$19x -17y=456$
5. $x=7$,$y=-2$
6. $ax+by=a-b$ , $bx-ay=a+b$
7. $2x+3y=0$, $124x+13y=0$
8. $y=11$ ,$y =-11$
Solution
Condition Algebraic interpretation $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ One unique solution only. $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ Infinite solution. $\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ No solution
One unique solution: (a), (d), (e), (f), (d)
Infinite solution :( b)
No solution: (c), (h)
Question 2.
True or False statement
a. Line $4x+5y=0$ and $11x+17y=0$ both passes through origin
b. Pair of lines $117x+14y=30$ , $65x+11y=19$ are consistent and have a unique solution
c. There are infinite solution for equation $17x+12y=30$
d. $x=0$ ,$y=0$ has one unique solution
e. Lines represented by $x-y=0$ and $x+y=0$ are perpendicular to each other
f. $2x+6y=12$ and $8x+24y=65$ are consistent pair of equation
g. $x+6y=12$ and $4x+24y=64$ are inconsistent pair of equation
Solution
1. True
2. True
3. True
4. True
5. True
6. False
7. True
## Multiple choice Questions
Question 3.
find the value of p for which the linear pair has infinite solution
$12x+14y=0$
$36x+py=0$
a. 14
b. 28
c. 56
d. 42
Solution (d)
For infinite solution
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}$
So $\frac {12}{36}=\frac {14}{p}$ => p=42
Question 4.
There are 10 students in XII class. Some are maths and some bio student. The no of bio students are 4 more then math’s students. Find the no of math’s and bio students
a. 1,9
b. 4,6
c. 2,8
d. 3,7
Solution (d)
Let x be math’s students
y be bio students
Then
$x+y=10$
$y=x+4$
Solving these linear pair through any method we get
x=3 and y=7
Question 5.
which of the below pair are consistent pair?
a. $x-3y=3$ , $3x-9y=2$
b. $51x +68y=110$ ,$3x+4y=99$
c. $2x+3y=10$ , $9x+11y=12$
d. None of these
Solution c
For consistent pair
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}$
Or
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Analyzing all of them ,we get c as the answer
Question 6.
There are two numbers. Two conditions are there for them
(i) Sum of these two numbers are 100
(ii) One number is four time another number.
What are these numbers?
a. 20,80
b. 30,70
c. 40,60
d. 25,75
Solution (a)
Let x and y are the number
$x+y=100$
$y=4x$
Solving them we get x=20 and y=80
Question 7.
The number of solution of the linear pair
$x+37y=123$
$21x-41y=125$
a. No solution
b. One solution
c. infinite many
d. None of these
Solution (d)
Question 8.
The sum of a 2 digit number and number obtained by reversing the order of the digits is 99. If the digits of the number differ by 3. Find the number
Solution 63 or 36
Question 9.
Rajdhani train covered the distance between Lucknow and Delhi at a uniform speed. It is observed that if rajdhani would have run slower by 10 km/hr,it would have taken 3 hours more to reach the destination and if rajdhani would have run faster by 10 km/hr,it would have taken 2 hours less. Find the distance Lucknow and Delhi?
Solution
Let x be the speed and t be the original timing ,then distance between Lucknow and Delhi
$Distance =speed \times time =xy$
Now from first observation
$xy=(x-10)(y+3)$ => $3x-10y-30=0$
From second observation
$xy=(x+10)(y-2)$ => $2x-10y+20=0$
Solving both we get
x=50km/hr
y=12hours
So distance between Lucknow and Delhi=$50 \times 12=600$ Km
Question 10.
Using substitution method solve the below equation
a. $x-2y+300=0$, $6x-y-70=0$
b. $5x-y=5, 3x-y=3$
Solution a. From Ist equation
$x=2y-300$
Substituting this in second equation
$6(2y-300)- y-70=0$ => $y=170$
Putting this Ist equation
$x=40$
b. Similarly,this can be solved. Answer is ( 1,0)
Question 11.
Using elimination method ,solve the following
a. $x+y-40=0$ ,$7x+3y=180$
b. $x+10y =68$ , $x+15y=98$
Solution a. $x+y-40=0$ ---(1)
$7x+3y=180$ ---(2)
Multiplying equation (1) by 7
$7x+7y-280=0$ ---(3)
Subtracting equation 2 from equation 7
$7x+7y-280=0$
$7x+3y=180$
We get
$4y=100$ => $y=25$
Substituting this in (1) ,we get $x=15$
b. ( 8,6)
Question 12.
Solve the below linear equation using cross-multiplication method
a. $(a-b)x + (a=b) y=a^2-2ab-b^2$ , $(a+b)(x+y)=a^2+ b^2$
b. $x+y=5$ ,$2x-3y=4$
Solution
3 Cross Multiplication method 1. Suppose the equation are $a_1x+b_1y+ c_1=0$ $a_2x +b_2y+c_2=0$ 2. This can be written as 3. This can be written as 4. Value of x and y can be find using the x => first and last expression y=> second and last expression
a. The equation can be written as
$(a-b)x + (a=b) y=a^2-2ab-b^2$
$(a+b)(x+y)=a^2+ b^2$
By cross multiplication we have $x=a+b$, $y=-2ab(a+b)^{-1}$
b. (19/5,6/5)
Question 13.
For what value of k, the following pair of linear equations has infinitely many solutions?
i. $kx + 4y - (k + 8) = 0$
$4x + ky + 4 = 0$
ii. $2x + 3y = 4$
$(k + 2)x + 6y = 3k + 2$
Question 14
For what value of k, will the following system of equations have no solutions?
i. $(3k + 1) x + 3y = 2$
$(k^2 + 1) x + (k - 2) y = 5$
ii. $3x + y = 1$
$(2k - 1) x + (k - 1) y = 2k + 1$
iii. $2x + ky = 11$
$5x -7y = 5$
Question 15.
For what value of k, does the pair of equations given below has a unique solution?
$2x + ky = 6$
$4x + 6y = 0$
Question 16.
Find the value of k for which the following system of linear equations has infinite solutions:
i. $x + (k + 1) y = 5$
$(k + 1) x + 9y = 8k - 1$
ii. $8x + 5y = 9$
$kx + 10y = 18$
iii. $2x + 3y = k$
$(k - 1) x + (k + 2) y = 3k$
Question 17
For what value of α, the system of equations
$\alpha x + 3y = \alpha - 3$
$12x + \alpha y = \alpha$
will have no solution?
-6
Question 18.
Find the value of k for which the system
$kx + 2y = 5$
$3x + y = 1$
has (i) a unique solution, and (ii) no solution.
k ≠6 , k=6
Question 19.
Find the value of k for which the system
$2x + ky = 1$
$3x - 5y = 7$
has (i) a unique solution, and (ii) no solution.
k ≠ -10/3 , k=10/3
Question 20
find the value of a and b for which the following system of equations has infinitely many solutions:
$2x - (2a + 5) y = 5$
$(2b + 1) x - 9y = 15$
(-1, 5/2)
Question 21.
Find the value of a for which the following system of equations has infinitely many solutions:
$2x + 3y - 7 = 0$
$(a - 1) x + (a + 1) y = (3a - 1)$
a=5
Question 22.
Replace p by an appropriate number so that the following system of equations has a unique solution.
$2x + 3y = 5$
$px +9y = 12$
## Summary
This pair of linear equations in two variables class 10 extra questions with solutions is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail. Go back to Class 10 Main Page using below links
### Practice Question
Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20 | HuggingFaceTB/finemath | |
# McGraw Hill My Math Grade 1 Chapter 9 Lesson 4 Answer Key Compare Shapes
All the solutions provided in McGraw Hill My Math Grade 1 Answer Key PDF Chapter 9 Lesson 4 Compare Shapes will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 1 Answer Key Chapter 9 Lesson 4 Compare Shapes
Explore and Explain
Teacher Directions: Use attribute block circles, squares. rectangles. and triangles. Place the shapes that have 4 sides on the left fence post. Trace the shapes. Place the shapes that have less than 4 vertices on the right fence post. Trace the shapes.
Explanation:
I placed the shapes that have 4 sides on the left fence post and placed the shapes that have less than 4 vertices on the right fence post.
See and Show
You can compare and sort two-dimensional shapes.
Circle the shapes that have straight sides.
Circle the shapes with more than 3 vertices.
Circle the shapes described.
Question 1.
shapes with 4 straight sides
Explanation:
I drew circles around the shapes that have 4 straight sides.
Question 2.
shapes with 3 vertices
Explanation:
I drew circles around the shapes with 3 vertices.
Question 3.
shapes with 4 sides the same length
Explanation:
I drew circles around the shapes with 4 sides the same length.
Question 4.
shapes with 3 sides and 3 vertices
Explanation:
I drew circles around the shapes that have 3 sides and 3 vertices.
Talk Math How do you compare two-dimensional shapes?
I compare the two-dimensional shapes based on the number of sides, vertices. That is the properties of the shapes.
Circle the shapes described.
Question 5.
shapes with 0 vertices
Explanation:
I drew circles around the shapes that have 0 vertices.
Question 6.
shapes with 4 sides
Explanation:
I drew circles around the shapes that have 4 sides.
Question 7.
shapes with straight sides
Explanation:
I drew circles around the shapes that have straight sides.
Question 8.
shapes with 0 vertices
Explanation:
I drew circles around the shapes that have 0 vertices.
Question 9.
shapes with 0 straight sides
Explanation:
I drew circles around the shapes that have 0 straight sides.
Question 10.
shapes that are not curved
Explanation:
I drew circles around the shapes that are not curved.
Problem Solving
Question 11.
Madeline sees these objects in her school. How many of the objects have more than three sides?
_____________ objects
Explanation:
3 objects
I drew circles around the shapes that have more than 3 straight sides.
HOT Problem Circle all of the same type of shapes. Explain.
Explanation:
I drew circles around the shapes that are same type.
### McGraw Hill My Math Grade 1 Chapter 9 Lesson 4 My Homework Answer Key
Practice
Circle the shapes described.
Question 1.
shapes with 0 vertices
Explanation:
I drew circles around the shapes that have 0 vertices.
Question 2.
shapes with 3 sides
Explanation:
I drew circles around the shapes that have 3 sides.
Question 3.
shapes with more than 2 sides
Explanation:
I drew circles around the shapes that have more than 2 sides.
Question 4.
shapes that are closed
Explanation:
I drew circles around the shapes that are closed.
Circle the shapes described.
Question 5.
shapes with more than 2 straight sides
Explanation:
I drew circles around the shapes that have more than 2 straight sides.
Question 6.
shapes with less than 4 vertices
Explanation:
I drew circle around the shape that have less than 4 vertices.
Question 7.
Draw a shape that has 4 vertices and 2 pair of sides that are different lengths.
Explanation:
A rectangle is a closed shape that has 4 vertices and 2 pair of sides that are different lengths.
Test Practice
Question 8.
Which shape has 3 sides and 3 vertices?
(A) circle
(B) square
(C) triangle
(D) rectangle | HuggingFaceTB/finemath | |
The transfer function model of an LTI system is shown in the following figure. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. $$i.e.,\: Transfer\: Function =\frac{Y(s)}{X(s)}$$. Sort by. These models are useful for analysis and design of control systems. In the field of signal processing, circuit theory, control theory etc., Fourier transformation and Laplace transformation are very important and powerful tools. Therefore, the transfer function of LTI system is equal to the ratio of $Y(s)$ and $X(s)$. Grüne, Bayreuth, GermanyAdvising EditorsE.D. Differential equation model is a time domain mathematical model of control systems. Consider the following electrical system as shown in the following figure. Control theory deals with the control of dynamical systems in engineered processes and machines. Control theory at it’s roots is the study of control functions to improve combined system/control performance metrics such as “regulation”, “tracking”,”disturbance rejections” and “sensitivity to parameter”. October 2012, issue 4; July 2012, issue 3; April 2012, issue 1-2. The input voltage applied to this circuit is $v_i$ and the voltage across the capacitor is the output voltage $v_o$. Here, we show a second order electrical system with a block having the transfer function inside it. Its usefulness against seismic motions and other pernitious perturbing actions The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Eduardo Sontag Distinguished University Professor, Northeastern University, Bioengineering and Electrical and Computer Engineering Faculty, Program in Therapeutic Science, Harvard Medical School Professor G. Halikias, Professor of Control Engineering. Apply basic laws to the given control system. Systems: Input-Output Properties, Classics in Applied Mathematics, SIAM, ISBN-13: 978-0898716702 (chapter on passivity) Romeo Ortega, Antonio Loria, Per Johan Nicklasson and Hebertt Sira-Ramirez, (1998), Passivity-Based Control of Euler-Lagrange Systems, Springer, ISBN 978-1-4471-3603-3. Professor Nabil Aouf, Professor of Robotics and Autonomous Systems, Centre Director. The control systems can be represented with a set of mathematical equations known as mathematical model. What are good resources to get into areas? A.E.Bashirov, K.Abuassba, Kalman filtering for wide band noise driven systems, Proceedings of the 5 th International Conference on Control and Optimization with Industrial Applications, 27-29 August, 2015, Baku, Azerbaijan, 3 pages. Get the differential equation in terms of input and output by eliminating the intermediate variable(s). Mathematics for Systems and Control. Design of Digital Control Systems {W-term only} Design of digital control systems, from frequency domain methods through state-variable methods. Underlying the Wolfram control systems solution is a powerful hybrid symbolic-numeric computation engine with numerics of any precision, high-performance symbolics, advanced visualizations and automated algorithm selection–everything to get accurate results efficiently. Linear systems can be regarded as a causal shift-invariant operator on a Hilbert space of signals, and by doing so this book presents an introduction to the common ground between operator theory and linear systems theory. The Transfer function of a Linear Time Invariant (LTI) system is defined as the ratio of Laplace transform of output and Laplace transform of input by assuming all the initial conditions are zero. eg., [2], Sect. Special Issue on Robust Stability and Control of Large-Scale Nonlinear Systems. Mathematics of Control, Signals, and Systems (MCSS) is an international journal devoted to mathematical control and system theory, including system theoretic aspects of signal processing. Watch the recordings here on Youtube! We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. New comments cannot be posted and votes cannot be cast. Mathematics of control systems in systems molecular biology. $$\Rightarrow\:v_i=RC\frac{\text{d}v_o}{\text{d}t}+LC\frac{\text{d}^2v_o}{\text{d}t^2}+v_o$$, $$\Rightarrow\frac{\text{d}^2v_o}{\text{d}t^2}+\left ( \frac{R}{L} \right )\frac{\text{d}v_o}{\text{d}t}+\left ( \frac{1}{LC} \right )v_o=\left ( \frac{1}{LC} \right )v_i$$. Missed the LibreFest? Design of control system means finding the mathematical model when we know the input and the output. And this block has an input $V_i(s)$ & an output $V_o(s)$. Abstract. Control theory is a branch of Applied Mathematics dealing with the use of feedback to influence the behaviour of a system in order to achieve a desired goal. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Click on the name of a member below to find out about their background, research interests and more. Let us discuss the first two models in this chapter. September 2013, issue 3; June 2013, issue 2; March 2013, issue 1; Volume 24 April - October 2012. For our Institute of Space Systems in Bremen we wish to recruit an Engineer Aerospace/Space, Control Engineering, Mathematics, Computer Science or similar Development of Technologies for Navigation Systems for Space Applications Control Signals Systems (1992) 5:233-261 Mathematics of Control, Signals, and Systems 9 1992 Springer-Verlag New York Inc. Nonlinear H ~176 Control Theory for Stable Plants* Joseph A. Ball1" and J. William Helton~ Abstract. Would like to see some problems solved to help my understanding. The control systems can be represented with a set of mathematical equations known as mathematical model. New mathematical results on the various aspects of consensus properties and issues in dynamic and control systems Active and semiactive control designs: modelling and stability. Special issue: Control, communication, and complexity. GATE CSE Discrete Mathematics Programming Languages Theory of Computation Operating Systems Digital Logic Computer Organization Previously, we got the differential equation of an electrical system as, $$\frac{\text{d}^2v_o}{\text{d}t^2}+\left ( \frac{R}{L} \right )\frac{\text{d}v_o}{\text{d}t}+\left ( \frac{1}{LC} \right )v_o=\left ( \frac{1}{LC} \right )v_i$$, $$s^2V_o(s)+\left ( \frac{sR}{L} \right )V_o(s)+\left ( \frac{1}{LC} \right )V_o(s)=\left ( \frac{1}{LC} \right )V_i(s)$$, $$\Rightarrow \left \{ s^2+\left ( \frac{R}{L} \right )s+\frac{1}{LC} \right \}V_o(s)=\left ( \frac{1}{LC} \right )V_i(s)$$, $$\Rightarrow \frac{V_o(s)}{V_i(s)}=\frac{\frac{1}{LC}}{s^2+\left ( \frac{R}{L} \right )s+\frac{1}{LC}}$$, $v_i(s)$ is the Laplace transform of the input voltage $v_i$, $v_o(s)$ is the Laplace transform of the output voltage $v_o$. These models are useful for analysis and design of control systems. Math. Analysis of control system means finding the output when we know the input and mathematical model. Follow these steps for differential equation model. Basic of Control Systems's Previous Year Questions with solutions of Control Systems from GATE ECE subject wise and chapter wise with solutions. International Journal of Systems Science and Applied Mathematics (IJSSAM) is a peer-reviewed and open access journal that provides an international forum for researchers, scholars and practitioners of systems science and applied mathematics to share experiences and communicate ideas. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Substitute, the current passing through capacitor $i=c\frac{\text{d}v_o}{\text{d}t}$ in the above equation. share. This thread is archived. 100% Upvoted. Have questions or comments? The journal encourages submissions from the research community where the priority will be on the novelty and the practical impact of the published findings. Dr V. Rakocevic, Reader in Networks and Communications save hide report. Budget R$24-45 BRL / hour. 20 comments. The transfer function model of this system is shown below. ... Engineering Mathematics. The simple answer is no. Mathematics of Control, Signals, and Systems features novel academic contributions which undergo peer review by experts in the field. Legal. Learning control systems. van Schuppen, CWI, Amsterdam, The NetherlandsFounding EditorsB.W. Process system engineers are entering a new era of control and optimization of complex process systems. Sontag, Boston, MA, USAJ. This circuit consists of resistor, inductor and capacitor. Although control theory has deep connections with classical areas of mathematics, such as the calculus of variations and the theory of differential equations , it did not become a field in its own right until the late 1950s and early 1960s. For more information, see more. Control theory, field of applied mathematics that is relevant to the control of certain physical processes and systems. Control theory does not deal directly with physical reality but with mathematical models. The above equation is a second order differential equation. Professor D. W. Stupples, Professor of Systems and Cryptography. The task of control theory is to study the mathematical quantification of these two basic problems and then to deduce applied mathematical methods whereby a concrete answer to optimization can be obtained. For example, the transfer function corresponding to a single linear controlled ordinary differential equation is a rational function (cf. Assumes a background in undergraduate control. Here, we represented an LTI system with a block having transfer function inside it. Managing EditorL. ME 661 Adaptive Control Systems{W-Term} Introduction to control of systems with undetermined or time-varying parameters. And this block has an input$X(s)$& output$Y(s)$. In mathematics, an autonomous system or autonomous differential equation is a system of ordinary differential equations which does not explicitly depend on the independent variable.When the variable is time, they are also called time-invariant systems.. best. What fields of mathematics are most widely used in systems and control theory? ito kazuhisa: Course description. Aims & scope. The above equation is a transfer function of the second order electrical system. Electrical Analogies of Mechanical Systems. from one par of the system to another… The reality of a control system is the physics not the blocks in frequency domain, but these block just try to let you understand it better and let you control it when its going to be unstable. What is Control Theory? $$v_i=Ri+L\frac{\text{d}i}{\text{d}t}+v_o$$. The objective is to develop a control model for controlling such systems using a control action in an optimum manner without delay or overshoot and ensuring control stability.. To do this, a controller with the requisite corrective behavior is required. Browse Top Mathematics Teachers Hire a Mathematics Teacher Browse Mathematics Jobs Post a Mathematics Project Learn more about Mathematics ... Control systems engineering - Mathematics - Linear Algebra. Analysis of control system means finding the output when we know the input and mathematical model. If$x(t)$and$y(t)$are the input and output of an LTI system, then the corresponding Laplace transforms are$X(s)$and$Y(s)$. Transfer function model is an s-domain mathematical model of control systems. Full members. The following mathematical models are mostly used. Various types of nonlinear equations or systems of equations arise in elementary and advanced control theory. All these electrical elements are connected in series. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Mathematics for systems and control theory. 52506: Engineer Aerospace/Space Engineering, Control Engineering, Mathematics, Computer Science, Electrical Engineering or similar - Technologies for Guidance & Control Systems for Space Applications Application Deadline: 03/02/2021 23:59 - Europe/Brussels The blocks in linear control systems, are just models to let you understand better about the physics, say voltages, currents or maybe fields, that travel(!) ). 1.2. [ "article:topic-guide", "license:ccby", "authorname:pwoolf" ], Assistant Professor (Chemical Engineering), 6.9: Blood Glucose Control in Diabetic Patients. Journal updates. One can distinguish two classes of systems for which control theory plays an indispensable role, namely man-made systems and biological systems. MCSS is an international journal devoted to mathematical control and system theory, including system theoretic aspects of signal processing. 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1. Chapter 6 Class 8 Squares and Square Roots
2. Serial order wise
3. Ex 6.4
Transcript
Ex 6.4, 2 Find the number of digits in the square root of each of the following numbers (without any calculation). (i) 64 Number of digits in 64 = n = 2. Since n is even, ∴ Number of digits in square root = 𝑛/2 = 2/2 = 1 Ex 6.4, 2 Find the number of digits in the square root of each of the following numbers (without any calculation). (ii) 144 Number of digits in 144 = n = 3. Since n is odd, ∴ Number of digits in square root = (𝑛 + 1)/2 = (3 + 1)/2 = 4/2 = 2 Ex 6.4, 2 Find the number of digits in the square root of each of the following numbers (without any calculation). (iii) 4489 Number of digits in 4489 = n = 4. Since n is even, ∴ Number of digits in square root = 𝑛/2 = 4/2 = 2 Ex 6.4, 2 Find the number of digits in the square root of each of the following numbers (without any calculation). (iv) 27225 Number of digits in 27225 = n = 5. Since n is odd, ∴ Number of digits in square root = (𝑛 + 1)/2 = (5 + 1)/2 = 6/2 = 3 Ex 6.4, 2 Find the number of digits in the square root of each of the following numbers (without any calculation). (v) 390625 Number of digits in 390625 = n = 6. Since n is even, ∴ Number of digits in square root = 𝑛/2 = 6/2 = 3
Ex 6.4 | HuggingFaceTB/finemath | |
# Introduction to Probability and Statistics Chapter 6 The
• Slides: 30
Introduction to Probability and Statistics Chapter 6 The Normal Probability Distribution
Continuous Random Variables • Continuous random variables can assume infinitely many values corresponding to points on a line interval. • Examples: – Heights, Weights – Lifetime of a particular product – Experimental laboratory error
Continuous Random Variables • A smooth curve describes the probability distribution of a continuous random variable. • The depth or density of the probability, which varies with x, may be described by a mathematical formula f (x ), called the probability distribution or probability density function for the random variable x.
Properties of Continuous Random Variable x • The area under the curve is equal to 1. • P(a < x < b) = area under the curve between a and b. • There is no probability attached to any single value of x. That is, P(x = a) = 0.
Properties of Continuous Random Variable x • Total probability is 1 • P( x = a) = 0 • P( x a) = P( x < a) (not true when x is discrete) • P( a < x < b) is the area between a and b under the density curve • P( x < a) is the area to the left of a • P( x > a) is the area to the right of a
Continuous Probability Distributions • There are many different types of continuous random variables a. Uniform; b. Exponential; c. Normal. • We try to pick a model that – Fits the data well – Allows us to make the best possible inferences using the data.
Normal Distribution • The formula that generates the normal probability density is: • Standard Normal: m = 0, s = 1.
Normal Distribution • The shape and location of the normal curve changes as the mean and standard deviation change. • Mean locates the center of the curve; • Standard deviation determines the shape: 1. Large values of standard deviation reduce height and increase spread. 2. Small values increase height and reduce spread.
The Standard Normal Distribution • To find P(a < x < b), we need to find the area under the appropriate normal curve. • To simplify the tabulation of these areas, we standardize each value of x by expressing it as a z-score, the number of standard deviations s it lies from the mean m.
Standard Normal (z) Distribution • • • z has Mean = 0; Standard deviation = 1 Symmetric about z = 0 Total probability Total area under curve is 1; P ( z > 0 ) =. 5 Area to the right of 0 is 0. 5; P ( z < 0 ) =. 5 Area to the left of 0 is 0. 5.
Using Table 3 Use Table 3 to calculate the probability: P(z < 1. 36) ? Area to the left of 1. 36 P ( z < 1. 36) =. 9131 Area to the left of 1. 36 =. 9131
Using Table 3 Use Table 3 to find the probability: Area to the right of 1. 36 P( z > 1. 36) ? P ( z > 1. 36) = 1 - P ( z 1. 36) = 1 -. 9131 =. 0869
Using Table 3 Use Table 3 to calculate the probability: (Area between) P(-1. 20 < z <1. 36) = P ( z < 1. 36) - P ( z <-1. 2) =. 9131 -. 1151 =. 7980 P(-1. 20 < z <1. 36)? Area between -1. 2 and 1. 36
Check Empirical Rule • within 3 standard deviations P(-3 < z <3) =. 9987 -. 0013=. 9974 Remember the Empirical Rule: Approximately 99. 7% of the measurements lie within 3 standard deviations of the mean. P(-1 < z <1) =. 8413 -. 1587 =. 6826 P(-2 < z < 2) =. 9772 -. 0228 =. 9544
z value with more than two decimals P( z < 1. 643) = P( z < 1. 64) =. 9495 P( z < 1. 6474) = P( z < 1. 65) =. 9505 P( z < 1. 645) ? 1. 1. 645 is halfway of 1. 64 and 1. 65 2. Look for areas of 1. 64 and 1. 65 in Table 3. 3. Since the value 1. 645 is halfway between 1. 64 and 1. 65, we average areas. 9495 and. 9505. 4. P( z 1. 645) = (. 9495+. 9505)/2 =. 9500
Extreme z values Using Table 3, calculate P(z<-5) = ? P(z>4) = ? • • P(z<-5) = 0 P(z>4) = 0
General Normal & Standard Normal • x is normal with mean m and standard deviation s. • Question: P(a < x < b) ? • i. e. area under the normal curve from a to b. • To simplify the tabulation of these areas, we standardize each value of x by expressing it as a z-score, the number of standard deviations s it lies from the mean m. z is standard normal
Example • x is normal with mean 0. 6 and standard deviation 2. z is standard normal • x is normal with mean 10. 2 and standard deviation 5. z is standard normal
Probabilities for General Normal Random Variable üTo find an area for a normal random variable x with mean m and standard deviation s, standardize or rescale the interval in terms of z. üFind the appropriate area using Table 3. Example: x has a normal distribution with m = 5 and s = 2. Find P(x > 7). 1 z
Example The weights of packages of ground beef are normally distributed with mean 1 pound and standard deviation. 10. What is the probability that a randomly selected package weighs between 0. 80 and 0. 85 pounds?
Area under General Normal Curve Studies show that gasoline use for compact cars sold in U. S. is normally distributed, with a mean of 25. 5 mpg and a standard deviation of 4. 5 mpg. What is the percentage of compacts get 30 mpg or more? 15. 87% of compacts get 30 mpg or more using Table 3
Using Table 3 üTo find an area to the left of a z-value, find the area directly from the table. e. g. P( z < 1. 36) üTo find an area to the right of a z-value, find the area in Table 3 and subtract from 1. or find the area with respect to the negative of the z-value; e. g. P( z >1. 36) = 1 -P( z <1. 36), P( z>1. 36) = P( z< -1. 36) üTo find the area between two values of z, find the two areas in Table 3, and subtract. e. g. P( -1. 20 < z < 1. 36 ) üTo find an area for a normal random variable x with mean m and standard deviation s, standardize or rescale the interval in terms of z. üFind the appropriate area using Table 3.
Working Backwards Find the value of z that has area. 25 to its left. 1. Look for the four digit area closest to. 2500 in Table 3. 2. What row and column does this value correspond to? 3. z = -. 67 4. What percentile does this value represent? 25 th percentile, or 1 st quartile (Q 1)
Working Backwards Find the value of z that has area. 05 to its right. 1. The area to its left will be 1 -. 05 =. 95 2. Look for the four digit area closest to. 9500 in Table 3. 3. Since the value. 9500 is halfway between. 9495 and. 9505, we choose z halfway between 1. 64 and 1. 65. 4. z = 1. 645
Example 1 Find the value of z, say z 0 , such that. 01 of the area is to its right. (tail area of. 01) using Table 3
Example 2 Find the value of z, say z 0 , such that. 95 of the area is within z 0 standard deviations of the mean. using Table 3
Example 3 The weights of packages of ground beef are normally distributed with mean 1 pound and standard deviation. 10. What is the weight of a package such that only 1% of all packages exceed this weight? using Table 3 99 th percentile
Exercise
Key Concepts I. Continuous Probability Distributions 1. Continuous random variables 2. Probability distributions or probability density functions a. Curves are smooth. b. The area under the curve between a and b represents the probability that x falls between a and b. c. P (x = a) = 0 for continuous random variables. II. The Normal Probability Distribution 1. Symmetric about its mean m. 2. Shape determined by its standard deviation s.
Key Concepts III. The Standard Normal Distribution 1. The normal random variable z has mean 0 and standard deviation 1. 2. Any normal random variable x can be transformed to a standard normal random variable using 3. Convert necessary values of x to z. 4. Use Table 3 in Appendix I to compute standard normal probabilities. 5. Several important z-values have tail areas as follows: Tail Area: . 005 . 01 . 025 . 05 . 10 z-Value: 2. 58 2. 33 1. 96 1. 645 1. 28 | HuggingFaceTB/finemath | |
# AP Statistics Curriculum 2007 EDA Center
(Difference between revisions)
Revision as of 16:46, 13 March 2008 (view source)IvoDinov (Talk | contribs) (→Other Measures of Centrality)← Older edit Revision as of 04:09, 7 April 2008 (view source)IvoDinov (Talk | contribs) (→Other Measures of Centrality: revised the Geometric/Harmonic mean sections)Newer edit → Line 46: Line 46: If you remove the student with the long jump distance of 106 and recalculate the median and mean, which one is altered less (therefore is more resistant)? Notice that the mean is very sensitive to outliers and atypical observations, and hence less resistant than the median. If you remove the student with the long jump distance of 106 and recalculate the median and mean, which one is altered less (therefore is more resistant)? Notice that the mean is very sensitive to outliers and atypical observations, and hence less resistant than the median. - ===Other Measures of Centrality=== + ===Resistant Mean-related Measures of Centrality=== The following two sample measures of population centrality estimate resemble the calculations of the [[AP_Statistics_Curriculum_2007_Distrib_MeanVar#Expectation_.28Mean.29 | mean]], however they are much more ''resistant'' to change in the presence of outliers. The following two sample measures of population centrality estimate resemble the calculations of the [[AP_Statistics_Curriculum_2007_Distrib_MeanVar#Expectation_.28Mean.29 | mean]], however they are much more ''resistant'' to change in the presence of outliers. Line 54: Line 54: ===='''Windsorized k-times mean'''==== ===='''Windsorized k-times mean'''==== The Windsorirized k-times mean is defined similarly by $\bar{y}_{w,k}={1\over n}( k\times y_{(k)}+\sum_{i=k+1}^{n-k-1}{y_{(i)}}+k\times y_{(n-k)})$, where $k\geq 0$ is the trim-factor and $y_{(i)}$ are the order statistics (small to large). In this case, before we compute the arithmetic average, we replace the ''k'' smallest and the ''k'' largest observations with the kth and (n-k)th largest observations, respectively. The Windsorirized k-times mean is defined similarly by $\bar{y}_{w,k}={1\over n}( k\times y_{(k)}+\sum_{i=k+1}^{n-k-1}{y_{(i)}}+k\times y_{(n-k)})$, where $k\geq 0$ is the trim-factor and $y_{(i)}$ are the order statistics (small to large). In this case, before we compute the arithmetic average, we replace the ''k'' smallest and the ''k'' largest observations with the kth and (n-k)th largest observations, respectively. + + ===Other Centrality Measures=== + The ''arithmetic'' mean answers the question, ''if all observations were equal, what would that value (''center'') have to be in order to achieve the same total?'' + : $n\times \bar{x}=\sum_{i=1}^n{x_i}$ + + In some situations, there is a need to think of the average in different terms, not in terms of arithmetic average. ====Harmonic Mean==== ====Harmonic Mean==== - In some situations, there is a need to think of the average in different terms, not in terms of arithmetic average. For instance if we study speeds (velocities) the ''arithmetic'' mean is inappropriate, however the [http://en.wikibooks.org/wiki/Statistics/Summary/Averages/Harmonic_Mean harmonic mean (computed differently)] gives the most intuitive answer to what is the "''middle''" for a process. + If we study speeds (velocities) the ''arithmetic'' mean is inappropriate, however the [http://en.wikibooks.org/wiki/Statistics/Summary/Averages/Harmonic_Mean harmonic mean (computed differently)] gives the most intuitive answer to what is the "''middle''" for a process. The harmonic mean answers the question ''if all the observations were equal, what would that value have to be in order to achieve the same sample sum of reciprocals?'' : ''Harmonic mean'': $\hat{\hat{x}}= \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \ldots + \frac{1}{x_n}}$ : ''Harmonic mean'': $\hat{\hat{x}}= \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \ldots + \frac{1}{x_n}}$ ====Geometric Mean==== ====Geometric Mean==== - The ''arithmetic'' mean answers the question, ''if all observations were equal, what would that value (''center'') have to be in order to achieve the same total?'' + In contrast, the [http://en.wikibooks.org/wiki/Statistics/Summary/Averages/Geometric_Mean geometric mean] answers the question, ''if all the observations were equal, what would that value have to be in order to achieve the same sample product?'' - : $n\times \bar{x}=\sum_{i=1}^n{x_i}$ + - + - In contrast, the [http://en.wikibooks.org/wiki/Statistics/Summary/Averages/Geometric_Mean geometric mean] answers the question, ''if all the observations were equal, what would that value have to be in order to achieve the same sample product ?'' + : ''Geometric mean'': $\tilde{x}^n={\prod_{i=1}^n x_i}$ : ''Geometric mean'': $\tilde{x}^n={\prod_{i=1}^n x_i}$
## General Advance-Placement (AP) Statistics Curriculum - Central Tendency
### Measurements of Central Tendency
There are three main features of all populations (or data samples) that are always critical in understanding and interpreting their distributions. These characteristics are Center, Spread and Shape. The main measure of centrality are mean, median and mode.
Suppose we are interested in the long-jump performance of some students. We can carry an experiment by randomly selecting 8 male statistics students and ask them to perform the standing long jump. In reality every student participated, but for the ease of calculations below we will focus on these eight students. The long jumps were as follows:
74 78 106 80 68 64 60 76
### Mean
The sample-mean is the arithmetic average of a finite sample of numbers. In the long-jump example, the sample-mean is calculated as follows:
$\overline{y} = {1 \over 8} (74+78+106+80+68+64+60+76)=75.75 in.$
### Median
The sample-median can be thought of as the point that divides a distribution in half (50/50). The following steps are used to find the sample-median:
• Arrange the data in ascending order
• If the sample size is odd, the median is the middle value of the ordered collection
• If the sample size is even, the median is the average of the middle two values in the ordered collection.
For the long-jump data above we have:
• Ordered data:
60 64 68 74 76 78 80 106
• $Median = {74+76 \over 2} = 75$.
### Mode(s)
The modes represent the most frequently occurring values (The numbers that appear the most). The term mode is applied both to probability distributions and to collections of experimental data.
For instance, for the Hot dogs data file, there appear to be 3 modes for the calorie variable! This is evident by the histogram of the Calorie content of all hotdogs, shown in the image below. Note the clear separation of the calories into 3 distinct sub-populations - the highest points in these three sub-populations are the three modes for these data.
### Resistance
A statistic is said to be resistant if the value of the statistic is relatively unchanged by changes in a small portion of the data. Referencing the formulas for the median, mean and mode which statistic seems to be more resistant?
If you remove the student with the long jump distance of 106 and recalculate the median and mean, which one is altered less (therefore is more resistant)? Notice that the mean is very sensitive to outliers and atypical observations, and hence less resistant than the median.
### Resistant Mean-related Measures of Centrality
The following two sample measures of population centrality estimate resemble the calculations of the mean, however they are much more resistant to change in the presence of outliers.
#### K-times trimmed mean
$\bar{y}_{t,k}={1\over n-2k}\sum_{i=k+1}^{n-k}{y_{(i)}}$, where $k\geq 0$ is the trim-factor (large k, yield less variant estimates of center), and y(i) are the order statistics (small to large). That is, we remove the smallest and the largest k observations from the sample, before we compute the arithmetic average.
#### Windsorized k-times mean
The Windsorirized k-times mean is defined similarly by $\bar{y}_{w,k}={1\over n}( k\times y_{(k)}+\sum_{i=k+1}^{n-k-1}{y_{(i)}}+k\times y_{(n-k)})$, where $k\geq 0$ is the trim-factor and y(i) are the order statistics (small to large). In this case, before we compute the arithmetic average, we replace the k smallest and the k largest observations with the kth and (n-k)th largest observations, respectively.
### Other Centrality Measures
The arithmetic mean answers the question, if all observations were equal, what would that value (center) have to be in order to achieve the same total?
$n\times \bar{x}=\sum_{i=1}^n{x_i}$
In some situations, there is a need to think of the average in different terms, not in terms of arithmetic average.
#### Harmonic Mean
If we study speeds (velocities) the arithmetic mean is inappropriate, however the harmonic mean (computed differently) gives the most intuitive answer to what is the "middle" for a process. The harmonic mean answers the question if all the observations were equal, what would that value have to be in order to achieve the same sample sum of reciprocals?
Harmonic mean: $\hat{\hat{x}}= \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \ldots + \frac{1}{x_n}}$
#### Geometric Mean
In contrast, the geometric mean answers the question, if all the observations were equal, what would that value have to be in order to achieve the same sample product?
Geometric mean: $\tilde{x}^n={\prod_{i=1}^n x_i}$
Alternatively: $\tilde{x}= \exp \left( \frac{1}{n} \sum_{i=1}^n\log(x_i) \right)$ | HuggingFaceTB/finemath | |
# Project #9607 - statistics
1. A coin is tossed 3 times.
(A) The sample space is____________________________________________________________
(B) Find the probability of getting exactly 1 heads by listing the number of outcomes of the only one head.
(C) Is this considered the binomial distribution? Why does it consider a binomial distribution? If yes, use the 4 requirements to identify it.
(D) Use the Binomial Probability formula to find probability from part (B).
(E) Find the probability of getting at most 1 head? What does getting at most 1 head mean? You can use binomial distribution formula sheet.
(F) Find the probability of getting at least 2 heads. What does getting at least 2 heads mean? You can use binomial distribution formula sheet.
(G) Find the mean, variance, and standard deviation for the number of head will be obtained. Remember tossing a coin getting a head is a binomial distribution.
2. Draw the normal curve and then find the area under the standard normal distribution.
a. to the left of -2.56
b. to the right of 2.56
c. between 0 and 2.56
8. Find the following probabilities using the table. Sketch the normal distribution shape and shade the region.
(A) p( z <2.18)
(B) Findp(z>2.18)
(C)Findp(1.46 <z<2.18)
9. The average height of LACC students is 60 inches. The standard deviation is 5 inches. The variable is normally distributed.
(A)Draw the normal distribution with labeling mean and standard deviation.
(B)Find the probability that a selected individual height will be greater than 50 inches. With the normal curve, shade the region that is greater than 50 inches.
(C)Find the probability that a selected individual height will be less than 65 inches. With the normal curve, shade the region that is less than 65 inches.
(D)If 25 students in English class will be selected, find the probability of the height will be less than 65 inches? What is the name of the theorem are you applying?
3. Estimate 90% confidence interval for the true population proportion.
In a survey at UCLA, there are 50 students out of 200 who like to drink cappuccino. Find the sample proportion, critical value, and then estimate the confidence interval (write out the result).
4. Estimate 90% confidence interval of the true mean.
In a survey at UCLA, the average hour of sleeping is 6 hours per week during the final week from a sample of 30 students and its standard deviation is 2 hours. Find critical value and then estimate the confidence interval (write out the result).
Subject Mathematics Due By (Pacific Time) 07/25/2013 12:00 am
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# Thevenin equivalent of circuit with no voltage sources I need to find the Thevenin Equivalent left of the >> symbols. Due to the lack of voltage sources I tried finding the Norton Equivalent first and came up with $Rn=R_O$, $v_{th}=i_S \cdot R_O$. Is that correct? If not how does one handle this case?
• It is trivial to go from Norton to Thevenin and vice versa. – Oldfart Apr 11 '18 at 15:51
• Also, I guarantee you that $R_N = R_O$ is incorrect. You haven't accounted for the VCCS, which is configured to act just like another resistor in parallel with $R_O$. – The Photon Apr 11 '18 at 15:55
• -oldfart; I know but doing that in this case yields a result which is independent of g, so i suppose i am missing something – Manouil Apr 11 '18 at 15:56
• what you've tried is correct but it's just an intermediate result: a Thevenin source connected to a dependent current source. Now use it to determine short circuit current and open circuit voltage – Curd Apr 11 '18 at 15:57
• Can you find the open-circuit output voltage and short-circuit output current of the circuit? – The Photon Apr 11 '18 at 15:58
## 1 Answer • To find Rth, open circuit all independent current sources, short all independent voltage sources.
• Connect a fictitious 1V source at the open ckt terminal.
• Find the current driven by 1V source.
• If I is the current driven by 1V source, then $\frac{V}{I} = \frac{1}{I}$ is the load seen by the voltage source. i.e., Rth. $$I = I_{Ro} + gv_x$$ vx = voltage across Ro = 1V $$\implies I = \frac{1}{Ro}+g = \frac{(1+gRo)}{Ro}$$ $$\therefore R_{th} = 1/I = \frac{Ro}{(1+gRo)}$$
• To find Vth, go back to the original ckt. Vth is the voltage drop across Ro which is same as vx.
• Write ohms law equation for voltage across Ro $$(i_s-gv_x)Ro= v_x$$ $$\implies v_x = i_s\frac{Ro}{(1+gRo)} = V_{th}$$ | HuggingFaceTB/finemath | |
# fixed point iteration of a single variable with unknown constants
Okay so given this simple looking fixed point iteration, i.e., $$x_{n+1}=g(x_n)$$
$$x_{n+1} = g(x) = -b - \frac{c}{x_n}$$
The idea is to find a region in the space of $$(b,c) \in \mathbb{R^2}$$ that will converge for all good starting points $$x_0$$ (ones that are relatively close to $$x$$) with the error being reduced by more than $$1/2$$ on each iteration, in other words the iteration is $$\mathcal{O}(1/2^n)$$.
I understand the brute force way of computing all the possible combinations, which my computer is working on, checking the error and then plotting the points that fit the criteria on a 2-D plot. It seems like there should be a way to figure this out more concisely, I envisage there is a way to take partials of $$g(x)$$ or optimize a potential of some sort or other and figure out the region analytically.
Step 1: If the iteration does converge, it would converge to some equilibrium point of $$x_{n+1} = -b - \dfrac{c}{x_{n}}$$, which is a root of $$x = -b - \dfrac{c}{x}$$.
That is, possible convergence points would be $$\dfrac{-b \pm \sqrt{b^2 - 4c}}{2}$$.
Step2: fix one of them, say $$X = \dfrac{-b + \sqrt{b^2 - 4c}}{2}$$, and consider $$\epsilon_{n} = x_{n} - X$$. Compute $$\epsilon_{n+1}$$. Obviously it depends on $$x_{n}$$, and you want $$|\epsilon_{n+1}| < |\dfrac{\epsilon_{n}}{{2}}|$$. This is a condition for $$x$$s you are looking for.
• I very well might misread your question. Are you interested in $b,c$ pairs for which an iteration converges fast for certain $x$, or, given $b,c$ you are interested in an initial $x$? – user58697 Jul 7 at 0:30 | HuggingFaceTB/finemath | |
# 1982 AHSME Problems/Problem 29
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Let $x,y,$ and $z$ be three positive real numbers whose sum is $1.$ If no one of these numbers is more than twice any other, then the minimum possible value of the product $xyz$ is
$\textbf{(A)}\ \frac{1}{32}\qquad \textbf{(B)}\ \frac{1}{36}\qquad \textbf{(C)}\ \frac{4}{125}\qquad \textbf{(D)}\ \frac{1}{127}\qquad \textbf{(E)}\ \text{none of these}$
## Solution
Suppose that the product $xyz$ is minimized at $(x,y,z)=(x_0,y_0,z_0).$ Without the loss of generality, let $x_0 \leq y_0 \leq z_0$ and fix $y=y_0.$
To minimize $xy_0z,$ we minimize $xz.$ Note that $x+z=1-y_0.$ By a corollary of the AM-GM Inequality (If two nonnegative numbers have a constant sum, then their product is minimized when they are as far as possible.), we get $z_0=2x_0.$ It follows that $y_0=1-3x_0.$
Recall that $x_0 \leq 1-3x_0 \leq 2x_0,$ so $\frac15 \leq x_0 \leq \frac14.$ This problem is equivalent to finding the minimum value of $$f(x)=xyz=x(1-3x)(2x)=2x^2(1-3x)$$ in the interval $I=\left[\frac15,\frac14\right].$ The graph of $y=f(x)$ is shown below: $[asy] /* Made by MRENTHUSIASM */ size(900,200); real f(real x) { return 2x^2 * (1-3x); } real xMin = -0.349; real xMax = 1/2; real yMin = -1/4; real yMax = 1/2; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("x",(xMax,0),(2,0)); label("y",(0,yMax),(0,2)); pair A[]; A[0] = (0,0); A[1] = (1/2,1/2); A[2] = (2/3,1/3); draw((1/5,-0.015)--(1/5,0.015),linewidth(1)); draw((1/4,-0.015)--(1/4,0.015),linewidth(1)); draw((1/3,-0.015)--(1/3,0.015),linewidth(1)); draw((1/5,yMin)--(1/5,yMax),dashed); draw((1/4,yMin)--(1/4,yMax),dashed); label("0",A[0],(-2,-2)); label("\frac15",(1/5,0),(0,-2),UnFill); label("\frac14",(1/4,0),(0,-2),UnFill); label("\frac13",(1/3,0),(0,-2)); draw(graph(f,xMin,xMax),red); dot(A[0],red+linewidth(5)); dot((1/3,0),red+linewidth(5)); [/asy]$ Since $f$ has a relative minimum at $x=0,$ and cubic functions have at most one relative minimum, we conclude that the minimum value of $f$ in $I$ is at either $x=\frac15$ or $x=\frac14.$ As $f\left(\frac14\right)=\frac{1}{32}\leq f\left(\frac15\right)=\frac{4}{125},$ the minimum value of $f$ in $I$ is $\boxed{\textbf{(A)}\ \frac{1}{32}}.$
~MRENTHUSIASM | HuggingFaceTB/finemath | |
# How Many Cups Is 1000 ML?
Last updated on January 23rd, 2023 at 09:58 pm
There are 4.22 cups in 1000 milliliters.
One cup and 1000 milliliters are both common measurements you might encounter, particularly if you’re baking or cooking.
Calculating how many cups are in 1000 milliliters can at first seem a bit tricky since these units are parts of two different measurement systems.
However, with a bit of simple math, you can easily calculate between cups and milliliters, and figuring out how many cups are in 1000 milliliters isn’t too difficult at all.
There are about 236.6 milliliters in a single cup.
In order to figure out how many cups are in a certain number of milliliters, then, all you would need to do is divide the number of milliliters by 236.6.
You can also round up to 237.
To figure out how many cups are in 1000 milliliters, divide 1000 by 236.6.
1000 / 236.6 = 4.22
The answer, which is 4.22, tells you that there are just over 4 cups in 1000 milliliters.
You can also reverse the math to figure out how many milliliters are in a certain number of cups.
For example, if you want to know how many milliliters are in 6 cups, simply multiply 236.6 by four.
The answer, which is 1419.6, tells you that there are just over 1400 milliliters in 6 cups.
Here is a conversion table for you to reference.
## Milliliters to cups conversion table
Recipes or other tasks that call for cup measurements are more common in the United States since this unit is a measurement in the imperial system.
CHECK OUT Household Items That Weigh 200 Grams
Milliliters, on the other hand, are part of the metric system, which is widely used around the world. | HuggingFaceTB/finemath | |
# Selection Sort
Selection sort is the last simplistic sorting algorithms covered here1. It works by repeatedly searching for the next minimum value and moving it to the correct position. It’s difficult to describe with prose; however, it’s the easiest sorting algorithm to implement. Examine the animated image below in conjunction with the supplied pseudo code to understand it’s inner working. Recall that bubble and insertion sort are $\Omega(n)-O(n^2)$ algorithms. When the data is pre-sorted the number of operations is linear to the input size. In the worst case scenario, reverse sorted data, the number of operations is $n^2$. Conversely, selection sort is $\Theta(n^2)$: the number of operations is the same regardless of the input data. One would assume that selection sort is inferior based on this data. However, recall that the RAM model weights all operations equally. The innovation of selection sort is that while the number of comparison operations is commiserate, the number of copy operation is minimized.
In this context, a copy is defined as copying data from one memory location to another. The pseudo code below demonstrates the three copy operations2 associated with swapping two array values.
int temp;
temp = A[i]; // 1st copy
A[i] = A[j]; // 2nd copy
A[j] = temp; // 3rd copy
Selection sort may outperform other simplistic sort algorithms owing to fewer copy operations. This is an important attribute to keep in mind for situations where moving data is expensive.
## Pseudo Code
\begin{algorithm}
\caption{Selection Sort}
\begin{algorithmic}
\REQUIRE $A$ = list of numbers
\ENSURE $A$ sorted in sequential order
\FUNCTION{selectionSort}{$A$}
\FOR{i $\gets$ 0 to ($\vert A \vert$ - 1)}
\STATE lowest $\gets$ i
\STATE
\FOR{j $\gets$ (i + 1) to ($\vert A \vert$ - 1)}
\IF{$A_{\text{lowest}} \gt A_j$ }
\STATE lowest $\gets$ j
\ENDIF
\ENDFOR
\STATE
\STATE swap $A_{\text{lowest}}$ and $A_i$
\ENDFOR
\ENDFUNCTION
\end{algorithmic}
\end{algorithm}
## Asymptotic Complexity
$\Theta(n^2)$
Full Repo
Relevant Files:
## Exercises
1. Assume line 6 of the selection sort pseudo code is changed to:
$\text{if} \space A_{\text{lowest}} \lt A_j$
How does this change the final state?
The data will be sorted in descending order rather than ascending order.
2. Using the programming language of your choice, implement a selection sort function that accepts:
See the C implementation provided in the links below. Your implementation may vary significantly and still be correct.
- selection_sort.h
- selection_sort.c
- selection_sort_test.c
3. At this point, you should have implemented bubble, insertion, and selection sort. Instrument each of your algorithms to calculate the total number of copy and comparison operations. Next, sort the data contained in the sort.txt file and determine the total number of copy and comparison for each algorithm.
Bubble Sort: copy = 71,844,390 comparisons = 49,988,559 total = 121,832,949 | HuggingFaceTB/finemath | |
# Inverse Functions
## Precalculus Quiz
- A function f(x) is one-to-one if it does not assign the same value to two different elements of its domain.
- If f(x) is a 1-1 function, then it has an inverse function f-1 defined by f-1(y)=x if f(x)=y, for all y in the range of f.
- The domain of f-1 is the range of f, and the range of f-1 is the domain of f.
- To find a formula for f-1, we can set y = f(x), solve for x in terms of y, and set f-1(y) = x.
Question 1: Which the functions f(x), g(x) and h(x) is a one-to-one function?
f(x) = -x5
g(x) = 2x2
h(x) = |x|
Questions 2-3: If , find a formula for f-1(x) and find the domain of f-1
f-1(x):
domain of f-1:
Questions 4-5: If f(x) = 3x3 - 8, find a formula for f-1(x) and find the domain of f-1
formula for f-1(x):
domain of f-1:
Questions 6-8: Let f(x) = x2 - 2x for x ≥ 1. Find the inverse f-1 and find the domain and the range for f-1.
formula for f-1(x):
domain of f-1:
range of f-1:
Questions 9-10: Let
Find the inverse f-1 and find the domain and the range for f-1.
formula for f-1(x):
domain of f-1:
Press the Submit button to see the results. | HuggingFaceTB/finemath | |
## New HL Resource
I’ve added a new resource to the HL Resources page that will be useful when studying. Note that (as with the resources you’ve already got), you will be able to find past paper questions that will be especially useful when studying for our end-of-year exam.
## Planes and the Cross Product
Here are the questions we considered in today’s class. See if you can answer these (with algebraic solutions) for tomorrow’s lesson. GeoGebra will be useful to check you answers, and to give you some insight into the question if you get stuck.
1. Find a vector normal to both
$\vec{a}=\begin{bmatrix}1\\2\\-2\end{bmatrix} \text{ and }\vec{b}=\begin{bmatrix}3\\4\\1\end{bmatrix}.$
2. a) Find a vector normal to the plane $\vec{r}=\begin{bmatrix}1\\4\\-2\end{bmatrix} +\lambda \begin{bmatrix}1\\1\\-1\end{bmatrix} + \mu \begin{bmatrix}-3\\1\\2\end{bmatrix}$ b) Using your answer to part a), can you find the distance of the point $$A(1, 1, 1)$$ to the given plane?
## Combined Class Postponed
We’ll join Mr Kilding’s group on Thursday this week, rather than today.
## Planes
Try to complete the following questions for our lesson on Monday.
1. Find a vector equation of the plane passing through $$A(1,2,3)$$, $$B(3,1,-2)$$, and $$C(4,4,4)$$.
2. Find a vector equation of the plane with Cartesian equation $$3x+2y-z=1$$.
3. Find a vector equation of the plane containing the line $\vec{r}=\begin{bmatrix}-1\\-1\\4\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}$ and passing through the point $$A(6,-3,2)$$.
## Vectors Homework Assignment
The attached Vectors Assignment contains a long answer question that you can work on during Tuesday’s lesson. Complete this question for Wednesday’s lesson. Your answers will be collected and marked. You can use GeoGebra to confirm you answers, but full algebraic solutions must be included.
## Lines in 3D
Try to complete the following questions for Monday’s lesson.
1. Find the distance of the point $$A(-1,1,2)$$ to the line $\displaystyle{\vec{r}=\begin{bmatrix}1\\-2\\3\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}}$
2. The lines below intersect at a point $$A$$. Use an algebraic method to find the coordinates of $$A$$, then verify your answer using GeoGebra.
$L_1: \vec{r}=\begin{bmatrix}-1\\-1\\4\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}$ $L_2: \vec{r}=\begin{bmatrix}-2\\0\\7\end{bmatrix}+\lambda \begin{bmatrix}-6\\4\\8\end{bmatrix}$
3. In 2D, any two non-parallel lines will have a point of intersection. In 3D, two lines can be non-parallel and have no point of intersection; such lines are called skew lines.
Show that $$L_1$$ from question 2 and $$L_3$$ below are skew lines.
$L_3: \vec{r}=\begin{bmatrix}-2\\5\\12\end{bmatrix}+\lambda \begin{bmatrix}-6\\4\\8\end{bmatrix}$
## Vector Equations of Lines 3
Find the distance of the point $$A(-3,1)$$ to the line
$\displaystyle{\vec{r}=\begin{bmatrix}1\\2\end{bmatrix}+\lambda \begin{bmatrix}-2\\3\end{bmatrix}}$
using the method outlined below.
1. Find an expression for an arbitrary point $$D$$ on the given line.
2. Using the expression you’ve produced, find a general expression for the vector $$\overrightarrow{AD}$$.
3. Where $$\vec{b}$$ is the direction vector of the given line, use the dot product $$\overrightarrow{AD}\cdot\vec{b}$$ to find the coordinates of the point on the line closest to $$A$$. Hence, find the distance from $$A$$ to the line.
## Vector Equations of Lines 2
Complete the following question for our lesson tomorrow. In the first instance, use the Cartesian equation to find your solution, then see of you can find a vector-based solution to the same problem.
Find the distance of the point $$A(1,3)$$ to the line
$\displaystyle{\vec{r}=\begin{bmatrix}4\\2\end{bmatrix}+\lambda \begin{bmatrix}4\\1\end{bmatrix}}$
## Vector Equations of Lines
Complete the following questions for tomorrow’s lesson.
1. Find a vector equation of the line passing through $$A(2,5)$$ and $$B(6,-1)$$.
2. Find a different vector equation for the same line.
## Applications of Vectors
As discussed in class last week, complete page 418 question 18 for tomorrow’s lesson.
For other examples of applications in physics, I suggest you also look at questions 28 and 29 on page 419. | HuggingFaceTB/finemath | |
# math
In right triangle, the side opposite to right angle is called what? ANSWER FAST PLZ!!! THX
1. 👍 0
2. 👎 0
3. 👁 100
1. hypotenuse
1. 👍 0
2. 👎 0
posted by Damon
2. thanks
1. 👍 0
2. 👎 0
## Similar Questions
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More Similar Questions | HuggingFaceTB/finemath | |
# How to calculate reaction moments and forces in a fixed axle with multiple point loads?
I need to calculate the maximum moment and reaction forces in an axle with both sides fixed in bearings, the axle is 1010mm long and has downwards forces of:
• 236kN at 403mm
• 86kN at 499mm
• 32kN at 595mm
• 12kN at 691mm
• 4kN at 787mm
I need the results to determine what bearings to use and to calculate the minimum axle diameter, I have had lessons in how to calculate these problems but those only explained a single point load or a distributed load. How would I go about calculating this?
EDIT: The A drum will be mounted on the axle to make a winch, the forces are based on the cable windings pulling on this drum.
• I’m voting to close this question because it looks like a homework exercise.
– NMech
May 10, 2021 at 12:34
• It is not a homework question, I'm designing a drum for a winch and these are the forces the individual cable windings exert on the axle of the winch. Sorry I didn't add that in the question but I really do not know how to solve this. May 10, 2021 at 13:07
• @David, you might want to add details in the comment to the body of the question. May 10, 2021 at 14:26
• @David, I assume you are primarily looking for the results. If so, you might want to give this tool a try: mechanicalc.com/calculators/beam-analysis May 10, 2021 at 17:24
• @Andrew Thank you! I wasn't aware such a calculator existed and this will definitely help me. May 10, 2021 at 22:07
If you know how to calculate the reactions (R, V & M) of a simply supported beam with a single concentrate load, you can repeat that calculation for every single load in a multi-load situation, then super-impose the results at the points of interest.
The graph below shows the process of a beam with two concentrate loads $$P_A$$ & $$P_B$$, and how to obtain the reactions $$R_L, R_R$$, the internal shear $$V$$ and internal moment $$M$$, along the span, by the method of superposition, which is valid for a linear elastic beam with any types/numbers of externally applied loads. Note the maximum moment always occurs at where the shear force changes sign and crosses the horizontal beam axis. (Note the graph is not made to scale)
Solution for a fixed end beam with a single concentrate load
• Thanks for the explanation, however in my situation the beam is not supported but fixed in both ends, as far as I know this takes away all degrees of freedom so I'm not sure if this applies? May 10, 2021 at 22:02
• Yes, it applies to beams with any kind of end support. It is tedious, but as long as you know how to calculate the reactions of the beam with a single load, the superposition is the most simple method to use without resorting to the structural analysis program. Note it is applicable to deflection calculation as well. Also, a spreadsheet can be utilized to eliminate the pain of repetition. If programmed correctly, it eliminates the mistake too.
– r13
May 10, 2021 at 22:38
• For a fixed end beam, the only addition to the graph above is the fixed-end moment at each end, again, the total is the sum of the parts.
– r13
May 10, 2021 at 22:44
• The solution for a fixed end beam with a single load is added FYI.
– r13
May 11, 2021 at 2:04 | HuggingFaceTB/finemath | |
+0
# Help
0
148
1
I need help
Nov 3, 2018
#1
0
C) You have to use your iterative algorithm about 6 times to get the accuracy of 3 decimal places as follows:
(5, -2.92, -2.765434415, -2.738480987, -2.733307664, -2.732297173)
So, -2.732 is ONLY one solution to your equation. There are two more: x= - 1 and x =sqrt(3) - 1, which you cannot get by using the iterative algorithm.
5) This is also an iterative algorithm. Using it 4 times gives the following results:
Nov 3, 2018
edited by Guest Nov 3, 2018
#1
0
C) You have to use your iterative algorithm about 6 times to get the accuracy of 3 decimal places as follows:
(5, -2.92, -2.765434415, -2.738480987, -2.733307664, -2.732297173)
So, -2.732 is ONLY one solution to your equation. There are two more: x= - 1 and x =sqrt(3) - 1, which you cannot get by using the iterative algorithm.
5) This is also an iterative algorithm. Using it 4 times gives the following results: | HuggingFaceTB/finemath | |
# Dice Game: Optimal Strategy
Recently, we played a simple dice game at school. The objective of the game is to gain as many points as possible (have the largest expected value of points). Each round consists of several turns. At each turn, the player rolls a dice. Then,
1. If the die is not a two, the player adds the number to his pot.
2. If the die is a two, the player loses everything and the round ends.
3. At the end of each roll, the player is given a chance to cash out, and receive the pot.
Finally, the player gets a free roll at the beginning (if the roll is a two, it is still added to the pot). What is the optimal strategy for winning?
I thought of two different ways to answer this question, but they gave different results.
1) Let $$s_{t}$$ be the current size of the pot. The expected value of taking the next turn is $$\mathbb{E}[s_{t+1}] = 5/6 * (s_{t} + 19 / 5)$$ where 19/5 is the expected value given that the roll is not a two, $$19 / 5 = (1+3+4+5+6)/ 5$$. Therefore, the time to stop, or when the expectation as its maximum, is when $$s_t$$ equals $$\mathbb{E}[s_{t+1}]$$, which is $$s_{t}= 19$$.
2) Instead of looking at the final score, I looked at what number of turns gives the best return. The equation I created was this: $$(19 / 5 n + 21 / 6) * (5/6) ^ n$$where n is the number of moves taken after the first free roll. The first part of the equation is the expected number of points gained from n moves given that each roll was not a two. The second part gives the probability of the person surviving until the nth decision. The maximum is obtained at around n = 4.38 with the equation equal to 9.38.
Here's where I try to compare the method 1 to method 2: Since the each roll contributes an expected value of (19 / 5), the expected number of decisions to make to reach 20 is $$19 / (19/5) - 1$$ or 4. That is not equal to the number of moves obtained from the second method, 4.38.
Are any of these methods valid arguments? Is the comparison valid? If not, are the two methods equivalent to each other?
The first argument is answering, "If I am facing a single decision on a single turn with $$s_t$$ in the pot, do I expect to make more by cashing out now or by cashing in after 1 more round?" dependent on the size of the pot.
The second argument is answering the question, "If I decide in advance that I'm going to play $$n$$ turns and then cash out, how much do I expect to make?" and then picking the $$n$$ that maximizes this. This throws away your ability to reevaluate each turn depending on new information. Staying in with $18 in the pot is a good idea, whether it's turn 1 or turn 7. But be careful, games like this get a bit weird if they go on for a while. For example, suppose you start with a dollar and flip a coin, heads you triple your current total and tails you lose it all. Expected value for the first round is +$0.50, so you should play. And then if you win, why not play again? And again? But if you keep playing indefinitely, you're almost sure to lose everything. How many rounds do you play? See, for comparison, the St. Petersburg Paradox | HuggingFaceTB/finemath | |
# Solved (Free): Two business schools, A and B, located in the same metropolitan area and they are competing for bragging rights. One of the points of competition is average salary of graduating seniors
#### ByDr. Raju Chaudhari
Apr 1, 2021
Two business schools, A and B, located in the same metropolitan area and they are competing for bragging rights. One of the points of competition is average salary of graduating seniors. 30 graduating seniors from A and 25 from B were surveyed. A's students had an average salary of \$62,000, and B's students had an average salary of \$67,000. Based on historical data, the population standard deviation is assumed to be \$10,000 for A and \$15,000 for B. Construct the hypotheses and conduct the appropriate tests that school B could use to claim that its students have a higher average graduating salary that A. Using a 5% level of significance, and the sample data provided, determine if school B can claim that its average graduating salary is greater than that of school A.
#### Solution
Given that the sample size $n_1 = 30$, $n_2 = 25$, sample mean $\overline{x}_1= 62000$,
$\overline{x}_2= 67000$, standard deviation $\sigma_1 = 10000$ and $\sigma_2 = 15000$.
##### Step 1 State the hypothesis testing problem
The hypothesis testing problem is
$H_0 : \mu_1 = \mu_2$ against $H_1 : \mu_1 < \mu_2$ ($\textit{left-tailed}$)
##### Step 2 Define test statistic
The test statistic is
\begin{aligned} Z=\frac{(\overline{x}_1 -\overline{x}_1)-(\mu_1 -\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}. \end{aligned}
The test statistic $Z$ follows standard normal distribution $N(0,1)$.
##### Step 3 Specify the level of significance
The significance level is $\alpha = 0.05$.
##### Step 4 Determine the critical value
As the alternative hypothesis is $\textit{left-tailed}$, the critical value of $Z$ $\text{is}$ $\text{-1.64}$.
The rejection region (i.e. critical region) is $\text{Z < -1.64}$.
##### Step 5 Computation
The test statistic for testing above hypothesis under the null hypothesis is
\begin{aligned} Z_{obs}&=\frac{(\overline{x}_1 -\overline{x}_1)-0}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}\\ &= \frac{62000-67000}{\sqrt{\frac{10000^2}{30}+\frac{15000^2}{25}}}\\ &= -1.424 \end{aligned}
##### Step 6 Decision
The rejection region (i.e. critical region) is $\text{Z < -1.64}$. The test statistic is $Z_{obs} =-1.424$ which falls $outside$ the critical region, we $\textit{fail to reject}$ the null hypothesis.
OR
$p$-value approach:
The test is $\text{left-tailed}$ test, so the p-value is the area to the $\text{left}$ of the test statistic ($Z_{obs}=-1.424$) is p-value = $0.0773$.
The p-value is $0.0773$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.
There is no sufficient evidence to conclude that school B has average graduating salary greater than that of school A. | HuggingFaceTB/finemath | |
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# Sony Placement Papers PDF Download 2017-2018 | Aptitude, Reasoning & Verbal Ability Model Papers
Sony Placement Papers PDF Download 2017-2018 | Aptitude, Reasoning & Verbal Ability Model Papers. Aspirants searching for Sony Placement Papers are at right place. The written exam conducted by Sony for selection is moderate to difficult. The exam has around 50 questions from Aptitude, Reasoning & Verbal Ability with a time allotment of around 60 minutes in total. Here we have provided the Sony Placement Papers for our readers to download and some sample questions gathered from various previous placement papers of Sony. Read the complete article and get an idea about the Sony Test Pattern.
## Latest Sony Placement Papers
The sample questions discussed below have been structured carefully so that it is comparable to the actual placement test of the Sony. The type of questions featured in them & their level of difficulty can be comparable to the actual selection tests aspirants will face in the selection process of Sony. All India Jobs also provides candidates with free Placement Papers PDF of Sony company to download.
You Can Also Check: Interview Questions
Candidates can download the Sony Placement Papers PDF from the link provided below and practice the Sony Placement Test question even in offline. By practicing regularly aspirants can easily secure good scores in Sony Recruitment Drive. We have gathered some sample questions from various Sony Placement Papers and discussed them below with answers and solutions. Practice them well and all the best for Sony Recruitment Test.
### Sony Aptitude Test Placement Papers
In this category, we have provided few questions from Sony Aptitude Test Placement Papers on topics like algebra, time & work, time, speed & distance, arithmetic, percentages, profit & loss, geometry etc.
1. Find out missing number 20, 19, 17, (.....), 10, 5
A.12
B.13
C.14
D.15
Solution:
The pattern is -1, -2, ...... Therefore, Missing number = 17 – 3 = 14.
2. 19, 2, 38, 3, 114, 4, (.....)
A.228
B.256
C.352
D.456
Solution:
The sequence is a combination of two series: I. 19 38, 114, (....) and II. 2, 3,4 The pattern followed in I is x 2, x 3, ..... Missing number = 114 x 4 = 456
3. 3, 6, 18, 72, (......)
A.144
B.216
C.288
D.360
Solution:
The pattern is x 2, x 3, x 4, ...... Missing number = 72 x 5 = 360
4. Fill in the blanks with suitable words Rajeev failed in the examination because none of his answers were ......... to the questions asked
A.allusive
B.revealing
C.pertinent
D.referential
5. There are ......... views on the issue of giving bonus to the employees
A.independent
B.divergent
C.modest
6. Man who has committed such an ........ crime must get the most severe punishment
A.injurious
B.uncharitable
C.unworthy
7. He has ....... people visiting him at his house because he fears it will cause discomfort to neighbours
A.curtailed
B.requested
C.stopped
D.warned
8. The following questions consist of three words two of which have a certain relationship to each other, followed by four words. Select the word that has the same relationship as the original pair of words. Errata : Books :: Flaws : ?
A.Manuscripts
B.Metals
C.Speech
D.Charter
Solution:
Errata comprises from the books. Similarly, Flaws are the defects in the metals.
9. Grain: Stock:: Stick : ?
A.Heap
B.Bundle
C.Collection
D.String
Solution:
Second is collection of the first.
10. Appraiser : Building :: Critic : ?
A.Book
B.Masterpiece
C.Judge
D.Gold
Solution:
11. Read the following information and answer the questions (11-15) given below it. There are five friends Sachin, Kunal, Mohit, Anuj and Rohan. Sachin ia shorter than Kunal but taller than Rohan. Mohit is tallest. Anuj is a little shorter than Kunal and little taller than Sachin. Who is the shortest?
A.Rohan
B.Sachin
C.Anuj
D.Kunal
Solution:
Let us denote the five boys by the first letter of their names,namely S, K, M, A and R. Then , R < S < K < M and S < A < K. R < S < A < K < M Rohan is shortest.
12. If they stand in the order of their heights, who will be in the middle?
A.Kunal
B.Rohan
C.Sachin
D.Anuj
Solution:
R < S < A < K < M Anuj is in the middle.
13. If they stand in the order of increasing heights, who will be the second?
A.Anuj
B.Sachin
C.Rohan
D.Kunal
Solution:
R < S < A < K < M. In the order of increasing heights i,e shortest to tallest, Sachin is second.
14. Who is taller than Anuj but shorter than Mohit?
A.Kunal
B.Rohan
C.Sachin
Solution:
R < S < A < K < M. Kunal is taller than Anuj but shorter than Mohit.
15. Who is the second tallest?
A.Sachin
B.Kunal
C.Anuj
D.Rohan
Solution:
R < S < A < K < M Kunal is second tallest.
16. Read the following information and answer the questions (16-20) given below it: (1)Seven students P, Q, R, S, T, U and v take a series of tests. (2)No two students get similar marks. (3)V always scores more than P. (4)P always scores more than Q. (5)Each time either R scores the highest and t gets the least or alternatively S scores the highest and U or Q scores the least. If S is ranked sixth and Q is ranked fifth, which of the following can be true?
A.V is ranked first or fourth
B.R is ranked second or third
C.P is ranked second or fifth
D.U is ranked third or fourth
Solution:
In terms of scores we have : V > P, P > Q i,e V > P > Q. If R scores the highest, we have R > ---------- > T. If S scores the highest, we have S > ----------- > Q or S > ---------- > U. If S is ranked sixth and Q is ranked fifth, we have _ > _ > _ > _ > Q > S > _ In this case, R will rank the highest and thus T will rank the least. we have R > _ > _ > _ > Q > S > T Also, the order V > P > Q will be maintained i,e V and P will have second,third or fourth places.
17. If R gets more, V should be ranked not lower than:
A.second
B.third
C.fourth
D.fifth
Solution:
R > _ > _ > _ > Q > S > T. Again, if R ranks most, T ranks Lowest and occupies seventh place. Since V always ranks above P and Q so in the maximum, P and Q will occupy fifth and sixth places. Thus, V will not rank lower than fourth.
18. If R gets more, V should be ranked not lower than:
A.second
B.third
C.fourth
D.fifth
Solution:
R > _ > _ > _ > Q > S > T. Again, if R ranks most, T ranks Lowest and occupies seventh place. Since V always ranks above P and Q so in the maximum, P and Q will occupy fifth and sixth places. Thus, V will not rank lower than fourth.
19. If S is ranked second, which of the following can be true?
A.U gets more than V
B.V gets more than S
C.P gets more than R
D.P gets more than V
Solution:
If S ranks second, R ranks first and T ranks lowest.The order V > P > Q will be followed. So, the arrangement will be R > S > _ > _ > _ > _ > T.
20. If V is ranked fifth, which of the following must be true?
A.S scores the highest
B.R is ranked second
C.T is ranked third
D.Q is ranked fourth
Solution:
If V ranks fifth, P and Q coming before it will occupy sixth and seventh places respectively i,e Q ranks least. So, S will score the highest.
21. Showing the man receiving the prize, Saroj said, “He is the brother of my uncle’s daughter.” Who is the man to Saroj ?
A.Son
B.Brother-in-law
C.Nephew
D.Cousin
Solution:
Brother of uncle’s daughter—Uncle’s son – Cousin. So, the man is Seema’s cousin.
22. Pointing to a man, a woman said, “His mother is the only daughter of my mother.” How is the woman related to the man?
A.Mother
B.Daughter
C.Sister
D.Grandmother
Solution:
Only daughter of my mother – Myself. So, the woman is man’s mother.
23. Pointing to a photograph, a person tells his friend, “She is the granddaughter of the elder brother of my father.” How is the girl in the photograph related to his man?
A.Niece
B.Sister
C.Aunt
D.Sister-in-law
Solution:
Brother of father – Uncle; Uncle’s grand daughter – daughter of uncle’s son – daughter of cousin -- niece
24. If in a certain language FASHION is coded as FOIHSAN, how is PROBLEM coded in that code?
A.ROBLEMP
B.PELBORM
C.PRBOELM
D.RPBOELM
Solution:
The first and last letters of the word remain as such and the remaining letters are written in a reverse order to obtain the code.
25. If in a certain language KINDLE is coded as ELDNIK, how is EXOTIC coded in that code?
A.EXOTLC
B.CXOTIE
C.COXITE
D.CITOXE
Solution:
The letters of the word are written in a reverse order to obtain the code.
26. If VICTORY is coded as YLFWRUB, how can SUCCESS be coded?
A.VXEEIVV
B.VXFFHVV
C.VYEEHVV
D.VYEFIVV
Solution:
Each letter of the word is moved three steps forward to obtain the code.
27. (Directions: 27-30). Each question given has a problem and two statements numbered 1 and 2 giving certain information. You have to decide if the information given in the statements is sufficient for answering the problem. Indicate your answer as A. statement 1 alone is sufficient, but statement 2 alone is not sufficient to answer the question B. statement 2 alone is sufficient, but statement 1 alone is not sufficient to answer the question C. both statements taken together are sufficient to answer the question, but neither statement alone is sufficient D. each statement alone is sufficient E. statements 1 and 2 together are not sufficient, and additional data is needed to answer the question How many ewes (female sheep) in a flock of 50 sheep are black? 1. There are 10 rams (male sheep) in the flock. 2. Forty percent of the animals are black.
A.A
B.C
C.D
D.E
Solution:
From (1) we know the ratio of male to female sheep, but nothing about the color distribution. So the answer cannot be A or D. From (2) we know that forty percent of the animals are black but nothing about whether they are male of female. So the answer cannot be B. Even putting the information together does not help because there is no way to tell what fraction of the female sheep re black.
28. Is the length of a side of equilateral triangle E less than the length of a side of square F? 1. The perimeter of E and the perimeter of F are equal. 2.The ratio of the height of triangle E to the diagonal of square F is 2v3 : 3v2.
A.A
B.B
C.C
D.D
Solution:
From (1) we can tell that a side of E is longer than a side of F, since 3 x side E = 4 x side F. Hence (1) is sufficient to answer the question and the answer must be either A or D. From (2) we could work out the ratio of the lengths of the sides (there is no need to do this, but since we are dealing with regular plane figures the geometry is quite simple), and although we cannot get the actual lengths, we can see from the ratio whether one is bigger than the other.
29. If a and b are both positive, what percent of b is a? 1. a = 3/11 2. b/a = 20
A.A
B.B
C.C
D.D
Solution:
Statement (1) tells us nothing about b and so the answer cannot be A or D. To find what percent a is of b we need to solve the expression (a/b) x 100. Statement (2) allows us to do just that: (a/b) = 1/20.
30. A wheel of radius 2 meters is turning at a constant speed. How many revolutions does it make in time T? 1. T = 20 minutes. 2. The speed at which a point on the circumference of the wheel is moving is 3 meters per minute.
A.A
B.B
C.C
D.D
Solution:
To find the number of revolutions we need to know the rate of turning and the time duration. Statement (1) gives us only the time, and so the answer cannot be A or D. Statement (2) tells us the rate at which a point on the circumference is moving, which, since we know the dimensions of the wheel, is sufficient to determine the number of rotations per minute. But since we do not know the time, B cannot be correct. But putting (1) and (2) together we have all we need, so the answer is C.
### Sony Reasoning Test Sample Papers
Reasoning Questions that show a similar pattern with the ones on Sony Reasoning Test can be attempted here. We provide a large question bank of all the Sony Reasoning Test Sample Papers.
1. In a certain code language, '+' means '/', '/' means '*', '*' means '-', '-' means '+'.
5 * 10 - 15 / 20 + 25 / 30
A. 210
B. 240
C. 355
D. 245
E. 360
Solution:
'+' means '/', '/' means '*', '*' means '-', '-' means '+'.
The given expression is
5 * 10 - 15 / 20 + 25 / 30
According to the given direction, the above expression becomes
5 - 10 + 15 * 20 / 25 * 30
Using BODMAS rule,
5 - 10 + 15 * 20 / 25 * 30
= -5 + 300 / 25 * 30
= -5 + 12 * 30
= 355.
2. 24 : 576 :: 32 : __
A. 1024
B. 992
C. 1228
D. 865
E. 1728
Solution:
24 : 576 :: 32 : __
24 : (24)2 :: 32 : (32)2
(32)2 = 1024 is the next number.
3. (A) The problem of child labour has reached epidemic proportions in India with the number of child labourers increasing from 12 million to over 40 million child workers.
(B) India has the worlds largest number of child labourers.
A. if statement (A) is the cause and statement (B) is its effect.
B. if statement (B) is the cause and statement (A) is its effect.
C. if both the statements (A) and (B) are independent causes.
D. if both the statements (A) and (B) are effects of independent causes.
E. if both the statements are effect of some common causes.
Solution:
The increase in child labour lead to India becoming the country with largest number of child labourers.
4. Study the following information carefully and answer the given questions:
A word and number arrangement machine when given an input line of words and numbers rearranges them following a particular rule in each step. The following is an illustration of input and rearrangement:
Input: 39 121 48 18 76 112 14 45 63 96
Step I: 14 39 121 48 18 76 112 45 63 96
Step II: 14 39 48 18 76 112 45 63 96 121
Step III: 14 18 39 48 76 112 45 63 96 121
Step IV: 14 18 39 48 76 45 63 96 112 121
Step V: 14 18 39 45 48 76 63 96 112 121
Step VI: 14 18 39 45 48 63 76 96 112 121
This is the final arrangement and step VI is the last step for this input.
Which of the following will be the third step for the following input?
Input: 45 78 97 132 28 16 146 54 99 112
A. 16 28 45 78 97 146 54 99 112 132
B. 16 28 45 97 78 54 99 112 132 146
C. 16 28 45 78 97 132 54 99 112 146
D. 16 28 45 97 78 132 99 54 112 146
E. None of these
Solution:
Step I: The smallest number becomes first and the remaining numbers shift one position rightward.
Step II: The largest number among given numbers becomes last and the remaining numbers shift one position leftward.
The steps are repeated alternately till all the numbers get arranged in ascending order and that will be that last step for that particular input.
Input: 45 78 97 132 28 16 146 54 99 112
Step I: 16 45 78 97 132 28 146 54 99 112
Step II: 16 45 78 97 132 28 54 99 112 146
Step III: 16 28 45 78 97 132 54 99 112 146
5. Should children be allowed to use internet in internet kiosks.
Arguments:
I. Yes, all the internet kiosks are using filter-software's, which do not allow adult software to be downloaded.
II. No, internet is a communication medium, which transmits information and does not distinguish between a child and a grown-up.
A. if only argument I is strong.
B. if only argument II is strong.
C. if either I or II is strong.
D. if neither I nor II is strong.
E. if both I and II are strong.
Solution:
One important reason to stop children from using kiosks is audit material available on the net. Hence, statement I is a strong argument. Statement II gives an irrelevant reason as its argument and is very weak.
6. Arrange the given words in alphabetical order and tick the one that comes in the middle ?
A. Verrigate
B. Vibrate
C. Vindictive
D. Trench
E. Wavering
Solution:
Trench, Verrigate, Vibrate, Vindictive, Wavering
Which of the statements A, B, C, D, and E mentioned above would represent an assumption implicit in the given paragraph or an inference which can be drawn from it?
A. Only C
B. Only E
C. Only B and D
D. Only A
E. Only D
Solution:
None of the given statements is an assumption implied in the given information. It is stated that a corridor of 2 km on either side of LOC has been declared as no military zone. Though the information provides
does not indicate the exact distance from LOC at which these roads are being laid, but from the fact that objections of the Indian government it can be understood that it is in violation of the bilateral agreement.
Hence, (E) can be inferred.
8. If it is possible to make a meaningful word using the sixth, seventh, ninth and tenth letters only once of the word "FUNDAMENTAL", what will be the first letter of the word? If no word can be formed, mark the answer as X, If two words can be formed, mark the answer as Y. If more than two words can be formed, mark the answer as Z.
A. T
B. M
C. X
D. Y
E. Z
Solution:
The sixth, seventh, ninth and tenth letters of the word FUNDAMENTAL are M, E, T, A
The words that can be formed from the letters are TEAM, TAME, MATE and MEAT.
9. If it is possible to make a meaningful word from the third, sixth, eight and eleventh letters of the word 'DISTINGUISH' using each letter only once, first letter of the word would be your answer. If more than one such word can be formed, your answer would be 'M' and if no such word can be formed, answer is 'X'.
A. N
B. S
C. H
D. M
E. X
Solution:
The 3rd, 6th, 8th and 11th letters are S, N, U and H respectively. The word that can be made is SHUN.
10. ASTN: ZTSO:: MSUB:?
A. CRRC
B. LTTA
C. NTVC
D. LTTC
Solution:
The 1st and 3rd letter of the group are each moved to 1 step backward while the 2nd and 4th letters are each moved one step forward to the corresponding letters the second group.
11. Choose the correct alternative that will continue the same pattern and fill in the blank spaces:
3, 6, 18, 72, (....)
A. 144
B. 216
C. 288
D. 360
Solution:
The pattern is *2, *3, *4, ...
Missing number = 72 * 5 = 360.
12. Obesity is a serious problem in this country. Research suggests that obesity can lead to a number of health problems including diabetes, asthma, and heart disease. Recent research has even indicated that there may be a relationship between obesity and some types of cancer. Major public health campaigns that increase awareness and propose simple lifestyle changes that will, with diligence and desire, eliminate or least mitigate the incidence of obesity are a crucial first step in battling this critical problem. This paragraph best supports the statement that
A. public health campaigns that raise consciousness and propose lifestyle changes are a productive way to fight obesity.
B. obesity is the leading cause of diabetes in our country.
C. people in our country watch too much television and do not exercise enough.
D. a decline in obesity would radically decrease the incidence of asthma.
E. fast-food restaurants and unhealthy school lunches contribute greatly to obesity.
Solution:
The support for this choice is in the last sentence, which states that major public health campaigns that increase awareness and propose lifestyle changes are important in our fight against obesity. Choice b can be ruled out because although the paragraph states that obesity can lead to diabetes, it doesn’t tell us that it is the leading cause of this disease. Choices c and e might sound reasonable and true, but they are not supported in the paragraph. And although we are told that obesity has been connected to asthma, this fact is not quantified in any way, so choice d is also not supported by the information given.
13. Statement:
Should India support all the international policies of United States of America?
Arguments:
I. No, many other powerful countries do not support the same.
II. Yes, this is the only way to gain access to USA developmental funds.
A. If only argument I is strong
B. If only argument II is strong
C. If either I or II is strong
D. If neither I nor II is strong
E. If both I and II are strong
Solution:
Argument I is not strong because India should take its stand according to its need. Argument II is not correct because of the word 'only'. Hence, argument II is not strong.
14. Here are some words translated from an artificial language.
slar means jump
slary means jumping
slarend means jumped
Which word could mean “playing”?
A. clargslarend
B. clargy
C. ellaclarg
D. slarmont
Solution:
According to this language, slar means jump. The suffix –ing is represented by –y. Since choice b is the only one that ends in the letter y, this is the only possible option.
15. Choose out the odd one.
A. Trunk
B. Tree
C. Fruit
D. Leaf
Solution:
All other parts are of the tree.
### Sony English Test Model Papers
In this section, we have discussed some sample questions from Sony English Test Model Papers. Contenders can practice these questions and get an idea about the questions asked in Sony English Placement Test.
1. The robbers ————————- by the police.
A. Have arrested
B. Have been arrested
C. Was arrested
Solution:
Have been arrested
2. One who cuts precious stones.
A. drover
B. oculist
C. philatelist
D. lapidist
3. VIVACIOUS
A. languid
B. open
C. strong
D. bright
E. lively
Solution:
The word vivacious (animated, cheerful) and languid (sluggish) are antonyms.
4. Patrice was a(n) ______ girl when she was teenager—long-limbed and constantly tripping over her own feet.
A. blithe
B. resolute
C. ungainly
D. preternatural
E. conducive
Solution:
Ungainly (adj.) means clumsy or awkward.
5. He holds a very important position. He is not well paid.
(A) Although he holds a very important position .....
(B) Unless he holds a very important .....
(C) Inspite of the fact that he holds a .....
A. A and B
B. B and C
C. A and C
D. All the three
E. None of these
Solution:
He deserves to be well-paid, though he is not. Hence, conjunctions like but, though, although, in spite of the fact that, etc. can be used to effectively put forth the idea while combining the two sentences.
6. I do not agree --------- you on this point.
A. to
B. in
C. at
D. with
7. Select the odd one out
A. Lungs
B. Cancer
C. Pancreas
D. Heart
8. AFFIDAVIT
A. document
B. protocol
C. affirmation
D. will
E. certificate
Solution:
Affidavit refers to a written statement for use as evidence in court, sworn on oath to be true. The word affirmation also means the same.
9. She went ahead with dogged determination, _____by the adversities which came her way and ultimately emerged successful.
(A) over-awed
(B) unintimidated
(C) dismayed
(D) undaunted
(E) unalarmed
(F) incapacitated
A. A and C
B. B and C
C. E and F
D. B and C
E. B and E
Solution:
She emerged successful in spite of the 'adversities' because she was conrageous or 'undaunted' or 'unintimidated'.
10. Child bereaved of one or both the parents
A. Desolate
B. Destitute
C. Orphan
D. Lout
11. ----- unit means a measurement
A. a
B. an
C. the
D. no article
12. You must have a through understanding of the subject. Only then you will be able to answer any question pertaining to it.
(A) Although you have .....
(B) Despite having .....
(C) Unless you have .....
A. Only A
B. Only B
C. Only C
D. B and C
E. None of these.
Solution:
Only 'C' which presents a condition is correct. A and B denote a contradiction.
13. Choose the correct spelt word out of the given alternatives.
A. Fabullous
B. Fablous
C. Fabuleous
D. Fabulous
14. On my request (a) / Shalu introduced me (b) / to his friend (c) / who it singer and a scientist (d) / No error (e)
A. on my request
B. shalu introduced me
C. to his friend
D. who it singer and a scientist
E. no error
15. You would not drive so fast; we have plenty of time.
A. will not
C. need not
D. do not
E. No correction required
Solution:
The sentence suggests that driving fast is not necessary. Hence the most appropriate modal verb to suit this context would be 'need not'. 'Will not' suggests simple future which is inapt. 'Had not' and 'do not' are illogical.
### Sony Placement Paper Syllabus:
#### Quantitative Aptitude Syllabus:
• Probability
• Permutations & Combinations
• Algebra
• Averages
• Time Speed & Distance
• Time & Work
• Profit & Loss
• Ratio & Proportion
• Simple & Compound Interest
• Percentage
• Number Series
• Mixtures & Alligations
• Simplification
• Number System
• Heights and Distances
• Geometry & Mensuration
• Data Sufficiency
• Logarithms
• Progressions
• LCM and HCL
• Pipes and Cisterns
• Partnership
• Boats and Streams
• Areas, Volumes
#### Reasoning Syllabus:
• Number Series
• Letter Series
• Analogies
• Puzzles
• Syllogisms
• Binary Logic
• Clocks & Calendars
• Cubes & Dice
• Classification
• Blood Relations
• Coding-Decoding
• Data Sufficiency
• Seating Arrangement
• Venn Diagrams
• Problem Solving
• Coded Inequalities
• Double Lineup
• Logical Deductions
• Routes & Networks
• Grouping & Selections
• Evaluating Course of Action
• Statements and Conclusions
• Mathematical and Computer Operations
• Critical Reasoning
• Inferences
• Situation Reaction Test
• Decision Making
• Symbols and Notations
• Direction Sense Test
• Logical Sequence Of Words
• Assertion and Reason
• Verification of Truth of the Statement
• Statements and Assumptions
• Data Interpretation
#### Verbal Ability Syllabus:
• Synonyms
• Antonyms
• Sentence Completion
• Spelling Test
• Passage Completion
• Sentence Arrangement
• Idioms and Phrases
• Para Completion
• Error Correction (Underlined Part)
• Fill in the blanks
• Synonyms
• Prepositions
• Active and Passive Voice
• Spotting Errors
• Substitution
• Transformation
• Sentence Improvement
• Joining Sentences
• Error Correction (Phrase in Bold)
• Articles
• Gerunds
• Identify the Errors
• Plural Forms
• Odd Words
• Prepositions
• Suffix
• Tense
• Homophones
• Identify the Sentences
• Nouns
• Prefix
• Sentence Pattern
• Tag Questions
Hope the syllabus and Sony placement papers provided here are helpful for aspirants while preparing for Sony Placement Test. Sometimes Sony company may change their syllabus and exam pattern. So Please check this article periodically for the latest syllabus.
## Top MNC's Registration Links 2017
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# Ounces and Pounds and Tons... Oh My!
Contributor: Rachel Lewis. Lesson ID: 12220
It doesn't take a ton of knowledge to weigh your options when converting weight measurements! No need to pound it into your head; just an ounce of time is needed to learn the simple way to convert!
categories
## Elementary
subject
Math
learning style
Visual
personality style
Lion, Beaver
Intermediate (3-5)
Lesson Type
Skill Sharpener
## Lesson Plan - Get It!
• Which is heavier: a slice of bread, a football, or a blue whale?
• Which tastes better?
Well, forget that part . . .
• How much do you weigh?
Your scale probably measures your weight in pounds. In the United States, we use three different units to measure weight. We use different units to measure objects with different sizes. Weight is measured using ounces, pounds, and tons.
• An ounce (oz.) is the smallest unit of mass. One ounce weighs about the same as one slice of bread.
• A pound (lb.) is the same as 16 ounces. We use pounds to measure our own weight. A football or a baseball weighs about one pound.
• We can use tons to weigh very heavy objects. In some countries, one ton equals 2,240 pounds, but in the United States, one ton is the same as 2,000 pounds. A small car weighs around one ton. A blue whale can weigh up to 150 tons!
Compare ounces, pounds, and tons, when you read US Standard Mass (Weight) by MathIsFun.com.
On a sheet of paper, create your own weight conversion chart.
• How many ounces are in one pound?
• How many pounds are in one ton? Include the following conversions:
• 1 ounce = 16 pounds
• 2,000 pounds = 1 ton
We can use conversions to change one unit to another.
If one pound equals 16 ounces, then how many ounces are in 1 pound 4 ounces?
• 16 ounces = 1 pound
• 16 ounces + 4 ounces = 20 ounces
• There are 20 ounces in 1 pound 4 ounces.
How can we convert 5 pounds into ounces?
We know there are 16 ounces in one pound, so we can multiply 5 pounds by 16:
• 5 pounds x 16 ounces = 80 ounces
• There are 80 ounces in 5 pounds.
• How do you think we would convert pounds to tons?
Take a moment to try an example:
• How do you express 6,000 pounds in tons?
If you said 3 tons, you’re right!
You can use repeated subtraction:
• There are 2,000 pounds in 1 ton:
• 6,000 - 2,000 = 4,000
• 4,000 - 2,000 = 2,000
• You can count up by 2,000:
• 2,000 + 2,000 = 4,000
• 4,000 + 2,000 = 6,000
• You can divide 6,000 by 2,000:
• 6,000 ÷ 2,000 = 3 tons
• Which method did you use to solve the problem?
• If you made a mistake, do you see where you went wrong?
• When would you use ounces in real life?
• When would you use pounds?
• Can you think of a time when you needed to weigh something on a scale? Did you use ounces or pounds?
Discuss your ideas with a parent or teacher.
Now, go to the Got It? section and use your conversion chart to complete a measurement activity.
## Elephango's Philosophy
We help prepare learners for a future that cannot yet be defined. They must be ready for change, willing to learn and able to think critically. Elephango is designed to create lifelong learners who are ready for that rapidly changing future. | HuggingFaceTB/finemath | |
# Cofactor matrix 4x4, evaluated by hand
I am currently studying linear algebra. So far I have studied method for aquiring the inverse of a matrix A. Now I would like to evaluate the inverse of a $4\times 4$ matrix using the following formula:
$$\mathbf{A}^{-1} = \frac{ \mathrm{adj}(\mathbf{A}) }{ \det(\mathbf{A}) } % A^(-1)=1/det(A) adj(A)$$
My question is in order to get the adjoint matrix I will need the cofactor matrix. In a $4\times 4$ matrix this means I could do it with row expansion in 2 dimensions. For instance, I take the entry A[a11], I cross out first row and first column. Next I choose B[a11] in the submatrix to get my minor. Now, do I have to multiply the minor of the submatrix with the original entry A[a11]? Does it matter what minor I choose in the submatrix in further calculation? For example, using A[a12] to get the minor of the main matrix, can I use any row or column in the submatrix?
I appreciate any help in this manner. Thank you.
-Daniel
-
That was quite incomprehensible... But you know how to compute a 3 by 3 determinant, right? That's what you have to do (16 times!), since each cofactor is a 3 by 3 determinant (namely the determinant that you get by crossing out one row and one column of $A$). – Hans Lundmark Oct 11 '12 at 7:48
If I understand your question correctly, when calculating the adjoint matrix, you do not need to multiply the determinant of the submatrix by the entry you're finding the cofactor for.
One reason this can be confusing is that you do multiply the submatrix determinants by the entries to find the determinant of the matrix.
- | HuggingFaceTB/finemath | |
# Order of Operations
What if you were asked to calculate the following:
${4}\hspace{0.33em}{+}\hspace{0.33em}{2}^{2}\times\hspace{0.33em}{7}\hspace{0.33em}{-}\hspace{0.33em}\left({{1}\hspace{0.33em}{+}\hspace{0.33em}{4}^{2}}\right)\hspace{0.33em}\div\hspace{0.33em}{2}$
Well person A could go left to right and do the 4 plus 2 squared to get 8 then multiply by 7 to get 56. then you might think that you should do the operations in the brackets first to get 17 and subtract that from 56 to get 39 then finally divide by 2 to get 19.5.
Person B may first do the operations in the brackets to get 17, then square the 2 to get 4 so that the problem now looked like:
${4}\hspace{0.33em}{+}\hspace{0.33em}{4}\times\hspace{0.33em}{7}\hspace{0.33em}{-}\hspace{0.33em}{17}\hspace{0.33em}\div\hspace{0.33em}{2}$
Then multiply the 4 and 7 to get 28 and divide the 17 by 2 to get 8.5. The problem would now look like:
${4}\hspace{0.33em}{+}\hspace{0.33em}{28}\hspace{0.33em}{-}\hspace{0.33em}{8}{.}{5}$
Then go left to right: add the 4 and 28 to get 32 then subtract the 8.5 to get 23.5. Who is right? Well mathematicians saw that different people would get different answers if there were no rules to dictate how this should be done. So they came up with a rule called order of operations.
This rule is easy to remember with an acronym. Depending on which dialect of maths you use (Australian or American), this acronym is either BODMAS (Australian) or PEDMAS (American). These stand for:
B: Brackets P: Parentheses
O: Order (Powers) E: Exponents (Powers)
D: Division D: Division
M: Multiplication M: Multiplication
S: Subtraction S: Subtraction
The main difference being that Americans call brackets, parentheses and Australians call exponents, order. The order of a number is just the exponent or the power of a number.
In actual fact, multiplication and division are equal in order and addition and subtraction are equal. So the acronym could be BOMDAS or BOMDSA but BODMAS is the accepted one. For equal operations, you would proceed left to right. Also, sometimes you get brackets within brackets. You would proceed with the inner most bracket first, and work outwards. By the way, for nested brackets, the convention used so that it is easier to distinguish which end bracket applies to which beginning bracket is {[( )]}. For my Australian friends these are called brackets “()”, square brackets “[]” and curly brackets “{}”. For my American friends, these are called parentheses, brackets, and braces.
$\left\{{\left[{\left({{4}\hspace{0.33em}{+}\hspace{0.33em}{28}\hspace{0.33em}{-}{9}}\right)\hspace{0.33em}{+}\hspace{0.33em}{7}}\right]\hspace{0.33em}{-}\hspace{0.33em}{8}}\right\}\hspace{0.33em}{=}\hspace{0.33em}\left\{{\left[{{23}\hspace{0.33em}{+}\hspace{0.33em}{7}}\right]\hspace{0.33em}{-}\hspace{0.33em}{8}}\right\}\hspace{0.33em}{=}\hspace{0.33em}\left\{{{30}\hspace{0.33em}{-}\hspace{0.33em}{8}}\right\}\hspace{0.33em}{=}\hspace{0.33em}{22}$
So it turns out, person B did the calculation correctly.
Please try this one and see if you get the answer 28:
$\left[{{\left({{3}\hspace{0.33em}{+}\hspace{0.33em}{4}}\right)}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}\times\hspace{0.33em}{3}}\right]$
Posted on Categories Algebra, Pre-VCE | HuggingFaceTB/finemath | |
## Saturday, October 29, 2011
### A Long Way to Go . . .
. . . for a very small joke.
TMAN: Dude, tell me that thing again, with the numbers and the countries.
SWELLSMAN: . . . . Okay, let’s try this from the top. First, pick a number between 1 and 10.
TMAN: Got it.
SWELLSMAN: Tell me what it is.
TMAN: How is that a trick?
SWELLSMAN: It isn’t. But we fucked the trick up the last time we tried so let’s just do a dry run. What’s the number?
TMAN: It’s two.
SWELLSMAN: Swell, it’s two. So multiply that by 9, and waddaya got?
TMAN: Eighteen.
SWELLSMAN: Okay. Now, take all the digits in that last answer and add them together . . . waddaya got now?
TMAN: Nine. One and Eight equals Nine.
SWELLSMAN: Perfect. This is the heart of the trick. Now subtract 5 from your answer. Waddaya got?
TMAN: Four.
SWELLSMAN: Great. Now count down the alphabet until you get to the letter that is that number. So “A” would be “one,” “B” would be “two,” etc.
Got it?
TMAN: Yeah.
SWELLSMAN: Okay. Now think of a country whose name begins with the same letter as the one you just counted down to.
Got it?
TMAN: Yeah.
SWELLSMAN: Good. Now think of an animal whose name begins with the last letter of the name of the country you just thought of.
Got it?
TMAN: Yeah.
SWELLSMAN: Good. Now think of a fruit that begins with the last letter of the name of the animal you just thought of.
Got it?
TMAN: Yeah.
SWELLSMAN: Good. It was “Denmark,” “kangaroo,” and “orange,” wasn’t it?
TMAN: Whoa. Not bad. How’d you do it?
SWELLSMAN: It’s not hard. Most of what you are doing when you do this trick is convincing people they have a lot more control over their answers than they really do.
For example, any number between 1 and 10 – when multiplied by 9 – yields a number whose digits add up to 9. 18, 27, 36, 45, etc. etc. So no matter what you ask the dupe to do, he’ll always end up with 9. Which means that when you tell him to subtract 5 from “whatever number he’s thinking of,” he’ll always end up with 4.
The fourth letter of the alphabet is “D,” and there are only a few countries that begin with ‘D” – Denmark is the one almost everyone goes with. Which means “K” is the first letter of the animal, and so almost everyone goes with Kangaroo.
And then “O” is the first letter of the fruit, and so that has gotta be Orange.
The key is to convince the dupe that he or she has any choice in the matter, despite the fact you are funneling him to a predetermined destination.
It is, in its way, a lot like voting. | HuggingFaceTB/finemath | |
# REVIEW. Write down this problem on your COMMUNICATOR Be prepared to share your work with the class. Prepare for my Unit 5 Test By completing my practice.
## Presentation on theme: "REVIEW. Write down this problem on your COMMUNICATOR Be prepared to share your work with the class. Prepare for my Unit 5 Test By completing my practice."— Presentation transcript:
REVIEW
Write down this problem on your COMMUNICATOR Be prepared to share your work with the class. Prepare for my Unit 5 Test By completing my practice test and making a 3x5 card of notes. Write down 5 Rational Numbers and 5 Irrational Numbers Defend your choices to your partner (justify your answers).
Try this CHALLENGE question: 0 It is the only Rational Number
Chapter 5’s Write them down on your 3x5 card. Notecards and Calculators can only be used on part of the test this time.
Benchmark Fractions MEMORIZE!! 0.50.25 0.75 0.125 0.375 0.625 0.875
Comparing Numbers When comparing numbers, they need to be in the same format (in this case: both decimal) Compare, using the symbol. >
Fractions Decimals 6.2
TERMINATING DECIMALS In your groups, discuss a way to sort these decimals into two groups. 0.888… 0.264 -6.15 0.758574 0.5 4.123123… -1.222… Types of Decimals REPEATING DECIMALS
0.3 Decimal Place Review What decimal place is this? tenth 5.347 thousandth 0.34 hundredth 0.347 thousandth
Do you see a pattern? Last time we converted repeating decimals to fractions, such as these examples… 0.555… Assign: x = 0.555… 0.444… Assign: x = 0.444… 0.151515… Assign: x = 0.151515… This Shortcut only works when the repeating numbers are right after the decimal. 0.1333… Assign: x = 0.1333… So this type of equation does not follow the pattern. Do you see a pattern?
RATIONAL NUMBERS Decimals that terminate or repeat. Examples: 0.356 and 0.555… IRRATIONAL NUMBERS Real numbers that are NOT RATIONAL The CRAZY Ones!
Perfect Squares MEMORIZE THEM 1 4 9 16 25 36 49 64 81 100 121 144
Square Roots UNDO Squares 2 INVERSE OPERATIONS
Square Roots of Perfect Squares 2 3 4 5 6 7 8 10 9
Square Roots UNDO Squares INVERSE OPERATIONS We can use this inverse operation to solve equations with exponents.
Find the Side, given the Area 6 ft.
Between which two Integers would you find… Square Roots jkkkk
Perfect Cubes MEMORIZE THEM 1 64 8 125 27
Cube Roots UNDO Cubes 3 3 INVERSE OPERATIONS
Cube Roots of Perfect Cubes 2 3 4 5
Cube Roots UNDO Squares INVERSE OPERATIONS We can use this inverse operation to solve equations with exponents.
Find the edge, given the Volume 2 m.
Cube Roots Foldable: Highlight the Perfect Cube Between which two Integers would you find…
CALCULATOR REVIEW SQUARE ROOTS CUBE ROOTS / ANY ROOTS # = 3
1, 2, 3, 4…0 … -4, -3, -2 -1 Decimals that repeat and terminate Decimals that do not terminate or repeat 0.444… 0.125 0.6543986204… X
No Calculator or Notecard on First 2 Pages Do the CIRCLED PROBLEMS on the PRACTICE TEST. When you finish, you can work on the rest of the problems. PRACTICE TEST
CIRCLED ANSWERS 3b. 7 in. 4b. 2 in. 6b. 321/100 6e. 215/99 8a & b
What’s your Score? Each answer part is worth 1 point 73 possible Now, finish the rest of the Practice Test!!!
SHOW YOUR WORK Unit 5 Practice TEST
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# Harmonic Motion and Waves
## Presentation on theme: "Harmonic Motion and Waves"— Presentation transcript:
Harmonic Motion and Waves
Simple Harmonic Motion(SHM)
Vibration (oscillation) Equilibrium position – position of the natural length of a spring Amplitude – maximum displacement
Period and Frequency Period (T) – Time for one complete cycle (back to starting point) Frequency (Hz) – Cycles per second F = 1 T = 1 T f
Period and Frequency A radio station has a frequency of M Hz. What is the period of the wave? 103.1 M Hz 1X106 Hz = X 108 Hz 1M Hz T = 1/f = 1/(1.031 X 108 Hz) = X 10-9 s
Hooke’s Law F = -kx F = weight of an object k = spring constant (N/m)
x = displacement when the object is placed on the spring
Hooke’s Law: Example 1 What is the spring constant if a kg mass causes the spring to stretch 6.0 cm? (ANS: 16 N/m)
Special Note: If a spring has a mass on it, and then is stretched further, equilibrium position is the starting length (with the mass on it)
Hooke’s Law: Example 2 A family of four has a combined mass of 200 kg. When they step in their 1200 kg car, the shocks compress 3.0 cm. What is the spring constant of the shocks? F = -kx k = -F/x k = -(200 kg)(9.8 m/s2)/(-0.03 m) k = 6.5 X 104 N/m (note that we did not include the mass of the car)
Hooke’s Law: Example 2a How far will the car lower if a 300 kg family borrows the car? F = -kx x = -F/k x = -(300 kg)(9.8 m/s2)/ 6.5 X 104 N/m x = 4.5 X 10-2 m = 4.5 cm
Forces on a Spring Extreme Position (Amplitude) Equilibrium position
Force at maximum Velocity = 0 Equilibrium position Force = 0 Velocity at maximum
Energy and Springs KE = ½ mv2 PE = ½ kx2 Maximum PE = ½ kA2
Law of conservation of Energy ½ kA2 = ½ mv2+ ½ kx2
All PE All KE Some KE and Some PE
Spring Energy: Example 1
A 0.50 kg mass is connected to a light spring with a spring constant of 20 N/m. Calculate the total energy if the amplitude is 3.0 cm. Maximum PE = ½ kA2 Maximum PE = ½ (20 N/m)(0.030 m2) Maximum PE = 9 X 10-3 Nm (J)
Spring Energy: Example 1a
What is the maximum speed of the mass? ½ kA2 = ½ mv2+ ½ kx2 ½ kA2 = ½ mv2 (x=0 at the origin) 9 X 10-3 J = ½ (0.50 kg)v2 v = 0.19 m/s
Spring Energy: Example 1b
What is the potential energy and kinetic energy at x = 2.0 cm? PE = ½ kx2 PE = ½ (20 N/m)(0.020 m2) = 4 X 10-3 J ½ kA2 = ½ mv2 + ½ kx2 ½ mv2 = ½ kA2 - ½ kx2 KE = 9 X 10-3 J - 4 X 10-3 J = 5 X 10-3 J
Spring Energy: Example 1c
At what position is the speed 0.10 m/s? (Ans: cm)
Spring Energy: Example 2a
A spring stretches m when a kg mass is suspended from it (diagrams a and b). Find the spring constant. (Ans: N/m)
Spring Energy: Example 2b
The spring is now stretched an additional m and allowed to oscillate (diagram c). What is the maximum velocity?
The maximum velocity occurs through the origin:
½ kA2 = ½ mv2+ ½ kx2 ½ kA2 = ½ mv2 (x=0 at the origin) kA2 = mv2 v2 = kA2/m v = \/kA2/m = \/(19.6 N/m)(0.100m)2/0.300kg v = m/s
Spring Energy: Example 2c
What is the velocity at x = m? ½ kA2 = ½ mv2+ ½ kx2 kA2 = mv2+ kx2 mv2 = kA2 - kx2 v2 = kA2 - kx2 m v2 = 19.6 N/m(0.100m2 – m2) = 0.49 m2/s2 0.300 kg v = m/s
Spring Energy: Example 2d
What is the maximum acceleration? The force is a maximum at the amplitude F = ma and F = kx ma = kx a = kx/m = (19.6 N/m)(0.100 m)/(0.300 kg) a = 6.53 m/s2
Trigonometry and SHM Ball rotates on a table
Looks like a spring from the side One rev(diameter) = 2pA T = 2p m k f = 1 T
Period depends only on mass and spring constant
Amplitude does not affect period vo = 2pAf or vo = 2pA T vo is the initial (and maximum) velocity
Period: Example 1 What is the period and frequency of a 1400 kg car whose shocks have a k of 6.5 X 104 N/m after it hits a bump? T = 2p m = 2p (1400 kg/6.5 X 104 N/m)1/2 k T = 0.92 s f = 1/T = 1/0.92 s = 1.09 Hz
Period: Example 2a An insect (m=0.30 g) is caught in a spiderweb that vibrates at 15 Hz. What is the spring constant of the web? T = 1/f = 1/15 Hz = s T = 2p m k T2 = (2p)2m k = (2p)2m = (2p)2(3.0 X 10-4 kg) = 2.7 N/m T (0.0667)2
Period: Example 2b What would be the frequency for a lighter insect, 0.10 g? Would it be higher or lower? T = 2p m k T = 2p (m/k)1/2 T = 2p (1.0 X 10-4 kg/2.7 N/m)1/2 = s f = 1/T = 1/0.038 s = 26 Hz
Cosines and Sines Imagine placing a pen on a vibrating mass
Draws a cosine wave
x = A cos2pt or x = A cos2pft T A = Amplitude t = time T = period f = frequency
x = A cos2pft v = -vosin2pft a = -aocos2pft
Velocity is the derivative of position Acceleration is the derivative of velocity
Cos: Example 1a A loudspeaker vibrates at 262 Hz (middle C). The amplitude of the cone of the speaker is 1.5 X 10-4 m. What is the equation to describe the position of the cone over time? x = A cos2pft x = (1.5 X 10-4 m) cos2p(262 s-1)t x = (1.5 X 10-4 m) cos(1650 s-1)t
Cos: Example 1b What is the position at t = 1.00 ms (1 X 10-3 s)
x = A cos2pft x = (1.5 X 10-4 m) cos2p(262 s-1) (1 X 10-3 s) x = (1.5 X 10-4 m) cos(1.65 rad) = -1.2 X 10-5 m
Cos: Example 1c What is the maximum velocity and acceleration?
vo = 2pAf vo = 2p(1.5 X 10-4 m)(262 s-1) = 0.25 m/s
F = ma kx = ma a = kx/m But we don’t know k or m a = k x Solve for k/m m T = 2p m k T2 = (2p)2m k = (2p)2 = (2p)2f2 m T2
a = k x m a = (2pf)2x = (2pf)2A a = [(2p)(262 Hz)]2(1.5 X 10-4 m) = 410 m/s2
Cos: Example 2a Find the amplitude, frequency and period of motion for an object vibrating at the end of a spring that follows the equation: x = (0.25 m)cos p t 8.0
x = A cos2pft x = (0.25 m)cos p t 8.0 Therefore A = 0.25 m 2pft = p t 2f = 1 f = 1/16 Hz T = 1/f = 16 s
Cos: Example 2b Find the position of the object after 2.0 seconds.
x = (0.25 m)cos p t 8.0 x = (0.25 m)cos p 4.0 x = 0.18 m
The Pendulum Pendulums follow SHM only for small angles (<15o)
The restoring force is at a maximum at the top of the swing. q Fr = restoring Force
Remember the circle (360o = 2p rad) q = x L
Fr = mgsinq at small angles sinq = q Fr = mgq q L x q mg Fr
Fr = mgq Fr = mgx (Look’s like Hook’s Law F = -kx) L k = mg T = 2p m k T = 2p mL mg
T = 2p L g f = 1 = g T 2p L The Period and Frequency of a pendulum depends only on its length
Swings and the Pendulum
To go fast, you need a high frequency Short length (tucking and extending your legs) f = g 2p L decrease the denominator
Example 1: Pendulum What would be the period of a grandfather clock with a 1.0 m long pendulum? T = 2p L g Ans: 2.0 s
Example 2: Pendulum Estimate the length of the pendulum of a grandfather clock that ticks once per second (T = 1.0 s). T = 2p L g Ans: m
Damped Harmonic Motion
Most SHM systems slowly stop For car shocks, a fluid “dampens” the motion
Resonance: Forced Vibrations
Can manually move a spring (sitting on a car and bouncing it) Natural or Resonant frequency (fo) When the driving frequency f = fo, maximum amplitude results Tacoma Narrows Bridge 1989 freeway collapse Shattering a glass by singing
Wave Medium Mechanical Waves Electromagnetic Waves Require a medium
Water waves Sound waves Medium moves up and down but wave moves sideways Electromagnetic Waves Do not require a medium EM waves can travel through the vacuum of space
Parts of a wave Crest Trough Amplitude Wavelength
Frequency (cycles/s or Hertz (Hz)) Velocity v = lf
Velocity of Waves in a String
Depends on: Tension (FT) [tighter string, faster wave] Mass per unit length (m/L) [heavier string, more inertia] v = FT m/L
Example 1: Strings A wave of wavelength 0.30 m is travelling down a 300 m long wire whose total mass is 15 kg. If the wire has a tension of 1000 N, what is the velocity and frequency? v = N = 140 m/s 15 kg/300 m v = lf f = v/l = 140 m/s/0.30 m = 470 Hz
Transverse and Longitudinal Waves
Transverse Wave – Medium vibrates perpendicular to the direction of wave EM waves Water waves Guitar String Longitudinal Wave – Medium vibrates in the same direction as the wave Sound
Longitudinal Waves: Velocity
Wave moving along a long solid rod Wire Train track vlong= E Elastic modulus r Wave moving through a liquid or gas vlong = B Bulk modulus
Ex. 1: Longitudinal Waves: Velocity
How fast would the sound of a train travel down a steel track? How long would it take the sound to travel 1.0 km? vlong= E = (2.0 X 1011/7800 kg/m3)1/2 r vlong = 5100 m/s (much fast than in air) v = x/t t = x/v = 1000m/5100m/s = 0.20 s
Earthquakes Both Transverse and Longitudinal waves are produced
S(Shear) –Transverse P(Pressure) – Longitudinal In a fluid, only p waves pass Center of earth is liquid iron
Energy Transported by Waves
Intensity = Power transported across a unit area perpendicular to the wave’s direction I = Power = P Area pr2 Comparing two distances: I1r12 = I2r22
Intensity: Example 1 The intensity of an earthquake wave is 1.0 X 106 W/m2 at a distance of 100 km from the source. What is the intensity 400 km from the source? I1r12 = I2r22 I2 = I1r12/r22 I2 = (1.0 X 106 W/m2)(100 km)2/(400 km)2 I2 = 6.2 X 104 W/m2
Reflection of a Wave Hard boundary inverts the wave
Exerts an equal and opposite force Loose rope returns in same direction
Continue in same direction if using another rope boundary
Constructive and Destructive Interference
Destructive Constructive Interference Interference
Constructive and Destructive Interference : Phases
Waves “in phase” “out of phase” in between
Resonance Standing Wave – a wave that doesn’t appear to move
Node – Point of destructive interference Antinode – Point of constructive interference (think “Antinode,Amplitude) “Standing waves are produced only at the natural (resonant) frequencies.”
Resonance: Harmonics Fundamental Lowest possible frequency
“first harmonic” L = ½ l First overtone (Second Harmonic) Second overtone (Third Harmonic)
Resonance: Equations L = nln n = 1, 2, 3….. 2 f = nv = nf1 2L v = lf
v = FT m/L
Example 1: Resonance A piano string is 1.10 m long and has a mass of 9.00 g. How much tension must the string be under to vibrate at 131 Hz (fund. freq.)? L = nln 2 l1 = 2L = m 1 v = lf = (2.20 m)(131 Hz) = 288 m/s
v = FT m/L v2 = FT FT = v2m = (288 m/s)2(0.009 kg) = 676 N L (1.10 m)
What are the frequencies of the first four harmonics of this string?
f1 = 131 Hz 1st Harmonic f2 = 262 Hz 2nd Harmonic 1st Overtone f3 = 393 Hz 3rd Harmonic 2nd Overtone f4 = 524 Hz 4th Harmonic 3rd Overtone
Hitting a Boundary Both reflection and refraction occur
Angle of incidence = angle of reflection q1 q2 q1 = q2 Reflected wave air water Refracted wave
Refraction Velocity of a wave changes when crossing between substances
Soldiers slow down marching into mud sin q1 = v1 sin q2 = v2
Example 1: Refraction An earthquake p-wave crosses a rock boundary where its speed changes from 6.5 km/s to 8.0 km/s. If it strikes the boundary at 30o, what is the angle of refraction? sin q1 = v1 sin 30o = km/s sin q2 = v2 sin q2 8.0 km/s q2 = 38o
Example 2: Refraction A sound wave travels through air at 343 m/s and strikes water at an angle of 50. If the refracted angle is 21.4o, what is the speed of sound in water? (Ans: m/s)
Diffraction Note bending of wave into “shadow region”
Diffraction Bending of waves around an object
Only waves diffract, not particles The smaller the obstacle, the more diffraction in the shadow region | HuggingFaceTB/finemath | |
# 155384 (number)
155,384 (one hundred fifty-five thousand three hundred eighty-four) is an even six-digits composite number following 155383 and preceding 155385. In scientific notation, it is written as 1.55384 × 105. The sum of its digits is 26. It has a total of 4 prime factors and 8 positive divisors. There are 77,688 positive integers (up to 155384) that are relatively prime to 155384.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 26
• Digital Root 8
## Name
Short name 155 thousand 384 one hundred fifty-five thousand three hundred eighty-four
## Notation
Scientific notation 1.55384 × 105 155.384 × 103
## Prime Factorization of 155384
Prime Factorization 23 × 19423
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 38846 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 155,384 is 23 × 19423. Since it has a total of 4 prime factors, 155,384 is a composite number.
## Divisors of 155384
8 divisors
Even divisors 6 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 291360 Sum of all the positive divisors of n s(n) 135976 Sum of the proper positive divisors of n A(n) 36420 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 394.188 Returns the nth root of the product of n divisors H(n) 4.26645 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 155,384 can be divided by 8 positive divisors (out of which 6 are even, and 2 are odd). The sum of these divisors (counting 155,384) is 291,360, the average is 36,420.
## Other Arithmetic Functions (n = 155384)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 77688 Total number of positive integers not greater than n that are coprime to n λ(n) 38844 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 14258 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 77,688 positive integers (less than 155,384) that are coprime with 155,384. And there are approximately 14,258 prime numbers less than or equal to 155,384.
## Divisibility of 155384
m n mod m 2 3 4 5 6 7 8 9 0 2 0 4 2 5 0 8
The number 155,384 is divisible by 2, 4 and 8.
## Classification of 155384
• Arithmetic
• Refactorable
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
## Base conversion (155384)
Base System Value
2 Binary 100101111011111000
3 Ternary 21220010222
4 Quaternary 211323320
5 Quinary 14433014
6 Senary 3155212
8 Octal 457370
10 Decimal 155384
12 Duodecimal 75b08
20 Vigesimal j894
36 Base36 3bw8
## Basic calculations (n = 155384)
### Multiplication
n×y
n×2 310768 466152 621536 776920
### Division
n÷y
n÷2 77692 51794.7 38846 31076.8
### Exponentiation
ny
n2 24144187456 3751620423663104 582941787910467751936 90579826772680121166823424
### Nth Root
y√n
2√n 394.188 53.7612 19.8542 10.9215
## 155384 as geometric shapes
### Circle
Diameter 310768 976306 7.58512e+10
### Sphere
Volume 1.57148e+16 3.03405e+11 976306
### Square
Length = n
Perimeter 621536 2.41442e+10 219746
### Cube
Length = n
Surface area 1.44865e+11 3.75162e+15 269133
### Equilateral Triangle
Length = n
Perimeter 466152 1.04547e+10 134566
### Triangular Pyramid
Length = n
Surface area 4.1819e+10 4.42133e+14 126871
## Cryptographic Hash Functions
md5 1c518080bed9a4b47f52c47aa0a4bc5a 23d5ae563f0c2e5e320cd690eb593a1d52aa1543 a84eba69be0080253d067b73d7b8751faa569829c7029cbf8775e65a0efc22d2 0a4357cfbcb1dc59e200e1085bfe9ae200a10e0703442633cf187c1bfa04c081641ac9febede9415a49fa6790915c9c3fca636b40504b5891f4742822c8af95f 192cf7173a48636138b66da8c1f26d70b9bc52ed | HuggingFaceTB/finemath | |
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Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
Impossible Sandwiches
Age 11 to 18
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
Iffy Logic
Age 14 to 18 Challenge Level:
Can you rearrange the cards to make a series of correct mathematical statements?
L-triominoes
Age 14 to 16 Challenge Level:
L triominoes can fit together to make larger versions of themselves. Is every size possible to make in this way?
Proofs with Pictures
Age 14 to 18
Some diagrammatic 'proofs' of algebraic identities and inequalities.
Age 11 to 16 Challenge Level:
Draw some quadrilaterals on a 9-point circle and work out the angles. Is there a theorem?
Janine's Conjecture
Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
For What?
Age 14 to 16 Challenge Level:
Prove that if the integer n is divisible by 4 then it can be written as the difference of two squares.
Leonardo's Problem
Age 14 to 18 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
Dalmatians
Age 14 to 18 Challenge Level:
Investigate the sequences obtained by starting with any positive 2 digit number (10a+b) and repeatedly using the rule 10a+b maps to 10b-a to get the next number in the sequence.
Parallel Universe
Age 14 to 16 Challenge Level:
An equilateral triangle is constructed on BC. A line QD is drawn, where Q is the midpoint of AC. Prove that AB // QD.
Magic Squares II
Age 14 to 18
An article which gives an account of some properties of magic squares.
Picturing Pythagorean Triples
Age 14 to 18
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
Mouhefanggai
Age 14 to 16
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
The Great Weights Puzzle
Age 14 to 16 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
Yih or Luk Tsut K'i or Three Men's Morris
Age 11 to 18 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
Multiplication Square
Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
Always Perfect
Age 14 to 16 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
Pythagorean Triples II
Age 11 to 16
This is the second article on right-angled triangles whose edge lengths are whole numbers.
Some Circuits in Graph or Network Theory
Age 14 to 18
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
Geometry and Gravity 2
Age 11 to 18
This is the second of two articles and discusses problems relating to the curvature of space, shortest distances on surfaces, triangulations of surfaces and representation by graphs.
To Prove or Not to Prove
Age 14 to 18
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
Age 14 to 18 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
Sprouts Explained
Age 7 to 18
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
Angle Trisection
Age 14 to 16 Challenge Level:
It is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square.
Pythagorean Triples I
Age 11 to 16
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
Unit Interval
Age 14 to 18 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
Find the Fake
Age 14 to 16 Challenge Level:
There are 12 identical looking coins, one of which is a fake. The counterfeit coin is of a different weight to the rest. What is the minimum number of weighings needed to locate the fake coin?
Converse
Age 14 to 16 Challenge Level:
Clearly if a, b and c are the lengths of the sides of an equilateral triangle then a^2 + b^2 + c^2 = ab + bc + ca. Is the converse true?
Age 14 to 18 Challenge Level:
Which of these roads will satisfy a Munchkin builder?
Cosines Rule
Age 14 to 16 Challenge Level:
Three points A, B and C lie in this order on a line, and P is any point in the plane. Use the Cosine Rule to prove the following statement.
More Number Sandwiches
Age 11 to 16 Challenge Level:
When is it impossible to make number sandwiches?
No Right Angle Here
Age 14 to 16 Challenge Level:
Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.
A Biggy
Age 14 to 16 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
Composite Notions
Age 14 to 16 Challenge Level:
A composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.
Age 14 to 16 Challenge Level:
Kyle and his teacher disagree about his test score - who is right?
Proximity
Age 14 to 16 Challenge Level:
We are given a regular icosahedron having three red vertices. Show that it has a vertex that has at least two red neighbours.
Russian Cubes
Age 14 to 16 Challenge Level:
I want some cubes painted with three blue faces and three red faces. How many different cubes can be painted like that?
Square Mean
Age 14 to 16 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
Natural Sum
Age 14 to 16 Challenge Level:
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
Calculating with Cosines
Age 14 to 18 Challenge Level:
If I tell you two sides of a right-angled triangle, you can easily work out the third. But what if the angle between the two sides is not a right angle?
Picture Story
Age 14 to 16 Challenge Level:
Can you see how this picture illustrates the formula for the sum of the first six cube numbers?
Age 11 to 18 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
A Long Time at the Till
Age 14 to 18 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
Pareq Exists
Age 14 to 16 Challenge Level:
Prove that, given any three parallel lines, an equilateral triangle always exists with one vertex on each of the three lines.
Proof: A Brief Historical Survey
Age 14 to 18
If you think that mathematical proof is really clearcut and universal then you should read this article.
Triangular Intersection
Age 14 to 16 Short Challenge Level:
What is the largest number of intersection points that a triangle and a quadrilateral can have?
Zig Zag
Age 14 to 16 Challenge Level:
Four identical right angled triangles are drawn on the sides of a square. Two face out, two face in. Why do the four vertices marked with dots lie on one line?
Diophantine N-tuples
Age 14 to 16 Challenge Level:
Can you explain why a sequence of operations always gives you perfect squares?
There's a Limit
Age 14 to 18 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely? | HuggingFaceTB/finemath | |
Q
#### Understanding Multiplying by 10s & 100s - Quiz 4
(11 Questions)
1 viewed last edited 4 years ago
Understanding Multiplying by 10s & 100s
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2
If 7 × 7 = 49 , then 7 × 70 = _______
0
2
If 3 × 3 = 9 , then 30 × 3 = _______
0
2
If 9 × 6 = 54 , then 9 × 600 = _______
0
2
If 5 × 6 = 30 , then 5 × 60 = _______
0
2
If 9 × 8 = 72 , then 9 × 800 = _______
0
2
If 6 × 4 = 24 , then 6 × 40 = _______
0
2
If 5 × 2 = 10 , then 500 × 2 = _______
0
2
If 8 × 5 = 40 , then 8 × 500 = _______
0
2
If 7 × 1 = 7 , then 7 × 10 = _______
0
2
If 7 × 4 = 28 , then 7 × 400 = _______
0
2
If 2 × 9 = 18 , then 20 × 9 = _______ | HuggingFaceTB/finemath | |
# Circle Parametric equation in $3D$ space?
What are the parametric equations of a circle in $x z$ plane with a rotation a round $z$-axis ? so if
$x = r * \cos(\theta)$
$z = r * \sin(\theta)$
what should $y =$ ??
• If I'm understanding your description properly, it sounds like $y=0$, so your parametric representation is: $(r\cos\theta,0,r\sin\theta)$ – G Tony Jacobs May 13 '18 at 16:02
• yes, when y = 0 the circle is in xz plane, now I want to rotate it around z-axis ? – Mostafa Said May 13 '18 at 16:13
• I see... it sounds as if you'll need one parameter, $\theta$, to trace out your circle, and another parameter, $t$, to make it rotate? – G Tony Jacobs May 13 '18 at 16:29
• You should start from the parametrization of a sphere in $\mathbb{R}^3$, and then fix the polar angle to the desired value – user438666 May 13 '18 at 16:30
• G Tony, --> thats exactly what i am trying to do, so $theta$ for the circle and t for the tilt – Mostafa Said May 13 '18 at 16:41
The circle
$$p = (r\cos(t),0,r\sin(t))$$
rotated around the $z$ axis is built with the rotation matrix
$$R(\theta) = \left( \begin{array}{ccc} \cos (\theta ) & -\sin (\theta ) & 0 \\ \sin (\theta ) & \cos (\theta ) & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$$
$$p\cdot R(\theta) = (r \cos (\theta ) \cos (t),-r \sin (\theta ) \cos (t),r \sin (t))$$
Attached a rotated circle (red) by $\frac{\pi}{3}$ • great visuals, thanks a lot :) – Mostafa Said May 13 '18 at 19:14
$HINT:$
You can see geometrically that :
any point $P(x,0,z)$ in the $XY-plane$,
if rotated about the $Z-axis$ by an angle $α$ with the $XZ- plane$ changes to :
$(x-x.cosα, x.sinα, z)$.
so if your original coordinates are:
$(r.cosθ,0,r.sinθ)$,(in the $XZ$-plane)
they will change to:
$x=r.cosθ-r.cosθ.cosα$
$y=r.cosθ.sinα$
$z=r.sinθ$
• thanks a lot, works :) could you tell where is the source of such equations ? – Mostafa Said May 13 '18 at 17:13
• you can take the coordinates of a point (x,0,z). Rotate it about z-axis in a rough diagram by an angle α, and use trignometric ratios to find the final directed distances from x,y,z axis. Hence finding the final coordinates.(sorry i cannot attach a photo yet) – kadoodle May 14 '18 at 14:36
• Thanks a lot, thats really helpful – Mostafa Said May 15 '18 at 14:44 | HuggingFaceTB/finemath | |
# Centre of the ring of quaternions
by Wingeer
Tags: centre, quaternions, ring
P: 79 1. The problem statement, all variables and given/known data What is the centre of the ring of the quaternions defined by: $$\mathbf{H}=\{ \begin{pmatrix} a & b \\ -\bar{b} & \bar{a} \end{pmatrix} | a,b \in \mathbf{C} \}$$? 2. Relevant equations The definition of the centre of a ring: The centre Z of a ring R is defined by $$Z(R)=\{A | AX=XA, \forall X \in R\}$$ 3. The attempt at a solution I figured that multiples of the 2x2 identity matrix must be in the centre. Also if we denote an element of H by: $$\begin{pmatrix} x & y \\ -\bar{y} & \bar{x} \end{pmatrix}$$ where $$x=x_1 + ix_2$$ and similarly for a,b and y that: 1. $$b\bar{y}=\bar{b}y$$ 2. $$y(a-\bar{a})=b(x-\bar{x})$$ 3. $$\bar{b}(x-\bar{x})=\bar{y}(a-\bar{a})$$ Then for instance we get from the first equation that: $$b_2x_1=a_1y_2$$ But I am not sure whether this approach really is any useful at all. Some hints would be greatly appreciated.
P: 79 Anyone? I actually have another question about the quaternions. I am asked to show that: $$\mathbf{H'} = \{ a+bi+cj+dk | a,b,c,d \in \mathbf{R} \}$$ with: i^2=j^2=k^2=-1, ij=k=-ji, ik=-j=-ki and jk=i=-kj. is isomorphic as rings to the quaternions defined in the previous post. I started by noticing that (where x,y are complex numbers): $$\begin{pmatrix} x & y \\ -\bar{x} & \bar{y} \end{pmatrix} = \begin{pmatrix} a+bi & c+di \\ c-di & a-bi \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} + \begin{pmatrix} bi & 0 \\ 0 & bi \end{pmatrix} + \begin{pmatrix} 0 & c \\ -c & 0 \end{pmatrix} + \begin{pmatrix} 0 & di \\ di & 0 \end{pmatrix}$$ And so we see that every element in H is a linear combination of these matrices which all are linearly independent as well. This means we have found a basis for H. So if we define a function f:H -> H' by: $$1= f \left( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right)$$ $$i=f \left( \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix}\right)$$ $$j=f \left( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\right)$$ $$k=f \left( \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\right)$$ We see that obviously f is both surjective and injective as these are the only values f are defined for. Therefore f is an bijection and H and H' are isomorphic. Do I have to mix ring homomorphisms in this? Or?
HW Helper Sci Advisor Thanks P: 24,454 You've have the right general ideas there. Let's call your matrices M1, Mi, Mj and Mk. By the way, I think you've Mi wrong, check it again. You need to define f for all matrices. But that's easy just define it to be the linear map H->H' defined by your mapping of the basis elements. That you have a bijection between H and H' isn't really in question, because you've mapped the basis for a four dimensional real vector space into the basis of another one. So it's a bijection. Now you have to worry whether it's a ring homomorphism. Ring addition is not a problem just because f is linear. It's multiplication you have to check. If f(xy)=f(x)f(y). For example is f(Mi*Mj)=f(Mi)f(Mj)?
P: 905
## Centre of the ring of quaternions
note that it is sufficient to check the 16 possible products of M1,Mi,Mj,Mk because of linearity and the distributive laws.
your formula for Mi is indeed wrong, as the lower right coordinate is not the complex conjugate of the upper left coordinate, so that matrix isn't even in H.
*****
with regard to your first problem, note that
[a 0]
[0 a] is not in Z(H) unless a is real, because:
[x+iy .0..][0 i]....[.0.. -y+ix]
[.0.. x-iy][i 0] = [b+ia ...0..]
whereas:
[0 i][x+iy .0.]....[...0.. y+ix]
[i 0][.0.. x-iy] = [-y+ix .0..], these two matrices aren't equal unless y = 0.
remember that an element of Z(H) has to commute with ALL of H, so if you find just ONE element of H a certain matrix doesn't commute with, that matrix cannot be in the center. so i suggest you find which matrices commute with your matrices Mi, Mj and Mk.
Related Discussions Introductory Physics Homework 1 Calculus & Beyond Homework 5 Introductory Physics Homework 0 Introductory Physics Homework 7 | HuggingFaceTB/finemath | |
Conversion formula
The conversion factor from grams to pounds is 0.0022046226218488, which means that 1 gram is equal to 0.0022046226218488 pounds:
1 g = 0.0022046226218488 lb
To convert 481 grams into pounds we have to multiply 481 by the conversion factor in order to get the mass amount from grams to pounds. We can also form a simple proportion to calculate the result:
1 g → 0.0022046226218488 lb
481 g → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 481 g × 0.0022046226218488 lb
M(lb) = 1.0604234811093 lb
The final result is:
481 g → 1.0604234811093 lb
We conclude that 481 grams is equivalent to 1.0604234811093 pounds:
481 grams = 1.0604234811093 pounds
Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 0.94301948024948 × 481 grams.
Another way is saying that 481 grams is equal to 1 ÷ 0.94301948024948 pounds.
Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that four hundred eighty-one grams is approximately one point zero six pounds:
481 g ≅ 1.06 lb
An alternative is also that one pound is approximately zero point nine four three times four hundred eighty-one grams.
Conversion table
grams to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from grams to pounds
grams (g) pounds (lb)
482 grams 1.063 pounds
483 grams 1.065 pounds
484 grams 1.067 pounds
485 grams 1.069 pounds
486 grams 1.071 pounds
487 grams 1.074 pounds
488 grams 1.076 pounds
489 grams 1.078 pounds
490 grams 1.08 pounds
491 grams 1.082 pounds | HuggingFaceTB/finemath | |
You are on page 1of 4
# Name ______________________ Due ___________
## Olympic Physics- Michael Phelps
Fill in the Table below by calculating the Average Speed to 4 decimal places.
Remember to include the units.
Michael Phelps Career Best Times
Event
Time
Min:sec.millisec
Time in
Seconds
Average Speed
100 m freestyle 47.51
200 m freestyle 1:42.96
400 m freestyle 3:47.79
100 m backstroke 53.01
200 m backstroke 1:54.65
100 m breaststroke 1:02.57
200 m breaststroke 2:15.06
100 m butterfly 49.82
200 m butterfly 1:51.51
1) Which event does Michael Phelps have the greatest speed?
2) Does Michael Phelps tend to have an increase or decrease in speed between the 100m and
200m laps? Why do you think this is the case?
3) What is the difference in speed between his slowest and fastest event?
4) Predict the time you think it would take Michel Phelps to complete
the 200m Individual Medley where he completes 50m laps of each
event(freestyle, backstroke, butterfly, breast stroke). And the 400m
Individual Medley where he completes 100m of each. Show your
work and justify why.
5) Fill in the chart below and compare your predicted results to his record below. Are they close?
Why or why not? What may cause the differences between your calculations and the actual
speed?
Event
Time
Min:sec.millisec
Time in Seconds Speed
200 m Individual
Medley
1:54.16
400 m Individual
Medley
4:03.84
Olympic Physics- Michael Phelps
Fill in the Table below by calculating the Average Speed to 4 decimal places.
Remember to include the units.
Michael Phelps Career Best Times
Event
Time
Min:sec.millisec
Time in
Seconds
Average Speed
100 m freestyle 47.51
47.51 s 2.1048 m/s
200 m freestyle 1:42.96
102.96 s 1.9425 m/s
400 m freestyle 3:47.79
227.79 s 1.7560 m/s
100 m backstroke 53.01
53.01 s 1.8864 m/s
200 m backstroke 1:54.65
114.65 s 1.7444 m/s
100 m breaststroke 1:02.57
62.57 s 1.5982 m/s
200 m breaststroke 2:15.06
135.06 s 1.4808 m/s
100 m butterfly 49.82
49.82 s 2.0072 m/s
200 m butterfly 1:51.51
111.51 s 1.7936 m/s
1) Which event does Michael Phelps have the greatest speed?
100m Freestyle
2) Does Michael Phelps tend to have an increase or decrease in speed between the 100m and
200m races? Why do you think this is the case?
He tends to have a decrease in speed between the 100m and 200m races. He may get tired and
there are also extra turns with the 200m race (3 turns) and only 1 turn with the 100m race which
can increase time.
3) What is the difference in speed between his slowest and fastest event?
4) Calculate the time and his speed you think it would take Michel
Phelps to complete the 200m Individual Medley where he completes
50m laps of each event(freestyle, backstroke, butterfly, breast
stroke). And the 400m Individual Medley where he completes 100m
of each. Show your work and justify why.
You may get a few different responses for this. Most students will base the
time on the individual races add up the times and account for the distances.
*I used the 200m times since it includes the extra turns.
200 freestyle 102.96 s
200 backstroke 114.65 s
200 butterfly 111.51 s
200 breast stroke 135.06 s
Total time 464.18 s
200 individual medley
200IM speed
400 individual medley
400IM speed
Some students may go further to account for him getting tired so they may add on a few extra seconds
or multiply by their estimated factor such as 1.1 or 1.05.
5) Fill in the chart below and compare your predicted results to his record below. Are they close?
Why or why not? What may cause the differences between your calculations and the actual
speed?
Event
Time
Min:sec.millisec
Time in Seconds Speed
200 m Individual
Medley
1:54.16 114.16 s 1.7519 m/s
400 m Individual
Medley
4:03.84 243.84 s 1.6404 m/s
The 200m IM he was faster than my estimated time. One reason may be that the turns are easier
with some strokes than others. He saved approximately 2 seconds less than my estimate.
The 400m IM predictions were less accurate. I estimated 232.09 seconds and his speed was 11.75
seconds over. Michael Phelps likely slowed throughout the race due to tiredness of the 400m race
and water disturbances which can decrease his speed. | HuggingFaceTB/finemath | |
## Thursday, June 10, 2010
### Long Odds on Drawing an Inside Straight
It's often said trying to draw to an inside straight is a fool's bet. Why is that?
1. Because there is only one card out of 52 that will satisfy your draw.
1. El wrongo, betting breath. You have 4 cards out of 52.
2. Trying to draw an outside straight will give you twice the chance to get the right card. But that's pretty relative, isn't it?
3. For an inside straight, your chances are 4 out of 52. If you're going for an inside straight flush, then it's one out of 52. I don't go for inside straights but the few times I have, it never came in.
4. err... If you have at least 4 cards in your hand already... plus the card you may have thrown back?
5. Let's say you're playing 5-card draw--that will make this comparatively simple.
If you're drawing to an inside straight, it means that you have four communally non-consecutive members of a series of five consecutive cards (e.g. 4,5,6,8), and one other card not matching this pattern. Out of the full deck of 52 cards, you've now seen 5--leaving 47 cards unseen. Some of these cards will be in other players' hands--but, since the chances of any given card being in their hands vs. still dealable are equal, this is not important.
We're dropping one card and drawing one replacement from the 47 possible remaining cards in the deck. There is one value (e.g. 7 in the above example)--and that value is still available in all four suits (since none of the cards we still hold have this value, nor does the card we've discarded). This leaves you with 4 chances for a successful inside straight draw out of a possible 47; so your odds of success are 4/47 = 0.085106. For even money, or anything close to it, this is a terrible bet; if you run this situation out many times, you'll win only 4/47ths of the time--so you'll lose lots and lots of money.
Now, dropping one card for the inside straight draw also gives you a shot at drawing a pair. You hold four cards that have possible pairs left to be drawn, and (assuming you didn't have a pair before your discard) each has three possible matches still dealable. Your odds of making one of these are 12/47 = 0.255319--just a bit better than one in four. If you think a pair is good in the hand (depending on the actual values of the cards you hold--high-end or low-end), then you actually have 16 outs out of 47 if you choose to drop the one card and "draw for the straight". That's 16/47 = 0.340425; so you'll make a hand fully 1/3 of the time. If your card are suited, your odds obviously improve even more; subtracting the outs we've already named (and assuming you don't already have a flush before the draw), there are 13 - 1 (straight card--actually straight-flush card) - 4 (cards in your hand) = 8 cards that will make your flush (independently of any of the above hands) if your four cards are suited. This increases you to 24 out out of 47 to make a hand--24/47, or 0.51, which is not bad at all. If you've got suited cards and an inside straight draw, you're fairly certain that a pair is sufficient to win, and you're getting even just even money on the bet, you should go for it.
Of course, the bet is the important part in the decision. Usually a pair isn't good in this situation, and usually you're not also drawing to a flush, so let's go with the straight-up inside straight draw odds of 4/47. If the ratio of the call you have to put in to the pot at stake is good enough, you'd still be obliged to call. For instance: Suppose you're drawing twice, rather than once; and that, in the first round, the pot was raised up to 50,000. In the second round, after your first draw, you're left with a possible inside straight draw--but your opponent, sensing you have nothing solid, bets an additional 1,000 to move you out of the pot. If you make the inside straight draw in this scenario many times, 4/47th of the time you'll win, and you'll make a gain of 49,000 each time; and 43/47th of the time you'll lose, losing 1,000 each time. Your average yield on the decision, then, is (49,000 * 4 - 43,000)/50 = 3,060; so the bet yields a net gain, and you should decidedly take it.
6. too much time on your hands
Leave your answer or, if you want to post a question of your own, send me an e-mail. Look in the about section to find my e-mail address. If it's new, I'll post it soon.
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Enter your Email and join hundreds of others who get their Question of the Day sent right to their mailbox | HuggingFaceTB/finemath | |
# PSEB 7th Class Maths Solutions Chapter 1 Integers Ex 1.1
Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 1 Integers Ex 1.1 Textbook Exercise Questions and Answers.
## PSEB Solutions for Class 7 Maths Chapter 1 Integers Ex 1.1
1. Use the appropriate symbol >, <, = to fill in the blanks (i) -3 -5
(ii) – 2 5-4
(iii) 8-4 – 3
(iv) – 6 5-0
(v) 5 8-3
(vi) 0 – 3.
2. Arrange the following integers in ascending order.
Question (i)
– 2, 12, – 43, 31, 7, – 35, – 10
Given positive integers are 12, 31, 7
Ascending order is 7 < 12 < 31
Given negative integers are – 2, – 43, – 35, -10
Ascending order is – 43 < – 35 < – 10 < -2.
Hence, all given integers in ascending order are :
– 43 < – 35 < -10 < – 2 < 7 < 12 < 31.
i.e. – 43, – 35, – 10, – 2, 7, 12, 31.
Question (ii)
– 20, 13, 4, 0, – 5, 5
Given positive integers are 13, 4, 5
Ascending order is 4 < 5 < 13
Given negative integers are – 20, – 5
Ascending order is -20 < – 5.
Hence all given integers in ascending order are :
– 20 < – 5 < 0 < 4 < 5 < 13
i.e. – 20, – 5, 0, 4, 5, 13 3. Arrange the following integers in descending order.
Question (i)
0, – 7, 19, – 23, – 3, 8, 46
Given positive integers are 19, 8, 46
Descending order is 46 > 19 > 8
Given negative integers are – 7, – 23, – 3
Descending order is – 3 > – 1 > – 23
Hence, all given integers in descending order are :
46 > 19 > 8 > 0 > – 3 > – 7 > – 23
i.e. 46, 19, 8, 0, – 3, – 7, – 23
Question (ii)
30, – 2, 0, – 6, – 20, 8.
Given positive integers are 30, 8
Descending order is 30 > 8
Given negative integers are – 2, – 6, – 20
Descending order is-2 > -6 > -20
Hence all given integers in descending order are :
30 > 8 > 0 > -2 > – 6 > – 20
i.e. 30, 8, 0, – 2, – 6, – 20
4. Evaluate :
Question (i)
30 – | -21 |
30 – | -21 | = 30 – 21
[∵ | – 21 | = 21]
= 9
Question (ii)
| -25 | – | -18 |
| -25 | – | – 18 | = 25 – 18
[∵ | -25 | = 25 and | – 18 | = 18]
= 7
Question (iii)
6 – | -4 |
6 – | -4 | = 6 – 4
[∵ | – 4 | = 4]
= 2
Question (iv)
| – 125 | + | 110 |
| – 125 | + | 110 | = 125 + 110
[∵ | -125 | = 125 and | 110 | = 110]
= 235 5. Fill in the blanks :
Question (i)
0 is greater than every …………… integer.
Negative
Question (ii)
Modulus of a negative integer is always ……………
Positive
Question (iii)
The smallest positive integer is …………… | HuggingFaceTB/finemath | |
# Proportion Formula
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Proportion is a fixed ratio between two substances. Proportion always maintains a fixed ratio between two fractions. For example, a / b = c / d. It can be written as a : b = c : d. This can also be called as the proportion formula. Proportion in math is defined as the similarity while comparison between two quantities or substances. Proportion is also used between two geometric objects where their dimensions can be compared.
Example 1: Find the x in the proportion 2 : y = 10 : 3.
Solution: Here the given proportion is 2 : y = 10 : 3.
It can be expressed as a fraction in the form.
2 / y = 10 / 3
Now multiply both sides of the equation by y.
This gives 2 = 10 y /3.
Now multiply both sides of the equation by 3.
This gives 10 y = 6.
Now divide both sides of the equation by 10.
So y = 6 / 10.
Hence the value of y for the given proportion = 3 / 5.
Example 2: Find the x in the proportion x : 5 = 1 : 15.
Solution: Here the given proportion is x : 5 = 1 :15.
It can be expressed as a fraction in the form.
This gives, x / 5 = 1 / 15
Now multiply both sides of the equation by 5.
This gives 5(x/ 5) = (1 / 15) (5).
x = 1 / 3.
Hence the value of x for the given proportion is= 1/3. | HuggingFaceTB/finemath | |
# math
posted by .
In an urn, there are several colored balls, with equal numbers of each color. We add 14 balls which are all of the same new color, that is different from those in the urn. It is calculated that the probability of drawing , without replacement, two balls of the same color is constant. How many balls are there in the urn initially?
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8. ### Pre Algebra Probability
Each of two urns contains green balls and red balls. urn 1 contains 10 green balls and 14 red balls. urn 2 contains 4 green balls and 11 red balls. if a ball is drawn from each urn what is p(red and red)?
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More Similar Questions | HuggingFaceTB/finemath | |
As a formula: where: w is the width of the rectangle h is the height of the rectangle Calculator. Viewed sideways it has a base of 20m and a height of 14m. How can I get the diagonal of square? You can use Pythagoras' theorem to find the diagonal from the length and width of the rectangle: d^2 = l^2 + w^2. And it's simple, like the saying, "The shortest distance between any two points is the hypotenuse!" Diagonal refers to the line segment that connects the two opposite corners or vertex of the rectangle. But because a polygon can’t have a negative number of sides, n must be 15. Diagonal of a Rectangle $$\ Diagonal\;of\;a\;Rectangle=\sqrt{l^{2}+b^{2}}$$ Area formula of a rectangle. A diamond is often used to describe a rhombus of 4 equal sides, sides parallel and angles equal (a square, Fig. The square root of 32 is equal to 4 square root of 2. It means the opposite vertex connection and the length is called length of the diagonal. Then find the square root for it, the resultant value is the length of diagonal. So, for example, if the square side is equal to 5 in, then the diagonal is 5√2 in ≈ 7.071 in. Formulas for Hexagon 1] Area of a Hexagon. We know ads can be annoying, but they’re what allow us to make all of wikiHow available for free. Fortunately, an easy formula exists to tell you exactly how many diagonals a polygon has. Then, we will find the area of one of the triangles and finally, we will multiply by 6 to find the total area of the polygon. A square always has equal lengths. When the parallelogram is specified from the lengths B and C of two adjacent sides together with the length D 1 of either diagonal, then the area can be found from Heron's formula. All four sides 2. In the bellow figure diagonal is AD and the length of the diagonal is square root of 2 multiple with the length. The diagonal side rectangles are calculated by the following formula: d 1i = â (a i 2 + h 2). Area of B = ½b × h = ½ × 20m × 14m = 140m 2 Such triangle has a half of the area of a square, its legs are square sides and hypotenuse equals to the length of the diagonal of a square. Cut all three squares out. So now we will discuss third formula. In strictly correct mathematical wording the formula above should be spoken as "s raised to the power of 2", meaning s is multiplied by itself. The three formulas are first one is area of rectangle formula, second one is perimeter of rectangle formula and third one formula for diagonal of rectangle. It only has an area, which is calculated by multiplying two sides together, or squaring one. This simple equation is used in many fields, including crystallography, chemistry, and art. Here, we will discuss some interesting facts about the box and how to calculate the volume and the surface area of a box with the help of mathematical formula. A regular polygon is a polygon where all the sides are the same length and all the angles are equal. 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How to Find the Area of a Square Using the Length of its Diagonal, https://www.mathsisfun.com/geometry/square.html, https://www.mathsisfun.com/pythagoras.html, http://www.coolmath.com/reference/squares#The_diagonal_of_a_square, http://www.sosmath.com/calculus/diff/der07/der07.html, encontrar el área de un cuadrado usando el largo de su diagonal, Calcolare l'Area di un Quadrato a Partire dalla Diagonale, Encontrar a Área de um Quadrado Usando o Comprimento de Sua Diagonal, вычислить площадь квадрата по длине диагонали, calculer l’aire d’un carré en utilisant la longueur de sa diagonale, Die Fläche eines Quadrats mittels der Länge seiner Diagonale berechnen, Memperoleh Luas Persegi Menggunakan Panjang Diagonal, De oppervlakte van een vierkant berekenen met behulp van de diagonaal, consider supporting our work with a contribution to wikiHow, The two shorter sides of the triangle are the sides of the square: each one has a length of. The formula for finding the area of a rhombus is A = (d 1 * d 2)/2, where d 1 and d 2 are the lengths of the diagonals. There are three different formulas for finding the area of a rhombus. Join any two opposite corners to make a diagonal. To find all possible diagonals of a simple polygon with just a few sides, you can easily count them. (a) 30 (b) 125 Start with the area formula A = s^2 and solve for s. (Hint: a square root will cancel out the exponent.). That is the length of the side of a square. The length of the diagonal in this case is equal to â(l2+w2). Note: Sometimes, base and height are used instead of length and width. Direct Proportional Adjustment Try Direct Proportional Adjustment of the 3-4-5 Triple. 1. If some base diagonal is denoted by the letter d 0i, then the volume diagonal ⦠To find the diagonal, square the width and height of the rectangle and add the squared values. The formula for the relationship between diagonals and sides of a parallologram is , where represents one diagonal, represents the other diagonal, represents a side, and represents the adjoining side. The area of the quadrangle is calculated using the formula of the sum of the length of the four sides and the diagonal. This will be useful later on. wikiHow's. wikiHow is where trusted research and expert knowledge come together. For example, enter the two side lengths. How do I find the diagonal of a square with different side lengths? Here, we will discuss some interesting facts about the box and how to calculate the volume and the surface area of a box with the help of mathematical formula. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. How do I find the area of a square if I know the length? Area of Parallelogram The parallelogram is a geometrical figure that is formed by the pair of parallel sides having opposite sides of equal length and the opposite angles of equal measure. If you know Altitude (height) and side s the formula is: So, in this problem, substitute the known values and solve for the missing diagonal. ABDC is a parallelogram with a side of length 11 units, and its diagonal lengths are 24 units and 20 units. Height = h = 12. To find the diagonal, square the width and height of the rectangle and add the squared values. Please help ASAP Find its area. Square 8, so you will get 64. This shape is not a square, since it does not have equal sides. The shape has nine diagonals, means the lines between the interior angles. The Surface area of a box formula . Diagonal of a Cube Formula. Question Based on Area, Perimeter, Diagonal. Heronâs formula gives its area as â[sâ
(s-a)â
(s-b)â
(s-c)]. When diagonals and angle between them are given. Diagonal of a Cube Formula. In the figure above, click 'reset'. For example, let's say a square has a diagonal that measures 10 cm. Explanation: . If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. % of people told us that this article helped them. For example, a square with a diagonal of 10cm has sides with length, If you need to find both the side length and the area from the diagonal, you can use this formula first, then quickly square the answer to get the area: Area. This diagonal divides the square into two right-triangles. This diagonal divides the square into two right-triangles. To compute the area of the diagonal cross section of the parallelepiped (S) proceed the same way, but note that the Pythagorean theorem in this case involved the legs different lengths - length (l) and width (w) volumetric shapes. Enter the length of four side (a, b, c, and d) and sum of the diagonal (A+B), and click "Calculate area of quadrangle", the area of the quadrangle is calculated from the length of four side and the sum of the diagonal, and it is displayed. The most common formula for the area of a square is simple: it's the length of the side squared, or s2. How do I determine the length of a side of a square given the area? Get the square root of 32. But sometimes you only know the length of the square's diagonal, running between opposite vertices. If diagonals and angle between those diagonals are given, the quadrilateral area formula for that case can be expressed as: In regular hexagons, there are nine diagonals in six equilateral triangles. In most of the cases, the box is an enclosed figure either a rectangle or a square. When the polygon gets a bit complicated, counting them can be very hard. Type that value into the diagonal of a square calculator to check it yourself! Calculate the diagonal of a trapezoid if given 1. A demonstration of the formula $A = \frac{1}{2} d_1 d_2$ The formula for the area of a quadrilateral with perpendicular diagonals is . Diagonal refers to the line segment that connects the two opposite corners or vertex of the rectangle. Since any diagonal of a parallelogram divides it into two congruent triangles, you can calculate the diagonal by knowing the sides of the parallelogram and the angle between them. Make sure all the sides are equal. Using the law of cosines 3. Include your email address to get a message when this question is answered. Since each side is s, the formula is Area = s x s = s 2. Area or midsegment Diagonal Of Square. How do I find the area of a circle inside a square? d = a√2. Draw a second square using that measurement as the length of the square. The perimeter, area, length of diagonals, as well as the radius of an inscribed circle and circumscribed circle will all be available in the blink of an eye. References. Area of A = a 2 = 20m × 20m = 400m 2 Part B is a triangle. The Relation Between Diagonal And Side of a square is, Diagonal = a 2 + a 2 = ( 2 a 2) = 2 a = 2 x side. There is no standard formula to find out the diagonals of irregular hexagons. Let the measure of this diagonal be d units. Thanks to all authors for creating a page that has been read 992,223 times. The area, A, of a rectangle is the product of its length, l, and width, w. A = l×w. Find length of diagonal of a parallelogram if given area, angle between the diagonals and other diagonal ( D d ) : diagonal of a parallelogram : = Digit 1 2 4 6 10 F A regular octagon is a geometric shape with 8 equal lengths and 8 equal angles. Now divide 64 by 2, you will get 32. Area of Regular Polygon . Enter the two side lengths and the rest will be calculated. Quadrilaterals can be classified into different types some of them are square, rectangle, parallelogram, rhombus and trapezoid. Draw a right triangle inside the shape with the diagonal as the hypotenuse. What is a regular octagon? Therefore, the diagonal of a square that is 12 feet would be around 17 feet. When the parallelogram is specified from the lengths B and C of two adjacent sides together with the length D 1 of either diagonal, then the area can be found from Heron's formula. Since each side is s, the formula is Area = s x s = s 2. Assuming it is a square or rectangle, you have to know the other dimension, too. So you have a 15-sided polygon (a pentadecagon, in case youâre curious). Review the basic formula for a square's area. To answer this question, we must find the diagonal of a rectangle that is by .Because a rectangle is made up of right angles, the diagonal of a rectangle creates a … A parallelogram is a 4-sided polygon that has two sets of parallel sides. Area of kite = product of diagonals . Find length of diagonal of a parallelogram if given area, angle between the diagonals and other diagonal ( D d ) : diagonal of a parallelogram : = Digit 1 2 4 6 10 F So you have a 15-sided polygon (a pentadecagon, in case you’re curious). How did I find the length of side of each square? A rectangle has two diagonal and they are congruent, that is both will measure the same length. Divide this by two to get the circle's radius, then use the standard area formula for circles: Area = pi * r^2. Here, the length of the base is equal to the length of perpendicular which is denoted by ‘a’ and hypotenuse is equal to the diagonal which is denoted by ‘d’. For example, if a square has sides of 7 inches, its diagonal d = 7√2 inches, or about 9.9 inches. Since each side is s, the formula is Area = s x s = s2. Solution (1) a= 11 //given (2) b=10 //AC=20, given, the diagonals of a parallelogram bisect each other so AO=10 A quadrilateral is a four-sided polygon, having the sum of interior angles equal to 360o. To calculate the diagonal of a square, multiply the length of the side by the square root of 2: d = aâ2. Area = length X width 10.5 X 10.5 = 110.25 sq ft. How do I find the surface area of a square? First two formulas i think everyone knows. By using this service, some information may be shared with YouTube. Review the basic formula for a square's area. To find the area of a square using the length of its diagonal, use the formula area = d^2 divided by 2, where d is the length of the diagonal. Diagonal of a Square $$Diagonal\ of \ square=a\sqrt{2}$$ Where, a is the length of the side of the square. Area formula The area of a square is given by the formula But since the width and height are by definition the same, the formula is usually written as where s is the length of one side. In this case, start with the formula A = s x s. Substitute (2x + 5) for each s, then simplify as much as you can. In strictly correct mathematical wording the formula above should be spoken as "s raised to the power of 2", meaning s is multiplied by itself. But if you've ever wondered whether you could find a square's area by using its diagonal(s), the answer is yes. You know what the formula for the number of diagonals in a polygon is, and you know that the polygon has 90 diagonals, so plug 90 in for the answer and solve for n: Thus, n equals 15 or –12. How do I find the area of the square in terms of x? If you've studied right triangles, you can find a new area formula that uses this diagonal as its only variable. The apothem of a regular polygon is a line segment from the centre of the polygon to the midpoint of one of its sides. The formula for the area of a square is simple -- it's side squared, or s^2. Area of a Rhombus Formula. A square's area is equal to its length times its width. In short, to find the area of a triangle, all you need to do is take the area of a rectangle formula (A = b h) and divide it by 2. The length is 12 ft if one of the sides is 12 ft. sinδ: The perimeter of a parallelogram. What is the area of a school lawn with a width of 69 meters? Draw a square on a piece of paper. Record your answer and fill in the bubbles on your answer document. The length of the diagonal can be found using the Pythagorean Theorem (a^2+b^2=c^2). If a kite has an area of 1.5 square feet and a diagonal with a length of 3 feet, what is the length of the other diagonal? Area of a kite The area of a kite is found in the same way as a rhombus, that is, by multiplying $$\frac{1}{2}$$ by the diagonal by the other diagonal . One side of the square is (2x+5) meters long. With the above equations, we can now derive various diagonal of a rectangle formulas that are used by this diagonal of a rectangle calculator: Given length and width: d … This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. The area is the amount of two-dimensional space taken up by an object is calculated using Area=1/2*(Diagonal)^2.To calculate Area of a Square when diagonal is given, you need Diagonal (d).With our tool, you need to enter the respective value for Diagonal and hit the calculate button. Example: What is the area of this circle? Area of a rectangle: A = w * l, Perimeter of a rectangle P = 2 * w + 2 * l, Circumcircle radius of a rectangle r = d/2. So, the missing diagonal ⦠For example, you can use it to calculate the area of landscape you can see when surveying, or when using perspective in photography or painting, by measuring the distance you've walked and imagining a grid with that distance as the diagonal. 144+144=c^2, 288=c^2, c=sqrt(288)=approximately 16.97. The perimeter is determined by the formula: $$P = \sqrt{\frac{32A}{\sqrt{2}+1}}$$ Diagonals. The height and base of the parallelogram should be perpendicular to each other. If your shape has four right angles, it is a rectangle. Task children with substituting the area and the known measure of the diagonal in the formula and finding the length of the missing diagonal. Area of a kite uses the same formula as the area of a rhombus. 1. The three formulas to find area depend on information you know about the rhombus. Example: What is the area of this rectangle? For a cube, we find the diagonal by using a three-dimensional version of the Pythagorean Theorem/distance formula: $\text{The formula of diagonal of cube} = s\sqrt{3}$ Where s is the side of a cube. ENTER THE TWO SIDE LENGTHS: Side 1 : clear: Side 2: clear: Area: Perimeter: Diagonal: Calculate Clear : Use the calculator above to calculate the properties of a rectangle. as a square has equal edges. Area formula using the diagonal. Length of square is 15 cm. In most of the cases, the box is an enclosed figure either a rectangle or a square. Diagonal of a Polygon Formula. Base = b = 20. Area = length × width. For a cube, we find the diagonal by using a three-dimensional version of the Pythagorean Theorem/distance formula: $\text{The formula of diagonal of cube} = s\sqrt{3}$ Where s is the side of a cube. Although it looks strange at first, you can substitute (2x + 5) for the side length in any formula. If you don't have a calculator, you can use 1.4 as an estimate for √2. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. Suppose that the diagonal is 8. A square is a 2D shape, so therefore does not have a surface area. This problem is trying to get you used to working with algebraic terms. So, the missing diagonal … Formula of rectangle circumscribed radius in terms of sine of the acute angle between the diagonals and the area of a rectangle: R = â 2A : sin β: 2: An angle between the diagonal and rectangle side An angle between the diagonal and rectangle side formulas. To learn how to find the length of a square's sides using the diagonal, scroll down! Let’s look at an example. The formula is: Area = w × h w = width h = height. Enter the length of four side(a, b, c, and d) and sum of the diagonal(A+B), and click "Calculate area of quadrangle", the area of the quadrangle is calculated from the length of four side and the sum of the diagonal, and it is displayed. Area of parallelogram(two sides and angle), Area of quadrilateral(four sides and angle), Christian era and japanese calendar conversion table, Quadrangle area(The four sides and the sum of the opposite angle). A rectangle has two diagonal and they are congruent, that is both will measure the same length. It is an irregular quadrilateral, either a trapezoid or an irregular shape with no name. Definition. Therefore, a2+ a2= d2. Just square the length of the diagonal and then divide that number by 2 to find the square's area. Last Updated: October 8, 2020 The area of the quadrangle is calculated using the formula of the sum of the length of the four sides and the diagonal. But because a polygon canât have a negative number of sides, n must be 15. Because a rectangle is made up of right angles, the diagonal of a rectangle creates a right triangle with two of the sides. Trace a copy of your first square so you have two of them. The area is the amount of two-dimensional space taken up by an object is calculated using Area=1/2*(Diagonal)^2.To calculate Area of a Square when diagonal is given, you need Diagonal (d).With our tool, you need to enter the respective value for Diagonal and hit the calculate button. The … The three formulas to find area depend on information you know about the rhombus. Differentation: Perimeter of a sector with given area (Edexcel A Level C2 Exercise 9C Q3) 2 Show that the rectangle of largest possible area, for a given perimeter, is a square. You know what the formula for the number of diagonals in a polygon is, and you know that the polygon has 90 diagonals, so plug 90 in for the answer and solve for n: Thus, n equals 15 or â12. Let us now discuss the quadrilateral formula in detail. If you have enough information to calculate the height and base of this right triangle (usually with trigonometry), you can use Pythagoras' Theorem to find the diagonal. Differentation: Perimeter of a sector with given area (Edexcel A Level C2 Exercise 9C Q3) 2 Show that the rectangle of largest possible area, for a given perimeter, is a square. [1] If the circle's diameter fits perfectly across the square, it must be equal to the side's square length. If we have the length of the diagonal, then the area can be calculated as: Area of Square = ½ × d 2 Square units Here, “d” is the length of any of the diagonal (in a square, diagonals are equal) Hereâs an example of using this formula for a kite with a long diagonal length of 4 and short diagonal length of 2. Khan Academy has a nifty drag tool that lets you see how the area of a triangle is found using the rectangle/parallelogram it's inscribed in. Review the basic formula for a square's area. There is one using the altitude and side, another using the side and angle, and one for the diagonals. Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. Let the measure of this diagonal be d units. We know w = 5 and h = 3, so: Area = 5 × 3 = 15. This can be easily understood visually - if you use the height to cut a triangle from the shape, then spin it 180° and move it to other side, it will perfectly match, and the shape will be that of a rectangle. With numerous diagonals, the task of calculating the lengths may seem daunting. What is the length of a square if one side of it is 12 feet? Note: Sometimes, base and height are used instead of length and width. This will be useful later on. Area formula The area of a square is given by the formula But since the width and height are by definition the same, the formula is usually written as where s is the length of one side. Area formula of a rectangle. Source: en.wikipedia.org. Because a right triangle is formed by the diagonal, we can use the Pythagorean Theorem, which is: To learn how to find the length of a square's sides using the diagonal, scroll down! A square has side lengths of 26.2 m, 21.4m, 27m, and 24.3m. {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/a\/a7\/Find-the-Area-of-a-Square-Using-the-Length-of-its-Diagonal-Step-3-Version-2.jpg\/v4-460px-Find-the-Area-of-a-Square-Using-the-Length-of-its-Diagonal-Step-3-Version-2.jpg","bigUrl":"\/images\/thumb\/a\/a7\/Find-the-Area-of-a-Square-Using-the-Length-of-its-Diagonal-Step-3-Version-2.jpg\/aid2751851-v4-728px-Find-the-Area-of-a-Square-Using-the-Length-of-its-Diagonal-Step-3-Version-2.jpg","smallWidth":460,"smallHeight":345,"bigWidth":"728","bigHeight":"546","licensing":" | open-web-math/open-web-math | |
# 303.9 minutes in seconds
## Result
303.9 minutes equals 18234 seconds
You can also convert 303.9 minutes to minutes and seconds.
## Conversion formula
Multiply the amount of minutes by the conversion factor to get the result in seconds:
303.9 min × 60 = 18234 s
## How to convert 303.9 minutes to seconds?
The conversion factor from minutes to seconds is 60, which means that 1 minutes is equal to 60 seconds:
1 min = 60 s
To convert 303.9 minutes into seconds we have to multiply 303.9 by the conversion factor in order to get the amount from minutes to seconds. We can also form a proportion to calculate the result:
1 min → 60 s
303.9 min → T(s)
Solve the above proportion to obtain the time T in seconds:
T(s) = 303.9 min × 60 s
T(s) = 18234 s
The final result is:
303.9 min → 18234 s
We conclude that 303.9 minutes is equivalent to 18234 seconds:
303.9 minutes = 18234 seconds
## Result approximation:
For practical purposes we can round our final result to an approximate numerical value. In this case three hundred three point nine minutes is approximately eighteen thousand two hundred thirty-four seconds:
303.9 minutes ≅ 18234 seconds
## Conversion table
For quick reference purposes, below is the minutes to seconds conversion table:
minutes (min) seconds (s)
304.9 minutes 18294 seconds
305.9 minutes 18354 seconds
306.9 minutes 18414 seconds
307.9 minutes 18474 seconds
308.9 minutes 18534 seconds
309.9 minutes 18594 seconds
310.9 minutes 18654 seconds
311.9 minutes 18714 seconds
312.9 minutes 18774 seconds
313.9 minutes 18834 seconds
## Units definitions
The units involved in this conversion are minutes and seconds. This is how they are defined:
### Minutes
The minute is a unit of time or of angle. As a unit of time, the minute (symbol: min) is equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). As a unit of angle, the minute of arc is equal to 1⁄60 of a degree, or 60 seconds (of arc). Although not an SI unit for either time or angle, the minute is accepted for use with SI units for both. The SI symbols for minute or minutes are min for time measurement, and the prime symbol after a number, e.g. 5′, for angle measurement. The prime is also sometimes used informally to denote minutes of time. In contrast to the hour, the minute (and the second) does not have a clear historical background. What is traceable only is that it started being recorded in the Middle Ages due to the ability of construction of "precision" timepieces (mechanical and water clocks). However, no consistent records of the origin for the division as 1⁄60 part of the hour (and the second 1⁄60 of the minute) have ever been found, despite many speculations.
### Seconds
The second (symbol: s) (abbreviated s or sec) is the base unit of time in the International System of Units (SI). It is qualitatively defined as the second division of the hour by sixty, the first division by sixty being the minute. The SI definition of second is "the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom". Seconds may be measured using a mechanical, electrical or an atomic clock. SI prefixes are combined with the word second to denote subdivisions of the second, e.g., the millisecond (one thousandth of a second), the microsecond (one millionth of a second), and the nanosecond (one billionth of a second). Though SI prefixes may also be used to form multiples of the second such as kilosecond (one thousand seconds), such units are rarely used in practice. The more common larger non-SI units of time are not formed by powers of ten; instead, the second is multiplied by 60 to form a minute, which is multiplied by 60 to form an hour, which is multiplied by 24 to form a day. The second is also the base unit of time in other systems of measurement: the centimetre–gram–second, metre–kilogram–second, metre–tonne–second, and foot–pound–second systems of units. | HuggingFaceTB/finemath | |
### Problem Statement
A binary string is a non-empty finite sequence of 0's and 1's. Given two binary strings, u and v, their concatenation, u * v, is defined to be the binary string obtained by appending v to the end of u. For example, if u = 01100 and v = 110, then u * v = 01100110.
Consider a function, h, that maps binary strings to other binary strings. Suppose that for every string u with digits a1, a2, ..., ak in that order, it is true that h(u) = h(a1) * h(a2) * ... * h(ak). Then, h is called a non-degenerate homomorphism. In general, this means that h is uniquely determined by the values of h(0) and h(1). For example, if h(0) = 001, and h(1) = 10, then,
• - h(110) = h(1) * h(1) * h(0) = 1010001.
• - h(00) = h(0) * h(0) = 001001.
• - h(0101) = h(0) * h(1) * h(0) * h(1) = 0011000110.
Create a class Homomorphism that contains a method count, which is a given a String u and a String v. The method should return the number of distinct non-degenerate homomorphisms, h, which satisfy h(u) = v. If there are infinitely many such non-degenerate homomorphisms, the method should return -1.
### Definition
Class: Homomorphism Method: count Parameters: String, String Returns: int Method signature: int count(String u, String v) (be sure your method is public)
### Notes
-Two non-degenerate homomorphisms, h and h' are considered distinct if and only if h(u) is different from h'(u) for at least one binary string, u.
### Constraints
-u and v will each contain between 1 and 50 characters inclusive.
-Each character in both u and v will be either '0' or '1'.
### Examples
0)
`"10"` `"11001"`
`Returns: 4`
Since h(0) and h(1) cannot be empty strings, there are precisely 4 legal non-degenerate homomorphisms: 1. h(1) = 1 and h(0) = 1001. 2. h(1) = 11 and h(0) = 001. 3. h(1) = 110 and h(0) = 01. 4. h(1) = 1100 and h(0) = 1.
1)
`"00"` `"11111"`
`Returns: 0`
If h is a valid non-degenerate homomorphism, then h(00) = h(0) * h(0), which implies that h(00) has an even length. Thus, there are no non-degenerate homomorphisms satisfying h(00) = 11111.
2)
`"11"` `"00"`
`Returns: -1`
Any non-degenerate homomorphism, h, satisfying h(1) = 0 will also satisfy h(11) = 00. This leaves no restrictions at all on h(0), so there are an infinite number of such non-degenerate homomorphisms.
3)
`"001"` `"1010001"`
`Returns: 1`
The unique non-degenerate homomorphism, h, satisfying h(001) = 1010001 is given by h(0) = 10 and h(1) = 001.
4)
`"101"` `"11111111111111111111111111111111111111111111111111"`
`Returns: 24`
#### Problem url:
http://www.topcoder.com/stat?c=problem_statement&pm=2363
#### Problem stats url:
http://www.topcoder.com/tc?module=ProblemDetail&rd=4770&pm=2363
dgarthur
#### Testers:
lbackstrom , brett1479
#### Problem categories:
Brute Force, String Manipulation | HuggingFaceTB/finemath | |
# Water flow down a 9% grade for 750' then fills tanks before traveling up 20' over 175'
I'm in hopes that someone far smarter than myself can tell me the answer to my questions.
I am setting up a water capture system for a spring. The spring drains into a pond where I am installing a 525 gallon tank that will essentially be underwater. That tank will feed a 2 inch pipe running 750 feet down a 9% grade where it will fill two consecutive 1700 gallon tanks. The 2 inch pipe will then continue it's run another 175 feet but now going up 20 feet in elevation in that 175 feet before entering the house. Once inside it will downsize to 1" pipe. The two 1700 gallon tanks will be sealed.
Will I have enough pressure from the long 750 foot drop or will I need to put in a pressure pump?
What will the pressure be once it enters the house?
Thank you to anyone who works on my problem and is able to answer my questions.
Be well, Mark
• Static pressure depends on head which is height difference. If you need to know dynamic pressure then you need to specify a flow rate. Your underwater tank will experience an uplift of 1 kg/L (10 N) if there is air in it. You may need to take that into consideration when anchoring it. Apr 25 at 18:09
• Assume the two 1700 gallons tank are full at all time, by gravity flow, you can figure out the flow rate Q1 at the inlet by the head H1, then Q1 becomes the flow rate at the downstream pipe, which has a negative flow rate due to the head H2. Now you can write the continuity equation to get the end flow rate Q2 and determine whether a pump is required to increase the flow rate for the service. Note, you need to include friction losses in the calculations.
– r13
Apr 25 at 23:17
I am not attempted to solve this problem but to express my understanding as a starter. If anything wrong please advise.
Equations:
$$v = \sqrt2g*H$$, Discharge Velocity
$$Q = V*A$$, Flow Rate
$$H_f = f*\frac{L}D*\frac{v^2}2g$$, Friction Head Loss, in which
$$f - f(Re, \frac{\epsilon}D)$$, The Moody, Darcy, or Stanton friction factor, and
$$\epsilon -$$ Roughness Factor of the Pipe
Assume the diagram below is correct, then you can calculate the discharge velocity and flow rate at the points of interest as shown. Then from the output flow rare $$Q_4$$ you can decide whether a pump is required to increase the flow. Note that from point 3 to point 4, the flow is in reversed direction. | HuggingFaceTB/finemath | |
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### Nim-interactive
##### Stage: 3 and 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter.
### Estimating Angles
##### Stage: 2, 3 and 4 Challenge Level:
How good are you at estimating angles?
### Tower Rescue
##### Stage: 2, 3, 4 and 5 Challenge Level:
Help the bee to build a stack of blocks far enough to save his friend trapped in the tower.
### The Triangle Game
##### Stage: 3 and 4 Challenge Level:
Can you discover whether this is a fair game?
### One, Three, Five, Seven
##### Stage: 3 and 4 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Pentanim
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### Nim
##### Stage: 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
### Nim-like Games
##### Stage: 2, 3 and 4 Challenge Level:
A collection of games on the NIM theme
### Connect Three
##### Stage: 3 and 4 Challenge Level:
Can you be the first to complete a row of three?
### Stop or Dare
##### Stage: 2, 3 and 4 Challenge Level:
All you need for this game is a pack of cards. While you play the game, think about strategies that will increase your chances of winning.
### Snail Trails
##### Stage: 3 Challenge Level:
This is a game for two players. You will need some small-square grid paper, a die and two felt-tip pens or highlighters. Players take turns to roll the die, then move that number of squares in. . . .
### Winning Lines
##### Stage: 2, 3 and 4
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Jam
##### Stage: 4 Challenge Level:
A game for 2 players
### Sprouts Explained
##### Stage: 2, 3, 4 and 5
This article invites you to get familiar with a strategic game called "sprouts". The game is simple enough for younger children to understand, and has also provided experienced mathematicians with. . . .
### Intersection Sudoku 1
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
### Ratio Sudoku 2
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
### 18 Hole Light Golf
##### Stage: 1, 2, 3 and 4 Challenge Level:
The computer starts with all the lights off, but then clicks 3, 4 or 5 times at random, leaving some lights on. Can you switch them off again?
### Square It
##### Stage: 3 and 4 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Learning Mathematics Through Games Series: 4. from Strategy Games
##### Stage: 1, 2 and 3
Basic strategy games are particularly suitable as starting points for investigations. Players instinctively try to discover a winning strategy, and usually the best way to do this is to analyse. . . .
### Spiralling Decimals
##### Stage: 2 and 3 Challenge Level:
Take turns to place a decimal number on the spiral. Can you get three consecutive numbers?
### Spiralling Decimals for Two
##### Stage: 2 and 3 Challenge Level:
Spiralling Decimals game for an adult and child. Can you get three decimals next to each other on the spiral before your partner?
### Have You Got It?
##### Stage: 3 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### Patience
##### Stage: 3 Challenge Level:
A simple game of patience which often comes out. Can you explain why?
### Behind the Rules of Go
##### Stage: 4 and 5
This article explains the use of the idea of connectedness in networks, in two different ways, to bring into focus the basics of the game of Go, namely capture and territory.
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Troublesome Triangles
##### Stage: 2 and 3 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .
### Diamond Mine
##### Stage: 3 Challenge Level:
Practise your diamond mining skills and your x,y coordination in this homage to Pacman.
##### Stage: 3 and 4 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### Twin Corresponding Sudokus II
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
##### Stage: 1, 2, 3 and 4 Challenge Level:
Advent Calendar 2010 - a mathematical game for every day during the run-up to Christmas.
### The Remainders Game
##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
### Twin Corresponding Sudoku III
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Charlie's Delightful Machine
##### Stage: 3 and 4 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### Dicey Operations
##### Stage: 3 Challenge Level:
Who said that adding, subtracting, multiplying and dividing couldn't be fun?
### Nice or Nasty
##### Stage: 2 and 3 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Diamond Collector
##### Stage: 3 Challenge Level:
Collect as many diamonds as you can by drawing three straight lines.
##### Stage: 3 Challenge Level:
A game for 2 or more people, based on the traditional card game Rummy. Players aim to make two `tricks', where each trick has to consist of a picture of a shape, a name that describes that shape, and. . . .
### Tangram Pictures
##### Stage: 1, 2 and 3 Challenge Level:
Use the tangram pieces to make our pictures, or to design some of your own!
### Seasonal Twin Sudokus
##### Stage: 3 and 4 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it? | HuggingFaceTB/finemath | |
# What is a general solution to the differential equation y'=2+2x^2+y+x^2y?
Jul 24, 2016
$y = C {e}^{x + {x}^{3} / 3} - 2$
#### Explanation:
$y ' = 2 + 2 {x}^{2} + y + {x}^{2} y$
this is separable,
$y ' = 2 \left(1 + {x}^{2}\right) + y \left(1 + {x}^{2}\right)$
$y ' = \left(2 + y\right) \left(1 + {x}^{2}\right)$
$\frac{1}{2 + y} y ' = \left(1 + {x}^{2}\right)$
$\int \setminus \frac{1}{2 + y} y ' \setminus \mathrm{dx} = \int \setminus \left(1 + {x}^{2}\right) \setminus \mathrm{dx}$
$\int \setminus \frac{d}{\mathrm{dx}} \left(\ln \left(2 + y\right)\right) \setminus \mathrm{dx} = \int \setminus \left(1 + {x}^{2}\right) \setminus \mathrm{dx}$
$\ln \left(2 + y\right) = x + {x}^{3} / 3 + C$
$2 + y = {e}^{x + {x}^{3} / 3 + C}$
$2 + y = C {e}^{x + {x}^{3} / 3}$
$y = C {e}^{x + {x}^{3} / 3} - 2$ | HuggingFaceTB/finemath | |
Engineering Questions with Answers - Multiple Choice Questions
# Computational Fluid Dynamics – Turbulence Modelling – Averaging Rules
1 - Question
These rules for averaging are used to average ___________
a) fluctuations in the turbulent flow
b) variation in results of turbulent flow
c) the coefficients in FVM
d) the coefficients in FDM
Explanation: The flow variables in a turbulent flow are divided into mean and fluctuating components. These fluctuating components in the turbulent flow are averaged for the further solution of the system. These rules are used for averaging.
2 - Question
According to the rules for averaging, which of these will sum up to zero?
a) The mean component of the flow variable
b) The fluctuating component of the flow variable
c) The flow variable
d) Integration of the flow variable
Explanation: The mean component of a flow variable is the overall average of the flow variable. So, when the average of the flow variable is its mean component, the average of the fluctuating component and hence its summation will be zero.
3 - Question
The average of the mean component will be ____________
a) equal to zero
b) equal to the mean component itself
c) equal to 1
d) equal to the fluctuating component
Explanation: The mean component is already found by taking the arithmetic mean (average) of the flow variables. So, if the average of only the mean component is taken, it will again be the same mean component itself.
4 - Question
The mean of the spatial partial derivative of a flow variable will be equal to ____________
a) 0
b) 1
c) the spatial partial derivative of the mean component
d) the mean component
Explanation: The mean of the flow variable will be equal to the mean variable. The mean of the flow variable’s spatial partial derivative will be equal to the spatial partial derivative of the mean component of that variable.
5 - Question
The mean of the summation of two flow variables will be equal to ____________
a) the summation of their mean components – the summation of the mean of their fluctuating components
b) the summation of their mean components + the summation of the mean of their fluctuating components
c) the summation of their fluctuating components
d) the summation of their mean components
Explanation: Consider two flow variables which can be decomposed as a=A+a’ and b=B+b’. The mean of their summation means a+b = A+a’+B+b’ But, a’ = 0 and b’ = 0. Therefore, a+b = A+B Also, A = A and B = B. Hence, a+b = A+B.
6 - Question
The mean of the space-based integral of a flow variable is equal to ____________
a) the summation of its mean component
b) the space-based integral of its fluctuating component
c) the space-based integral of its mean component
d) the summation of its fluctuating components
Explanation: As the mean of the fluctuating component is zero and the mean of the mean component is the mean component itself, the mean of the space-based integral of a flow variable is equal to the space-based integral of its mean component alone.
7 - Question
The mean of the product of the mean component of one variable and the fluctuating component of another variable is ____________
a) 1
b) 0
c) the product of their mean components
d) the product of their fluctuating components
Explanation: The mean of a fluctuating component is zero. The mean of a mean component is a variable. So, the mean of the product of the mean component of one variable and the fluctuating component of another variable will become zero.
8 - Question
The mean of the product of a flow variable and the mean component of another flow variable is ____________
a) the product of their mean components
b) the product of their fluctuating components
c) the mean of the product of their mean components
d) the mean of the product of their fluctuating components
Explanation: Consider two flow variables which can be decomposed as a=A+a’ and b=B+b’. The mean of the product of one flow variable and the mean component of another flow variable is represented as aB=(A+a’)B aB=AB+a’B As a’B=0 and AB=AB, aB=AB.
9 - Question
Consider a vector flow variable which can be decomposed as a⃗ =A⃗ +a.diva⃗ ¯¯¯¯¯¯¯¯¯¯ will be equal to ____________
a) div A⃗
b) diva¯¯¯¯¯¯¯¯¯¯¯¯
c) diva⃗ ¯¯¯¯¯¯¯¯¯
d) divA⃗ ¯¯¯¯¯¯¯¯¯¯¯
Explanation: From the given problem,
diva⃗ ¯¯¯¯¯¯¯¯¯¯=diva⃗ ¯¯¯=divA⃗ +a¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=divA⃗ ¯¯¯¯=divA⃗ .
10 - Question
Consider two flow variables which can be decomposed as a=A+a’ and b=B+b’. What is ab?
a) 0
b) 1
c) AB
d) a’b’ | HuggingFaceTB/finemath | |
Tools
Text
# Define a Vector – Linear Algebra
Lesson 8 Chapter 3
Let's talk about how we define a vector with Numpy. Assume we would like to define a column vector with has a size of . Take a careful look to the code below and the shape of the arrays:
# Import Numpy library
import numpy as np
# Rank-1 array
v = np.array([0,8])
print('Shape: ', v.shape)
# Rank-2 array (row vector)
v = np.array([[0,8]])
print('Shape: ', v.shape)
# Rank-2 array (column vector)
v = np.array([,])
print('Shape: ', v.shape)
In the above code, we defined the same arrays in terms of numeric values with different ranks and shape. At line 7, we defined a rank-1 array (has only one dimension). At 11, we defined a rank-2 array which is a row vector (1 row and multiple columns). For defining vectors, the preference is how we did at line 15 which results in a rank-2 array and a column vector (multiple rows and 1 column). The output of the above code is as below:
Shape: (2,)
Shape: (1, 2)
Shape: (2, 1)
Remember we do NOT usually need to define vectors as we did in lines 11 or 15. That approach seemed to be a little bit complicated using all those sorts of nested Python lists! Now let's do it the easy way:
import numpy as np
# Rank-1 array
v = np.array([0,8])
print('Shape: ', v.shape)
# Rank-1 array (row vector)
row_v = v.reshape(1,-1)
print('Shape: ', row_v.shape)
# Rank-1 array (column vector)
column_v = v.reshape(-1,1)
print('Shape: ', column_v.shape)
What I did above? (1) I used "-1" as it indicates all rows (columns). (2) I used the Numpy "reshape" method which simply changes the shape of the array to the desired shape (details later in this tutorial). (3) I used "1" indexing which indicates one!
Let me explain the line 8 of the above code for further illustration. (1) "-1" is the total columns which are the total elements of the vector , equals 2. (2) Numpy "reshape" method changes the shape to (1,2) which means the new vector ( ) has 2 columns and only one row! It is worth to emphasize is a row vector as it only has one row.
NOTE: In simple words, (1) (1,-1) means put only one row and place all elements in columns and (-1,1) means put only one column and place all elements in rows. Check the below figure. | HuggingFaceTB/finemath | |
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# Center of Mass and Linear Momentum MCQ Quiz - Objective Question with Answer for Center of Mass and Linear Momentum - Download Free PDF
Last updated on Feb 2, 2023
## Latest Center of Mass and Linear Momentum MCQ Objective Questions
#### Center of Mass and Linear Momentum Question 1:
Kinetic energy will become/remain________ when the momentum of the body is doubled.
1. two times
2. four times
3. unchanged
4. half
5. None of the above
Option 2 : four times
#### Center of Mass and Linear Momentum Question 1 Detailed Solution
CONCEPT:
• Kinetic energy (KE): The energy possessed by a body by virtue of its motion is called kinetic energy.
$$KE = \frac{1}{2}m{v^2}$$
Where m = mass of the body and v = velocity of the body
• Momentum (p): It is the product of the mass and velocity of an object.
p = m v
• The relationship between the kinetic energy (KE) and Linear momentum (p) is
$$p = \sqrt {2mKE}$$
Or KE = P2/2m
EXPLANATION:
KE = P2/2m
• Since the mass of an object is constant, therefore we can say that Kinetic energy is directly proportional to the square of the Momentum.
• Hence, when momentum is doubled, kinetic energy will become four times. Sp option 2 is correct.
#### Center of Mass and Linear Momentum Question 2:
A bullet of mass 25 g is fired horizontally with a velocity of 120 m/s from a pistol of 2·5 kg. The recoil velocity of the pistol is
1. - 0·6 m/s
2. 0·6 m/s
3. - 1·2 m/s
4. 1·2 m/s
Option 3 : - 1·2 m/s
#### Center of Mass and Linear Momentum Question 2 Detailed Solution
Concept:
Momentum:
• It is defined as the product of mass and velocity.
• Formula, P = mv, where m = mass, v = velocity
• Its SI unit is kg m/s.
Conservation of linear momentum:
• It states that if two objects collide, then the total momentum before and after the collision will be the same if there is no external force acting on the colliding objects.
• Formula, m1u+ m2u2 = m1v1 + m2 v2
Calculation:
Given,
The mass of the bullet, m = 25g
The velocity of the bullet, v = 120 m/s
The mass of the pistol, M = 2.5 kg
From the conservation of the momentum,
mv + MV= 0
$$V=-\frac{mv}{M}$$
$$V=-\frac{25\times 10^{-3}\times 120}{2.5}=-1.2 m/s$$
Hence, the recoil velocity is -1.2 m/s.
#### Center of Mass and Linear Momentum Question 3:
A heavy particle of rest mass M while moving along the positive z - direction, decays into two identical light particles with rest mass m (where M > 2m). The maximum value of the momentum that any one of the lighter particles can have in a direction perpendicular to the z-direction, is
1. $$\rm\frac{1}{2}c\sqrt{M^2−4m^2}$$
2. $$\rm\frac{1}{2}c\sqrt{M^2−2m^2}$$
3. $$\rm c\sqrt{M^2−4m^2}$$
4. $$\frac{1}{2}$$Mc
Option 1 : $$\rm\frac{1}{2}c\sqrt{M^2−4m^2}$$
#### Center of Mass and Linear Momentum Question 3 Detailed Solution
Concept:
We know,
E = mc2 where m is the mass of the particle and c is the speed of light.
The invariant mass (or rest mass) is an invariant for all frames of reference, not just in inertial frames in flat spacetime, but also accelerated frames traveling through curved spacetime.
The energy of mass M,
E = $$\sqrt{P^2c^2 + M^2c^2}$$
where p is the momentum and c is the speed of light.
Calculation:
Let P be the momentum of heavy mass M . And let P1 be the momentum of the light particles of mass m in the direction perpendicular to z and P2 be the momentum in z -direction. According to the conservation of momentum, the Momentum of mass M,
P = P2 + P2 = 2 P2
⇒ P2 = P/2
The energy of mass M,
E = $$\sqrt{P^2c^2 + M^2c^2}$$
The momentum of a mass m,
P = $$\sqrt{P^2_1+P^2_2}$$
$$\sqrt{P^2_1+ {P^2\over 4}}$$
The energy of mass m,
E21 = (P21 + $${P^2\over 4}$$)c2 + m2c2
As energy is conserved
E = E1 + E2
= 2E1
⇒ E1 = E/2
∴ E21 = $$E^2\over 4$$
= (P12 + $${P^2 \over 4}$$)c2 + m2c4
⇒ 4 (P12 + $${P^2 \over 4}$$)c2 + 4 m2c4 = P2c2 + M2c4
∴ P1 = $$\rm\frac{1}{2}c\sqrt{M^2−4m^2}$$
The correct answer is option (1).
#### Center of Mass and Linear Momentum Question 4:
Two bodies of unequal masses are dropped from a cliff. At any instant, they have equal
1. Momenta
2. Accelerations
3. Potential energies
4. Kinetic energies
Option 2 : Accelerations
#### Center of Mass and Linear Momentum Question 4 Detailed Solution
Key Points:
• Acceleration is the rate at which speed and direction of velocity vary over time.
• $$a = \frac{dv}{dt}$$
• Acceleration is a vector quantity.
• One of the vector quantities is acceleration.
• Given that the two bodies' masses differ.
• Let's calculate the body's travel time to the ground given that the height is the same, h.
• Knowing this,
• $$h = {1 \over 2}gt^2$$
• $$t= { \sqrt{2h} \over g}$$
• We can infer from the foregoing that time is independent of mass.
• As a result, both bodies have an equal amount of time to reach the ground.
• Due to gravity, they also experience the same acceleration.
• Equal time is given to each.
• Due to gravity, they also experience the same acceleration.
• Momentum:
• The relationship between a particle's mass (m) and velocity is known as its momentum (v).
• P = mv
• Momentum is vector quantity.
Potential energy:
• The energy a body acquires as a result of its position or configuration is known as potential energy.
• The equation PE = mgh describes the potential energy attributable to an object's height.
• Where m is the body's mass, g is its gravitational acceleration, and h is its height
• Kinetic energy:
• Kinetic energy is the energy that an object produces as result of its motion.
• $$K.E = \frac 12 mv^2$$
• where m is the body's mass and v is its velocity.
#### Center of Mass and Linear Momentum Question 5:
A ball of mass M moving with a speed of 2 m/s hits another ball of mass 1 kg moving in the same direction with a speed of 1 m/s. If the kinetic energy of center of mass is $$\frac{4}{3}$$ Joule, then the magnitude of M is
1. 1 kg
2. 0.25 kg
3. 0.50 kg
4. 2 kg
Option 3 : 0.50 kg
#### Center of Mass and Linear Momentum Question 5 Detailed Solution
CONCEPT:
As we know;
$$v_{CM} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$$
CALCULATION:
Given: mass $$m_1 = M$$
$$m_2 = 1$$ kg
As we have;
$$v_{CM} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$$
⇒ $$v_{CM} = \frac{2M+1}{M+1}$$
The kinetics of the center of mass is $$\frac{4}{3}$$ Joule, therefore we have;
$$\frac{4}{3} =\frac{1}{2}(M+1) (\frac{2M+1}{M+1})^2$$
Now solving the above equation we have;
M = 0.50 kg
Hence option 3) is the correct answer.
## Top Center of Mass and Linear Momentum MCQ Objective Questions
#### Center of Mass and Linear Momentum Question 6
If two unequal masses possess the same kinetic energy, then the heavier mass has:
1. equal momentum
2. lesser momentum
3. greater momentum
4. Can't say
Option 3 : greater momentum
#### Center of Mass and Linear Momentum Question 6 Detailed Solution
CONCEPT:
• Kinetic energy (K.E): The energy possessed by a body by the virtue of its motion is called kinetic energy.
The expression for kinetic energy is given by:
$$KE = \frac{1}{2}m{v^2}$$
Where m = mass of the body and v = velocity of the body
• Momentum (p): The product of mass and velocity is called momentum.
Momentum (p) = mass (m) × velocity (v)
The relationship between the kinetic energy and Linear momentum is given by:
As we know,
$$KE = \frac{1}{2}m{v^2}$$
Divide numerator and denominator by m, we get
$$KE = \frac{1}{2}\frac{{{m^2}{v^2}}}{m} = \frac{1}{2}\frac{{\;{{\left( {mv} \right)}^2}}}{m} = \frac{1}{2}\frac{{{p^2}}}{m}\;$$ [p = mv]
$$\therefore KE = \frac{1}{2}\frac{{{p^2}}}{m}\;$$
$$p = \sqrt {2mKE}$$
EXPLANATION:
Given that:
K.E1 = K.E2= K.E (let say)
The relation between the momentum and the kinetic energy is given by:
$$P = \sqrt {2m\;K.E}$$
But as K.E is same
∴ $$P \propto \sqrt m$$
Or, $$\frac{{{P_1}}}{{{P_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}}$$
Here, if m1 > m2 then p> p2
• If two unequal masses possess the same kinetic energy, then the heavier mass has greater momentum. So option 3 is correct.
#### Center of Mass and Linear Momentum Question 7
A sphere of mass 2kg strikes another sphere of mass 3 kg at rest with a velocity of 5 m/s. if they move together after collision. What is their common velocity?
1. 5 m/s
2. 6 m/s
3. 1 m/s
4. 2 m/s
Option 4 : 2 m/s
#### Center of Mass and Linear Momentum Question 7 Detailed Solution
Concept:
Momentum:
momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.
• The unit of momentum (P) is kg m/s.
• Dimension: [MLT-1]
Law of conservation of Momentum:
A conservation law states that the total linear momentum of a closed system remains constant through time, regardless of other possible changes within the system.
P1 = P2
m1 v1 = m2 v2
Where, P1 = initial momentum of system, P2 = final momentum of system, m1 = mass of first object, v1 = velocity of first object, m2 = mass of second object and v2 = velocity of second object.
Calculation:
Given: m1 = 2 kg m2 = 3 kg u1 = 5 m/s u2 = 0 m/s
Let the common velocity of the combined body be V m/s
Mass of combined body M = 2 + 3 = 5 kg
Applying conservation of momentum:
m1 v1 + m2 v2 = M V
⇒ (2 × 5) + (3 × 0) = 5 V
⇒ 10 + 0 = 5 V
V = 2 m/s
Hence the combined velocity of both the spheres is 2 m/s.
#### Center of Mass and Linear Momentum Question 8
A 30 N of force is acting on a body moving on a straight line with initial momentum 10 kg m s-1. Find the final momentum after 3 seconds.
1. 100 Kg m s-1
2. 90 kg m s-1
3. 120 kg m s-1
4. 110 kg m s-1
Option 1 : 100 Kg m s-1
#### Center of Mass and Linear Momentum Question 8 Detailed Solution
Concept:
Second Law of Motion:
• The rate of change of momentum is directly proportional to the applied force.
$$⇒ F = \dfrac{Δ p}{t}$$
⇒ F × t = Δ p
Where F is force, t is time, and Δ p is change in momentum.
Calculation:
Given: Initial momentum (pi) = 10 kg m s-1, Force (F) = 30 N, and time (t) = 3 second
Let final momentum be pf
• Change in momentum is
⇒ Δ p = F × t
⇒ Δ p = 30 N × 3 s = 90 N s = 90 kg m s-1
• As we know the change in momentum is equal to the difference between the final momentum and initial momentum i.e.,
⇒ Δ p = pf - pi
⇒ pf = Δ p + pi
⇒ pf = 90 kg m s-1 + 10 kg m s-1 = 100 kg m s-1
• So, the correct option is 100 kg m s-1
#### Center of Mass and Linear Momentum Question 9
If the velocity of a body is doubled, its momentum ________.
1. remains same
2. doubles
3. becomes half
4. becomes 4 times
Option 2 : doubles
#### Center of Mass and Linear Momentum Question 9 Detailed Solution
CONCEPT:
• Momentum (P): The product of mass and velocity is called momentum.
• The SI unit of momentum is kg m/s.
• Momentum (P) = Mass (m) × Velocity (v)
EXPLANATION:
Since, P = m v
Let the initial velocity of the body be v
Since the Mass of the body is constant
So, P1 = m v ----(i)
According to the question
The new velocity of the body = 2v
New Momentum (P2) = m × 2v
⇒ P= 2mv ----(ii)
On dividing (ii) by (i), we get
$$\frac {P_2} {P_1} = \frac{2mv}{mv}$$
⇒ P2 = 2 P1
∴ The momentum will be doubled.
Key Points
• If the velocity of a body is doubled then its momentum doubles because velocity is directly proportional to momentum. So option 2 is correct.
#### Center of Mass and Linear Momentum Question 10
A 2,000 kg truck moving at 10 m/s strikes a car waiting at a traffic light. After collision, the two move together at 8 m/s. The mass of the car is _____.
1. 250 kg
2. 500 kg
3. 750 kg
4. 1,000 kg
Option 2 : 500 kg
#### Center of Mass and Linear Momentum Question 10 Detailed Solution
The correct answer is 500 kg.
CONCEPT:
• The type of collision in which there is a loss of kinetic energy and after the collision both the colliding particles move together is called perfectly inelastic collision.
• In this type of collision the coefficient of restitution is equal to 0.
• Momentum of the system remains constant in this collision.
• Momentum before collision (P1) = Momentum after the collision (P2)
P1 = m1u1 + m2u2
P2 = m1v1 + m2v2
CALCULATION:
Here, speed of the car is 0 m/s
⇒ After collision speed is 8 m/s
According to "Principle of conservation of momentum"
⇒ So, 8 = (m1v1 + m2v2)/(m1 + m2) where m1 is the mass of truck, v1 is the velocity of truck, v2 is the velocity of car and m2 is the mass of car.
⇒ 8 = (2000 × 10 + m2 × 0)/(2000 + m2)
⇒ 16000 + 8 m2 = 20000
Therefore, mass of the car is 500 kg.
#### Center of Mass and Linear Momentum Question 11
If the momentum of a particle increases by 30%, then increase in its kinetic energy is:
1. 30%
2. 60%
3. 69%
4. 80%
Option 3 : 69%
#### Center of Mass and Linear Momentum Question 11 Detailed Solution
CONCEPT:
• Momentum: A property of a body in motion that is equal to the product of the body's mass and velocity is called momentum.
P = mv
where P is the momentum of the body, m is the mass of the body, and v is the velocity of the body.
• Momentum Conservation: When in a system, there is no external force then the total momentum (P) of the system will be conserved.
• Kinetic energy: The energy in a body due to its motion, is known as kinetic energy.
The kinetic energy in terms of momentum is given by:
$$K=\frac{P^2}{2m}$$
where K is the kinetic energy of the body, P is the momentum of the body and m is the mass of the body.
CALCULATION:
$$K=\frac{P^2}{2M}$$
Let initial value of P = 100, m = 100
After increase of 30% P = 130
Initial kinetic energy $$K_i=\frac{P_i^2}{2m}=\frac{100^2}{2\times 100}=50$$
Final kinetic energy $$K_f=\frac{P_f^2}{2m}=\frac{130^2}{2\times 100}=169/2$$
% Increase in Kinetic energy $$\frac{(K_f-K_i)}{K_i}\times100 = \frac{(\frac{169}{2}-50)}{50}\times100 = 69$$%
So the correct answer is option 3.
#### Center of Mass and Linear Momentum Question 12
What will be the magnitude of recoil speed of a tank which weighs 0.4 tonnes and fires a projectile of mass 3 kg with a velocity of 24 m/s?
1. 0
2. 0.18
3. 18
Option 2 : 0.18
#### Center of Mass and Linear Momentum Question 12 Detailed Solution
CONCEPT:
• Momentum: Momentum is the product of the mass and the velocity of an object or particle. It is a vector quantity.
• The standard unit of momentum magnitude is the kilogram-meter per second (kg·m/s).
P = m v
Where P = momentum, m = mass of the body, v = velocity of body.
• Conservation of Linear Momentum: Conservation of Linear Momentum states that a body in motion retains its total momentum (product of mass and vector velocityunless an external force is applied to it.
• In mathematical term
Initial momentum = Final momentum
P= P2
We can say that
M1V1 + M2V= 0
Where M1 = mass of projectile, V1 = velocity of projectile, M2 = mass of tank and V2 = Unknown recoil speed of tank
CALCULATION:
Given that M1 = 3 kg, V= 24 m/s, M2 = 0.4 tonnes = 400 kg
Substituting these values in the formula:
M1V1 + M2V= 0
(3 × 24) + (400 × V2) = 0
V2 = -0.18 m/s,
The tank recoils in the backward direction with a magnitude of 0.18 m/s.
So option 2 is correct.
#### Center of Mass and Linear Momentum Question 13
What is the momentum of an object of mass $$\frac{1}{2}m$$ and moving with a velocity of 2V?
1. $$\frac{1}{2}mv$$
2. mv2
3. (mv)2
4. mv
Option 4 : mv
#### Center of Mass and Linear Momentum Question 13 Detailed Solution
CONCEPT:
• Momentum (p): The product of mass and velocity is called momentum. It is a vector quantity.
• The SI unit of momentum is kg m/s.
Momentum (p) = Mass (m') × Velocity (v')
EXPLANATION:
Given that:
The mass of an object (m') is ½ m
The velocity (v') of an object is 2v
We know that momentum (p) = mass × velocity = ((m/2) × 2 v) = m v
• The momentum of the object is mv kg m/s. Hence option 4 is correct.
#### Center of Mass and Linear Momentum Question 14
Two bodies of 2 kg and 4 kg are moving with velocities 20 m/s and 10 m/s respectively towards each other under mutual gravitational attraction. Find the velocity of their centre of mass in m/s.
1. 5
2. 6
3. 8
4. zero
Option 4 : zero
#### Center of Mass and Linear Momentum Question 14 Detailed Solution
Concept:
position of the centre of mass:
$${x_{cm}} = \frac{{{m_1}{x_1} + {m_2}{x_2} + \ldots + {m_n}{x_n}}}{{{m_1} + {m_2} + \ldots + {m_n}}}$$
The velocity of the centre of mass:
$${V_{cm}} = \frac{{{m_1}{v_1} + {m_2}{v_2} + \ldots + {m_n}{v_n}}}{{{m_1} + {m_2} + \ldots + {m_n}}}$$
Acceleration of the centre of mass:
$${a_{cm}} = \frac{{{m_1}{a_1} + {m_2}{a_2} + \ldots + {m_n}{a_n}}}{{{m_1} + {m_2} + \ldots + {m_n}}}$$
Calculations:
$${\vec V_{cm}} = \frac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}$$
$$= \frac{{2 \times 20 - 4 \times 10}}{{2 + 4}}$$ (Negative because both are moving in the opposite direction)
= 0
#### Center of Mass and Linear Momentum Question 15
If the momentum of a body is tripled keeping the mass as constant, its Kinetic Energy will ________.
1. remain the same
2. become nine times its original value
3. become six times its original value
4. become three times its original value
Option 2 : become nine times its original value
#### Center of Mass and Linear Momentum Question 15 Detailed Solution
CONCEPT:
• Kinetic energy: The energy of an object due to its motion is called kinetic energy.
• It is denoted by KE.
• Mathematically kinetic energy can be written as
• Kinetic energy (KE) = 1/2 mv2
• Momentum: The product of mass and velocity is called momentum or linear momentum.
• It is denoted by P.
• Mathematically it can be written as
• Momentum (P) = m × v
• The relationship between kinetic energy and momentum is given by:
$$KE=\frac{P^2}{2m}$$
Where m is mass of the object and v is the velocity
CALCULATION:
Given - Final momentum (P2) = 3P (initial momentum)
• The relationship between kinetic energy and momentum is given by:
$$KE=\frac{P^2}{2m}$$
• When the momentum of an object is tripled keeping the mass as constant:
$$\Rightarrow {KE}_2=\frac{P^2_1}{2m}$$
$$\Rightarrow {KE}_2=\frac{(3P)^2}{2m}=\frac{9P^2}{2m} = 9(KE)$$ $$[\because KE=\frac{P^2}{2m}]$$
• Hence, if the momentum of a body is tripled, it's Kinetic energy will become nine times its original value. So option 2 is correct. | HuggingFaceTB/finemath | |
# The third equation is redundant since if we add up
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Unformatted text preview: p = ⇡ says ⇡1 ⇡2 ⇡3 0 .7 @.3 .2 which translates into three equations .2 .5 .4 1 .1 .2A = ⇡1 .4 .7⇡1 + .3⇡2 + .2⇡3 .2⇡1 + .5⇡2 + .4⇡3 .1⇡1 + .2⇡2 + .4⇡3 = ⇡1 = ⇡2 = ⇡3 ⇡2 ⇡3 20 CHAPTER 1. MARKOV CHAINS Note that the columns of the matrix give the numbers in the rows of the equations. The third equation is redundant since if we add up the three equations we get ⇡1 + ⇡2 + ⇡3 = ⇡1 + ⇡2 + ⇡3 If we replace the third equation by ⇡1 + ⇡2 + ⇡3 = 1 and subtract ⇡1 from each side of the first equation and ⇡2 from each side of the second equation we get .3⇡1 + .3⇡2 + .2⇡3 = 0 .2⇡1 .5⇡2 + .4⇡3 = 0 ⇡1 + ⇡2 + ⇡3 = 1 (1.10) At this point we can solve the equations by hand or using a calculator. By hand. We note that the third equation implies ⇡3 = 1 substituting this in the first two gives .2 = .5⇡1 ⇡1 ⇡2 and .1⇡2 .4 = .2⇡1 + .9⇡2 Multiplying the first equation by .9 and adding .1 times the second gives 2.2 = (0.45 + 0.02)⇡1 or ⇡1 = 22/47 Multiplying the first equation by .2 and adding 0.16...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
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Categories
# Unit digit | Algebra | AMC 8, 2014 | Problem 22
Try this beautiful problem from Algebra about unit digit from AMC-8, 2014. You may use sequential hints to solve the problem.
Try this beautiful problem from Algebra about unit digit from AMC-8, 2014.
## Unit digit | AMC-8, 2014|Problem 22
A $2$-digit number is such that the product of the digits plus the sum of the digits is equal to the number. What is the unit digit of the number?
• 7
• 9
• 5
### Key Concepts
Algebra
Multiplication
integer
Answer:$9$
AMC-8, 2014 problem 22
Challenges and Thrills in Pre College Mathematics
## Try with Hints
Let the ones digit place be y and ten’s place be x
Therefore the number be $10x+y$
Can you now finish the problem ……….
Given that the product of the digits plus the sum of the digits is equal to the number
can you finish the problem……..
Let the ones digit place be y and ten’s place be x
Therefore the number be $10x+y$
Now the product of the digits=$xy$
Given that the product of the digits plus the sum of the digits is equal to the number
Therefore $10x+y=(x\times y)+(x+y)$
$\Rightarrow 9x=xy$
$\Rightarrow y=9$
Therefore the unit digit =$y$=9 | HuggingFaceTB/finemath | |
# Question Video: Comparing the Number of Sides of 2D Shapes Mathematics • Kindergarten
Which shape is the odd one out? Hint: Think about the sides.
02:00
### Video Transcript
Which shape is the odd one out? Hint: think about the sides.
We’re shown three different shapes. And we have to find the odd one out. We’re given a hint to think about the sides. What do you notice about the sides of these shapes? Let’s look at the first shape. Let’s count how many sides it has. One, two, three, four, five, six. The first shape has six sides.
Can you spot another shape which could have six sides? These two shapes look very similar. Let’s count the number of sides. One, two, three, four, five, six. Both of these shapes have six sides. If both of these shapes have the same number of sides, the circle must be the odd one out. A circle doesn’t have six sides. The other two shapes are hexagons. Hexagons have six sides.
So the shape which is the odd one out is the circle. | HuggingFaceTB/finemath | |
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Homework9
# Homework9 - Homework 9 Answers 95.657 Electromagnetic...
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Homework 9 Answers, 95.657 Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Jackson 4.10 Two concentric conducting spheres of inner and outer radii a and b , respectively, carry charges ± Q . The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant ε / ε 0 ), as shown in the figure. (a) Find the electric field everywhere between the spheres. SOLUTION We can split the region where we want to know the electric field into two regions, the left hemisphere and the right hemisphere, solve for the field in each region separately and then apply the boundary conditions to get the final solution. In the left region, there are no dielectrics and no charges. So the electric potential obeys the Laplace equation: 2 L = 0 If we align the z -axis pointing to the right, the problem has azimuthal symmetry. The general solution to the Laplace equation in spherical coordinates for azimuthal symmetry is: L r , , = l A l r l B l r l 1 P l cos The potential on the surface of a conductor is always constant: L r = a = C C = l A l a l B l a l 1 P l cos Because this equation must hold for all values of the independent polar variable and because the Legendre polynomials are orthogonal, all coefficients must equate independently: C = A 0 B 0 a and B l =− A l a 2 l 1 for l > 0 a b + Q -Q
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The solution now becomes: L r , , = C B 0 a B 0 r l = 1 A l r a l a r l 1 P l cos The other surface is also a conductor and must have a constant potential as well: L r = b = D D = C B 0 a B 0 b l = 1 A l b a l a b l 1 P l cos Because this equation must hold for all values of the independent polar variable and because the Legendre polynomials are orthogonal, all coefficients must equate independently: B 0 = ab D C a b and A l = 0 for l > 1 The solution now becomes: L r , , = C ab D C a b 1 a 1 r At this point C and D are undetermined, so we can redefine them to simplify the equation: L r , , = D C r This leads to a total electric field of: E L = C r 2 r We could have guessed this form of the solution based on the symmetry of the problem, but it is often safer and more instructive to go through all the steps. The right hemisphere is a separate region and can be now solved separately. In the region between the shells, we can tell that there are no free charges and no bound charges (the bound charges, or
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Question
# The value of derivative at the point x=−2 on the curve y=x2 is equal to .
A
-4
B
4
C
-2
D
2
Solution
## The correct option is A -4We know that as per the first principle, the derivative of a function is given by f′(x)=limh→0f(x+h)−f(x)h f(x+h)=(x+h)2=x2+h2+2hxf(x)=x2 So f′(x)=limh→0x2+h2+2hx−x2h=limh→0h2+2hxh=limh→0h(h+2x)h=2x At x = -2, f'(x)= 2 × (-2) = -4
 Suggest corrections
 | HuggingFaceTB/finemath | |
# Prove that:(its urgent plz help someone who's intelligent enough !!)
1
by vivek2001
2015-02-18T11:13:03+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Let x = tan A , y = Tan B , z = Tan C
(x - y)/(1+x y) = tan (A-B)
(y-z)//(1+y z) = tan (B-C)
(z-x)/(1+z x) = tan (C - A)
LHS = (A-B) + (B-C) + (C-A) = 0
RHS is also similar to LHS.
If we assume x³ = tan P , y³ = tan Q , z³ = tan R
RHS = tan⁻¹ (tan (P-Q) ) + tan⁻¹ [tan (Q-R) ] + tan⁻¹ [ tan (R-P) ] = 0
LHS = RHS
thanx sir......... !!! :) | HuggingFaceTB/finemath | |
# Thoughts on 1/7 and other rational numbers (Part 3)
In Part 2 of this series, I discussed the process of converting a fraction into its decimal representation. In this post, I consider the reverse: starting with a decimal representation, and ending with a fraction.
Let me say at the onset that the process I’m about to describe appears to be a dying art. When I show this to my math majors who want to be high school teachers, roughly half have either not seen it before or else have no memory of seeing it before. (As always, I hold my students blameless for the things that they were simply not taught at a younger age, and part of my job is repairing these odd holes in their mathematical backgrounds so that they’ll have their best chance at becoming excellent high school math teachers.) I’m guessing that this algorithm is a dying art because of the ease and convenience of modern calculators.
So let me describe how I describe this procedure to my students. To begin, suppose that we’re given a repeating decimal like $0.\overline{432} = 0.432432432\dots$. How do we change this into a decimal? Let’s call this number $x$.
I’m now about to do something that, if you don’t know what’s coming next, appears to make no sense. I’m going to multiply $x$ by $1000$. Students often give skeptical, quizzical, and/or frustrated looks about this non-intuitive next step… they’re thinking, “How would I ever have thought to do that on my own?” To allay these concerns, I explain that this step comes from the patented Bag of Tricks. Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students. Multiplying by $1000$ on the next step is absolutely not obvious, unless you happen to know via clairvoyance what’s going to come next.
Anyway, let’s write down $x$ and also $1000x$. $1000x = 432.432432\dots$ $x = 0.432432\dots$
Notice that the decimal parts of both $x$ and $1000x$ are the same. Subtracting, the decimal parts cancel, leaving $999x = 432$
or $x = \displaystyle \frac{432}{999} = \displaystyle \frac{16}{37}$
In my experience, most students — even senior math majors who have taken a few theorem-proof classes and hence are no dummies — are a little stunned when they see this procedure for the first time. To make this more real and believable to them, I then ask them to pop out their calculators to confirm that this actually worked. (Indeed, many students need this confirmation to be psychologically sure that it really did work.) Then I ask my students, why did we multiply by $1000$? They’ll usually give the correct answer: so that the decimal parts will cancel. My follow-up question is, what should we do if the decimal is $0.\overline{24}$? They’ll usually respond that we should multiply by $100$ or, in general, by $10^n$, where $n$ is the length of the repeating block.
This strategy, of course, works for $0.\overline{142857}$, eventually yielding $0.\overline{142587} = \displaystyle \frac{142857}{999999} = \displaystyle \frac{1}{7}$
after cancellation. The same procedure works for decimal expansions with a delay, like $x = 0.72\overline{3}$. This time, I’ll ask them how we should go about changing this into a fraction. I usually get at least one of three responses. I love getting multiple responses, as this gives the students a chance to came the “different” answers, compare the different strategies, and
Answer #1. Multiply $x$ by $1000$ since the repeating pattern starts at the 3rd decimal place. $1000x = 723.333\dots$ $x = 0.7233\dots$ $\therefore 999x = 722.61$ $x =\displaystyle\frac{722.61}{999} = \displaystyle\frac{72261}{99900} = \displaystyle \frac{217}{300}$
Answer #2. Multiply $x$ by $10$ since the repeating block has length 1. $10x = 7.23333\dots$ $x = 0.7233\dots$ $\therefore 9x = 6.51$ $x = \displaystyle \frac{6.51}{9} = \displaystyle\frac{651}{900} = \displaystyle\frac{217}{300}$
Answer #3. First multiply $x$ by 100 to get rid of the delay. Then multiply $100 x$ by an extra $10$ since the repeating block has length 1. $1000x = 723.333\dots$ $100x = 72.33\dots$ $\therefore 900x = 651$ $x = \displaystyle\frac{651}{900} = \displaystyle\frac{217}{300}$ The above discussion concerned repeating decimals. For completeness, converting terminating decimals into a fraction is easy. For example, $0.124 = \displaystyle \frac{1}{10} + \frac{2}{100} + \frac{4}{1000} = \displaystyle \frac{124}{1000} = \displaystyle \frac{31}{250}$ One more thought. The concept behind Part 2 of this series shows that a rational number of the form $k/n$, where both $k$ and $n$ are integers, must have either a terminating decimal expansion or else a repeating decimal expansion (possibly with a delay). In this post, we went the other direction. Therefore, we have the basis for the following theorem.
Theorem. A number $x$ is rational if and only if it has either a terminating decimal expansion or else a repeating decimal expansion.
The contrapositive of this theorem is perhaps intuitively obvious.
Theorem. A number $x$ is irrational if and only if it has a non-terminating and non-repeating decimal expansion.
In my experience, most students absolutely believe both of these theorems. For example, most students believe that $\sqrt{2}$ has a decimal expansion that neither terminates nor repeats. That said, most math majors are surprised to discover that it does take quite a bit of work — like a formal write-up of Parts 2 and 3 of this series — to actually prove this statement from middle-school mathematics.
## 3 thoughts on “Thoughts on 1/7 and other rational numbers (Part 3)”
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# Fraction Practice
The fraction practice offered in this set of worksheets will help build a strong foundation in math skills. Kids will color in fractions, add them, and subtract them, as well as decide which fractions are greatest, least or equivalent. Understanding fractions is an essential math skill for third graders to master, and when it comes to math, practice makes perfect.
### Find the Fraction
Each shape on this third grade math worksheet is divided into equal parts, and some of the parts are shaded. Kids identify and write the fraction represented.
### "Color-fill" Fractions
Each geometric shape on this third grade math worksheet is divided into equal parts. Kids color each shape to represent a certain fraction.
### Word Up! Fraction Word Problems
Eating fractions is a great way to advance kids' understanding of fractions. This worksheet asks kids to solve fraction word problems about food.
### Portion Control: Color the Fractions
Hunting for a worksheet to help with fractions? This printable uses geometry to help your child get a handle on fractions.
### Crazy Coconut Fractions
On this third grade math worksheet, kids add simple fractions that have the same denominator.
### Hula Subtraction
On this third grade math worksheet, kids subtract simple fractions that have the same denominator.
### Equivalent Fractions: Find the Partner
This colorful worksheet helps children grasp the concept of equivalent fractions.
### Comparing Fractions: Least and Greatest
Children determine which fraction is greatest, and which is least, in a row of fractions with unlike denominators.
### Gardening with Fractions
In this 4th grade worksheet, your child will practice reducing fractions as she colors each plot of plants according to the directions given.
### Color by Fraction
Here's a challenging math worksheet for your 4th grader: your child must color the fractions according to their value, giving each value its own color.
### Fraction Practice: Color It!
To complete this second grade math worksheet, kids color parts of shapes to show the fractions 1/2, 1/3, and 1/4.
### Probability Darts 4
Building on math skills with fractions, addition, and degrees, kids will figure out how much of the dart board each panel takes up then answer questions.
### Tropical Fruit Fractions
This third grade math worksheet asks questions about a set of different kinds of fruit. Kids respond to the questions by counting to determine the fractions.
### Greater Than or Less Than? Comparing Fractions
Covering a variety of fraction skills such as converting to like fractions, this worksheet is sure to challenge your fifth grader's fraction savvy.
### Baking with Fractions
In this 5th grade math worksheet, your child will work with fractions as she calculates how much of each ingredient she needs for a recipe.
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# Shors Algorithm Osama Awwad Department of Computer Science
• Slides: 20
Shor’s Algorithm Osama Awwad Department of Computer Science Western Michigan University 12 December 2021
Overview n n Shor's algorithm is a quantum algorithm for factoring a number N in O((log N)3) time and O(log N) space, named after Peter Shor. The algorithm is significant because it implies that RSA, a popular public-key cryptography method, might be easily broken, given a sufficiently large quantum computer
Overview n n RSA uses a public key N which is the product of two large prime numbers One way to crack RSA encryption is by factoring N, but with classical algorithms, factoring becomes increasingly time-consuming as N grows large; more specifically no classical algorithm is known that can factor in polynomial time. Shor's algorithm can crack RSA in polynomial time.
Overview n n Like many quantum computer algorithms, Shor's algorithm is probabilistic It gives the correct answer with high probability, and the probability of failure can be decreased by repeating the algorithm.
Overview n n n Shor's algorithm was discovered in 1994 by Peter Shor, but the classical part was known before. it is credited to G. L. Miller. Seven years later, in 2001. it was demonstrated by a group at IBM, which factored 15 into 3 and 5, using a quantum computer with 7 qubits.
Shor’s Algorithm §Shor’s algorithm shows (in principle, ) that a quantum computer is capable of factoring very large numbers in polynomial time. The algorithm is dependant on §Modular Arithmetic §Quantum Parallelism §Quantum Fourier Transform
Shor’s Algorithm - Periodicity § An important result from Number Theory: F(a) = xa mod N is a periodic function § Choose N = 15 and x = 7 and we get the following: 7 0 mod 15 = 1 1 7 mod 15 = 7 2 7 mod 15 = 4 3 7 mod 15 = 13 7 4 mod 15 = 1. . .
Shor’s Algorithm - In Depth Analysis To Factor an odd integer N (Let’s choose 15) : 1. Determine if the number n is a prime, a even number, or an integer power of a prime number. 2 If it is we 2 will not use Shor's algorithm. There are efficient classical methods for determining if a integer n belongs to one of the above groups, and providing factors for it if it is. This step would be performed on a classical computer. 2. Choose an integer q such that N 2< q < 2 N 2 let’s pick 256 3. Choose a random integer x such that GCD(x, N) = 1 let’s pick 7
Shor’s Algorithm - In Depth Analysis 4. Create two quantum registers (these registers must also be entangled so that the collapse of the input register corresponds to the collapse of the output register) q Input register: must contain enough qubits to represent numbers as large as q-1. up to 255, so we need 8 qubits q Output register: must contain enough qubits to represent numbers as large as N-1. up to 14, so we need 4 qubits
Shor’s Algorithm - Preparing Data 5. Load the input register with an equally weighted superposition of all integers from 0 to q-1. 0 to 255 6. Load the output register with all zeros. The total state of the system at this point will be: 1 255 √ 256 a=0 Input Register ∑ |a, 000> Output Register Note: the comma here denotes that the registers are entangled
Shor’s Algorithm - Modular Arithmetic 7. Apply the transformation x a mod N to each number in the input register, storing the result of each computation in the output register. Note that we are using decimal numbers here only for simplicity. Input Register 7 a Mod 15 Output Register |0> 7 0 Mod 15 1 |1> 7 1 Mod 15 7 |2> 7 2 Mod 15 4 |3> 7 3 Mod 15 13 |4> 7 4 Mod 15 1 |5> 7 5 Mod 15 7 |6> 7 6 Mod 15 4 |7> 7 7 Mod 15 13 . .
Shor’s Algorithm - Superposition Collapse 8. Now take a measurement on the output register. This will collapse the superposition to represent just one of the results of the transformation, let’s call this value c. Our output register will collapse to represent one of the following: |1>, |4>, |7>, or |13 For sake of example, lets choose |1>
Shor’s Algorithm - Entanglement Now things really get interesting ! 9. Since the two registers are entangled, measuring the output register will have the effect of partially collapsing the input register into an equal superposition of each state between 0 and q-1 that yielded c (the value of the collapsed output register. ) Since the output register collapsed to |1>, the input register will partially collapse to: 1 √ 64 1 1 1 |0> +√ 64 |4> +√ 64 |8> +√ 64 |12>, . . . 1 √ 64 The probabilities in this case are since our register is now in an equal superposition of 64 values (0, 4, 8, . . . 252)
Shor’s Algorithm - QFT We now apply the Quantum Fourier transform on the partially collapsed input register. The fourier transform has the effect of taking a state |a> and transforming it into a state given by: 1 √q q-1 ∑ |c> * e 2 iac / q c=0
Shor’s Algorithm - QFT 1 √ 64 ∑ |a> , |1> a A 1 √ 256 255 ∑ |c> * e 2 iac / 256 c=0 Note: A is the set of all values that 7 a mod 15 yielded 1. In our case A = {0, 4, 8, …, 252} So the final state of the input register after the QFT is: 1 √ 64 ∑ a A 1 √ 256 255 ∑ |c> * e 2 iac / 256, |1> c=0
Shor’s Algorithm - QFT The QFT will essentially peak the probability amplitudes at integer multiples of q/r , where r is the desired period in our case r is 4. |0>, |64>, |128>, |192>, … So we no longer have an equal superposition of states, the probability amplitudes of the above states are now higher than the other states in our register. Measure the state of register one, call this value m, this integer m has a very high probability of being a multiple of q/r With our knowledge of q, and m, there are methods of calculating the period (one method is the continuous fraction expansion of the ratio between q and m. )
Shor’s Algorithm - The Factors : ) 10. Now that we have the period, the factors of N can be determined by taking the greatest common divisor of N with respect to x ^ (P/2) + 1 and x ^ (P/2) - 1. The idea here is that this computation will be done on a classical computer. We compute: Gcd(7 4/2 + 1, 15) = 5 Gcd(7 4/2 - 1, 15) = 3 We have successfully factored 15!
What to do if Shor's algorithm failed to produce factors of n §The QFT comes up short and reveals the wrong period. This probability is actually dependant on your choice of q. The larger the q, the higher the probability of finding the correct probability. §The period of the series ends up being odd §Fourier transform could be measured to be 0, making the post processing in the next step impossible. §The algorithm will sometimes find factors of 1 and n, which is not useful either. If either of these cases occur, we go back to the beginning and pick a new x.
References n n n Peter W. Shor “Polynomial-Time Algorithms for Prime Factorization and Discrete Logarithms on a Quantum Computer “, SIAM Journal on Computing (1997) www. eecis. udel. edu/~saunders/courses/879 -03 s/ http: //en. wikipedia. org/wiki/Shor's_algorithm
Q&A Thank You | HuggingFaceTB/finemath | |
# $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2b\cos(\theta)}\left(\frac{-2ar\sin(\theta)}{\sqrt{2ar\cos(\theta)-r^{2}}}\right) rdrd\theta$
How can I calculate this double integral?
Calculate $$\xi=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2b\cos(\theta)}\left(\frac{-2ar\sin(\theta)}{\sqrt{2ar\cos(\theta)-r^{2}}}\right) rdrd\theta$$where $$a\in \mathbb{R}^{+}$$.
My attempt: We can see \begin{align*}\xi=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}-2a\sin(\theta)\int_{0}^{2b\cos(\theta)}\frac{r^{2}}{\sqrt{2ar\cos(\theta)-r^{2}}}drd\theta \end{align*}
but I think this double-integral is very difficult, but maybe can I solve it using change the order of the following double integral, but I don't sure how can I to do it.
This double-integral has a relation with that problem (using stokes-theorem): $C$ be curve of intersection of hemisphere $x^2+y^2+z^2=2ax$ and cylinder $x^2+y^2=2bx$ ; to evaluate $\int_C(y^2+z^2)dx+(x^2+z^2)dy+(x^2+y^2)dz$
Context of problem: Using of the information in the link, we can see that $$\boxed{\oint_{C}\vec{F}\cdot d \vec{r}=\int \int_{S}(\nabla \times \vec{F}) \cdot d S=\int \int_{S}(\nabla \times \vec{F})\cdot \vec{n}dS=\int \int_{D}(\nabla \times \vec{F}) \cdot \left(\frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y} \right)dA.}$$
Note that $$\begin{eqnarray*} \Gamma(x,y)&:=&[(2y-2z)\vec{i}+(2z-2x)\vec{j}+(2x-2y)\vec{k}]\cdot \left( \frac{x-a}{\sqrt{2ax-x^{2}-y^{2}}}\vec{i}+\frac{y}{\sqrt{2ax-x^{2}-y^{2}}}\vec{j}+k \right)\\ &=&\frac{(2y-2z)(x-a)}{\sqrt{2ax-x^{2}-y^{2}}}+\frac{y(2z-2x)}{\sqrt{2ax-x^{2}-y^{2}}}+(2x-2y)\\ &=& \frac{(2y-2\sqrt{a^{2}-y^{2}-(x-a)^{2}})(x-a)}{\sqrt{2ax-x^{2}-y^{2}}}+\frac{y(2\sqrt{a^{2}-y^{2}-(x-a)^{2}}-2x)}{\sqrt{2ax-x^{2}-y^{2}}}+(2x-2y)\\ &:=&\Gamma_{1}(x,y)+\Gamma_{2}(x,y)+\Gamma_{3}(x,y) \end{eqnarray*}$$ So $$\begin{eqnarray*} \int \int_{D} (\nabla \times \vec{F})\cdot \left(\frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y} \right)dA&=&\int \int_{D} \Gamma(x,y)dA\\ &=&\underbrace{\int \int_{D} \Gamma_{1}(x,y)dA}_{I_{1}}+\underbrace{\int \int_{D} \Gamma_{2}(x,y)dA}_{I_{2}}+\underbrace{\int \int_{D} \Gamma_{3}(x,y)dA}_{I_{3}} \end{eqnarray*}$$
Note that calculate $$I_{3}$$ is easy, because $$\begin{eqnarray*}I_{3}&=&\int \int_{D} \Gamma_{3}(x,y)dA\\ &=& \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2b\cos(\theta)}\Gamma_{3}(r\cos(\theta),r\sin(\theta))rdrd\theta\\ &=&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2b\cos(\theta)}(2r\cos(\theta)-2r\sin(\theta)) rdrd\theta\\ &=&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2b\cos(\theta)}2r^{2}(\cos(\theta)-\sin(\theta)drd\theta\\ &=&\frac{2}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[r^{3} \right]_{0}^{2b\cos(\theta)}(\cos(\theta)-\sin(\theta))d\theta\\ &=&\frac{2}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(2b\cos(\theta))^{3}(\cos(\theta)-\sin(\theta))d\theta\\ &=&\frac{2\cdot 2^{3}b^{3}}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{3}(\theta)(\cos(\theta)-\sin(\theta))d\theta\\ &=&\frac{2^{4}b^{3}}{3}\cdot \frac{3 \pi}{8}\\ &=&2\pi b^{3} \implies \boxed{I_{3}=2\pi b^{3}} \end{eqnarray*}$$
but my problem is how can I calculate $$I_{1}+I_{2}$$?
Add information, we can see $$\boxed{r(x,y)=x\vec{i}+y\vec{j}+\left(\sqrt{a^{2}-y^{2}-(x-a)^{2}}\right)\vec{k}}$$
and $$\boxed{\frac{\partial r}{\partial x}=\vec{i}+\frac{\partial}{\partial x}\left(\sqrt{a^{2}-y^{2}-(x-a)^{2}} \right)\vec{k}=\vec{i}-\frac{x-a}{\sqrt{2ax-x^{2}-y^{2}}\vec{k}}}$$ and $$\boxed{\frac{\partial r}{\partial y}=\vec{j}+\frac{\partial}{\partial y}\left(\sqrt{a^{2}-y^{2}-(x-a)^{2}} \right)\vec{k}=\vec{j}-\frac{y}{\sqrt{2ax-x^{2}-y^{2}}}\vec{k}}$$ so, $$\boxed{\frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y}=\frac{x-a}{\sqrt{2ax-x^{2}-y^{2}}}\vec{i}+\frac{y}{\sqrt{2ax-x^{2}-y^{2}}}\vec{j}+k}$$Now, using information of link, we can see $$\boxed{D=\left\{(r,\theta): \frac{-\pi}{2}\leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq 2b\cos(\theta) \right\}}$$and finally $$\boxed{\vec{F}(x,y,z):=(y^{2}+z^{2})\vec{i}+(x^{2}+z^{2})\vec{j}+(x^{2}+y^{2})\vec{k}}.$$ Note that $$\boxed{\nabla \times \vec{F}=(2y-2z)\vec{i}+(2z-2x)\vec{j}+(2x-2y)\vec{k}}$$
• @RachidAtmai I will write a little more the context of my problem and how I got to that integral – mathproof Sep 7 '20 at 22:44
• argh! instead of editing I deleted my comment...sorry about that. yes please post the context. If $r=0$ then you have an issue with your integral (my comment was about the integral being improper...). – Rachid Atmai Sep 7 '20 at 22:47
• I already added the full context of my problem. When trying to use polar coordinates to calculate $I_ {1} + I_ {2}$ the problem becomes very ugly and from there comes that integral that I placed at the beginning that I don't know how to integrate. – mathproof Sep 7 '20 at 23:03
• What is the relationship between $a$ and $b$ – Ninad Munshi Sep 7 '20 at 23:24
• $0<b<a$ that is the relationship. – mathproof Sep 7 '20 at 23:28
Under the interchange $$\theta \leftrightarrow -\theta$$
$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2b\cos(-\theta)}\frac{-2ar^2\sin(-\theta)}{\sqrt{2ar\cos(-\theta)-r^2}}drd\theta$$
$$= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{2b\cos \theta}\frac{2ar^2\sin \theta}{\sqrt{2ar\cos \theta-r^2}}drd\theta = -I$$
Thus $$I = 0$$ | HuggingFaceTB/finemath | |
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# n is an integer from 1 to 50, what is the probability that n(n+1) is n
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Re: n is an integer from 1 to 50, what is the probability that n(n+1) is n [#permalink]
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siddharthsinha123 wrote:
n is an integer from 1 to 50, what is the probability that n(n+1) is not divisible by 4?
A. 12/25
B. 13/25
C. 14/25
D. 15/25
E. 16/25
We can use the following equation:
1 = P(n(n+1) is divisible by 4) + P(n(n+1) is NOT divisible by 4)
Thus:
P(n(n+1) is NOT divisible by 4) = 1 - P(n(n+1) is divisible by 4)
Let’s determine the probability that n(n+1) is divisible by 4. If n(n+1) is divisible by 4, then either n is divisible by 4 or n +1 is divisible by 4. Calculating the number of values of n divisible by 4 is the same as calculating the number of multiples of 4 between 1 and 50 inclusive. To calculate this, we can use this formula:
(largest multiple of 4 - smallest multiple of 4)/4 + 1
(48 - 4)/4 + 1
44/4 + 1 = 11 + 1 = 12
Thus, there are 12 multiples of 4 between 1 and 50 inclusive. That is, n can be any one of these 12 multiples of 4 so that n(n + 1) will be divisible by 4.
Similarly, if (n + 1) is a multiple of 4, n(n + 1) also will be divisible by 4. Since we know that there are 12 values of n that are multiples of 4, there must be another 12 values of n such that (n + 1) is a multiple of 4. Let’s expand on this idea:
When n = 3, n + 1 = 4, and thus n(n+1) is a multiple of 4.
When n = 23, n + 1 = 24, and thus n(n+1) is a multiple of 4.
When n = 47, n + 1 = 48, and thus n(n+1) is a multiple of 4.
We can see that there are 12 values of n that are multiples of 4, and 12 more values of n for (n + 1) to be a multiple of 4. Since there are 50 total integers from 1 to 50, inclusive, the probability of selecting a value of n so that n(n+1) is a multiple of 4 is:
24/50 = 12/25
Thus, the probability that n(n+1) IS NOT a multiple of 4 is 1 - 12/25 = 13/25.
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Re: n is an integer from 1 to 50, what is the probability that n(n+1) is n [#permalink]
(N) (N + 1) will be divisible by 4 if the integer N that is picked is either of the form:
N = 4k
or
N = 4k - 1
Find how many multiples of 4 there are from 1 to 50
Then double that amount because for each corresponding multiple there will be one value that is (-1) below that multiple
Ex: (3 and 4) ——- (7 and 8) ........ (47 and 48)
Count = (48 - 4)/4 + (1)
Count = 12
12 * 2 = 24 integers N such that (N) (N + 1) will be divisible by 4
Thus:
50 integers - (24 multiples of 4) =
26 Numbers that will NOT make (N) (N + 1) divisible by 4
Probability = 26/50 = 13/25
13/25
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Proof that traveling salesman problem is NP Hard
• Last Updated : 04 Jun, 2020
Pre-requisite: Travelling Salesman Problem, NP Hard
Given a set of cities and the distance between each pair of cities, the travelling salesman problem finds the path between these cities such that it is the shortest path and traverses every city once, returning back to the starting point.
Problem – Given a graph G(V, E), the problem is to determine if the graph has a TSP consisting of cost at most K.
Explanation –
In order to prove the Travelling Salesman Problem is NP-Hard, we will have to reduce a known NP-Hard problem to this problem. We will carry out a reduction from the Hamiltonian Cycle problem to the Travelling Salesman problem.
Every instance of the Hamiltonian Cycle problem consists of a graph G =(V, E) as the input can be converted to a Travelling Salesman problem consisting of graph G’ = (V’, E’) and the maximum cost, K. We will construct the graph G’ in the following way:
For all the edges e belonging to E, add the cost of edge c(e)=1. Connect the remaining edges, e’ belonging to E’, that are not present in the original graph G, each with a cost c(e’)= 2.
And, set .
The new graph G’ can be constructed in polynomial time by just converting G to a complete graph G’ and adding corresponding costs. This reduction can be proved by the following two claims:
• Let us assume that the graph G contains a Hamiltonian Cycle, traversing all the vertices V of the graph. Now, these vertices form a TSP with Since it uses all the edges of the original graph having cost c(e)=1. And, since it is a cycle, therefore, it returns back to the original vertex.
• We assume that the graph G’ contains a TSP with cost, . The TSP traverses all the vertices of the graph returning to the original vertex. Now since none of the vertices are excluded from the graph and the cost sums to n, therefore, necessarily it uses all the edges of the graph present in E, with cost 1, hence forming a hamiltonian cycle with the graph G. Thus we can say that the graph G’ contains a TSP if graph G contains Hamiltonian Cycle. Therefore, any instance of the Travelling salesman problem can be reduced to an instance of the hamiltonian cycle problem. Thus, the TSP is NP-Hard.
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My Personal Notes arrow_drop_up | HuggingFaceTB/finemath | |
```Mean, variance, standard
deviation and expectation
Section 5-3
Find the mean, variance, standard
deviation, and expected value for a
discrete random variable
Objectives
Once we know that a probability
distribution exist, we can describe it using
various descriptive statistics
◦ Visually using a graph, table, or formula
◦ Algebraically, we can find the mean, variance,
and standard deviation
Introduction
m= population mean since ALL possible
values are considered
m x p(x) x1 p(x1) x2 p(x2 ) ...xn p(xn )
Mean is also known as “Expected Value”
Mean should be rounded to one more
decimal place than the outcome x.
Always simplify fractions
Mean of a general discrete
probability distribution
Variance
( x p( x)) m
2
2
2
Standard Deviation
( x
2
p( x)) m
2
Variance & standard deviation
Example: Find the mean &
standard deviation
x
1
2
3
4
5
P(x)
0.3
0.1
0.1
0.2
0.3
In a study of brand
recognition of Sony,
groups of four
consumers are
interviewed. If x is the
number of people in
the group who
recognize the Sony
brand name, then x
can be 0, 1, 2, 3, or 4
and the corresponding
probabilities are
0.0016,0.0564,
0.1432, 0.3892, and
0.4096
``` | HuggingFaceTB/finemath | |
# Math formula for dating age Free online sexual dating for teenagers
When finding the age of an organic organism we need to consider the half-life of carbon 14 as well as the rate of decay, which is –0.693.
For example, say a fossil is found that has 35% carbon 14 compared to the living sample. We can use a formula for carbon 14 dating to find the answer.
Radiocarbon dating can be used on samples of bone, cloth, wood and plant fibers.
The half-life of a radioactive isotope describes the amount of time that it takes half of the isotope in a sample to decay.
After 5,730 years, the amount of carbon 14 left in the body is half of the original amount.
If the amount of carbon 14 is halved every 5,730 years, it will not take very long to reach an amount that is too small to analyze.
In the case of radiocarbon dating, the half-life of carbon 14 is 5,730 years.
This half life is a relatively small number, which means that carbon 14 dating is not particularly helpful for very recent deaths and deaths more than 50,000 years ago.
Libby invented carbon dating for which he received the Nobel Prize in chemistry in 1960.
There are also social theories for age differences in relationships as well as suggested reasons for 'alternative' age-hypogamous relationships.
Age-disparity relationships have been documented for most of recorded history and have been regarded with a wide range of attitudes dependant on sociocultural norms and legal systems. Relationships with age disparity of all kinds have been observed with both men and women as the older or younger partner.
Ofcourse at the end of the day, if you love that person, and if the sex is good, go for it no matter what anyone saids about anything.
Here is a simple way of calculating your age range, and proven to be quite useful and accurate.(ex) your age=30 (not that I am, just that its a simple number to calculate mind you! so if you're 30 you can't date anyone younger than 22. That person's age / 2 7 = your age ==There are also good things in life as you grow older...
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Archaeologists use the exponential, radioactive decay of carbon 14 to estimate the death dates of organic material.
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# How to derive van 't Hoff equation for Henry's law constant?
While studying about Henry's law, I encountered this equation:
$$\frac{\mathrm d\ln H}{\mathrm d(1/T)}= \frac{-\Delta_\text{sol}H }{R},$$
where $$\displaystyle\mathrm d(1/T) = -\frac{\mathrm dT}{T^2}$$ and the $$\displaystyle\frac{\Delta H}{RT^2}$$ reminds me of some kind of $$\Delta G$$ equation.
I didn't know how to derive it, so I searched about it and got nothing. I tried to substitute $$H$$ (Henry's law constant) by $$\gamma\cdot f_i$$ where $$\gamma$$ is the activity coefficient at infinite dilution and $$f_i$$ is the fugacity of pure species.
But still I have no idea what to do to obtain the formula that I'm familiar with, so I welcome any idea about how to derive or how to reverse engineer the formula.
Assuming Henry's coefficient $$\mathcal{H}_i$$ is defined by the following relation:
$$\mathcal{H}_i = \gamma_i^\inf \cdot p_i^\mathrm{sat}$$
Where $$\gamma_i^\inf$$ is the activity coefficient for an infinite diluted solution and $$p_i^\mathrm{sat}$$ is the vapor pressure of pure component $$i$$.
One can try to extrapolate Henry's coefficient at another temperature by assessing the following ratio (this is a common approach in Chemistry, not always valid):
$$\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\gamma_i^\inf(T_1) \cdot p_i^\mathrm{sat}(T_1)}{\gamma_i^\inf(T_0) \cdot p_i^\mathrm{sat}(T_0)}$$
The RHS maybe rearranged in order to have the well know form of ratio of mixed equilibrium constants $$K_i$$ considering that bulk concentration are close and ratio of activities also simplifies (over a small range of temperature).
$$\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\frac{\gamma_i(T_1)\cdot p_i(T_1)}{\eta_i(T_1)\cdot x_i(T_1)}}{\frac{\gamma_i(T_0)\cdot p_i(T_0)}{\eta_i(T_0)\cdot x_i(T_0)}} =\frac{K_i(T_0)}{K_i(T_0)}$$
That is, we are stating that ratio of Henry's constants approximates to the ratio of equilibrium constant because we assume that only ratio of pressure is significant and other ratios tend to unity. This is, off course, not always acceptable, but in some restrained conditions it might hold.
Where the concerned reaction is the converse of gas solubilization:
$$\ce{X_{i,(aq)} <=> X_{i,(\mathrm{g})}}{\quad \Delta_\mathrm{R}H = -\Delta_\mathrm{sol}H}$$
And its mixed equilibrium constant:
$$K_i = \frac{\gamma_i\cdot p_i}{\eta_i \cdot x_i}$$
Then one may apply van 't Hoff relation using enthalpy $$\Delta_\mathrm{sol}H$$ of the concerned reaction, from its integrated form, it comes:
$$\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \exp\left[\frac{\Delta_\mathrm{sol}H}{R}\left(\frac{1}{T_1}-\frac{1}{T_0}\right)\right]$$
This formulae is generally a good approximation around $$\pu{20°C}$$ over a small range of temperature. Do not forget that $$\Delta_\mathrm{sol}H$$ is not a constant but rather a function of temperature.
If you wish to get your definition, you just have to differentiate the last relation with respect to temperature:
$$\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}T} \approx \frac{\Delta_\mathrm{R}H}{RT^2} = -\frac{\Delta_\mathrm{sol}H}{RT^2}$$
Or equivalently: $$\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}\left(\frac{1}{T}\right)} \approx -\frac{\Delta_\mathrm{R}H}{R} = \frac{\Delta_\mathrm{sol}H}{R}$$
• Can you please explain how you changed the right side of second equation? May 24, 2016 at 16:15
• assuming that ratio of $x's$ and ratio of activity coefficients are 1 right? so what's the formula used for $K$ here ? May 24, 2016 at 16:28
• How the equation for $Ki$ has obtained ? I know $K$ is the ratio of $fi$ to $fi$ (standard condition)، I understand the numerator,But Can't understand the denominator. May 25, 2016 at 10:17
• This is the expression of a mixed (polyphasic) equilibrium constant. $x_i$ stands for the bulk dissolved gas concentration at equilibrium and $\eta_i$ is its activity coefficient. It is just the ratio of product/reactant as you always write them for aqueous solutions. May 25, 2016 at 11:35
• If you feel uncomfortable with this mixed constant, try convert it into $K_c$ or $K_p$. You will get some $RT$ terms that simply because of the ration of constants. May 25, 2016 at 11:44 | HuggingFaceTB/finemath | |
## 39.21 Descent in terms of groupoids
Cartesian morphisms are defined as follows.
Definition 39.21.1. Let $S$ be a scheme. Let $f : (U', R', s', t', c') \to (U, R, s, t, c)$ be a morphism of groupoid schemes over $S$. We say $f$ is cartesian, or that $(U', R', s', t', c')$ is cartesian over $(U, R, s, t, c)$, if the diagram
$\xymatrix{ R' \ar[r]_ f \ar[d]_{s'} & R \ar[d]^ s \\ U' \ar[r]^ f & U }$
is a fibre square in the category of schemes. A morphism of groupoid schemes cartesian over $(U, R, s, t, c)$ is a morphism of groupoid schemes compatible with the structure morphisms towards $(U, R, s, t, c)$.
Cartesian morphisms are related to descent data. First we prove a general lemma describing the category of cartesian groupoid schemes over a fixed groupoid scheme.
Lemma 39.21.2. Let $S$ be a scheme. Let $(U, R, s, t, c)$ be a groupoid scheme over $S$. The category of groupoid schemes cartesian over $(U, R, s, t, c)$ is equivalent to the category of pairs $(V, \varphi )$ where $V$ is a scheme over $U$ and
$\varphi : V \times _{U, t} R \longrightarrow R \times _{s, U} V$
is an isomorphism over $R$ such that $e^*\varphi = \text{id}_ V$ and such that
$c^*\varphi = \text{pr}_1^*\varphi \circ \text{pr}_0^*\varphi$
as morphisms of schemes over $R \times _{s, U, t} R$.
Proof. The pullback notation in the lemma signifies base change. The displayed formula makes sense because
$(R \times _{s, U, t} R) \times _{\text{pr}_1, R, \text{pr}_1} (V \times _{U, t} R) = (R \times _{s, U, t} R) \times _{\text{pr}_0, R, \text{pr}_0} (R \times _{s, U} V)$
as schemes over $R \times _{s, U, t} R$.
Given $(V, \varphi )$ we set $U' = V$ and $R' = V \times _{U, t} R$. We set $t' : R' \to U'$ equal to the projection $V \times _{U, t} R \to V$. We set $s'$ equal to $\varphi$ followed by the projection $R \times _{s, U} V \to V$. We set $c'$ equal to the composition
\begin{align*} R' \times _{s', U', t'} R' & \xrightarrow {\varphi , 1} (R \times _{s, U} V) \times _ V (V \times _{U, t} R) \\ & \xrightarrow {} R \times _{s, U} V \times _{U, t} R \\ & \xrightarrow {\varphi ^{-1}, 1} V \times _{U, t} (R \times _{s, U, t} R) \\ & \xrightarrow {1, c} V \times _{U, t} R = R' \end{align*}
A computation, which we omit shows that we obtain a groupoid scheme over $(U, R, s, t, c)$. It is clear that this groupoid scheme is cartesian over $(U, R, s, t, c)$.
Conversely, given $f : (U', R', s', t', c') \to (U, R, s, t, c)$ cartesian then the morphisms
$U' \times _{U, t} R \xleftarrow {t', f} R' \xrightarrow {f, s'} R \times _{s, U} U'$
are isomorphisms and we can set $V = U'$ and $\varphi$ equal to the composition $(f, s') \circ (t', f)^{-1}$. We omit the proof that $\varphi$ satisfies the conditions in the lemma. We omit the proof that these constructions are mutually inverse. $\square$
Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. Then we obtain a groupoid scheme $(X, X \times _ Y X, \text{pr}_1, \text{pr}_0, c)$ over $S$. Namely, $j : X \times _ Y X \to X \times _ S X$ is an equivalence relation and we can take the associated groupoid, see Lemma 39.13.3.
Lemma 39.21.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of schemes over $S$. The construction of Lemma 39.21.2 determines an equivalence
$\begin{matrix} \text{category of groupoid schemes} \\ \text{cartesian over } (X, X \times _ Y X, \ldots ) \end{matrix} \longrightarrow \begin{matrix} \text{ category of descent data} \\ \text{ relative to } X/Y \end{matrix}$
Proof. This is clear from Lemma 39.21.2 and the definition of descent data on schemes in Descent, Definition 35.31.1. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | open-web-math/open-web-math | |
# Kinematics in one dimention, where do two balls meet?
1. Sep 9, 2011
### feihong47
1. You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but your friend down on the ground throws another ball upward at the same speed v, releasing her ball at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location (a) at the halfway point between window and ground, (b) above this point, or (c) below this point?
Someone said the solution is b. Subjectively this may be find and dandy, but does any have proof of this using equations? I tried to find the time of each ball to reach halfway and then compare. Using the dropped ball as t1, i get t1 = SQRT(h/g), where h is the height of the building. Using t2 as the thrown ball, i have to solve t2 as a quadratic formula (1/2 h = vt-1/2gt2), in which case i have t2 in terms of v, g, and h. however I had no way to compare the two time intervals.
2. Sep 9, 2011
### wukunlin
you want to find the intersection of two displacement vs time curves
the first one is the ball dropping from the upper floor:
$$d = - \frac{1}{2}gt^2$$
the other one is the ball being tossed up
$$d = vt - \frac{1}{2}gt^2 - h$$
where h is the height between the floors
can you see how to solve it now?
3. Sep 9, 2011
### feihong47
i set the two equations equal to each other and i get h=vt. not sure how to use that, so i solve for t and plug it back in EQ 1 and i get d = -4.9h2/v2 ... doesn't really help yet
another thing is -- for ball being tossed up, i set up the equation as d=vt−1/2gt2 but without - h... how did you set yours up like that to make sense?
Last edited: Sep 9, 2011
4. Sep 9, 2011
### vela
Staff Emeritus
How are v and h related?
5. Sep 9, 2011
### wukunlin
about the -h:
the expressions for d need to have the same origin, in the equations I have written, d=0 is the release point at the upper floor, therefore the equation for the ball being tossed up, d must be equal to -h at time t=0.
if you like you can define the d=0 at any other point, your answer won't change qualitatively
6. Sep 9, 2011
### feihong47
ok, for some reason I dont see it..
i understand the intersection of the two curves is the point where the two balls meet.
therefore I set the two equations equal to each other and get :
vt = h
I solved for t = h/v, and then substituting into EQ 1, and I get
$$d = - \frac{1}{2}g(\frac{h}{v})^2$$
but how does that indicate where exactly the balls are?
7. Sep 9, 2011
### wukunlin
what is v in terms of h?
8. Sep 9, 2011
### feihong47
thanks! got it now so d = -h/4, so that indicates it's above the halfway line (rather than the bottome 1/4) right?
9. Sep 10, 2011
### wukunlin
precisely :)
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# 62.06 kg to lbs - 62.06 kilograms to pounds
Do you want to learn how much is 62.06 kg equal to lbs and how to convert 62.06 kg to lbs? Here it is. You will find in this article everything you need to make kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to underline that all this article is dedicated to only one amount of kilograms - exactly one kilogram. So if you need to know more about 62.06 kg to pound conversion - read on.
Before we go to the more practical part - this is 62.06 kg how much lbs calculation - we want to tell you some theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 62.06 kg to lbs? 62.06 kilograms it is equal 136.8188797972 pounds, so 62.06 kg is equal 136.8188797972 lbs.
## 62.06 kgs in pounds
We will start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, that is International System of Units (in short form SI).
Sometimes the kilogram is written as kilogramme. The symbol of this unit is kg.
The kilogram was defined first time in 1795. The kilogram was described as the mass of one liter of water. First definition was simply but hard to use.
Then, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was prepared of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was switched by another definition.
Today the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It is also divided to 100 decagrams and 1000 grams.
## 62.06 kilogram to pounds
You learned a little bit about kilogram, so now we can move on to the pound. The pound is also a unit of mass. We want to highlight that there are not only one kind of pound. What does it mean? For example, there are also pound-force. In this article we are going to to centre only on pound-mass.
The pound is used in the Imperial and United States customary systems of measurements. Of course, this unit is in use also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is defined as 0.45359237 kilograms. One avoirdupois pound is divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of the pound 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 62.06 kg?
62.06 kilogram is equal to 136.8188797972 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 62.06 kg in lbs
The most theoretical part is already behind us. In next section we will tell you how much is 62.06 kg to lbs. Now you learned that 62.06 kg = x lbs. So it is high time to get the answer. Just see:
62.06 kilogram = 136.8188797972 pounds.
It is a correct result of how much 62.06 kg to pound. You can also round it off. After rounding off your outcome will be exactly: 62.06 kg = 136.532 lbs.
You learned 62.06 kg is how many lbs, so see how many kg 62.06 lbs: 62.06 pound = 0.45359237 kilograms.
Obviously, this time you may also round it off. After it your result will be exactly: 62.06 lb = 0.45 kgs.
We are also going to show you 62.06 kg to how many pounds and 62.06 pound how many kg outcomes in tables. Have a look:
We are going to begin with a chart for how much is 62.06 kg equal to pound.
### 62.06 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
62.06 136.8188797972 136.5320
Now look at a table for how many kilograms 62.06 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
62.06 0.45359237 0.45
Now you know how many 62.06 kg to lbs and how many kilograms 62.06 pound, so it is time to move on to the 62.06 kg to lbs formula.
### 62.06 kg to pounds
To convert 62.06 kg to us lbs you need a formula. We will show you a formula in two different versions. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 136.8188797972 outcome in pounds
The first version of a formula will give you the most exact result. In some cases even the smallest difference could be considerable. So if you need an accurate result - first formula will be the best solution to convert how many pounds are equivalent to 62.06 kilogram.
So let’s go to the another formula, which also enables calculations to learn how much 62.06 kilogram in pounds.
The another formula is down below, let’s see:
Amount of kilograms * 2.2 = the outcome in pounds
As you can see, the second formula is simpler. It could be better option if you want to make a conversion of 62.06 kilogram to pounds in fast way, for instance, during shopping. Just remember that final result will be not so exact.
Now we are going to learn you how to use these two formulas in practice. But before we are going to make a conversion of 62.06 kg to lbs we want to show you easier way to know 62.06 kg to how many lbs totally effortless.
### 62.06 kg to lbs converter
An easier way to know what is 62.06 kilogram equal to in pounds is to use 62.06 kg lbs calculator. What is a kg to lb converter?
Converter is an application. Calculator is based on first formula which we gave you in the previous part of this article. Due to 62.06 kg pound calculator you can effortless convert 62.06 kg to lbs. Just enter number of kilograms which you need to convert and click ‘calculate’ button. The result will be shown in a second.
So let’s try to calculate 62.06 kg into lbs using 62.06 kg vs pound converter. We entered 62.06 as an amount of kilograms. This is the result: 62.06 kilogram = 136.8188797972 pounds.
As you see, our 62.06 kg vs lbs calculator is intuitive.
Now we are going to our main issue - how to convert 62.06 kilograms to pounds on your own.
#### 62.06 kg to lbs conversion
We are going to start 62.06 kilogram equals to how many pounds calculation with the first version of a formula to get the most exact outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 136.8188797972 the outcome in pounds
So what need you do to know how many pounds equal to 62.06 kilogram? Just multiply amount of kilograms, this time 62.06, by 2.20462262. It is equal 136.8188797972. So 62.06 kilogram is exactly 136.8188797972.
It is also possible to round it off, for example, to two decimal places. It is 2.20. So 62.06 kilogram = 136.5320 pounds.
It is high time for an example from everyday life. Let’s convert 62.06 kg gold in pounds. So 62.06 kg equal to how many lbs? As in the previous example - multiply 62.06 by 2.20462262. It gives 136.8188797972. So equivalent of 62.06 kilograms to pounds, when it comes to gold, is exactly 136.8188797972.
In this example you can also round off the result. Here is the outcome after rounding off, this time to one decimal place - 62.06 kilogram 136.532 pounds.
Now we can go to examples calculated with short formula.
#### How many 62.06 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 136.532 the result in pounds
So 62.06 kg equal to how much lbs? And again, you need to multiply amount of kilogram, this time 62.06, by 2.2. See: 62.06 * 2.2 = 136.532. So 62.06 kilogram is exactly 2.2 pounds.
Let’s make another conversion with use of shorer formula. Now calculate something from everyday life, for example, 62.06 kg to lbs weight of strawberries.
So convert - 62.06 kilogram of strawberries * 2.2 = 136.532 pounds of strawberries. So 62.06 kg to pound mass is exactly 136.532.
If you know how much is 62.06 kilogram weight in pounds and can calculate it using two different formulas, we can move on. Now we want to show you all results in tables.
#### Convert 62.06 kilogram to pounds
We realize that outcomes shown in tables 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 quickly make a comparison 62.06 kg equivalent to lbs results.
Start with a 62.06 kg equals lbs chart for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
62.06 136.8188797972 136.5320
And now look 62.06 kg equal pound chart for the second version of a formula:
Kilograms Pounds
62.06 136.532
As you see, after rounding off, if it comes to how much 62.06 kilogram equals pounds, the outcomes are not different. The bigger number the more significant difference. Remember it when you need to make bigger amount than 62.06 kilograms pounds conversion.
#### How many kilograms 62.06 pound
Now you know how to convert 62.06 kilograms how much pounds but we want to show you something more. Are you interested what it is? What about 62.06 kilogram to pounds and ounces calculation?
We want to show you how you can convert it step by step. Begin. How much is 62.06 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, this time 62.06, by 2.20462262. So 62.06 * 2.20462262 = 136.8188797972. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To check how much 62.06 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is exactly 327396192 ounces.
So final outcome is exactly 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then final outcome will be equal 2 pounds and 33 ounces.
As you see, calculation 62.06 kilogram in pounds and ounces quite easy.
The last calculation which we want to show you is calculation of 62.06 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters you need another formula. Before we give you it, look:
• 62.06 kilograms meters = 7.23301385 foot pounds,
• 62.06 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 62.06 foot pounds to kilograms meters you have to multiply 62.06 by 0.13825495. It gives 0.13825495. So 62.06 foot pounds is 0.13825495 kilogram meters.
It is also possible to round off this result, for instance, to two decimal places. Then 62.06 foot pounds will be exactly 0.14 kilogram meters.
We hope that this conversion was as easy as 62.06 kilogram into pounds conversions.
We showed you not only how to do a conversion 62.06 kilogram to metric pounds but also two another conversions - to know how many 62.06 kg in pounds and ounces and how many 62.06 foot pounds to kilograms meters.
We showed you also other way to do 62.06 kilogram how many pounds calculations, it is with use of 62.06 kg en pound calculator. It will be the best option for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you can make 62.06 kilogram equal to how many pounds calculation - on your own or with use of our 62.06 kgs to pounds converter.
So what are you waiting for? Convert 62.06 kilogram mass to pounds in the way you like.
Do you need to make other than 62.06 kilogram as pounds calculation? For instance, for 5 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so simply as for 62.06 kilogram equal many pounds.
### How much is 62.06 kg in pounds
We want to sum up this topic, that is how much is 62.06 kg in pounds , we prepared for you an additional section. Here you can find the most important information about how much is 62.06 kg equal to lbs and how to convert 62.06 kg to lbs . Let’s see.
What is the kilogram to pound conversion? The conversion kg to lb is just multiplying 2 numbers. How does 62.06 kg to pound conversion formula look? . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 62.06 kilogram to pounds? The exact answer is 136.8188797972 lbs.
There is also another way to calculate how much 62.06 kilogram is equal to pounds with another, shortened type of the formula. Have a look.
The number of kilograms * 2.2 = the result in pounds
So in this case, 62.06 kg equal to how much lbs ? The answer is 136.8188797972 lb.
How to convert 62.06 kg to lbs in a few seconds? It is possible to use the 62.06 kg to lbs converter , which will do whole mathematical operation for you and you will get an accurate 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. | HuggingFaceTB/finemath | |
# Running Sum of 1d Array- Day 44(Python)
After writing a three-day series of Knapsack problem, I decided to write something super easy today. It is an “Easy” tagged question in Leetcode. Let us look into the question.
1480. Running Sum of 1d Array
Given an array `nums`. We define a running sum of an array as `runningSum[i] = sum(nums…nums[i])`.
Return the running sum of `nums`.
Example 1:
`Input: nums = [1,2,3,4]Output: [1,3,6,10]Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].`
Example 2:
`Input: nums = [1,1,1,1,1]Output: [1,2,3,4,5]Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].`
Example 3:
`Input: nums = [3,1,2,10,1]Output: [3,4,6,16,17]`
Constraints:
This is an easy question. All we need to do is run a loop that starts with the second number in the list. The loop takes the number in the previous position and current position in the list, and store the result in the current position.
`class RunningSumFinder: def runningSum(self, nums: List[int]) -> List[int]: for i in range(1, len(nums)): nums[i] = nums[i-1] + nums[i] return nums`
Complexity analysis.
Time Complexity
We are traversing through the list once. Hence the time complexity is O(N).
Space Complexity.
We are not using any extra data structure to store any intermediate results. Hence the space complexity is O(1).
I would love to hear your feedback about my posts. Do let me know if you have any comments or feedback.
## More from Annamariya Tharayil
Software Engineer. Find me @ www.linkedin.com/in/annamariya-jt | HuggingFaceTB/finemath | |
## 4674
4,674 (four thousand six hundred seventy-four) is an even four-digits composite number following 4673 and preceding 4675. In scientific notation, it is written as 4.674 × 103. The sum of its digits is 21. It has a total of 4 prime factors and 16 positive divisors. There are 1,440 positive integers (up to 4674) that are relatively prime to 4674.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 21
• Digital Root 3
## Name
Short name 4 thousand 674 four thousand six hundred seventy-four
## Notation
Scientific notation 4.674 × 103 4.674 × 103
## Prime Factorization of 4674
Prime Factorization 2 × 3 × 19 × 41
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 4674 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 4,674 is 2 × 3 × 19 × 41. Since it has a total of 4 prime factors, 4,674 is a composite number.
## Divisors of 4674
1, 2, 3, 6, 19, 38, 41, 57, 82, 114, 123, 246, 779, 1558, 2337, 4674
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 10080 Sum of all the positive divisors of n s(n) 5406 Sum of the proper positive divisors of n A(n) 630 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 68.3667 Returns the nth root of the product of n divisors H(n) 7.41905 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 4,674 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 4,674) is 10,080, the average is 630.
## Other Arithmetic Functions (n = 4674)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 1440 Total number of positive integers not greater than n that are coprime to n λ(n) 360 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 635 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 1,440 positive integers (less than 4,674) that are coprime with 4,674. And there are approximately 635 prime numbers less than or equal to 4,674.
## Divisibility of 4674
m n mod m 2 3 4 5 6 7 8 9 0 0 2 4 0 5 2 3
The number 4,674 is divisible by 2, 3 and 6.
• Arithmetic
• Abundant
• Polite
• Square Free
## Base conversion (4674)
Base System Value
2 Binary 1001001000010
3 Ternary 20102010
4 Quaternary 1021002
5 Quinary 122144
6 Senary 33350
8 Octal 11102
10 Decimal 4674
12 Duodecimal 2856
20 Vigesimal bde
36 Base36 3lu
## Basic calculations (n = 4674)
### Multiplication
n×i
n×2 9348 14022 18696 23370
### Division
ni
n⁄2 2337 1558 1168.5 934.8
### Exponentiation
ni
n2 21846276 102109494024 477259775068176 2230712188668654624
### Nth Root
i√n
2√n 68.3667 16.7197 8.26841 5.41923
## 4674 as geometric shapes
### Circle
Diameter 9348 29367.6 6.86321e+07
### Sphere
Volume 4.27715e+11 2.74528e+08 29367.6
### Square
Length = n
Perimeter 18696 2.18463e+07 6610.03
### Cube
Length = n
Surface area 1.31078e+08 1.02109e+11 8095.61
### Equilateral Triangle
Length = n
Perimeter 14022 9.45971e+06 4047.8
### Triangular Pyramid
Length = n
Surface area 3.78389e+07 1.20337e+10 3816.31
## Cryptographic Hash Functions
md5 1f5795e7b93f423c397e6f7aaff80133 ce2575945491168e19b00715da76c5bbdf8f0084 508d5031072e26a70f2cfb41f81d33a7909023dcec0f9f0aab4e83bb4faeae49 d8cc4e1b87f2f53ce4429e169e7b881d84fb35f2d8edfd57d7340cba90f1f6af0b0d3e70e92690a78a51fb7bc21ee987255b5239c07ccda1c4e096fe158f1101 56e2fec8ffd5296373a9cb33c63a81428af33208 | HuggingFaceTB/finemath |
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