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## NY,NY one year ago Which trigonometric expression does not simplify to 1? (see link)
1. NY,NY
2. Vocaloid
well, let's try each answer one by one and see what we get, shall we?
3. NY,NY
ok.
4. Vocaloid
|dw:1434685599699:dw|
5. anonymous
help me please my question is at the top of the feed
6. Vocaloid
does that make sense?
7. NY,NY
I dont understand how cos^2/sin^2 became just cos^2.
8. Vocaloid
I distributed sin^2(x)
9. Vocaloid
|dw:1434685846593:dw|
10. Vocaloid
see how sin^2(x) cancels out?
11. Vocaloid
|dw:1434685945486:dw|
12. Vocaloid
|dw:1434685962311:dw|
13. NY,NY
Oh yes that makes more sense.
14. NY,NY
Thank you for explaining that.
15. Vocaloid
ok, so we know that #1 is not the right answer because we get 1 in the end
16. Vocaloid
let's move on to number 2
17. NY,NY
Ok.
18. Vocaloid
|dw:1434686053840:dw|
19. Vocaloid
|dw:1434686204867:dw|
20. Vocaloid
does this make sense so far?
21. NY,NY
yes, because you used the Pythagorean identity at the end, right?
22. Vocaloid
yup! ok, we know that #2 is out, because we ended up with 1, so let's move on to #3
23. NY,NY
Right.
24. Vocaloid
|dw:1434686335785:dw|
25. Vocaloid
now, I believe #3 is the right answer, but let's check #4 just in case
26. Vocaloid
|dw:1434686426089:dw|
27. Vocaloid
|dw:1434686523673:dw|
28. Vocaloid
whew, that took a while, but do you understand everything so far? :)
29. Vocaloid
the important thing to remember, is that when you have csc, sec, cot, or tan, you can re-write them in terms of cos or sin, which usually helps you cross things out and simplify :)
30. NY,NY
Im a little confused with the second step you took.
31. Vocaloid
which one? #3 or #4?
32. NY,NY
#4.
33. Vocaloid
can you tell me exactly which part is confusing?
34. NY,NY
the step that the arrow is pointing to |dw:1434687552340:dw|
35. NY,NY
Im confused how that concluded from the previous step, where you substituted the reciprocals.
36. Vocaloid
|dw:1434687726888:dw|
37. Vocaloid
which leaves me with|dw:1434687748727:dw|
38. Vocaloid
does it make it more clear? or is there something you're still unsure about?
39. NY,NY
Ok, yes that makes sense.
40. NY,NY
So that leaves us with choice 3.
41. Vocaloid
yup ~
42. NY,NY
Ok, thank you for helping.
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## NY,NY one year ago Which trigonometric expression does not simplify to 1? (see link). 1. NY,NY. 2. Vocaloid. well, let's try each answer one by one and see what we get, shall we?. 3. NY,NY. ok.. 4. Vocaloid. |dw:1434685599699:dw|. 5. anonymous. help me please my question is at the top of the feed. 6. Vocaloid. does that make sense?. 7. NY,NY. I dont understand how cos^2/sin^2 became just cos^2.. 8. Vocaloid. I distributed sin^2(x). 9. Vocaloid. |dw:1434685846593:dw|. 10. Vocaloid. see how sin^2(x) cancels out?. 11. Vocaloid. |dw:1434685945486:dw|. 12. Vocaloid. |dw:1434685962311:dw|. 13. NY,NY. Oh yes that makes more sense.. 14. NY,NY. Thank you for explaining that.. 15. Vocaloid. ok, so we know that #1 is not the right answer because we get 1 in the end. 16. Vocaloid. let's move on to number 2. 17. NY,NY. Ok.. 18. Vocaloid. |dw:1434686053840:dw|. 19. Vocaloid. |dw:1434686204867:dw|. 20. Vocaloid. does this make sense so far?. 21. NY,NY. yes, because you used the Pythagorean identity at the end, right?. 22.
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Vocaloid. yup! ok, we know that #2 is out, because we ended up with 1, so let's move on to #3. 23. NY,NY. Right.. 24. Vocaloid. |dw:1434686335785:dw|. 25. Vocaloid. now, I believe #3 is the right answer, but let's check #4 just in case. 26. Vocaloid. |dw:1434686426089:dw|. 27. Vocaloid. |dw:1434686523673:dw|. 28. Vocaloid. whew, that took a while, but do you understand everything so far? :). 29. Vocaloid. the important thing to remember, is that when you have csc, sec, cot, or tan, you can re-write them in terms of cos or sin, which usually helps you cross things out and simplify :). 30. NY,NY. Im a little confused with the second step you took.. 31. Vocaloid. which one? #3 or #4?. 32. NY,NY. #4.. 33. Vocaloid. can you tell me exactly which part is confusing?. 34. NY,NY. the step that the arrow is pointing to |dw:1434687552340:dw|. 35. NY,NY. Im confused how that concluded from the previous step, where you substituted the reciprocals.. 36. Vocaloid. |dw:1434687726888:dw|. 37. Vocaloid. which leaves me with|dw:1434687748727:dw|. 38. Vocaloid. does it make it more clear? or is there something you're still unsure about?. 39. NY,NY. Ok, yes that makes sense.. 40. NY,NY. So that leaves us with choice 3.. 41. Vocaloid. yup ~. 42. NY,NY. Ok, thank you for helping.
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https://goprep.co/ex-12.3-q7-the-perimeter-of-a-triangular-field-is-420-m-and-i-1nkc90
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Q. 74.5( 8 Votes )
# The perimeter of
Given that the sides are in ratio of 6:7:8
Let the sides of the triangle be 6x, 7x and 8x.
Perimeter(Δ) = 420
6x + 7x + 8x = 420
21x = 420
x = 20
Therefore the sides are 120m, 140m and 160m.
a = 120, b = 140, c = 160
s = (a + b + c)/2
s = (120 + 140 + 160)/2 = 420/2 = 210.
Area(Δ) = √s(s-a)(s-b)(s-c)
Area(Δ) = √210(210-120)(210-140)(210-160)
Area(Δ) = √210×90×70×50
Area(Δ) = 2100√15 m2
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Q. 74.5( 8 Votes ). # The perimeter of. Given that the sides are in ratio of 6:7:8. Let the sides of the triangle be 6x, 7x and 8x.. Perimeter(Δ) = 420. 6x + 7x + 8x = 420. 21x = 420. x = 20. Therefore the sides are 120m, 140m and 160m.. a = 120, b = 140, c = 160. s = (a + b + c)/2. s = (120 + 140 + 160)/2 = 420/2 = 210.. Area(Δ) = √s(s-a)(s-b)(s-c). Area(Δ) = √210(210-120)(210-140)(210-160). Area(Δ) = √210×90×70×50. Area(Δ) = 2100√15 m2. Rate this question :.
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http://docplayer.net/21763825-C-complex-numbers-1-complex-arithmetic.html
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# C. Complex Numbers. 1. Complex arithmetic.
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1 C. Complex Numbers. Complex arithmetic. Most people think that complex numbers arose from attempts to solve quadratic equations, but actually it was in connection with cubic equations they first appeared. Everyone knew that certain quadratic equations, like x 2 + = 0, or x 2 +2x+5 = 0, had no solutions. The problem was with certain cubic equations, for example x 6x+2 = 0. This equation was known to have three real roots, given by simple combinations of the expressions () A = + 7, B = 7; one of the roots for instance is A+B: it may not look like a real number, but it turns out to be one. What was to be made of the expressions A and B? They were viewed as some sort of imaginary numbers which had no meaning in themselves, but which were useful as intermediate steps in calculations that would ultimately lead to the real numbers you were looking for (such as A+B). This point of view persisted for several hundred years. But as more and more applications for these imaginary numbers were found, they gradually began to be accepted as valid numbers in their own right, even though they did not measure the length of any line segment. Nowadays we are fairly generous in the use of the word number : numbers of one sort or another don t have to measure anything, but to merit the name they must belong to a system in which some type of addition, subtraction, multiplication, and division is possible, and where these operations obey those laws of arithmetic one learns in elementary school and has usually forgotten by high school the commutative, associative, and distributive laws. To describe the complex numbers, we use a formal symbol i representing ; then a complex number is an expression of the form (2) a + bi, a, b real numbers. If a = 0 or b = 0, they are omitted (unless both are 0); thus we write a+0i = a, 0+bi = bi, 0+0i = 0. The definition of equality between two complex numbers is () a+bi = c+di a = c, b = d. This shows that the numbers a and b are uniquely determined once the complex number a+bi is given; we call them respectively the real and imaginary parts of a+bi. (It would be more logical to call bi the imaginary part, but this would be less convenient.) In symbols, (4) a = Re(a+bi), b = Im(a+bi)
2 2 8.0 NOTES Addition and multiplication of complex numbers are defined in the familiar way, making use of the fact that i 2 = : (5a) (5b) Addition Multiplication (a+bi)+(c+di) = (a+c)+(b+d)i (a+bi)(c+di) = (ac bd)+(ad+bc)i Division is a little more complicated; what is important is not so much the final formula but rather the procedure which produces it; assuming c+di 0, it is: (5c) Division a+bi c+di = a+bi c+di c di c di = ac+bd c 2 +d 2 + bc ad c 2 +d 2 i This division procedure made use of complex conjugation: if z = a + bi, we define the complex conjugate of z to be the complex number (6) z = a bi (note that z z = a 2 +b 2 ). The size of a complex number is measured by its absolute value, or modulus, defined by (7) z = a+bi = a 2 +b 2 ; (thus : z z = z 2 ). Remarks. For the sake of computers, which do not understand what a formal expression is, one can define a complex number to be just an ordered pair (a,b) of real numbers, and define the arithmetic operations accordingly; using (5b), multiplication is defined by (a,b)(c,d) = (ac bd, ad+bc). Then if we let i represent the ordered pair (0,), and a the ordered pair (a,0), it is easy to verify using the above definition of multiplication that i 2 = (0,)(0,) = (,0) = and (a,b) = (a,0)+(b,0)(0,) = a+bi, and we recover the human way of writing complex numbers. Since it is easily verified from the definition that multiplication of complex numbers is commutative: z z 2 = z 2 z, it does not matter whether the i comes before or after, i.e., whether we write z = x + yi or z = x + iy. The former is used when x and y are simple numbers because it looks better; the latter is more usual when x and y represent functions (orvaluesoffunctions), tomaketheistandoutclearlyortoavoidhavingtouseparentheses: 2. Polar representation. 2+i, 5 2πi; cos π 2 +i sin π 2, x(t)+iy(t). Complex numbers are represented geometrically by points in the plane: the number a+ib is represented by the point (a,b) in Cartesian coordinates. When the points of the plane represent complex numbers in this way, the plane is called the complex plane. By switching to polar coordinates, we can write any non-zero complex number in an alternative form. Letting as usual x = rcosθ, y = rsinθ, we get the polar form for a non-zero complex number: assuming x+iy 0, (8) x+iy = r(cosθ +isinθ). When the complex number is written in polar form, we see from (7) that r = x + iy. (absolute value, modulus)
3 C. COMPLEX NUMBERS We call θ the polar angle or the argument of x+iy. In symbols, one sometimes sees θ = arg (x+iy) (polar angle, argument). The absolute value is uniquely determined by x+iy, but the polar angle is not, since it can be increased by any integer multiple of 2π. (The complex number 0 has no polar angle.) To make θ unique, one can specify 0 θ < 2π principal value of the polar angle. This so-called principal value of the angle is sometimes indicated by writing Arg (x + iy). For example, Arg ( ) = π, arg ( ) = ±π,±π,±5π,.... Changing between Cartesian and polar representation of a complex number is essentially the same as changing between Cartesian and polar coordinates: the same equations are used. Example. Give the polar form for: i, +i, i, +i. Solution. i = i sin π 2 +i = 2(cos π 4 +i sin π 4 ) +i = 2(cos 2π +i sin 2π ) i = 2(cos π 4 +i sin π 4 ) The abbreviation cisθ is sometimes used for cosθ+isinθ; for students of science and engineering, however, it is important to get used to the exponential form for this expression: (9) e iθ = cosθ +isinθ Euler s formula. Equation (9) should be regarded as the definition of the exponential of an imaginary power. A good justification for it however is found in the infinite series e t = + t! + t2 2! + t! If we substitute iθ for t in the series, and collect the real and imaginary parts of the sum (remembering that i 2 =, i = i, i 4 =, i 5 = i,..., and so on, we get ) e iθ = ( θ2 2! + θ4 4!... + i = cosθ + isinθ, in view of the infinite series representations for cosθ and sinθ. ) (θ θ! + θ5 5!... Since we only know that the series expansion for e t is valid when t is a real number, the above argument is only suggestive it is not a proof of (9). What it shows is that Euler s formula (9) is formally compatible with the series expansions for the exponential, sine, and cosine functions. Using the complex exponential, the polar representation (8) is written (0) x+iy = re iθ The most important reason for polar representation is that multiplication and division of complex numbers is particularly simple when they are written in polar form. Indeed, by using Euler s formula (9) and the trigonometric addition formulas, it is not hard to show
4 4 8.0 NOTES () e iθ e iθ = e i(θ+θ ). This gives another justification for the definition (9) it makes the complex exponential follow the same exponential addition law as the real exponential. The law () leads to the simple rules for multiplying and dividing complex numbers written in polar form: (2a) multiplication rule re iθ r e iθ = rr e i(θ+θ ) ; to multiply two complex numbers, you multiply the absolute values and add the angles. (2b) reciprocal rule re iθ = r e iθ ; (2c) division rule r = r e iθ r e i(θ θ ) ; to divide by a complex number, divide by its absolute value and subtract its angle. re iθ The reciprocal rule (2b) follows from (2a), which shows that r e iθ re iθ =. The division rule follows by writing reiθ r e iθ = r e iθ re iθ and using (2b) and then (2a). Using (2a), we can raise x+iy to a positive integer power by first using x+iy = re iθ ; the special case when r = is called DeMoivre s formula: () (x+iy) n = r n e inθ ; DeMoivre s formula: (cosθ+isinθ) n = cosnθ+isinnθ. Example 2. Express a) (+i) 6 in Cartesian form; b) +i +i in polar form. Solution. a) Change to polar form, use (), then change back to Cartesian form: (+i) 6 = ( 2e iπ/4 ) 6 = ( 2) 6 e i6π/4 = 8e iπ/2 = 8i. +i b) Changing to polar form, = 2eiπ/ +i 2e = iπ/6 eiπ/6, using the division rule (2c). You can check the answer to (a) by applying the binomial theorem to (+i) 6 and collecting the real and imaginary parts; to (b) by doing the division in Cartesian form (5c), then converting the answer to polar form.. Complex exponentials Because of the importance of complex exponentials in differential equations, and in science and engineering generally, we go a little further with them. Euler s formula (9) defines the exponential to a pure imaginary power. The definition of an exponential to an arbitrary complex power is: (4) e a+ib = e a e ib = e a (cosb+isinb). We stress that the equation (4) is a definition, not a self-evident truth, since up to now no meaning has been assigned to the left-hand side. From (4) we see that (5) Re(e a+ib ) = e a cosb, Im(e a+ib ) = e a sinb.
5 The complex exponential obeys the usual law of exponents: (6) e z+z = e z e z, as is easily seen by combining (4) and (). C. COMPLEX NUMBERS 5 The complex exponential is expressed in terms of the sine and cosine by Euler s formula (9). Conversely, the sin and cos functions can be expressed in terms of complex exponentials. There are two important ways of doing this, both of which you should learn: (7) cosx = Re(e ix ), sinx = Im(e ix ) ; (8) cosx = 2 (eix +e ix ), sinx = 2i (eix e ix ). The equations in (8) follow easily from Euler s formula (9); their derivation is left for the exercises. Here are some examples of their use. Example. Express cos x in terms of the functions cosnx, for suitable n. Solution. We use (8) and the binomial theorem, then (8) again: cos x = 8 (eix +e ix ) = 8 (eix +e ix +e ix +e ix ) = 4 cosx+ 4 cosx. As a preliminary to the next example, we note that a function like e ix = cosx+isinx is a complex-valued function of the real variable x. Such a function may be written as u(x)+iv(x), u, v real-valued and its derivative and integral with respect to x are defined to be (9a,b) a) D(u+iv) = Du+iDv, b) (u+iv)dx = udx+i vdx. From this it follows by a calculation that (20) D(e (a+ib)x = (a+ib)e (a+ib)x, and therefore e (a+ib)x dx = a+ib e(a+ib)x. Example 4. Calculate e x cos2xdx by using complex exponentials. Solution. The usual method is a tricky use of two successive integration by parts. Using complex exponentials instead, the calculation is straightforward. We have ( e x cos2x = Re e (+2i)x), by (4) or (5); therefore ( ) e x cos2xdx = Re e (+2i)x dx, by (9b). Calculating the integral, e (+2i)x dx = +2i e(+2i)x by (20); ( = 5 2 ) (e 5 i x cos2x+ie x sin2x ), using (4) and complex division (5c). According to the second line above, we want the real part of this last expression. Multiply using (5b) and take the real part; you get 5 ex cos2x+ 2 5 ex sin2x.
6 6 8.0 NOTES In this differential equations course, we will make free use of complex exponentials in solving differential equations, and in doing formal calculations like the ones above. This is standard practice in science and engineering, and you need to get used to it. 4. Finding n-th roots. To solve linear differential equations with constant coefficients, you need to be able find the real and complex roots of polynomial equations. Though a lot of this is done today with calculators and computers, one still has to know how to do an important special case by hand: finding the roots of z n = α, where α is a complex number, i.e., finding the n-th roots of α. Polar representation will be a big help in this. Let s begin with a special case: the n-th roots of unity: the solutions to z n =. To solve this equation, we use polar representation for both sides, setting z = re iθ on the left, and using all possible polar angles on the right; using the exponential law to multiply, the above equation then becomes r n e inθ = e (2kπi), k = 0,±,±2,.... Equating the absolute values and the polar angles of the two sides gives from which we conclude that ( ) r =, θ = 2kπ n r n =, nθ = 2kπ, k = 0,±,±2,...,, k = 0,,...,n. In the above, we get only the value r =, since r must be real and non-negative. We don t need any integer values of k other than 0,...,n since they would not produce a complex number different from the above n numbers. That is, if we add an, an integer multiple of n, to k, we get the same complex number: θ = 2(k +an)π n = θ +2aπ; and e iθ = e iθ, since e 2aπi = (e 2πi ) a =. We conclude from ( ) therefore that (2) the n-th roots of are the numbers e 2kπi/n, k = 0,...,n. 2π i e πi e This shows there are n complex n-th roots of unity. They all lie on the unit circle in the complex plane, since they have absolute value ; they are evenly spaced around the unit circle, starting with ; the angle between two consecutive ones is 2π/n. These facts are illustrated on the right for the case n = 6. 4πi π i e e 5
7 C. COMPLEX NUMBERS 7 From (2), we get another notation for the roots of unity (ζ is the Greek letter zeta ): (22) the n-th roots of are,ζ,ζ 2,...,ζ n, where ζ = e 2πi/n. We now generalize the above to find the n-th roots of an arbitrary complex number w. We begin by writing w in polar form: w = re iθ ; θ = Arg w, 0 θ < 2π, i.e., θ is the principal value of the polar angle of w. Then the same reasoning as we used above shows that if z is an n-th root of w, then (2) z n = w = re iθ, so z = n re i(θ+2kπ)/n, k = 0,,...,n. Comparing this with (22), we see that these n roots can be written in the suggestive form (24) n w = z 0, z 0 ζ, z 0 ζ 2,..., z 0 ζ n, where z 0 = n re iθ/n. As a check, we see that all of the n complex numbers in (24) satisfy z n = w : (z 0 ζ i ) n = z0ζ n ni = z0 n i, since ζ n =, by (22); = w, by the definition (24) of z 0 and (2). Example 5. Find in Cartesian form all values of a) b) 4 i. Solution. a) According to (22), the cube roots of are,ω, and ω 2, where ω = e 2πi/ = cos 2π +isin 2π ω 2 = e 2πi/ = cos 2π +isin 2π = 2 +i 2 = 2 i 2. The greek letter ω ( omega ) is traditionally used for this cube root. Note that for the polar angle of ω 2 we used 2π/ rather than the equivalent angle 4π/, in order to take advantage of the identities cos( x) = cosx, sin( x) = sinx. Note that ω 2 = ω. Another way to do this problem would be to draw the position of ω 2 and ω on the unit circle, and use geometry to figure out their coordinates. b) To find 4 i, we can use (24). We know that 4 =,i,, i (either by drawing the unit circle picture, or by using (22)). Therefore by (24), we get 4 i = z 0, z 0 i, z 0, z 0 i, where z 0 = e πi/8 = cos π 8 +isin π 8 ; = a+ib, b+ia, a ib, b ia, where z 0 = a+ib = cos π 8 +isin π 8.
8 8 8.0 NOTES Example 6. Solve the equation x 6 2x +2 = 0. Solution. Treating this as a quadratic equation in x, we solve the quadratic by using the quadratic formula, the two roots are + i and i (check this!), so the roots of the original equation satisfy either x = +i, or x = i. This reduces the problem to finding the cube roots of the two complex numbers ± i. We begin by writing them in polar form: +i = 2e πi/4, i = 2e πi/4. (Once again, note the use of the negative polar angle for i, which is more convenient for calculations.) The three cube roots of the first of these are (by (2)), 6 2 e πi/2 = 6 ( 2 cos π 2 +isin π e πi/4 = 6 ( 2 cos π 4 +isin π e 7πi/2 = 6 ( 2 ) cos 7π 7π isin 2 2 ), since ), since The second cube root can also be written as 6 ( ) +i 2 2 π 2 + 2π = π 4 ; π 2 2π = 7π 2. = +i 2. This gives three of the cube roots. The other three are the cube roots of i, which may be found by replacing i by i everywhere above (i.e., taking the complex conjugate). The cube roots can also according to (24) be described as z, z ω, z ω 2 and z 2, z 2 ω, z 2 ω 2, where z = 6 2 e πi/2, z 2 = 6 2 e πi/2. Exercises: Section 2E
9 M.I.T. 8.0 Ordinary Differential Equations 8.0 Notes and Exercises c Arthur Mattuck and M.I.T. 988, 992, 996, 200, 2007, 20
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# C. Complex Numbers. 1. Complex arithmetic.. Save this PDF as:. Size: px. Start display at page:. ## Transcription. 1 C. Complex Numbers. Complex arithmetic. Most people think that complex numbers arose from attempts to solve quadratic equations, but actually it was in connection with cubic equations they first appeared. Everyone knew that certain quadratic equations, like x 2 + = 0, or x 2 +2x+5 = 0, had no solutions. The problem was with certain cubic equations, for example x 6x+2 = 0. This equation was known to have three real roots, given by simple combinations of the expressions () A = + 7, B = 7; one of the roots for instance is A+B: it may not look like a real number, but it turns out to be one. What was to be made of the expressions A and B? They were viewed as some sort of imaginary numbers which had no meaning in themselves, but which were useful as intermediate steps in calculations that would ultimately lead to the real numbers you were looking for (such as A+B). This point of view persisted for several hundred years. But as more and more applications for these imaginary numbers were found, they gradually began to be accepted as valid numbers in their own right, even though they did not measure the length of any line segment. Nowadays we are fairly generous in the use of the word number : numbers of one sort or another don t have to measure anything, but to merit the name they must belong to a system in which some type of addition, subtraction, multiplication, and division is possible, and where these operations obey those laws of arithmetic one learns in elementary school and has usually forgotten by high school the commutative, associative, and distributive laws. To describe the complex numbers, we use a formal symbol i representing ; then a complex number is an expression of the form (2) a + bi, a, b real numbers. If a = 0 or b = 0, they are omitted (unless both are 0); thus we write a+0i = a, 0+bi = bi, 0+0i = 0. The definition of equality between two complex numbers is () a+bi = c+di a = c, b = d. This shows that the numbers a and b are uniquely determined once the complex number a+bi is given; we call them respectively the real and imaginary parts of a+bi. (It would be more logical to call bi the imaginary part, but this would be less convenient.) In symbols, (4) a = Re(a+bi), b = Im(a+bi). 2 2 8.0 NOTES Addition and multiplication of complex numbers are defined in the familiar way, making use of the fact that i 2 = : (5a) (5b) Addition Multiplication (a+bi)+(c+di) = (a+c)+(b+d)i (a+bi)(c+di) = (ac bd)+(ad+bc)i Division is a little more complicated; what is important is not so much the final formula but rather the procedure which produces it; assuming c+di 0, it is: (5c) Division a+bi c+di = a+bi c+di c di c di = ac+bd c 2 +d 2 + bc ad c 2 +d 2 i This division procedure made use of complex conjugation: if z = a + bi, we define the complex conjugate of z to be the complex number (6) z = a bi (note that z z = a 2 +b 2 ). The size of a complex number is measured by its absolute value, or modulus, defined by (7) z = a+bi = a 2 +b 2 ; (thus : z z = z 2 ). Remarks. For the sake of computers, which do not understand what a formal expression is, one can define a complex number to be just an ordered pair (a,b) of real numbers, and define the arithmetic operations accordingly; using (5b), multiplication is defined by (a,b)(c,d) = (ac bd, ad+bc). Then if we let i represent the ordered pair (0,), and a the ordered pair (a,0), it is easy to verify using the above definition of multiplication that i 2 = (0,)(0,) = (,0) = and (a,b) = (a,0)+(b,0)(0,) = a+bi, and we recover the human way of writing complex numbers. Since it is easily verified from the definition that multiplication of complex numbers is commutative: z z 2 = z 2 z, it does not matter whether the i comes before or after, i.e., whether we write z = x + yi or z = x + iy. The former is used when x and y are simple numbers because it looks better; the latter is more usual when x and y represent functions (orvaluesoffunctions), tomaketheistandoutclearlyortoavoidhavingtouseparentheses: 2. Polar representation. 2+i, 5 2πi; cos π 2 +i sin π 2, x(t)+iy(t). Complex numbers are represented geometrically by points in the plane: the number a+ib is represented by the point (a,b) in Cartesian coordinates. When the points of the plane represent complex numbers in this way, the plane is called the complex plane. By switching to polar coordinates, we can write any non-zero complex number in an alternative form. Letting as usual x = rcosθ, y = rsinθ, we get the polar form for a non-zero complex number: assuming x+iy 0, (8) x+iy = r(cosθ +isinθ). When the complex number is written in polar form, we see from (7) that r = x + iy. (absolute value, modulus). 3 C. COMPLEX NUMBERS We call θ the polar angle or the argument of x+iy. In symbols, one sometimes sees θ = arg (x+iy) (polar angle, argument). The absolute value is uniquely determined by x+iy, but the polar angle is not, since it can be increased by any integer multiple of 2π. (The complex number 0 has no polar angle.) To make θ unique, one can specify 0 θ < 2π principal value of the polar angle. This so-called principal value of the angle is sometimes indicated by writing Arg (x + iy). For example, Arg ( ) = π, arg ( ) = ±π,±π,±5π,.... Changing between Cartesian and polar representation of a complex number is essentially the same as changing between Cartesian and polar coordinates: the same equations are used. Example. Give the polar form for: i, +i, i, +i. Solution. i = i sin π 2 +i = 2(cos π 4 +i sin π 4 ) +i = 2(cos 2π +i sin 2π ) i = 2(cos π 4 +i sin π 4 ) The abbreviation cisθ is sometimes used for cosθ+isinθ; for students of science and engineering, however, it is important to get used to the exponential form for this expression: (9) e iθ = cosθ +isinθ Euler s formula. Equation (9) should be regarded as the definition of the exponential of an imaginary power. A good justification for it however is found in the infinite series e t = + t! + t2 2! + t! If we substitute iθ for t in the series, and collect the real and imaginary parts of the sum (remembering that i 2 =, i = i, i 4 =, i 5 = i,..., and so on, we get ) e iθ = ( θ2 2! + θ4 4!... + i = cosθ + isinθ, in view of the infinite series representations for cosθ and sinθ. ) (θ θ! + θ5 5!... Since we only know that the series expansion for e t is valid when t is a real number, the above argument is only suggestive it is not a proof of (9). What it shows is that Euler s formula (9) is formally compatible with the series expansions for the exponential, sine, and cosine functions. Using the complex exponential, the polar representation (8) is written (0) x+iy = re iθ The most important reason for polar representation is that multiplication and division of complex numbers is particularly simple when they are written in polar form. Indeed, by using Euler s formula (9) and the trigonometric addition formulas, it is not hard to show. 4 4 8.0 NOTES () e iθ e iθ = e i(θ+θ ). This gives another justification for the definition (9) it makes the complex exponential follow the same exponential addition law as the real exponential. The law () leads to the simple rules for multiplying and dividing complex numbers written in polar form: (2a) multiplication rule re iθ r e iθ = rr e i(θ+θ ) ; to multiply two complex numbers, you multiply the absolute values and add the angles. (2b) reciprocal rule re iθ = r e iθ ; (2c) division rule r = r e iθ r e i(θ θ ) ; to divide by a complex number, divide by its absolute value and subtract its angle. re iθ The reciprocal rule (2b) follows from (2a), which shows that r e iθ re iθ =. The division rule follows by writing reiθ r e iθ = r e iθ re iθ and using (2b) and then (2a). Using (2a), we can raise x+iy to a positive integer power by first using x+iy = re iθ ; the special case when r = is called DeMoivre s formula: () (x+iy) n = r n e inθ ; DeMoivre s formula: (cosθ+isinθ) n = cosnθ+isinnθ. Example 2. Express a) (+i) 6 in Cartesian form; b) +i +i in polar form. Solution. a) Change to polar form, use (), then change back to Cartesian form: (+i) 6 = ( 2e iπ/4 ) 6 = ( 2) 6 e i6π/4 = 8e iπ/2 = 8i. +i b) Changing to polar form, = 2eiπ/ +i 2e = iπ/6 eiπ/6, using the division rule (2c). You can check the answer to (a) by applying the binomial theorem to (+i) 6 and collecting the real and imaginary parts; to (b) by doing the division in Cartesian form (5c), then converting the answer to polar form.. Complex exponentials Because of the importance of complex exponentials in differential equations, and in science and engineering generally, we go a little further with them. Euler s formula (9) defines the exponential to a pure imaginary power. The definition of an exponential to an arbitrary complex power is: (4) e a+ib = e a e ib = e a (cosb+isinb). We stress that the equation (4) is a definition, not a self-evident truth, since up to now no meaning has been assigned to the left-hand side. From (4) we see that (5) Re(e a+ib ) = e a cosb, Im(e a+ib ) = e a sinb.. 5 The complex exponential obeys the usual law of exponents: (6) e z+z = e z e z, as is easily seen by combining (4) and (). C. COMPLEX NUMBERS 5 The complex exponential is expressed in terms of the sine and cosine by Euler s formula (9). Conversely, the sin and cos functions can be expressed in terms of complex exponentials. There are two important ways of doing this, both of which you should learn: (7) cosx = Re(e ix ), sinx = Im(e ix ) ; (8) cosx = 2 (eix +e ix ), sinx = 2i (eix e ix ). The equations in (8) follow easily from Euler s formula (9); their derivation is left for the exercises. Here are some examples of their use. Example. Express cos x in terms of the functions cosnx, for suitable n. Solution. We use (8) and the binomial theorem, then (8) again: cos x = 8 (eix +e ix ) = 8 (eix +e ix +e ix +e ix ) = 4 cosx+ 4 cosx. As a preliminary to the next example, we note that a function like e ix = cosx+isinx is a complex-valued function of the real variable x. Such a function may be written as u(x)+iv(x), u, v real-valued and its derivative and integral with respect to x are defined to be (9a,b) a) D(u+iv) = Du+iDv, b) (u+iv)dx = udx+i vdx. From this it follows by a calculation that (20) D(e (a+ib)x = (a+ib)e (a+ib)x, and therefore e (a+ib)x dx = a+ib e(a+ib)x. Example 4. Calculate e x cos2xdx by using complex exponentials. Solution. The usual method is a tricky use of two successive integration by parts. Using complex exponentials instead, the calculation is straightforward. We have ( e x cos2x = Re e (+2i)x), by (4) or (5); therefore ( ) e x cos2xdx = Re e (+2i)x dx, by (9b). Calculating the integral, e (+2i)x dx = +2i e(+2i)x by (20); ( = 5 2 ) (e 5 i x cos2x+ie x sin2x ), using (4) and complex division (5c). According to the second line above, we want the real part of this last expression. Multiply using (5b) and take the real part; you get 5 ex cos2x+ 2 5 ex sin2x.. 6 6 8.0 NOTES In this differential equations course, we will make free use of complex exponentials in solving differential equations, and in doing formal calculations like the ones above. This is standard practice in science and engineering, and you need to get used to it. 4. Finding n-th roots. To solve linear differential equations with constant coefficients, you need to be able find the real and complex roots of polynomial equations. Though a lot of this is done today with calculators and computers, one still has to know how to do an important special case by hand: finding the roots of z n = α, where α is a complex number, i.e., finding the n-th roots of α. Polar representation will be a big help in this. Let s begin with a special case: the n-th roots of unity: the solutions to z n =. To solve this equation, we use polar representation for both sides, setting z = re iθ on the left, and using all possible polar angles on the right; using the exponential law to multiply, the above equation then becomes r n e inθ = e (2kπi), k = 0,±,±2,.... Equating the absolute values and the polar angles of the two sides gives from which we conclude that ( ) r =, θ = 2kπ n r n =, nθ = 2kπ, k = 0,±,±2,...,, k = 0,,...,n. In the above, we get only the value r =, since r must be real and non-negative. We don t need any integer values of k other than 0,...,n since they would not produce a complex number different from the above n numbers. That is, if we add an, an integer multiple of n, to k, we get the same complex number: θ = 2(k +an)π n = θ +2aπ; and e iθ = e iθ, since e 2aπi = (e 2πi ) a =. We conclude from ( ) therefore that (2) the n-th roots of are the numbers e 2kπi/n, k = 0,...,n. 2π i e πi e This shows there are n complex n-th roots of unity. They all lie on the unit circle in the complex plane, since they have absolute value ; they are evenly spaced around the unit circle, starting with ; the angle between two consecutive ones is 2π/n. These facts are illustrated on the right for the case n = 6. 4πi π i e e 5. 7 C. COMPLEX NUMBERS 7 From (2), we get another notation for the roots of unity (ζ is the Greek letter zeta ): (22) the n-th roots of are,ζ,ζ 2,...,ζ n, where ζ = e 2πi/n. We now generalize the above to find the n-th roots of an arbitrary complex number w. We begin by writing w in polar form: w = re iθ ; θ = Arg w, 0 θ < 2π, i.e., θ is the principal value of the polar angle of w. Then the same reasoning as we used above shows that if z is an n-th root of w, then (2) z n = w = re iθ, so z = n re i(θ+2kπ)/n, k = 0,,...,n. Comparing this with (22), we see that these n roots can be written in the suggestive form (24) n w = z 0, z 0 ζ, z 0 ζ 2,..., z 0 ζ n, where z 0 = n re iθ/n. As a check, we see that all of the n complex numbers in (24) satisfy z n = w : (z 0 ζ i ) n = z0ζ n ni = z0 n i, since ζ n =, by (22); = w, by the definition (24) of z 0 and (2). Example 5. Find in Cartesian form all values of a) b) 4 i. Solution. a) According to (22), the cube roots of are,ω, and ω 2, where ω = e 2πi/ = cos 2π +isin 2π ω 2 = e 2πi/ = cos 2π +isin 2π = 2 +i 2 = 2 i 2. The greek letter ω ( omega ) is traditionally used for this cube root. Note that for the polar angle of ω 2 we used 2π/ rather than the equivalent angle 4π/, in order to take advantage of the identities cos( x) = cosx, sin( x) = sinx. Note that ω 2 = ω. Another way to do this problem would be to draw the position of ω 2 and ω on the unit circle, and use geometry to figure out their coordinates. b) To find 4 i, we can use (24). We know that 4 =,i,, i (either by drawing the unit circle picture, or by using (22)). Therefore by (24), we get 4 i = z 0, z 0 i, z 0, z 0 i, where z 0 = e πi/8 = cos π 8 +isin π 8 ; = a+ib, b+ia, a ib, b ia, where z 0 = a+ib = cos π 8 +isin π 8.. 8 8 8.0 NOTES Example 6. Solve the equation x 6 2x +2 = 0. Solution. Treating this as a quadratic equation in x, we solve the quadratic by using the quadratic formula, the two roots are + i and i (check this!), so the roots of the original equation satisfy either x = +i, or x = i. This reduces the problem to finding the cube roots of the two complex numbers ± i. We begin by writing them in polar form: +i = 2e πi/4, i = 2e πi/4. (Once again, note the use of the negative polar angle for i, which is more convenient for calculations.) The three cube roots of the first of these are (by (2)), 6 2 e πi/2 = 6 ( 2 cos π 2 +isin π e πi/4 = 6 ( 2 cos π 4 +isin π e 7πi/2 = 6 ( 2 ) cos 7π 7π isin 2 2 ), since ), since The second cube root can also be written as 6 ( ) +i 2 2 π 2 + 2π = π 4 ; π 2 2π = 7π 2. = +i 2. This gives three of the cube roots. The other three are the cube roots of i, which may be found by replacing i by i everywhere above (i.e., taking the complex conjugate). The cube roots can also according to (24) be described as z, z ω, z ω 2 and z 2, z 2 ω, z 2 ω 2, where z = 6 2 e πi/2, z 2 = 6 2 e πi/2. Exercises: Section 2E. 9 M.I.T. 8.0 Ordinary Differential Equations 8.0 Notes and Exercises c Arthur Mattuck and M.I.T. 988, 992, 996, 200, 2007, 20. ### COMPLEX NUMBERS. a bi c di a c b d i. a bi c di a c b d i For instance, 1 i 4 7i 1 4 1 7 i 5 6i. COMPLEX NUMBERS _4+i _-i FIGURE Complex numbers as points in the Arg plane i _i +i -i A complex number can be represented by an expression of the form a bi, where a b are real numbers i is a symbol with. ### THE COMPLEX EXPONENTIAL FUNCTION. Math 307 THE COMPLEX EXPONENTIAL FUNCTION (These notes assume you are already familiar with the basic properties of complex numbers.) We make the following definition e iθ = cos θ + i sin θ. 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We re thrilled that you ve decided to make us part of your homeschool curriculum. This lesson. ### Core Maths C1. Revision Notes. Core Maths C Revision Notes November 0 Core Maths C Algebra... Indices... Rules of indices... Surds... 4 Simplifying surds... 4 Rationalising the denominator... 4 Quadratic functions... 4 Completing the. ### MTH304: Honors Algebra II. MTH304: Honors Algebra II This course builds upon algebraic concepts covered in Algebra. Students extend their knowledge and understanding by solving open-ended problems and thinking critically. Topics. ### a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4). ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x. ### Mathematics Pre-Test Sample Questions A. { 11, 7} B. { 7,0,7} C. { 7, 7} D. { 11, 11}. Mathematics Pre-Test Sample Questions 1. 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NCDPI staff are continually updating and improving these tools to better serve teachers. Algebra. ### 2 Complex Functions and the Cauchy-Riemann Equations. 2 Complex Functions and the Cauchy-Riemann Equations 2.1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z). ### The program also provides supplemental modules on topics in geometry and probability and statistics.. Algebra 1 Course Overview Students develop algebraic fluency by learning the skills needed to solve equations and perform important manipulations with numbers, variables, equations, and inequalities. Students. ### Continued Fractions and the Euclidean Algorithm. Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction. ### Homework 5 Solutions. Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which. ### Student Performance Q&A:. Student Performance Q&A: AP Calculus AB and Calculus BC Free-Response Questions The following comments on the free-response questions for AP Calculus AB and Calculus BC were written by the Chief Reader,. ### MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education). MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education) Accurately add, subtract, multiply, and divide whole numbers, integers,. ### Precalculus with Geometry and Trigonometry. 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Fields 4 1.3. The field of complex numbers. 6 1.4.
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# Tent map
Graph of tent map function
In mathematics, the tent map with parameter μ is the real-valued function fμ defined by
${\displaystyle f_{\mu }:=\mu \min\{x,\,1-x\},}$
the name being due to the tent-like shape of the graph of fμ. For the values of the parameter μ within 0 and 2, fμ maps the unit interval [0, 1] into itself, thus defining a discrete-time dynamical system on it (equivalently, a recurrence relation). In particular, iterating a point x0 in [0, 1] gives rise to a sequence ${\displaystyle x_{n}}$ :
${\displaystyle x_{n+1}=f_{\mu }(x_{n})={\begin{cases}\mu x_{n}&\mathrm {for} ~~x_{n}<{\frac {1}{2}}\\\\\mu (1-x_{n})&\mathrm {for} ~~{\frac {1}{2}}\leq x_{n}\end{cases}}}$
where μ is a positive real constant. Choosing for instance the parameter μ=2, the effect of the function fμ may be viewed as the result of the operation of folding the unit interval in two, then stretching the resulting interval [0,1/2] to get again the interval [0,1]. Iterating the procedure, any point x0 of the interval assumes new subsequent positions as described above, generating a sequence xn in [0,1].
The ${\displaystyle \mu =2}$ case of the tent map is a non-linear transformation of both the bit shift map and the r=4 case of the logistic map.
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• ✪ ChaosBook.org chapter Charting the state space - Tent map
• ✪ ChaosBook.org chapter Stretch, fold, prune - Tent map symbolic dynamics
• ✪ Lecture - 9 The Logistic Map and Period doubling
• ✪ [NJIT Math 491 - 695] Chaos - Lecture 4: Maps! What are they good for!?
• ✪ ChaosBook.org chapter Charting the state space - Bernoulli map
## Behaviour
Orbits of unit-height tent map
Bifurcation diagram for the tent map. Higher density indicates increased probability of the x variable acquiring that value for the given value of the μ parameter.
The tent map with parameter μ=2 and the logistic map with parameter r=4 are topologically conjugate,[1] and thus the behaviours of the two maps are in this sense identical under iteration.
Depending on the value of μ, the tent map demonstrates a range of dynamical behaviour ranging from predictable to chaotic.
• If μ is less than 1 the point x = 0 is an attractive fixed point of the system for all initial values of x i.e. the system will converge towards x = 0 from any initial value of x.
• If μ is 1 all values of x less than or equal to 1/2 are fixed points of the system.
• If μ is greater than 1 the system has two fixed points, one at 0, and the other at μ/(μ + 1). Both fixed points are unstable i.e. a value of x close to either fixed point will move away from it, rather than towards it. For example, when μ is 1.5 there is a fixed point at x = 0.6 (because 1.5(1 − 0.6) = 0.6) but starting at x = 0.61 we get
${\displaystyle 0.61\to 0.585\to 0.6225\to 0.56625\to 0.650625\ldots }$
• If μ is between 1 and the square root of 2 the system maps a set of intervals between μ − μ2/2 and μ/2 to themselves. This set of intervals is the Julia set of the map i.e. it is the smallest invariant sub-set of the real line under this map. If μ is greater than the square root of 2, these intervals merge, and the Julia set is the whole interval from μ − μ2/2 to μ/2 (see bifurcation diagram).
• If μ is between 1 and 2 the interval [μ − μ2/2, μ/2] contains both periodic and non-periodic points, although all of the orbits are unstable (i.e. nearby points move away from the orbits rather than towards them). Orbits with longer lengths appear as μ increases. For example:
${\displaystyle {\frac {\mu }{\mu ^{2}+1}}\to {\frac {\mu ^{2}}{\mu ^{2}+1}}\to {\frac {\mu }{\mu ^{2}+1}}{\mbox{ appears at }}\mu =1}$
${\displaystyle {\frac {\mu }{\mu ^{3}+1}}\to {\frac {\mu ^{2}}{\mu ^{3}+1}}\to {\frac {\mu ^{3}}{\mu ^{3}+1}}\to {\frac {\mu }{\mu ^{3}+1}}{\mbox{ appears at }}\mu ={\frac {1+{\sqrt {5}}}{2}}}$
${\displaystyle {\frac {\mu }{\mu ^{4}+1}}\to {\frac {\mu ^{2}}{\mu ^{4}+1}}\to {\frac {\mu ^{3}}{\mu ^{4}+1}}\to {\frac {\mu ^{4}}{\mu ^{4}+1}}\to {\frac {\mu }{\mu ^{4}+1}}{\mbox{ appears at }}\mu \approx 1.8393}$
• If μ equals 2 the system maps the interval [0,1] onto itself. There are now periodic points with every orbit length within this interval, as well as non-periodic points. The periodic points are dense in [0,1], so the map has become chaotic. In fact, the dynamics will be non-periodic if and only if ${\displaystyle x_{0}}$ is irrational. This can be seen by noting what the map does when ${\displaystyle x_{n}}$ is expressed in binary notation: It shifts the binary point one place to the right; then, if what appears to the left of the binary point is a "one" it changes all ones to zeroes and vice versa (with the exception of the final bit "one" in the case of a finite binary expansion); starting from an irrational number, this process goes on forever without repeating itself. The invariant measure for x is the uniform density over the unit interval.[2] The autocorrelation function for a sufficiently long sequence {${\displaystyle x_{n}}$} will show zero autocorrelation at all non-zero lags.[3] Thus ${\displaystyle {x_{n}}}$ cannot be distinguished from white noise using the autocorrelation function. Note that the r=4 case of the logistic map and the ${\displaystyle \mu =2}$ case of the tent map are homeomorphic to each other: Denoting the logistically evolving variable as ${\displaystyle y_{n}}$, the homeomorphism is
${\displaystyle x_{n}={\tfrac {2}{\pi }}\sin ^{-1}(y_{n}^{1/2}).}$
• If μ is greater than 2 the map's Julia set becomes disconnected, and breaks up into a Cantor set within the interval [0,1]. The Julia set still contains an infinite number of both non-periodic and periodic points (including orbits for any orbit length) but almost every point within [0,1] will now eventually diverge towards infinity. The canonical Cantor set (obtained by successively deleting middle thirds from subsets of the unit line) is the Julia set of the tent map for μ = 3.
### Numerical errors
Time series of the Tent map for the parameter m=2.0 which shows numerical error: "the plot of time series (plot of x variable with respect to number of iterations) stops fluctuating and no values are observed after n=50". Parameter m= 2.0, initial point is random.
## Magnifying the orbit diagram
Magnification near the tip shows more details.
• A closer look at the orbit diagram shows that there are 4 separated regions at μ ≈ 1. For further magnification, 2 reference lines (red) are drawn from the tip to suitable x at certain μ (e.g., 1.10) as shown.
Further magnification shows 8 separated regions.
• With distance measured from the corresponding reference lines, further detail appears in the upper and lower part of the map. (total 8 separated regions at some μ)
## Asymmetric tent map
The asymmetric tent map is essentially a distorted, but still piecewise linear, version of the ${\displaystyle \mu =2}$ case of the tent map. It is defined by
${\displaystyle v_{n+1}={\begin{cases}v_{n}/a&\mathrm {for} ~~v_{n}\in [0,a]\\\\(1-v_{n})/(1-a)&\mathrm {for} ~~v_{n}\in [a,1]\end{cases}}}$
for parameter ${\displaystyle a\in [0,1]}$. The ${\displaystyle \mu =2}$ case of the tent map is the present case of ${\displaystyle a={\tfrac {1}{2}}}$. A sequence {${\displaystyle v_{n}}$} will have the same autocorrelation function [3] as will data from the first-order autoregressive process ${\displaystyle w_{n+1}=(2a-1)w_{n}+u_{n+1}}$ with {${\displaystyle u_{n}}$} independently and identically distributed. Thus data from an asymmetric tent map cannot be distinguished, using the autocorrelation function, from data generated by a first-order autoregressive process.
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To install click the Add extension button. That's it.. The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.. 4,5. Kelly Slayton. Congratulations on this excellent venture… what a great idea!. Alexander Grigorievskiy. I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.. Live Statistics. English Articles. Improved in 24 Hours. Languages. Recent. Show all languages. What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.. .. Leo. Newton. Brights. Milds. # Tent map. Graph of tent map function. In mathematics, the tent map with parameter μ is the real-valued function fμ defined by. ${\displaystyle f_{\mu }:=\mu \min\{x,\,1-x\},}$. the name being due to the tent-like shape of the graph of fμ. For the values of the parameter μ within 0 and 2, fμ maps the unit interval [0, 1] into itself, thus defining a discrete-time dynamical system on it (equivalently, a recurrence relation). In particular, iterating a point x0 in [0, 1] gives rise to a sequence ${\displaystyle x_{n}}$ :. ${\displaystyle x_{n+1}=f_{\mu }(x_{n})={\begin{cases}\mu x_{n}&\mathrm {for} ~~x_{n}<{\frac {1}{2}}\\\\\mu (1-x_{n})&\mathrm {for} ~~{\frac {1}{2}}\leq x_{n}\end{cases}}}$. where μ is a positive real constant. Choosing for instance the parameter μ=2, the effect of the function fμ may be viewed as the result of the operation of folding the unit interval in two, then stretching the resulting interval [0,1/2] to get again the interval [0,1]. Iterating the procedure, any point x0 of the interval assumes new subsequent positions as described above, generating a sequence xn in [0,1].. The ${\displaystyle \mu =2}$ case of the tent map is a non-linear transformation of both the bit shift map and the r=4 case of the logistic map.. • 1/5. Views:. 726. 489. 16 492. 416. 363. • ✪ ChaosBook.org chapter Charting the state space - Tent map. • ✪ ChaosBook.org chapter Stretch, fold, prune - Tent map symbolic dynamics. • ✪ Lecture - 9 The Logistic Map and Period doubling. • ✪ [NJIT Math 491 - 695] Chaos - Lecture 4: Maps! What are they good for!?. • ✪ ChaosBook.org chapter Charting the state space - Bernoulli map. ## Behaviour. Orbits of unit-height tent map.
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Bifurcation diagram for the tent map. Higher density indicates increased probability of the x variable acquiring that value for the given value of the μ parameter.. The tent map with parameter μ=2 and the logistic map with parameter r=4 are topologically conjugate,[1] and thus the behaviours of the two maps are in this sense identical under iteration.. Depending on the value of μ, the tent map demonstrates a range of dynamical behaviour ranging from predictable to chaotic.. • If μ is less than 1 the point x = 0 is an attractive fixed point of the system for all initial values of x i.e. the system will converge towards x = 0 from any initial value of x.. • If μ is 1 all values of x less than or equal to 1/2 are fixed points of the system.. • If μ is greater than 1 the system has two fixed points, one at 0, and the other at μ/(μ + 1). Both fixed points are unstable i.e. a value of x close to either fixed point will move away from it, rather than towards it. For example, when μ is 1.5 there is a fixed point at x = 0.6 (because 1.5(1 − 0.6) = 0.6) but starting at x = 0.61 we get. ${\displaystyle 0.61\to 0.585\to 0.6225\to 0.56625\to 0.650625\ldots }$. • If μ is between 1 and the square root of 2 the system maps a set of intervals between μ − μ2/2 and μ/2 to themselves. This set of intervals is the Julia set of the map i.e. it is the smallest invariant sub-set of the real line under this map. If μ is greater than the square root of 2, these intervals merge, and the Julia set is the whole interval from μ − μ2/2 to μ/2 (see bifurcation diagram).. • If μ is between 1 and 2 the interval [μ − μ2/2, μ/2] contains both periodic and non-periodic points, although all of the orbits are unstable (i.e. nearby points move away from the orbits rather than towards them). Orbits with longer lengths appear as μ increases. For example:. ${\displaystyle {\frac {\mu }{\mu ^{2}+1}}\to {\frac {\mu ^{2}}{\mu ^{2}+1}}\to {\frac {\mu }{\mu ^{2}+1}}{\mbox{ appears at }}\mu =1}$. ${\displaystyle {\frac {\mu }{\mu ^{3}+1}}\to {\frac {\mu ^{2}}{\mu ^{3}+1}}\to {\frac {\mu ^{3}}{\mu ^{3}+1}}\to {\frac {\mu }{\mu ^{3}+1}}{\mbox{ appears at }}\mu ={\frac {1+{\sqrt {5}}}{2}}}$. ${\displaystyle {\frac {\mu }{\mu ^{4}+1}}\to {\frac {\mu ^{2}}{\mu ^{4}+1}}\to {\frac {\mu ^{3}}{\mu ^{4}+1}}\to {\frac {\mu ^{4}}{\mu ^{4}+1}}\to {\frac {\mu }{\mu ^{4}+1}}{\mbox{ appears at }}\mu \approx 1.8393}$. • If μ equals 2 the system maps the interval [0,1] onto itself. There are now periodic points with every orbit length within this interval, as well as non-periodic points. The periodic points are dense in [0,1], so the map has become chaotic. In fact, the dynamics will be non-periodic if and only if ${\displaystyle x_{0}}$ is irrational. This can be seen by noting what the map does when ${\displaystyle x_{n}}$ is expressed in binary notation: It shifts the binary point one place to the right; then, if what appears to the left of the binary point is a "one" it changes all ones to zeroes and vice versa (with the exception of the final bit "one" in the case of a finite binary expansion); starting from an irrational number, this process goes on forever without repeating itself. The invariant measure for x is the uniform density over the unit interval.[2] The autocorrelation function for a sufficiently long sequence {${\displaystyle x_{n}}$} will show zero autocorrelation at all non-zero lags.[3] Thus ${\displaystyle {x_{n}}}$ cannot be distinguished from white noise using the autocorrelation function. Note that the r=4 case of the logistic map and the ${\displaystyle \mu =2}$ case of the tent map are homeomorphic to each other: Denoting the logistically evolving variable as ${\displaystyle y_{n}}$, the homeomorphism is. ${\displaystyle x_{n}={\tfrac {2}{\pi }}\sin ^{-1}(y_{n}^{1/2}).}$. • If μ is greater than 2 the map's Julia set becomes disconnected, and breaks up into a Cantor set within the interval [0,1]. The Julia set still contains an infinite number of both non-periodic and periodic points (including orbits for any orbit length) but almost every point within [0,1] will now eventually diverge towards infinity. The canonical Cantor set (obtained by successively deleting middle thirds from subsets of the unit line) is the Julia set of the tent map for μ = 3.. ### Numerical errors. Time series of the Tent map for the parameter m=2.0 which shows numerical error: "the plot of time series (plot of x variable with respect to number of iterations) stops fluctuating and no values are observed after n=50". Parameter m= 2.0, initial point is random.. ## Magnifying the orbit diagram. Magnification near the tip shows more details.. • A closer look at the orbit diagram shows that there are 4 separated regions at μ ≈ 1. For further magnification, 2 reference lines (red) are drawn from the tip to suitable x at certain μ (e.g., 1.10) as shown.. Further magnification shows 8 separated regions.. • With distance measured from the corresponding reference lines, further detail appears in the upper and lower part of the map. (total 8 separated regions at some μ). ## Asymmetric tent map. The asymmetric tent map is essentially a distorted, but still piecewise linear, version of the ${\displaystyle \mu =2}$ case of the tent map. It is defined by. ${\displaystyle v_{n+1}={\begin{cases}v_{n}/a&\mathrm {for} ~~v_{n}\in [0,a]\\\\(1-v_{n})/(1-a)&\mathrm {for} ~~v_{n}\in [a,1]\end{cases}}}$. for parameter ${\displaystyle a\in [0,1]}$. The ${\displaystyle \mu =2}$ case of the tent map is the present case of ${\displaystyle a={\tfrac {1}{2}}}$. A sequence {${\displaystyle v_{n}}$} will have the same autocorrelation function [3] as will data from the first-order autoregressive process ${\displaystyle w_{n+1}=(2a-1)w_{n}+u_{n+1}}$ with {${\displaystyle u_{n}}$} independently and identically distributed. Thus data from an asymmetric tent map cannot be distinguished, using the autocorrelation function, from data generated by a first-order autoregressive process.
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# Thread: Help with problem solving
1. ## Help with problem solving
A sculptor commissioned to design a monument for the long reach city council, has chosen a parabolic shape that stand 15m high with supports at 45° as shown in the diagram. For aesthetic reason the sculptor chooses the shape given by the function
Y = 15 -
π (pie)
ie. y equals 15 minus x² over π (pie)
Find the length of the support beam marked A. Provide justification for your answer.
2. do you mean
$E_1 \rightarrow y=\frac{15-x^2}{\pi}$
or
$E_2 \rightarrow y=15-\frac{x^2}{\pi}$
3. $\tan^{-1}(45^{\circ})=1$
so the equation of the line is $y=x$
setting $y=x$ equal to $y=15-\frac{x^2}{\pi}$
$x=15-\frac{x^2}{\pi} \iff x^2+\pi x=15 \pi$
so I will solve by completeing the square so we will add $\frac{\pi^2}{4}$ to both sides
$x^2+\pi x +\frac{\pi^2}{4}=15 \pi +\frac{\pi^2}{4} \iff (x-\frac{\pi}{2})^2=\frac{\pi(60+\pi)}{4}$
$x=\frac{\pi}{2} +\frac{\sqrt{\pi(60+\pi)}}{2}$
So this is half the distance from the diagram so if we multiply it by 2 we will get the length of the line.
$L=\pi+\sqrt{\pi(60+\pi)}$
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# Thread: Help with problem solving. 1. ## Help with problem solving. A sculptor commissioned to design a monument for the long reach city council, has chosen a parabolic shape that stand 15m high with supports at 45° as shown in the diagram. For aesthetic reason the sculptor chooses the shape given by the function. Y = 15 -. π (pie). ie. y equals 15 minus x² over π (pie). Find the length of the support beam marked A. Provide justification for your answer.. 2. do you mean. $E_1 \rightarrow y=\frac{15-x^2}{\pi}$.
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or. $E_2 \rightarrow y=15-\frac{x^2}{\pi}$. 3. $\tan^{-1}(45^{\circ})=1$. so the equation of the line is $y=x$. setting $y=x$ equal to $y=15-\frac{x^2}{\pi}$. $x=15-\frac{x^2}{\pi} \iff x^2+\pi x=15 \pi$. so I will solve by completeing the square so we will add $\frac{\pi^2}{4}$ to both sides. $x^2+\pi x +\frac{\pi^2}{4}=15 \pi +\frac{\pi^2}{4} \iff (x-\frac{\pi}{2})^2=\frac{\pi(60+\pi)}{4}$. $x=\frac{\pi}{2} +\frac{\sqrt{\pi(60+\pi)}}{2}$. So this is half the distance from the diagram so if we multiply it by 2 we will get the length of the line.. $L=\pi+\sqrt{\pi(60+\pi)}$.
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Mathematics
# $\displaystyle \int { \frac { { x }^{ 2 } }{ { \left( 2+{ 3x }^{ 2 } \right) }^{ 5/2 } } dx }$ is equal to
$\dfrac { 1 }{ 5 } { \left[ \dfrac { { x }^{ 2 } }{ 2+3{ x }^{ 2 } } \right] }^{ 3/2 }+C$
Its FREE, you're just one step away
Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105
#### Realted Questions
Q1 Subjective Medium
$-2 \int x^2 e^{-2x} dx = e^{-2x} (ax^2 + bx + c) + d$, then
1 Verified Answer | Published on 17th 09, 2020
Q2 Single Correct Medium
$\displaystyle\int \frac{\sqrt{2+x^{10}}}{x^{16}}dx$
• A. $\displaystyle \frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 3 }{ 2 } }$
• B. $\displaystyle { (\frac { 2 }{ { x }^{ 30 } } +1) }^{ \dfrac { 3 }{ 2 } }$
• C. $\displaystyle \frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 1 }{ 2 } }$
• D. $\displaystyle -\frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 3 }{ 2 } }$
1 Verified Answer | Published on 17th 09, 2020
Q3 Single Correct Hard
If $f(x) = \dfrac {x + 2}{2x + 3}$, then $\displaystyle \int \left (\dfrac {f(x)}{x^{2}}\right )^{1/2} dx = \dfrac {1}{\sqrt {2}}g \left (\dfrac {1 + \sqrt {2f(x)}}{1 - \sqrt {2f(x)}}\right ) - \sqrt {\dfrac {2}{3}}h \left (\dfrac {\sqrt {3f(x)} + \sqrt {2}}{\sqrt {3f(x)} - \sqrt {2}}\right ) + c$, where
• A. $g(x) = \log |x|, h(x) = \tan^{-1}x$
• B. $g(x) = h(x) = \tan^{-1}x$
• C. $g(x) = \log|x|, h(x) = \log |x|$
• D. $g(x) = \tan^{-1} x, h(x) = \log |x|$
1 Verified Answer | Published on 17th 09, 2020
Q4 Subjective Medium
Evaluate the following:
$\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{tan \,xdx}{1 + m^2 \,tan^2 x}$
1 Verified Answer | Published on 17th 09, 2020
Q5 Single Correct Medium
$\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \dfrac { 2x }{ 1-{ x }^{ 2 } } \right) dx } } =\dfrac{\pi}{a}-\ln a$. Find $a$.
• A. $1$
• B. $-1$
• C. None of these
• D. $2$
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Mathematics. # $\displaystyle \int { \frac { { x }^{ 2 } }{ { \left( 2+{ 3x }^{ 2 } \right) }^{ 5/2 } } dx }$ is equal to. $\dfrac { 1 }{ 5 } { \left[ \dfrac { { x }^{ 2 } }{ 2+3{ x }^{ 2 } } \right] }^{ 3/2 }+C$. Its FREE, you're just one step away. Single Correct Medium Published on 17th 09, 2020. Questions 203525. Subjects 9. Chapters 126. Enrolled Students 105. #### Realted Questions. Q1 Subjective Medium. $-2 \int x^2 e^{-2x} dx = e^{-2x} (ax^2 + bx + c) + d$, then. 1 Verified Answer | Published on 17th 09, 2020. Q2 Single Correct Medium. $\displaystyle\int \frac{\sqrt{2+x^{10}}}{x^{16}}dx$. • A. $\displaystyle \frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 3 }{ 2 } }$. • B. $\displaystyle { (\frac { 2 }{ { x }^{ 30 } } +1) }^{ \dfrac { 3 }{ 2 } }$. • C. $\displaystyle \frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 1 }{ 2 } }$. • D. $\displaystyle -\frac { 1 }{ 30 } \times { (\frac { 2 }{ { x }^{ 10 } } +1) }^{ \dfrac { 3 }{ 2 } }$. 1 Verified Answer | Published on 17th 09, 2020. Q3 Single Correct Hard.
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If $f(x) = \dfrac {x + 2}{2x + 3}$, then $\displaystyle \int \left (\dfrac {f(x)}{x^{2}}\right )^{1/2} dx = \dfrac {1}{\sqrt {2}}g \left (\dfrac {1 + \sqrt {2f(x)}}{1 - \sqrt {2f(x)}}\right ) - \sqrt {\dfrac {2}{3}}h \left (\dfrac {\sqrt {3f(x)} + \sqrt {2}}{\sqrt {3f(x)} - \sqrt {2}}\right ) + c$, where. • A. $g(x) = \log |x|, h(x) = \tan^{-1}x$. • B. $g(x) = h(x) = \tan^{-1}x$. • C. $g(x) = \log|x|, h(x) = \log |x|$. • D. $g(x) = \tan^{-1} x, h(x) = \log |x|$. 1 Verified Answer | Published on 17th 09, 2020. Q4 Subjective Medium. Evaluate the following:. $\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{tan \,xdx}{1 + m^2 \,tan^2 x}$. 1 Verified Answer | Published on 17th 09, 2020. Q5 Single Correct Medium. $\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \dfrac { 2x }{ 1-{ x }^{ 2 } } \right) dx } } =\dfrac{\pi}{a}-\ln a$. Find $a$.. • A. $1$. • B. $-1$. • C. None of these. • D. $2$.
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# You asked: How many volts is 2 amps?
Contents
Current Voltage Power
2 Amps 20 Volts 40 Watts
2 Amps 22.5 Volts 45 Watts
2 Amps 25 Volts 50 Watts
2 Amps 27.5 Volts 55 Watts
## How many amps are in a Volt?
A “volt” is a unit of electric potential, also known as electromotive force, and represents “the potential difference between two points of a conducting wire carrying a constant current of 1 ampere, when the power dissipated between these points is equal to 1 watt.” Stated another way, a potential of one volt appears …
## What does 2 amps mean?
The ampere, often shortened to “amp” or A, is the base unit of electric current in the International System of Units. … A current of 2 Amps can be written as 2A. The bigger the current the more electricity is flowing.
## How many amps is 11.1 volts?
11.1V Ohm to Amps
Assumed Ω = 20, the result of the conversion is 0.56 amps (rounded to two decimal digits).
## How many amps is 220 volts?
13.64 Amps, if you use 220 V.
## Is it better to charge a battery at 2 amps or 10 amps?
Slow charging using 10 amp chargers or less is generally accepted to be better. This is because repeatedly fast-charging a car battery may lead to overcharging and will severely reduce its performance. The short answer: 10 amp chargers are generally preferable than 2 amp chargers.
IT IS INTERESTING: Does xiaomi Redmi have Wi Fi calling?
## Is a 2 amp charger good?
Don’t worry about any problems using a 2-amp charger with a device designed to be used with a 1-amp charger. In some cases, it may speed up charging of the smaller device. In other cases, the smaller device will limit its power-sipping to 1 amp.
## How many amps are in a 12 volt?
Equivalent Volts and Amps Measurements
Voltage Current Power
12 Volts 0.8333 Amps 10 Watts
12 Volts 1.25 Amps 15 Watts
12 Volts 1.667 Amps 20 Watts
12 Volts 2.083 Amps 25 Watts
## How many volts is 70 amps?
The 70A ohm to volts formula is V = 70 × Ω. Replace Ω with your individual resistance in ohm. Assumed Ω = 8, the result of the conversion is 560 volts (rounded to two decimal places).
## How many volts are in 13 amps?
The 13A watts to volts formula is V = W / 13. Replace W with your actual power in watt. For example with W = 60, you obtain the result of 4.62 volts (rounded to two decimal digits).
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# You asked: How many volts is 2 amps?. Contents. Current Voltage Power. 2 Amps 20 Volts 40 Watts. 2 Amps 22.5 Volts 45 Watts. 2 Amps 25 Volts 50 Watts. 2 Amps 27.5 Volts 55 Watts. ## How many amps are in a Volt?. A “volt” is a unit of electric potential, also known as electromotive force, and represents “the potential difference between two points of a conducting wire carrying a constant current of 1 ampere, when the power dissipated between these points is equal to 1 watt.” Stated another way, a potential of one volt appears …. ## What does 2 amps mean?. The ampere, often shortened to “amp” or A, is the base unit of electric current in the International System of Units. … A current of 2 Amps can be written as 2A. The bigger the current the more electricity is flowing.. ## How many amps is 11.1 volts?. 11.1V Ohm to Amps. Assumed Ω = 20, the result of the conversion is 0.56 amps (rounded to two decimal digits).. ## How many amps is 220 volts?. 13.64 Amps, if you use 220 V.. ## Is it better to charge a battery at 2 amps or 10 amps?. Slow charging using 10 amp chargers or less is generally accepted to be better. This is because repeatedly fast-charging a car battery may lead to overcharging and will severely reduce its performance. The short answer: 10 amp chargers are generally preferable than 2 amp chargers.
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IT IS INTERESTING: Does xiaomi Redmi have Wi Fi calling?. ## Is a 2 amp charger good?. Don’t worry about any problems using a 2-amp charger with a device designed to be used with a 1-amp charger. In some cases, it may speed up charging of the smaller device. In other cases, the smaller device will limit its power-sipping to 1 amp.. ## How many amps are in a 12 volt?. Equivalent Volts and Amps Measurements. Voltage Current Power. 12 Volts 0.8333 Amps 10 Watts. 12 Volts 1.25 Amps 15 Watts. 12 Volts 1.667 Amps 20 Watts. 12 Volts 2.083 Amps 25 Watts. ## How many volts is 70 amps?. The 70A ohm to volts formula is V = 70 × Ω. Replace Ω with your individual resistance in ohm. Assumed Ω = 8, the result of the conversion is 560 volts (rounded to two decimal places).. ## How many volts are in 13 amps?. The 13A watts to volts formula is V = W / 13. Replace W with your actual power in watt. For example with W = 60, you obtain the result of 4.62 volts (rounded to two decimal digits).
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Top
# Number, Ranking and Time Sequence Test
1.
Sandhya's birthday falls on 15th August and Minu's birthday falls on 25th June. If Minu's birthday was on Wednesday, what was the day on Sandhya's birthday in the same year?
Total days from 25th June to 15th August = 5 + 31 + 15 = 51 days
51/7 = 7 weeks and 2 days
August 15 = Wednesday + 2 = Friday
Enter details here
2.
In the following question, select the odd number pair from the given alternatives.
No answer description available for this question.
Enter details here
3.
How many days will be there from 26th january 2008, to 15th may 2008(both days are included)?
Number of days = (6 + 29 + 31 + 30 + 15) = 111
2008 is a leap year, number of days in february = 29
Enter details here
4.
Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there be three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue?
No answer description available for this question.
Enter details here
5.
How many numbers amogst the numbers 9 to 54 are there which are exactly divisible by 9 but not by 3 ?
Enter details here
6.
If the position of the first, sixth digits of the number 2796543018 are interchanged, similarly the positions of the the second and seventh digits are interchanged and so on, which of the following will be the left of seventh digit from the left end?
The new number formed is 4 3 0 1 8 2 7 9 6 5
The seventh digit from the left is 7, the 3rd digit from the left of 7 is 1
Enter details here
7.
If tuesday falls on 4th of month, then which day will fall three days after the 24th ?
The 4th day is tuesday, then 11, 18, 25th also tuesdays
Then 3 days after 24th is 27th, which is thursday
Enter details here
8.
If the seventh day of a month is three days earlier than Friday, what day will it be on the nineteenth day of the month?
No answer description available for this question.
Enter details here
9.
In a row of trees, one tree is fifth from either end of the row. How many trees are there in the row?
No answer description available for this question.
Enter details here
10.
If the day before yesterday was Saturday, what day will fall on the day after tomorrow?
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Top. # Number, Ranking and Time Sequence Test. 1.. Sandhya's birthday falls on 15th August and Minu's birthday falls on 25th June. If Minu's birthday was on Wednesday, what was the day on Sandhya's birthday in the same year?. Total days from 25th June to 15th August = 5 + 31 + 15 = 51 days. 51/7 = 7 weeks and 2 days. August 15 = Wednesday + 2 = Friday. Enter details here. 2.. In the following question, select the odd number pair from the given alternatives.. No answer description available for this question.. Enter details here. 3.. How many days will be there from 26th january 2008, to 15th may 2008(both days are included)?. Number of days = (6 + 29 + 31 + 30 + 15) = 111. 2008 is a leap year, number of days in february = 29. Enter details here. 4.. Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there be three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue?. No answer description available for this question.. Enter details here. 5.
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How many numbers amogst the numbers 9 to 54 are there which are exactly divisible by 9 but not by 3 ?. Enter details here. 6.. If the position of the first, sixth digits of the number 2796543018 are interchanged, similarly the positions of the the second and seventh digits are interchanged and so on, which of the following will be the left of seventh digit from the left end?. The new number formed is 4 3 0 1 8 2 7 9 6 5. The seventh digit from the left is 7, the 3rd digit from the left of 7 is 1. Enter details here. 7.. If tuesday falls on 4th of month, then which day will fall three days after the 24th ?. The 4th day is tuesday, then 11, 18, 25th also tuesdays. Then 3 days after 24th is 27th, which is thursday. Enter details here. 8.. If the seventh day of a month is three days earlier than Friday, what day will it be on the nineteenth day of the month?. No answer description available for this question.. Enter details here. 9.. In a row of trees, one tree is fifth from either end of the row. How many trees are there in the row?. No answer description available for this question.. Enter details here. 10.. If the day before yesterday was Saturday, what day will fall on the day after tomorrow?.
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Teaching mathematics
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Free course
# 1.4 Using concrete resources
It is important to give learners of all ages a variety of concrete experiences connected with proportional reasoning to encourage them to compare, make conjectures, find relationships and generalise their learning.
Young learners may start with weighing and measuring objects in the classroom, and discussing comparisons between amounts.
## Activity 6 Sharing money
When asked to share out £24 between two cousins in the ratio of 3:1, learners can use counters to represent the pounds (Figure 7). How can you use counters to demonstrate the solution to this problem? Write responses in the box below.
Figure 7 Using counters to share out in a ratio
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Interactive feature not available in single page view (see it in standard view).
### Discussion
Grouping counters in groups of 1 and 3 is one way to work on this problem.
By grouping counters in this way, learners share out all the counters in the ratio of 1:3.
They can then look at how many of each colour there are altogether. Noticing that there are 6 groups of 3 and 6 groups of 1 would result in finding a total of 18 blue counters and 6 yellow counters, representing the £18 and £6 the cousins will each get.
Another way to approach this problem using counters is to create boxes (on paper or on mini-whiteboards) to represent the ratio. In this case, since the ratio is 3:1, you will need 3 boxes for the first cousin and 1 box for the second cousin.
The counters, which represent pounds, are then shared out equally between the 4 boxes. The first cousin is given 3 boxes, so receiving £18, and the second cousin is given 1 box, so receiving £6 (Figure 8).
Figure 8 Using ratio boxes with counters
This approach can also be used with pens on mini-whiteboards. Learners need to create the boxes to represent the ratio and then put a mark in every box until they get to the total, in this case 24 (Figure 9).
Figure 9 Using ratio boxes on mini-whiteboards
This imagery can then support learners in moving towards a more abstract concept of sharing amounts into a ratio. In other words, it can help them to understand that in a ratio of 3:1 there are 4 equal parts in total which the £24 needs to be divided equally between.
Note: When moving on to ratio problems involving fractions and decimals, using manipulative resources can become problematic. It is sensible to move onto more abstract methods with your learners before introducing these kinds of problems.
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Teaching mathematics. Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.. Free course. # 1.4 Using concrete resources. It is important to give learners of all ages a variety of concrete experiences connected with proportional reasoning to encourage them to compare, make conjectures, find relationships and generalise their learning.. Young learners may start with weighing and measuring objects in the classroom, and discussing comparisons between amounts.. ## Activity 6 Sharing money. When asked to share out £24 between two cousins in the ratio of 3:1, learners can use counters to represent the pounds (Figure 7). How can you use counters to demonstrate the solution to this problem? Write responses in the box below.. Figure 7 Using counters to share out in a ratio. To use this interactive functionality a free OU account is required. Sign in or register.. Interactive feature not available in single page view (see it in standard view).. ### Discussion. Grouping counters in groups of 1 and 3 is one way to work on this problem.. By grouping counters in this way, learners share out all the counters in the ratio of 1:3.
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They can then look at how many of each colour there are altogether. Noticing that there are 6 groups of 3 and 6 groups of 1 would result in finding a total of 18 blue counters and 6 yellow counters, representing the £18 and £6 the cousins will each get.. Another way to approach this problem using counters is to create boxes (on paper or on mini-whiteboards) to represent the ratio. In this case, since the ratio is 3:1, you will need 3 boxes for the first cousin and 1 box for the second cousin.. The counters, which represent pounds, are then shared out equally between the 4 boxes. The first cousin is given 3 boxes, so receiving £18, and the second cousin is given 1 box, so receiving £6 (Figure 8).. Figure 8 Using ratio boxes with counters. This approach can also be used with pens on mini-whiteboards. Learners need to create the boxes to represent the ratio and then put a mark in every box until they get to the total, in this case 24 (Figure 9).. Figure 9 Using ratio boxes on mini-whiteboards. This imagery can then support learners in moving towards a more abstract concept of sharing amounts into a ratio. In other words, it can help them to understand that in a ratio of 3:1 there are 4 equal parts in total which the £24 needs to be divided equally between.. Note: When moving on to ratio problems involving fractions and decimals, using manipulative resources can become problematic. It is sensible to move onto more abstract methods with your learners before introducing these kinds of problems.. TM_1.
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# Thread: Integral with partial fractions
1. ## Integral with partial fractions
so I got
A= -3 , B = 5 C = 5
and i got ∫-3/x-4 + 5x+5/x²+9 dx for my equation
my answer came out to be
I dont know why it is wrong.
2. I didn't check the details but one mistake is that it should be $\displaystyle -3\ln(x-4)$, not $\displaystyle -3\ln(x+4)$
3. Originally Posted by emurphy
so I got
A= -3 , B = 5 C = 5
and i got ∫-3/x-4 + 5x+5/x²+9 dx for my equation
my answer came out to be
I dont know why it is wrong.
As well as the mistake already mentioned (which may well be just a typo), your calculation of $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$ is very wrong. Please show all the details of your calculation so that it can be reviewed.
4. Oh yeah that is pretty bad. Use parentheses! You wrote $\displaystyle 3/x-4 + 5x+5/x^2 + 9$ and integrated it as $\displaystyle 3/(x-4) + 5x+5/(x^2 + 9)$ when in fact you should have written $\displaystyle 3/(x-4) + (5x+5)/(x^2 + 9)$. These expressions are completely different from one another.
5. For $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$
I separated the fraction into 2
∫5x+5 dx + ∫1/x²+9 dx
∫5x+5 dx = 5x²/2+5x
∫1/x²+9 dx = ∫1/(1/9)(x²+1) dx = 9 ∫ 1/x²+1
= 9tan^-1(x)
6. Originally Posted by emurphy
For $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$
I separated the fraction into 2
∫5x+5 dx + ∫1/x²+9 dx
∫5x+5 dx = 5x²/2+5x
∫1/x²+9 dx = ∫1/(1/9)(x²+1) dx = 9 ∫ 1/x²+1
= 9tan^-1(x)
NO! You cannot seperate the integrand like that!! Using this logic you would solve $\displaystyle \int \frac{x}{x} \, dx$ as $\displaystyle \int x \, dx + \int \frac{1}{x} \, dx$ which is clearly so wrong that I don't need to say anything more.
7. Originally Posted by mr fantastic
NO! You cannot seperate the integrand like that!! Using this logic you would solve $\displaystyle \int \frac{x}{x} \, dx$ as $\displaystyle \int x \, dx + \int \frac{1}{x} \, dx$ which is clearly so wrong that I don't need to say anything more.
Adding to this, as a diagnostic of where you're at could you please do the following two integrals:
$\displaystyle I_1 = \int \frac{2x}{x^2 + 9} \, dx$
$\displaystyle I_2 = \int \frac{5}{x^2 + 9} \, dx$
Edit: If you cannot do these there is little point in you attempting questions like the one you posted until you have thoroughly reviewed the basic integration techniques this questions assumes and have developed a more solid understanding of integration.
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# Thread: Integral with partial fractions. 1. ## Integral with partial fractions. so I got. A= -3 , B = 5 C = 5. and i got ∫-3/x-4 + 5x+5/x²+9 dx for my equation. my answer came out to be. I dont know why it is wrong.. 2. I didn't check the details but one mistake is that it should be $\displaystyle -3\ln(x-4)$, not $\displaystyle -3\ln(x+4)$. 3. Originally Posted by emurphy. so I got. A= -3 , B = 5 C = 5. and i got ∫-3/x-4 + 5x+5/x²+9 dx for my equation. my answer came out to be. I dont know why it is wrong.. As well as the mistake already mentioned (which may well be just a typo), your calculation of $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$ is very wrong. Please show all the details of your calculation so that it can be reviewed.. 4. Oh yeah that is pretty bad. Use parentheses! You wrote $\displaystyle 3/x-4 + 5x+5/x^2 + 9$ and integrated it as $\displaystyle 3/(x-4) + 5x+5/(x^2 + 9)$ when in fact you should have written $\displaystyle 3/(x-4) + (5x+5)/(x^2 + 9)$. These expressions are completely different from one another.. 5.
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For $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$. I separated the fraction into 2. ∫5x+5 dx + ∫1/x²+9 dx. ∫5x+5 dx = 5x²/2+5x. ∫1/x²+9 dx = ∫1/(1/9)(x²+1) dx = 9 ∫ 1/x²+1. = 9tan^-1(x). 6. Originally Posted by emurphy. For $\displaystyle \int \frac{5x + 5}{x^2 + 9} \, dx$. I separated the fraction into 2. ∫5x+5 dx + ∫1/x²+9 dx. ∫5x+5 dx = 5x²/2+5x. ∫1/x²+9 dx = ∫1/(1/9)(x²+1) dx = 9 ∫ 1/x²+1. = 9tan^-1(x). NO! You cannot seperate the integrand like that!! Using this logic you would solve $\displaystyle \int \frac{x}{x} \, dx$ as $\displaystyle \int x \, dx + \int \frac{1}{x} \, dx$ which is clearly so wrong that I don't need to say anything more.. 7. Originally Posted by mr fantastic. NO! You cannot seperate the integrand like that!! Using this logic you would solve $\displaystyle \int \frac{x}{x} \, dx$ as $\displaystyle \int x \, dx + \int \frac{1}{x} \, dx$ which is clearly so wrong that I don't need to say anything more.. Adding to this, as a diagnostic of where you're at could you please do the following two integrals:. $\displaystyle I_1 = \int \frac{2x}{x^2 + 9} \, dx$. $\displaystyle I_2 = \int \frac{5}{x^2 + 9} \, dx$. Edit: If you cannot do these there is little point in you attempting questions like the one you posted until you have thoroughly reviewed the basic integration techniques this questions assumes and have developed a more solid understanding of integration.
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https://learn.careers360.com/school/question-please-solve-rd-sharma-class-12-chapter-19-definite-integrals-exercise-19-point-5-question-3-maths-textbook-solution-2/?question_number=3.0
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please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19.5 question 3 maths textbook solution
8
Hint:
To solve the given statement, we have to use the formula of addition limits.
Given:
$\int_{1}^{3}\left ( 3x-2 \right )dx$
Solution:
We have,
$\int_{1}^{3}\left ( 3x-2 \right )dx$
$\int_{a}^{b}\left ( fx\right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$
Where, $h=\frac{b-a}{n}$
Here,
$a=1,b=3,f(x)=(3x-2)\\ h=\frac{2}{n}$
Thus, we have
\begin{aligned} I &=\int_{1}^{3}(3 x-2) d x \\ I &=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \\ \end{aligned}
\begin{aligned} &=\lim _{h \rightarrow 0} h[1+(3(1+h)-2)+\{3(1+2 h)-2\}+\ldots\{3(1+(n-1) h)-2\}] \\ &=\lim _{h \rightarrow 0} h[n+3 h(1+2+3+\ldots(n-1))] \\ \end{aligned}\begin{aligned} &=\lim _{h \rightarrow 0} h\left[n+3 h\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[n+\frac{6}{n} \frac{n(n-1)}{2}\right] \\ \end{aligned} $\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n} \\ n \rightarrow \infty \end{array}\right]$
\begin{aligned} &=\lim _{n \rightarrow \infty} 2+\frac{6}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ &=2+6=8 \end{aligned}
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please solve RD sharma class 12 chapter 19 Definite Integrals exercise 19.5 question 3 maths textbook solution. 8. Hint:. To solve the given statement, we have to use the formula of addition limits.. Given:. $\int_{1}^{3}\left ( 3x-2 \right )dx$. Solution:. We have,. $\int_{1}^{3}\left ( 3x-2 \right )dx$.
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$\int_{a}^{b}\left ( fx\right )dx=\lim_{h\rightarrow 0}h\left [ f(a)+f(a+h)+f(a+2h)+.....f(a+(n-1)h) \right ]$. Where, $h=\frac{b-a}{n}$. Here,. $a=1,b=3,f(x)=(3x-2)\\ h=\frac{2}{n}$. Thus, we have. \begin{aligned} I &=\int_{1}^{3}(3 x-2) d x \\ I &=\lim _{h \rightarrow 0} h[f(1)+f(1+h)+f(1+2 h)+\ldots f(1+(n-1)) h] \\ \end{aligned}. \begin{aligned} &=\lim _{h \rightarrow 0} h[1+(3(1+h)-2)+\{3(1+2 h)-2\}+\ldots\{3(1+(n-1) h)-2\}] \\ &=\lim _{h \rightarrow 0} h[n+3 h(1+2+3+\ldots(n-1))] \\ \end{aligned}\begin{aligned} &=\lim _{h \rightarrow 0} h\left[n+3 h\left(\frac{n(n-1)}{2}\right)\right] \\ &=\lim _{n \rightarrow \infty} \frac{2}{n}\left[n+\frac{6}{n} \frac{n(n-1)}{2}\right] \\ \end{aligned} $\left[\begin{array}{l} \mathrm{Q} h \rightarrow 0 \& h=\frac{2}{n} \\ n \rightarrow \infty \end{array}\right]$. \begin{aligned} &=\lim _{n \rightarrow \infty} 2+\frac{6}{n^{2}} \cdot n^{2}\left(1-\frac{1}{n}\right) \\ &=2+6=8 \end{aligned}.
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Is the number of members of Club X greater than the number : DS Archive
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# Is the number of members of Club X greater than the number
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Is the number of members of Club X greater than the number of members of Club Y ?
(1) Of the members of Club X, 20 percent are also members of Club Y.
(2) Of the members of Club Y, 30 percent are also members of Club X.
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03 Oct 2004, 03:35
C.
of course each one independently will not give anything.
for both looking at the intersection or members comon to X and Y
from (1) . intersection = 0.2 X
from (2). intersection = 0.3 Y
0.2X = 0.3 Y
this gives that memebres of X are greater than Y
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Is the number of members of Club X greater than the number : DS Archive. Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack. It is currently 03 Dec 2016, 16:03. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. # Events & Promotions. ###### Events & Promotions in June. Open Detailed Calendar. # Is the number of members of Club X greater than the number. Author Message. Intern. Joined: 25 Sep 2004. Posts: 21. Followers: 0. Kudos [?]: 98 [0], given: 0. Is the number of members of Club X greater than the number [#permalink]. ### Show Tags. 03 Oct 2004, 03:10.
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00:00. Difficulty:. (N/A). Question Stats:. 0% (00:00) correct 0% (00:00) wrong based on 0 sessions. ### HideShow timer Statistics. This topic is locked. If you want to discuss this question please re-post it in the respective forum.. Is the number of members of Club X greater than the number of members of Club Y ?. (1) Of the members of Club X, 20 percent are also members of Club Y.. (2) Of the members of Club Y, 30 percent are also members of Club X.. Manager. Joined: 07 Sep 2004. Posts: 60. Followers: 1. Kudos [?]: 1 [0], given: 0. ### Show Tags. 03 Oct 2004, 03:35. C.. of course each one independently will not give anything.. for both looking at the intersection or members comon to X and Y. from (1) . intersection = 0.2 X. from (2). intersection = 0.3 Y. 0.2X = 0.3 Y. this gives that memebres of X are greater than Y. Display posts from previous: Sort by.
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# Reasoning Booster on Direction Test for Teaching Exams
By Ashish Kumar|Updated : December 31st, 2021
Verbal reasoning comprises of a very important concept that is - Direction. It is one of the easiest topics to score in most of the Teaching Exams. The Concept behind the Directions is the same that we use in our day-to-day life. This direction sense test is nothing but a precise of sensing the direction. To solve the direction sense test first you need to make a sketch of the data provided. We have clearly described the direction Sense test Tricks, tips and Shortcuts here.
Verbal reasoning comprises a very important concept which is - Direction. It is one of the easiest topics to score in most Teaching Exams. The Concept behind the Directions is the same that we use in our day-to-day life. This direction sense test is nothing but a precise of sensing the direction. To solve the direction sense test first you need to make a sketch of the data provided. We have clearly described the direction Sense test Tricks, tips and Shortcuts here.
### Short Tricks & Basic Concepts on Direction Test
Four main Directions - North, South, East, West
Four Cardinal Direction - North-East, North-West, South-East, South-West
• At the time of sunrise if a man stands facing the east, his shadow will be towards the west.
• At the time of sunset the shadow of an object is always in the east.
• If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.
• At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow.
The above is the Directional Diagram. The direction sense consists of four different types of problems. Let’s see the types of Direction problems with solutions.
1. A man starting from his home walks 5 km towards East, and then he turns left and goes 4 km. At last he turn to his left and walks 5 km. Now find the distance between the man and his home and also find at which direction he is facing?
Sol.
From the above diagram we can find he is 4 km from his house and facing the North Direction.
2. A man starting from his home moves 4 km towards East, then he turns right and moves 3 km. Now what will be the minimum distance covered by him to come back to his home?
Sol.
The minimum distance covered by him = √42 + 32 = 5 km
3. After Sunrise, Prakash while going to college suddenly met with Lokesh at a crossing point. Lokesh's Shadow was exactly to right of Prakash. If they were facing each other on which direction was Prakash facing?
Sol. Always Sun rises in the East Direction. So Shadow falls towards West
4. Prem started from his home and moved 5 km to reach the crossing point of the palace. In which direction was Prem going, if the road opposite to his direction goes to the hospital. The road to the right of Prem goes to the station. If the road which goes to station is just opposite to the road of the IT-Park, then in which direction is Prem which leads to the IT- Park?
Sol.
From the above diagram its shows that the road which goes towards the IT-Park is left of Prem.
That’s how Direction questions are solved easily. We hope these concepts have made the chapter easy for you all.
Thanks!
The most comprehensive exam prep app
#DreamStriveSucceed
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# Reasoning Booster on Direction Test for Teaching Exams. By Ashish Kumar|Updated : December 31st, 2021. Verbal reasoning comprises of a very important concept that is - Direction. It is one of the easiest topics to score in most of the Teaching Exams. The Concept behind the Directions is the same that we use in our day-to-day life. This direction sense test is nothing but a precise of sensing the direction. To solve the direction sense test first you need to make a sketch of the data provided. We have clearly described the direction Sense test Tricks, tips and Shortcuts here.. Verbal reasoning comprises a very important concept which is - Direction. It is one of the easiest topics to score in most Teaching Exams. The Concept behind the Directions is the same that we use in our day-to-day life. This direction sense test is nothing but a precise of sensing the direction. To solve the direction sense test first you need to make a sketch of the data provided. We have clearly described the direction Sense test Tricks, tips and Shortcuts here.. ### Short Tricks & Basic Concepts on Direction Test. Four main Directions - North, South, East, West. Four Cardinal Direction - North-East, North-West, South-East, South-West. • At the time of sunrise if a man stands facing the east, his shadow will be towards the west.. • At the time of sunset the shadow of an object is always in the east.. • If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.. • At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow.. The above is the Directional Diagram. The direction sense consists of four different types of problems. Let’s see the types of Direction problems with solutions.. 1. A man starting from his home walks 5 km towards East, and then he turns left and goes 4 km. At last he turn to his left and walks 5 km.
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Now find the distance between the man and his home and also find at which direction he is facing?. Sol.. From the above diagram we can find he is 4 km from his house and facing the North Direction.. 2. A man starting from his home moves 4 km towards East, then he turns right and moves 3 km. Now what will be the minimum distance covered by him to come back to his home?. Sol.. The minimum distance covered by him = √42 + 32 = 5 km. 3. After Sunrise, Prakash while going to college suddenly met with Lokesh at a crossing point. Lokesh's Shadow was exactly to right of Prakash. If they were facing each other on which direction was Prakash facing?. Sol. Always Sun rises in the East Direction. So Shadow falls towards West. 4. Prem started from his home and moved 5 km to reach the crossing point of the palace. In which direction was Prem going, if the road opposite to his direction goes to the hospital. The road to the right of Prem goes to the station. If the road which goes to station is just opposite to the road of the IT-Park, then in which direction is Prem which leads to the IT- Park?. Sol.. From the above diagram its shows that the road which goes towards the IT-Park is left of Prem.. That’s how Direction questions are solved easily. We hope these concepts have made the chapter easy for you all.. Thanks!. The most comprehensive exam prep app. #DreamStriveSucceed.
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https://math.stackexchange.com/questions/2190775/how-to-prove-gcd-with-prime-factor-decomposition
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How to prove $gcd$ with prime factor decomposition?
So we know that $n=\prod_{p\in\mathbb{P}} p^{v_{n}(p)}$ is formula of factor decomposition.
How to show that $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ using factor decomposition?
I know that for example when we have $gcd(36,360)$, we can write $36$ as
$36 = 2^{p_1}3^{p_2}5^{p_3}$ which is $36 = 2^{2}3^{2}5^{0}$
and $360$ as
$360 = 2^{q_1}3^{q_2}5^{q_3}$ which is $360 = 2^{3}3^{2}5^{1}$
To get $gcd$ we now need to take smallest $p$ and $q$ for every pair with same base. In this case these are 2,2,0, therefore our $gcd(36,360) = 36$ which is obvious.
So, I can show that this is valid $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ through an example with numbers, but I need proof, and I don't know how to prove that.
That's just because $m|n$ iff $\nu_m(p) \leq \nu_n(p)$ for all relevant $p$, and then use the definition of greatest common divisor.
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How to prove $gcd$ with prime factor decomposition?. So we know that $n=\prod_{p\in\mathbb{P}} p^{v_{n}(p)}$ is formula of factor decomposition.. How to show that $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ using factor decomposition?. I know that for example when we have $gcd(36,360)$, we can write $36$ as. $36 = 2^{p_1}3^{p_2}5^{p_3}$ which is $36 = 2^{2}3^{2}5^{0}$. and $360$ as.
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$360 = 2^{q_1}3^{q_2}5^{q_3}$ which is $360 = 2^{3}3^{2}5^{1}$. To get $gcd$ we now need to take smallest $p$ and $q$ for every pair with same base. In this case these are 2,2,0, therefore our $gcd(36,360) = 36$ which is obvious.. So, I can show that this is valid $gcd(m,n) = \prod_{p\in\mathbb{P}} p^{min(v_{n}(p),v_{n}(p))}$ through an example with numbers, but I need proof, and I don't know how to prove that.. That's just because $m|n$ iff $\nu_m(p) \leq \nu_n(p)$ for all relevant $p$, and then use the definition of greatest common divisor.
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http://www.math-math.com/2016/04/goldbach-conjecture-explained-in-5.html
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Goldbach Conjecture Explained in 5 Minutes
Goldbach Conjecture Explained in 5 Minutes
In 1742, the German mathematician Christian Goldbach, in a discussion with the mathematician Leonhard Euler, made a simple statement..
Every even integer greater than 2 can be written as the sum of two prime numbers.
Mathematicians have tried to prove this ever since. None have. It's a great example of how a simple statement in mathematics can be amazingly difficult to prove. Computers have checked billions of numbers and shown it to be true for every number tested, but that's not the same as a proof. A proof would show it to be true for all even numbers, period.
It's easy to prove that every integer can be written as a product of primes. This is called the prime decomposition of an integer. So for any integer m we have.. m=(p1^s1)*(p2^s2)*......*(pn^sn)
where p1,p2,...,pn are prime numbers and the s1,s2,s3,...,sn are just integer powers and represent the simple fact that primes may be repeated. By collecting like primes together and raising them to a power we make sure that p1,p2,p3,...,pn are distinct primes with no duplication.
The condition that m be even simply means that one of these primes must be 2. Since the order of multiplication does not matter we can make p1=2. So m now looks like this..
m=(2^s1)*(p2^s2)*...*(pn^sn)
Goldbach's Conjecture says that when m is even there exists two prime numbers, let's call them g1 and g2, such that..
m=(2^s1)*(p2^s2)*...*(pn^sn)=g1+g2
Are we on our way to a proof? No, but it's still fun to try!
Perhaps there's some deep clue in the fact that Goldbach's Conjecture only works if p1=2. In other words, if 2 does not appear in the prime decomposition of an integer then Goldbach's Conjecture does not work. What's so special about the number 2?
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Goldbach Conjecture Explained in 5 Minutes. Goldbach Conjecture Explained in 5 Minutes. In 1742, the German mathematician Christian Goldbach, in a discussion with the mathematician Leonhard Euler, made a simple statement... Every even integer greater than 2 can be written as the sum of two prime numbers.. Mathematicians have tried to prove this ever since. None have. It's a great example of how a simple statement in mathematics can be amazingly difficult to prove. Computers have checked billions of numbers and shown it to be true for every number tested, but that's not the same as a proof. A proof would show it to be true for all even numbers, period.. It's easy to prove that every integer can be written as a product of primes. This is called the prime decomposition of an integer. So for any integer m we have.. m=(p1^s1)*(p2^s2)*......*(pn^sn). where p1,p2,...,pn are prime numbers and the s1,s2,s3,...,sn are just integer powers and represent the simple fact that primes may be repeated.
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By collecting like primes together and raising them to a power we make sure that p1,p2,p3,...,pn are distinct primes with no duplication.. The condition that m be even simply means that one of these primes must be 2. Since the order of multiplication does not matter we can make p1=2. So m now looks like this... m=(2^s1)*(p2^s2)*...*(pn^sn). Goldbach's Conjecture says that when m is even there exists two prime numbers, let's call them g1 and g2, such that... m=(2^s1)*(p2^s2)*...*(pn^sn)=g1+g2. Are we on our way to a proof? No, but it's still fun to try!. Perhaps there's some deep clue in the fact that Goldbach's Conjecture only works if p1=2. In other words, if 2 does not appear in the prime decomposition of an integer then Goldbach's Conjecture does not work. What's so special about the number 2?.
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https://math.stackexchange.com/questions/2929894/determinants-of-block-tridiagonal-matrices-when-off-diagonal-blocks-are-not-mm
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Determinants of block tridiagonal matrices when off-diagonal blocks are not m*m matrices
We know that if we have the block-tridiagonal matrix $$M(z)$$ with blocks $$A_i$$, $$B_i$$ and $$C_{i−1}$$ (i =1,...,n) which are all $$m \times m$$ matrices $$M(z)=\begin{bmatrix} A_1 & B_1 & \cdots &\frac{1}{z}C_0\\ C_1 & \ddots & \ddots& \\ \vdots & \ddots & \ddots& B_{n-1}\\ zB_n & & C_{n-1}& A_n \end{bmatrix}$$ where the off-diagonal blocks are nonsingular, it can be associated with a transfer matrix, $$T=\begin{bmatrix} -B_n^{-1}A_n& -B_n^{-1}C_{n-1} \\ I_m & 0 \end{bmatrix} \cdots \begin{bmatrix} -B_1^{-1}A_1& -B_1^{-1}C_0 \\ I_m & 0 \end{bmatrix}$$ where $$I_m$$ is the $$m×m$$ unit matrix and $$\det(T)=\prod_{i=1}^\infty \det[B_i^{-1}C_{i-1}]$$
But what will the transfer matrix $$T$$ and the determinant $$\det(T)$$ be if $$B_i$$ and $$C_{i}$$ are $$m \times n$$ matrices ($$m$$ is not equal to $$n$$) instead of $$m\times m$$ matrices? What's the new formula for that?
Thank you!!
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Determinants of block tridiagonal matrices when off-diagonal blocks are not m*m matrices. We know that if we have the block-tridiagonal matrix $$M(z)$$ with blocks $$A_i$$, $$B_i$$ and $$C_{i−1}$$ (i =1,...,n) which are all $$m \times m$$ matrices $$M(z)=\begin{bmatrix} A_1 & B_1 & \cdots &\frac{1}{z}C_0\\ C_1 & \ddots & \ddots& \\ \vdots & \ddots & \ddots& B_{n-1}\\ zB_n & & C_{n-1}& A_n \end{bmatrix}$$ where the off-diagonal blocks are nonsingular, it can be associated with a transfer matrix, $$T=\begin{bmatrix} -B_n^{-1}A_n& -B_n^{-1}C_{n-1} \\ I_m & 0 \end{bmatrix} \cdots \begin{bmatrix} -B_1^{-1}A_1& -B_1^{-1}C_0 \\ I_m & 0 \end{bmatrix}$$ where $$I_m$$ is the $$m×m$$ unit matrix and $$\det(T)=\prod_{i=1}^\infty \det[B_i^{-1}C_{i-1}]$$. But what will the transfer matrix $$T$$ and the determinant $$\det(T)$$ be if $$B_i$$ and $$C_{i}$$ are $$m \times n$$ matrices ($$m$$ is not equal to $$n$$) instead of $$m\times m$$ matrices? What's the new formula for that?.
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Thank you!!.
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https://math.stackexchange.com/questions/1290948/special-representation-of-a-number
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# Special representation of a number
How can I check, if a number $n$ can be representated by
$$pq+rs$$
where $p,q,r,s$ are pairwise different prime numbers with the same number of digits.
For example,
$$105153899965560312960 = 3022993637\times 6003631993 + 9069920719\times 9592692301$$
has such a representation.
My questions :
• Is such a representation (if it exists), always unique ?
• How can I find the primes $p,q,r,s$ if a representation exists ?
• How can I check if a representation exists ?
• There is a very large number of counterexamples to your uniqueness conjecture, of which the smallest is $$11\cdot13 + 19\cdot 29 = 11\cdot 43 + 13\cdot 17$$ – MJD May 20 '15 at 13:12
• So, the first part is answered. Thank you. And I conjectured the uniqueness only for numbers large enough. It would be interesting, which is the largest counterexample (I think there is a largest). – Peter May 20 '15 at 13:14
You said in a comment:
I conjectured the uniqueness only for numbers large enough. It would be interesting, which is the largest counterexample (I think there is a largest).
I think your intuition is backwards here, and I think it will be instructive if I explain to you why; this is also why I was able to immediately guess that there would be counterexamples. Short summary: There so many 4-tuples compared with the number of possible sums that it is not possible for every 4-tuple to get its own sum.
Suppose our primes have $d$ digits and that there are $D$ primes with $d$ digits. The prime number theorem tells us that $D$ is around $O\left(\frac{10^d}{d}\right)$.
Then there are $O(D^4) = O\left(\frac{10^{4d}}{d^4}\right)$ quadruples $(p,q,r,s)$ of distinct $d$-digit primes. The quantity $n=pq+rs$ has around $2d$ digits and there are $10^{2d}$ of these numbers.
There is no reason to think that the quantities $pq+rs$ will be distributed among these $10^{2d}$ possibilities in any way other than randomly. (Additive properties of prime numbers are almost always random unless there is some obvious reason they cannot be. To take an example I made up on the spot, the mod-3 remainder of $p+q$ is equal to 0, 1, or 2 with probability close to $\frac12, \frac14, \frac14$ just as one would expect.)
So we should expect that what we are doing here is essentially throwing $O\left(\frac{10^{4d}}{d^4}\right)$ balls at random into only $10^{2d}$ bins . The number of balls greatly exceeds the number of bins when $d$ is large, so it is not at all surprising that some balls end up in the same bin; that is, that some tuples yield the same value for $pq+rs$. As $d$ increases, we should expect this to be vastly more likely, not less likely, and this argues strongly against the possibility of a maximal counterexample.
Indeed, one would expect the opposite to be true: let $N$ be given. Then we should expect to find many sets, each containing $N$ 4-tuples, for which all the 4-tuples in the set have the same $pq+rs$ value, no matter how large $N$ is large. If we want to find $n$ that can be represented as $n=pq+rs$ in one million different ways, we should expect that there will be many such.
(The tuples won't be distributed uniformly over the space of sums—for example, only a few tuples map to an odd sum—but this will tend to increase, not decrease, the number of collisions.)
• That is convincing. +1 for the well explanation. – Peter May 20 '15 at 13:37
• Experiments bear out this theory. There are 20 ways to represent 3150 in the form $n=pq+rs$, and this is the most ways for any $n$ under 10,000. But if we start looking at larger primes, there are over ten thousand such $n$ below 200,000; for example there are 20 or more ways to represent each of 170280, 170430, 170436, 170520, 170790, 170940, 170970, 171318, 171780, 171990, 172410… – MJD May 20 '15 at 13:49
(Just to elaborate on MJD's answer.)
Note that if $p, q, r, s$ are all $d$ digits long, which means that they all lie in the interval $[10^{d-1}, 10^d)$, then $pq + rs$ lies in the interval $[2\times10^{2d-2}, 2\times10^d)$. For numbers $N$ that lie close to the endpoints of this interval there will be few representations as $pq + rs$, while for numbers "well inside" this interval there ought to be many.
Let's dispose of the case of $1$-digit primes: in that case $\{p, q, r, s\} = \{2, 3, 5, 7\}$, and the only possible values of $pq + rs$ are $\{29, 31, 41\}$. For larger lengths, $p, q, r, s$ are all odd numbers, so $pq + rs$ is even.
I computed the number of representations as $pq + rs$ for $p, q, r, s$ being 3-digit and 4-digit primes, respectively. For 3-digit primes numbers with such a representation lie in the interval $[20000, 2000000]$ (actually $[22030,1948418]$), and for 4-digit primes they lie in the interval $[2000000, 200000000]$ (actually $[2062436,198303900]$). Plotting them gives:
Far from the representation as $pq+rs$ being unique,
• all even numbers in $[60198, 1217132]$ have at least $2$ representations,
• all even numbers in $[90376, 1038516]$ and in $[3667846, 161023932]$ have at least $10$ representations,
• all even numbers in $[12346780, 95078484]$ have at least $1000$ representations, etc.
There are even numbers with over $12000$ representations as you can see, and as you go to larger $N$ this will only increase.
I think there is nothing special about prime numbers here (once you start adding them): if you take merely "odd numbers" you'd probably see something similar.
• The remarkable resemblance of the two plots is interesting though, and I'm curious what the general shape (of the "upper" and "lower" curves) are. – ShreevatsaR Jun 28 '15 at 16:33
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# Special representation of a number. How can I check, if a number $n$ can be representated by. $$pq+rs$$. where $p,q,r,s$ are pairwise different prime numbers with the same number of digits.. For example,. $$105153899965560312960 = 3022993637\times 6003631993 + 9069920719\times 9592692301$$. has such a representation.. My questions :. • Is such a representation (if it exists), always unique ?. • How can I find the primes $p,q,r,s$ if a representation exists ?. • How can I check if a representation exists ?. • There is a very large number of counterexamples to your uniqueness conjecture, of which the smallest is $$11\cdot13 + 19\cdot 29 = 11\cdot 43 + 13\cdot 17$$ – MJD May 20 '15 at 13:12. • So, the first part is answered. Thank you. And I conjectured the uniqueness only for numbers large enough. It would be interesting, which is the largest counterexample (I think there is a largest). – Peter May 20 '15 at 13:14. You said in a comment:. I conjectured the uniqueness only for numbers large enough. It would be interesting, which is the largest counterexample (I think there is a largest).. I think your intuition is backwards here, and I think it will be instructive if I explain to you why; this is also why I was able to immediately guess that there would be counterexamples. Short summary: There so many 4-tuples compared with the number of possible sums that it is not possible for every 4-tuple to get its own sum.. Suppose our primes have $d$ digits and that there are $D$ primes with $d$ digits. The prime number theorem tells us that $D$ is around $O\left(\frac{10^d}{d}\right)$.. Then there are $O(D^4) = O\left(\frac{10^{4d}}{d^4}\right)$ quadruples $(p,q,r,s)$ of distinct $d$-digit primes. The quantity $n=pq+rs$ has around $2d$ digits and there are $10^{2d}$ of these numbers.. There is no reason to think that the quantities $pq+rs$ will be distributed among these $10^{2d}$ possibilities in any way other than randomly. (Additive properties of prime numbers are almost always random unless there is some obvious reason they cannot be.
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To take an example I made up on the spot, the mod-3 remainder of $p+q$ is equal to 0, 1, or 2 with probability close to $\frac12, \frac14, \frac14$ just as one would expect.). So we should expect that what we are doing here is essentially throwing $O\left(\frac{10^{4d}}{d^4}\right)$ balls at random into only $10^{2d}$ bins . The number of balls greatly exceeds the number of bins when $d$ is large, so it is not at all surprising that some balls end up in the same bin; that is, that some tuples yield the same value for $pq+rs$. As $d$ increases, we should expect this to be vastly more likely, not less likely, and this argues strongly against the possibility of a maximal counterexample.. Indeed, one would expect the opposite to be true: let $N$ be given. Then we should expect to find many sets, each containing $N$ 4-tuples, for which all the 4-tuples in the set have the same $pq+rs$ value, no matter how large $N$ is large. If we want to find $n$ that can be represented as $n=pq+rs$ in one million different ways, we should expect that there will be many such.. (The tuples won't be distributed uniformly over the space of sums—for example, only a few tuples map to an odd sum—but this will tend to increase, not decrease, the number of collisions.). • That is convincing. +1 for the well explanation. – Peter May 20 '15 at 13:37. • Experiments bear out this theory. There are 20 ways to represent 3150 in the form $n=pq+rs$, and this is the most ways for any $n$ under 10,000. But if we start looking at larger primes, there are over ten thousand such $n$ below 200,000; for example there are 20 or more ways to represent each of 170280, 170430, 170436, 170520, 170790, 170940, 170970, 171318, 171780, 171990, 172410… – MJD May 20 '15 at 13:49. (Just to elaborate on MJD's answer.). Note that if $p, q, r, s$ are all $d$ digits long, which means that they all lie in the interval $[10^{d-1}, 10^d)$, then $pq + rs$ lies in the interval $[2\times10^{2d-2}, 2\times10^d)$. For numbers $N$ that lie close to the endpoints of this interval there will be few representations as $pq + rs$, while for numbers "well inside" this interval there ought to be many.. Let's dispose of the case of $1$-digit primes: in that case $\{p, q, r, s\} = \{2, 3, 5, 7\}$, and the only possible values of $pq + rs$ are $\{29, 31, 41\}$. For larger lengths, $p, q, r, s$ are all odd numbers, so $pq + rs$ is even.. I computed the number of representations as $pq + rs$ for $p, q, r, s$ being 3-digit and 4-digit primes, respectively. For 3-digit primes numbers with such a representation lie in the interval $[20000, 2000000]$ (actually $[22030,1948418]$), and for 4-digit primes they lie in the interval $[2000000, 200000000]$ (actually $[2062436,198303900]$). Plotting them gives:. Far from the representation as $pq+rs$ being unique,. • all even numbers in $[60198, 1217132]$ have at least $2$ representations,. • all even numbers in $[90376, 1038516]$ and in $[3667846, 161023932]$ have at least $10$ representations,. • all even numbers in $[12346780, 95078484]$ have at least $1000$ representations, etc.. There are even numbers with over $12000$ representations as you can see, and as you go to larger $N$ this will only increase.. I think there is nothing special about prime numbers here (once you start adding them): if you take merely "odd numbers" you'd probably see something similar.. • The remarkable resemblance of the two plots is interesting though, and I'm curious what the general shape (of the "upper" and "lower" curves) are. – ShreevatsaR Jun 28 '15 at 16:33.
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# Torus in 3-space
## Definition
The torus in $\R^3$ is constructed as follows:
• Take a base circle centered at a point which we shall call the origin. We shall call the radius of the base circle the average radius.
• Fix a length smaller than the average radius, called the tube radius.
• For each point on the torus, consider the circle centered at that point of radius equal to the tube radius, in the plane perpendicular to the plane of the base circle, and containing that center and the origin
• The union of all such circles is termed the torus
The torus can thus be thought of as the trace of a circle whose center is itself moving on a base circle, such that the plane of the circle always contains the origin of the base circle.
Some further terminology:
• The inner radius is the average radius minus the tube radius. It is the shortest possible distance between the origin and points on the torus. The set of points at this minimum distance forms a circle, called the inner rim or inner circle.
• The outer radius is the average radius plus the tube radius. It is the maximum distance between the origin and points on the torus. The set of points at this maixmum distance form a circle, called the outer rim or outer circle.
• Consider two planes parallel to the plane containing the base circle, with distance from it equal to the tube radius. Both these planes are tangent to the torus, meeting it at circles. These circles are termed the top circle and bottom circle (interchangeably). The radius of these circles is equal to the average radius.
## Equational descriptions
### Cartesian parametric equation
Let $r_1$ denote the average radius and $r_2$ denote the tube radius. Suppose the base circle is in the $xy$-plane and the origin of the torus is the origin. Then the parametric equations aer in terms of two angles, $\alpha$ and $\beta$, where:
• $x = r_1 \cos \alpha + r_2 \cos \alpha \cos \beta$
• $y = r_1 \sin \alpha + r_2 \sin \alpha \cos \beta$
• $z = r_2 \sin \beta$
Both $\alpha$ and $\beta$ are modulo $2\pi$.
## Abstract structure
Topologically, and even differentially, the torus is isomorphic to the direct product of circles, viz $S^1 \times S^1$. One coordinate describes the position of the center on the base circle, and the other coordinate describes the position of the point on that particular small circle. The Cartesian parametric equations given above make this explicit.
However, the metric structure on the torus is very different from that on $S^1 \times S^1$ with the direct product metric. The latter is actually the flat torus, which cannot be embedded in $\R^3$, but needs to be embedded in $\R^4$.
## Structure and symmetry
The isometry group of the torus is the semidirect product of the following two groups:
• The group $S^1$, which acts on the first coordinate by rotating the base circle.
• The group of order two which reflects the whole configuration about the $xy$-plane
In fact, this is precisely the group of isometries of $\R^3$ which fix the origin and the preserve the $z$-axis (though they may not preserve direction).
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# Torus in 3-space. ## Definition. The torus in $\R^3$ is constructed as follows:. • Take a base circle centered at a point which we shall call the origin. We shall call the radius of the base circle the average radius.. • Fix a length smaller than the average radius, called the tube radius.. • For each point on the torus, consider the circle centered at that point of radius equal to the tube radius, in the plane perpendicular to the plane of the base circle, and containing that center and the origin. • The union of all such circles is termed the torus. The torus can thus be thought of as the trace of a circle whose center is itself moving on a base circle, such that the plane of the circle always contains the origin of the base circle.. Some further terminology:. • The inner radius is the average radius minus the tube radius. It is the shortest possible distance between the origin and points on the torus. The set of points at this minimum distance forms a circle, called the inner rim or inner circle.. • The outer radius is the average radius plus the tube radius. It is the maximum distance between the origin and points on the torus. The set of points at this maixmum distance form a circle, called the outer rim or outer circle.. • Consider two planes parallel to the plane containing the base circle, with distance from it equal to the tube radius. Both these planes are tangent to the torus, meeting it at circles. These circles are termed the top circle and bottom circle (interchangeably). The radius of these circles is equal to the average radius.. ## Equational descriptions.
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### Cartesian parametric equation. Let $r_1$ denote the average radius and $r_2$ denote the tube radius. Suppose the base circle is in the $xy$-plane and the origin of the torus is the origin. Then the parametric equations aer in terms of two angles, $\alpha$ and $\beta$, where:. • $x = r_1 \cos \alpha + r_2 \cos \alpha \cos \beta$. • $y = r_1 \sin \alpha + r_2 \sin \alpha \cos \beta$. • $z = r_2 \sin \beta$. Both $\alpha$ and $\beta$ are modulo $2\pi$.. ## Abstract structure. Topologically, and even differentially, the torus is isomorphic to the direct product of circles, viz $S^1 \times S^1$. One coordinate describes the position of the center on the base circle, and the other coordinate describes the position of the point on that particular small circle. The Cartesian parametric equations given above make this explicit.. However, the metric structure on the torus is very different from that on $S^1 \times S^1$ with the direct product metric. The latter is actually the flat torus, which cannot be embedded in $\R^3$, but needs to be embedded in $\R^4$.. ## Structure and symmetry. The isometry group of the torus is the semidirect product of the following two groups:. • The group $S^1$, which acts on the first coordinate by rotating the base circle.. • The group of order two which reflects the whole configuration about the $xy$-plane. In fact, this is precisely the group of isometries of $\R^3$ which fix the origin and the preserve the $z$-axis (though they may not preserve direction).
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https://www.teacherspayteachers.com/Product/Candy-Corn-Ten-Frame-Fun-389956
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## Main Categories
Total:
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# Candy Corn Ten Frame Fun
Common Core Standards
Product Rating
File Type
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
187 KB|10 pages
Share
Product Description
Candy Corn Ten Frame Fun
Grab some candy corn, and have some math fun with your kiddos. This ten frame math activity can be used with your whole class or as a math center. It meets a variety of common core state standards for addition and subtraction, depending on how it is used in the classroom. Each activity reinforces the number and quantity for 10. There are 4 recording sheets: Adding to 10, Subtracting from 10, Using Addition to Solve Subtraction from 10, and Turn Around Facts for 10
Kindergarten Standards:
K.OA.1 Represent addition and subtraction with objects...
K.OA.2 Solve addition and subtraction word problems and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
K.OA.4 For any number from 1 to 9, find the number that makes 10 when added to the given number, e.g., by using objects or drawings, and record the answer with a drawing or equation.
1.OA.1 Represent and solve problems involving addition and subtraction.
1.OA.3 Apply properties of operations as strategies to add and subtract. (For example, if 8 + 3 = 11 is known, then 3 + 8 = 11 is also known).
1.OA.4 Understand subtraction as an unknown-addend problem. (For example, subtract 10 - 8 by finding the number that makes 10 when added to 8.)
Directions:
1. You will need candy corn in two colors for each child.
2. Give each child a small baggy of candy corn containing 10 pieces of each color.
3. Use the appropriate printable to match your instruction.
4. Instruct children to "Build, Color, and Record their Answers".
Build: Use 2 different colors of real candy corn to fill up a ten frame.
Color: Color the candy on the ten frame recording sheet to match the ten frame.
Record: Write a number sentence (equation) to match the ten frame.
I hope you enjoy this resource!
Graphics by ScrappinDoodles
Total Pages
10 pages
N/A
Teaching Duration
N/A
Report this Resource
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## Main Categories. Total:. \$0.00. Whoops! Something went wrong.. # Candy Corn Ten Frame Fun. Common Core Standards. Product Rating. File Type. PDF (Acrobat) Document File. Be sure that you have an application to open this file type before downloading and/or purchasing.. 187 KB|10 pages. Share. Product Description. Candy Corn Ten Frame Fun. Grab some candy corn, and have some math fun with your kiddos. This ten frame math activity can be used with your whole class or as a math center. It meets a variety of common core state standards for addition and subtraction, depending on how it is used in the classroom. Each activity reinforces the number and quantity for 10. There are 4 recording sheets: Adding to 10, Subtracting from 10, Using Addition to Solve Subtraction from 10, and Turn Around Facts for 10. Kindergarten Standards:. K.OA.1 Represent addition and subtraction with objects.... K.OA.2 Solve addition and subtraction word problems and add and subtract within 10, e.g., by using objects or drawings to represent the problem.. K.OA.4 For any number from 1 to 9, find the number that makes 10 when added to the given number, e.g., by using objects or drawings, and record the answer with a drawing or equation.. 1.OA.1 Represent and solve problems involving addition and subtraction.. 1.OA.3 Apply properties of operations as strategies to add and subtract. (For example, if 8 + 3 = 11 is known, then 3 + 8 = 11 is also known).
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1.OA.4 Understand subtraction as an unknown-addend problem. (For example, subtract 10 - 8 by finding the number that makes 10 when added to 8.). Directions:. 1. You will need candy corn in two colors for each child.. 2. Give each child a small baggy of candy corn containing 10 pieces of each color.. 3. Use the appropriate printable to match your instruction.. 4. Instruct children to "Build, Color, and Record their Answers".. Build: Use 2 different colors of real candy corn to fill up a ten frame.. Color: Color the candy on the ten frame recording sheet to match the ten frame.. Record: Write a number sentence (equation) to match the ten frame.. I hope you enjoy this resource!. Graphics by ScrappinDoodles. Total Pages. 10 pages. N/A. Teaching Duration. N/A. Report this Resource. \$1.00. More products from Fantastic First Grade. \$1.00.
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http://www.nzmaths.co.nz/resource/power-2
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Te Kete Ipurangi
Communities
Schools
### Te Kete Ipurangi user options:
Level Five > Number and Algebra
# The Power of 2
Keywords:
Achievement Objectives:
Purpose:
This is a level 5 algebra strand activity from the Figure It Out series.
Specific Learning Outcomes:
find a rule to describe a non linear number pattern
Required Resource Materials:
FIO, Level 4, Algebra, Book Two, The power of 2, pages 18-19
geoboard or square dot paper (see Copymaster)
newspaper, scissors, calculator
circular plastic lids (three different sizes)
classmate
Activity:
#### Activity One
In this activity, the students should initially use geoboards or square dot paper to help them work out the area of the striped squares.
The area of the striped square in figure 1 (which is the shape in iv) is the area of the surrounding 4 by 4 square, that is, 16 square units, minus the area of the four corner triangles. The area of each triangle is 2 square units (one-half of a 2 by 2 square). So the striped area is equal to 16 square units minus 4 x 2 square units. We usually write 8 square units as 8 units2, so in this case, the area is 16 units2 – 8 units2 = 8 units2.
The area of the striped square in figure 2 (which is the shape in iii) is the area of the striped square from figure 1 minus the area of the four corner triangles. The area of each of these triangles is 1 unit2 (two halves of a 1 by 1 square). So the striped area in figure 2 is equal to 8 units2 – 4 x 1 = 4 units2.
The students need to repeat this process for the other striped squares and write a rule connecting successive striped square areas. They will see that the striped area in figure 1 is double the striped area in figure 2. This relationship can also be seen clearly by folding squares of paper, as illustrated in the diagram below.
Square A is twice the area of square B. Square A is the square that encloses square B.
So a simple rule is: the area of a square is twice the area of the square it encloses or, as given in the Answers, the area of a square is double the area of the enclosed square.
#### Activity Two
In this activity, the students repeatedly fold and then cut pieces of paper. Each cut doubles the number of pieces of paper. These results can be shown in a table:
So, for 10 cuts, there will be 210 pieces of paper. This is 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1 024. We express this as “2 to the power of 10 equals 1 024”.
Pressing the following sequence of keys on most calculators will give the value of 210 as 1 024.
Where this sequence of keys doesn’t work, experiment until you find a sequence that does. For example, the following sequence works with some calculators:
Scientific calculators can also be used to find the value of powers of numbers. We can express these algebraically as yx, which we say as “any number, y, raised to the power of any number, x”. So the value of 57 can be found by pressing the following keys:
Note that some scientific calculators use the key xy instead of yx.
#### Activity Three
This activity explores a simplified version of a mathematical puzzle that was invented in 1883 by the French mathematician Edouard Lucas (1842–1891). The puzzle is known as the Tower of Hanoi and sometimes as the Tower of Brahma. It was inspired by a Hindu legend that tells of the mental discipline demanded of young
priests. The legend says that at the beginning of time, the priests in the temple were given 64 gold discs, each one a little smaller than the one beneath it. The priests were to transfer the 64 gold discs from one of three poles to another, via the second pole where necessary, in such a way that a disc could never be placed on top of a smaller disc. The legend goes on to say that when this task was finished, the temple would
crumble into dust and the world would end.
The students are initially asked to see if they can work out the minimum number of moves to transfer only 3 lids (discs) rather than the 64 in the legend. To understand how this puzzle works, the students will find it helpful to try the puzzle with just 1 lid and then 2 lids. The following diagrams show the moves for 1, 2, and 3 lids.
If you look closely at the diagrams, you may notice a symmetrical pattern. Each diagram has a centre of rotational symmetry, marked with a .
The pattern for the number of moves is shown in this table:
The pattern in the table shows that there are 2(number of lids), – 1 moves for any number of lids. When there are x lids, the number of moves can be expressed algebraically as 2x – 1.
If we return to the Hindu legend associated with this puzzle, we see that to transfer 64 gold discs, 264 – 1 moves will be required. Altogether, this is 18 446 744 073 709 551 615 moves. If the priests worked continuously for 24 hours a day, 7 days a week, making one move every second, the complete transfer would take slightly more than 580 billion years. This is more than current estimates for the age of the universe, so the legend may well be correct in asserting that the world will end when the priests finish their task!
Activity One
1. a. ii. 2 units2
iii. 4 units2
iv. 8 units2
v. 16 units2
b. 32 units2
2. A possible rule is: double the area of the enclosed square.
Activity Two
1. 32. (2 x 2 x 2 x 2 x 2)
2. At least 7. (26 = 64 and 27 = 128)
3. Yes. 20 cuts gives 1 048 576 pieces, which is more than one million pieces. There are several ways that he may have used a calculator. One way is to press
the following sequence of keys (pressing the “equals” key 19 times):
On some calculators, this is
Activity Three
1. a. 7 moves
b. It takes 3 moves to shift a stack of 2 lids and 15 moves to shift a stack of 4 lids.
2. It will take 31 moves to shift a stack of 5 lids. A rule for this is 25 – 1. The table below shows how the rule works:
AttachmentSize
SquareDotPaperCM.pdf113.84 KB
## A Helping Hand
This is a level 4 algebra activity from the Figure It Out theme series.
## Which Wheels Where?
This is a level 4 algebra strand link activity from the Figure It Out series.
## Tiling Teasers
This is a level 3 algebra strand link activity from the Figure It Out series.
## Pizza Order
This is a level 3 algebra strand link activity from the Figure It Out series.
## Digit Chains
This is a level 3 algebra strand link activity from the Figure It Out series.
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Te Kete Ipurangi. Communities. Schools. ### Te Kete Ipurangi user options:. Level Five > Number and Algebra. # The Power of 2. Keywords:. Achievement Objectives:. Purpose:. This is a level 5 algebra strand activity from the Figure It Out series.. Specific Learning Outcomes:. find a rule to describe a non linear number pattern. Required Resource Materials:. FIO, Level 4, Algebra, Book Two, The power of 2, pages 18-19. geoboard or square dot paper (see Copymaster). newspaper, scissors, calculator. circular plastic lids (three different sizes). classmate. Activity:. #### Activity One. In this activity, the students should initially use geoboards or square dot paper to help them work out the area of the striped squares.. The area of the striped square in figure 1 (which is the shape in iv) is the area of the surrounding 4 by 4 square, that is, 16 square units, minus the area of the four corner triangles. The area of each triangle is 2 square units (one-half of a 2 by 2 square). So the striped area is equal to 16 square units minus 4 x 2 square units. We usually write 8 square units as 8 units2, so in this case, the area is 16 units2 – 8 units2 = 8 units2.. The area of the striped square in figure 2 (which is the shape in iii) is the area of the striped square from figure 1 minus the area of the four corner triangles. The area of each of these triangles is 1 unit2 (two halves of a 1 by 1 square). So the striped area in figure 2 is equal to 8 units2 – 4 x 1 = 4 units2.. The students need to repeat this process for the other striped squares and write a rule connecting successive striped square areas. They will see that the striped area in figure 1 is double the striped area in figure 2. This relationship can also be seen clearly by folding squares of paper, as illustrated in the diagram below.. Square A is twice the area of square B. Square A is the square that encloses square B.. So a simple rule is: the area of a square is twice the area of the square it encloses or, as given in the Answers, the area of a square is double the area of the enclosed square.. #### Activity Two. In this activity, the students repeatedly fold and then cut pieces of paper. Each cut doubles the number of pieces of paper. These results can be shown in a table:. So, for 10 cuts, there will be 210 pieces of paper. This is 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1 024. We express this as “2 to the power of 10 equals 1 024”.. Pressing the following sequence of keys on most calculators will give the value of 210 as 1 024.. Where this sequence of keys doesn’t work, experiment until you find a sequence that does. For example, the following sequence works with some calculators:. Scientific calculators can also be used to find the value of powers of numbers. We can express these algebraically as yx, which we say as “any number, y, raised to the power of any number, x”. So the value of 57 can be found by pressing the following keys:. Note that some scientific calculators use the key xy instead of yx.. #### Activity Three. This activity explores a simplified version of a mathematical puzzle that was invented in 1883 by the French mathematician Edouard Lucas (1842–1891). The puzzle is known as the Tower of Hanoi and sometimes as the Tower of Brahma. It was inspired by a Hindu legend that tells of the mental discipline demanded of young. priests. The legend says that at the beginning of time, the priests in the temple were given 64 gold discs, each one a little smaller than the one beneath it. The priests were to transfer the 64 gold discs from one of three poles to another, via the second pole where necessary, in such a way that a disc could never be placed on top of a smaller disc. The legend goes on to say that when this task was finished, the temple would. crumble into dust and the world would end.. The students are initially asked to see if they can work out the minimum number of moves to transfer only 3 lids (discs) rather than the 64 in the legend. To understand how this puzzle works, the students will find it helpful to try the puzzle with just 1 lid and then 2 lids. The following diagrams show the moves for 1, 2, and 3 lids.. If you look closely at the diagrams, you may notice a symmetrical pattern. Each diagram has a centre of rotational symmetry, marked with a .. The pattern for the number of moves is shown in this table:.
|
The pattern in the table shows that there are 2(number of lids), – 1 moves for any number of lids. When there are x lids, the number of moves can be expressed algebraically as 2x – 1.. If we return to the Hindu legend associated with this puzzle, we see that to transfer 64 gold discs, 264 – 1 moves will be required. Altogether, this is 18 446 744 073 709 551 615 moves. If the priests worked continuously for 24 hours a day, 7 days a week, making one move every second, the complete transfer would take slightly more than 580 billion years. This is more than current estimates for the age of the universe, so the legend may well be correct in asserting that the world will end when the priests finish their task!. Activity One. 1. a. ii. 2 units2. iii. 4 units2. iv. 8 units2. v. 16 units2. b. 32 units2. 2. A possible rule is: double the area of the enclosed square.. Activity Two. 1. 32. (2 x 2 x 2 x 2 x 2). 2. At least 7. (26 = 64 and 27 = 128). 3. Yes. 20 cuts gives 1 048 576 pieces, which is more than one million pieces. There are several ways that he may have used a calculator. One way is to press. the following sequence of keys (pressing the “equals” key 19 times):. On some calculators, this is. Activity Three. 1. a. 7 moves. b. It takes 3 moves to shift a stack of 2 lids and 15 moves to shift a stack of 4 lids.. 2. It will take 31 moves to shift a stack of 5 lids. A rule for this is 25 – 1. The table below shows how the rule works:. AttachmentSize. SquareDotPaperCM.pdf113.84 KB. ## A Helping Hand. This is a level 4 algebra activity from the Figure It Out theme series.. ## Which Wheels Where?. This is a level 4 algebra strand link activity from the Figure It Out series.. ## Tiling Teasers. This is a level 3 algebra strand link activity from the Figure It Out series.. ## Pizza Order. This is a level 3 algebra strand link activity from the Figure It Out series.. ## Digit Chains. This is a level 3 algebra strand link activity from the Figure It Out series.
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https://socratic.org/questions/1-s-1-r-1-b-solve-for-b-1#510758
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# 1/s+1/r=1/b. solve for b?
## rational expressions
##### 1 Answer
Nov 24, 2017
$b = \frac{r s}{r + s}$
#### Explanation:
Given
$\frac{1}{s} + \frac{1}{r} = \frac{1}{b}$
Solve for $b$
1) Get $b$ out of the denominator by multiplying all the terms on both sides by $b$ and letting the denominator cancel.
After you have multiplied and canceled, you will have this:
$\frac{b}{s} + \frac{b}{r} = \frac{1}{1}$
2) Clear the first fraction by multiplying all the terms on both sides by $s$ and letting that denominator cancel.
$\frac{b}{1} + \frac{b s}{r} = \frac{s}{1}$
3) Clear the second fraction by multiplying all the terms on both sides by $r$ and letting that denominator cancel.
$\frac{b r}{1} + \frac{b s}{1} = \frac{r s}{1}$
This is the same as
$b r + b s = r s$
4) Factor out $b$
$b \left(r + s\right) = r s$
5) Divide both sides by $\left(r + s\right)$ to isolate $b$
$b = \frac{r s}{r + s}$$\leftarrow$ answer
Answer:
$b = \frac{r s}{r + s}$
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# 1/s+1/r=1/b. solve for b?. ## rational expressions. ##### 1 Answer. Nov 24, 2017. $b = \frac{r s}{r + s}$. #### Explanation:. Given. $\frac{1}{s} + \frac{1}{r} = \frac{1}{b}$. Solve for $b$. 1) Get $b$ out of the denominator by multiplying all the terms on both sides by $b$ and letting the denominator cancel.. After you have multiplied and canceled, you will have this:. $\frac{b}{s} + \frac{b}{r} = \frac{1}{1}$.
|
2) Clear the first fraction by multiplying all the terms on both sides by $s$ and letting that denominator cancel.. $\frac{b}{1} + \frac{b s}{r} = \frac{s}{1}$. 3) Clear the second fraction by multiplying all the terms on both sides by $r$ and letting that denominator cancel.. $\frac{b r}{1} + \frac{b s}{1} = \frac{r s}{1}$. This is the same as. $b r + b s = r s$. 4) Factor out $b$. $b \left(r + s\right) = r s$. 5) Divide both sides by $\left(r + s\right)$ to isolate $b$. $b = \frac{r s}{r + s}$$\leftarrow$ answer. Answer:. $b = \frac{r s}{r + s}$.
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https://www.online-sciences.com/physics/laws-of-circular-motion-centripetal-acceleration-tangential-linear-velocity-centripetal-force/
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# Laws of circular motion (Centripetal Acceleration, Tangential linear Velocity & Centripetal Force)
### Centripetal Acceleration
Changing the direction of velocity leads to the existence of acceleration called the centripetal acceleration ( a ) which is the acceleration acquired by an object moving in a circular path due to a continuous change in the direction of its velocity .
When a force ( F ) acts on a body of mass ( m ) moving at speed ( v ) in a circular path of radius ( r ) normally to the direction of its motion , The magnitude of velocity ( v ) remains constant along its path , The direction of velocity changes from one point to another on its path .
#### Finding the value of the centripetal acceleration
If a body moves from point ( A ) to point ( B ) , the velocity ( v ) changes in direction but maintains a constant magnitude .
a = v² / r
Laws of circular motion
##### Factors affecting the centripetal acceleration ( a ) :
The tangential velocity : Centripetal acceleration is directly proportional to square of the tangential velocity at constant radius of the circular path .
Slope = a / v² = 1 / r
The radius of circular path : Centripetal acceleration is inversely proportional to the radius of the circular path at constant tangential velocity .
Slope = a r = v²
#### Tangential linear Velocity
Finding the tangential linear velocity : If the body complete one circular revolution in an interval of time ( T ) which is called the periodic time , Periodic time is the time taken by the body to make one complete revolution .
The tangential linear velocity = Distance ( circle circumference ) / Periodic time
v = 2 πr / T
##### Factors affecting the tangential linear velocity ( v ) :
The radius of circular path : Tangential velocity is directly proportional to the radius of the circular path at constant periodic time .
Slope = v / r = 2 π / T
The periodic time : Tangential velocity is inversely proportional to the periodic time at constant radius of circular path .
Slope = v T = 2 π r
v = 2 π r / T
##### Calculating the angular velocity :
If a body moves at tangential velocity ( v ) in a circle of radius ( r ) from point ( A ) to point ( B ) , covering a distance ( Δ l ) corresponding to an angle ( Δθ ) , during time interval ( Δ t ) .
Then the value ( Δθ / Δ t ) is known as the angular velocity ( ω ) .
 ω = Δθ / Δ t
It is known that the value of the angle in radian equals the ratio between the arc length and the radius of the path .
Δθ = Δl / rÂ
ω = ( Δl / Δ t ) × ( 1 / r ) = v / r     ∴ v = ω r
Tangential ( linear ) velocity = Angular velocity × Radius of the path
Since v = 2 π r / T  , then : ω r = 2 π r / T
∴ ω = 2 π / T
#### The Centripetal Force
Finding the Centripetal Force ( F ) , According to Newton’s Second Law , the force is given by the relation :Â
F = m a  , a = v² / r
F = m a = m v² / r
##### Factors affecting the centripetal force ( F ) :
The object mass : Centripetal force is directly proportional to the object mass at constant tangential velocity and radius of circular path .
Slope = F / m = v² / r
The tangential velocity : Centripetal force is directly proportional to square of the tangential velocity at constant radius of circular path and object mass .
Slope = F / v² = m / r
F = m v² / r
The radius of circular path : Centripetal force is inversely proportional to the radius of the circular path at constant tangential velocity and object mass .
Slope = F r = m v²
##### Designing curved roads
It is necessary to calculate the centripetal force when designing the curved roads and railways in order to keep cars and trains moving along this curved path without skidding out .
If a car moves in a curved slippery road , the frictional forces may not be sufficient to turn the car around the curve , So , the car skids out the road .
Engineers define certain velocities for vehicles when moving in curves , If the vehicle velocity exceeds the predetermined velocity , the vehicle needs more more centripetal force to be kept in this curved path where F α v² .
It is forbidden for trucks and trailers to move on some dangerous curves , As the vehicle mass increases , it needs more centripetal force where F α m .
Slowing down in dangerous curves is a must to avoid accidents , As the radius of curve decreases the car needs more centripetal force to turn around , where F α 1/r .
When moving a bucket half filled with water in a vertical circular path at sufficient speed , the water does not spill out from the opening of the bucket because :
The centripetal force acting on water is normal to the direction of its motion , This force changes the direction of velocity without changing its magnitude to keep water inside the bucket rotating in a circular path .
We can make benefit of skidding objects out the circular path when the centripetal force is not sufficient to keep them rotating in the circular path :
1. Making candy floss .
2. Rotating barrels in amusement park .
3. Drying cloths in automatic washing machines : water droplets adhere to clothes by certain forces , when the dryer rotates at great velocity , these adhesive forces will not be sufficient to hold these droplets , They separate from the cloths and move tangential to the circular path .
On using electric sharpener , the glowing metal splints blow in straight lines at tangent velocities .
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# Laws of circular motion (Centripetal Acceleration, Tangential linear Velocity & Centripetal Force). ### Centripetal Acceleration. Changing the direction of velocity leads to the existence of acceleration called the centripetal acceleration ( a ) which is the acceleration acquired by an object moving in a circular path due to a continuous change in the direction of its velocity .. When a force ( F ) acts on a body of mass ( m ) moving at speed ( v ) in a circular path of radius ( r ) normally to the direction of its motion , The magnitude of velocity ( v ) remains constant along its path , The direction of velocity changes from one point to another on its path .. #### Finding the value of the centripetal acceleration. If a body moves from point ( A ) to point ( B ) , the velocity ( v ) changes in direction but maintains a constant magnitude .. a = v² / r. Laws of circular motion. ##### Factors affecting the centripetal acceleration ( a ) :. The tangential velocity : Centripetal acceleration is directly proportional to square of the tangential velocity at constant radius of the circular path .. Slope = a / v² = 1 / r. The radius of circular path : Centripetal acceleration is inversely proportional to the radius of the circular path at constant tangential velocity .. Slope = a r = v². #### Tangential linear Velocity. Finding the tangential linear velocity : If the body complete one circular revolution in an interval of time ( T ) which is called the periodic time , Periodic time is the time taken by the body to make one complete revolution .. The tangential linear velocity = Distance ( circle circumference ) / Periodic time. v = 2 Ï€r / T. ##### Factors affecting the tangential linear velocity ( v ) :. The radius of circular path : Tangential velocity is directly proportional to the radius of the circular path at constant periodic time .. Slope = v / r = 2 Ï€ / T. The periodic time : Tangential velocity is inversely proportional to the periodic time at constant radius of circular path .. Slope = v T = 2 Ï€ r. v = 2 Ï€ r / T. ##### Calculating the angular velocity :. If a body moves at tangential velocity ( v ) in a circle of radius ( r ) from point ( A ) to point ( B ) , covering a distance ( Δ l ) corresponding to an angle ( Δθ ) , during time interval ( Δ t ) .. Then the value ( Δθ / Δ t ) is known as the angular velocity ( ω ) ..  ω = Δθ / Δ t. It is known that the value of the angle in radian equals the ratio between the arc length and the radius of the path .. Δθ = Δl / rÂ. ω = ( Δl / Δ t ) × ( 1 / r ) = v / r     ∴ v = ω r. Tangential ( linear ) velocity = Angular velocity × Radius of the path.
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Since v = 2 Ï€ r / T  , then : ω r = 2 Ï€ r / T. ∴ ω = 2 Ï€ / T. #### The Centripetal Force. Finding the Centripetal Force ( F ) , According to Newton’s Second Law , the force is given by the relation :Â. F = m a  , a = v² / r. F = m a = m v² / r. ##### Factors affecting the centripetal force ( F ) :. The object mass : Centripetal force is directly proportional to the object mass at constant tangential velocity and radius of circular path .. Slope = F / m = v² / r. The tangential velocity : Centripetal force is directly proportional to square of the tangential velocity at constant radius of circular path and object mass .. Slope = F / v² = m / r. F = m v² / r. The radius of circular path : Centripetal force is inversely proportional to the radius of the circular path at constant tangential velocity and object mass .. Slope = F r = m v². ##### Designing curved roads. It is necessary to calculate the centripetal force when designing the curved roads and railways in order to keep cars and trains moving along this curved path without skidding out .. If a car moves in a curved slippery road , the frictional forces may not be sufficient to turn the car around the curve , So , the car skids out the road .. Engineers define certain velocities for vehicles when moving in curves , If the vehicle velocity exceeds the predetermined velocity , the vehicle needs more more centripetal force to be kept in this curved path where F α v² .. It is forbidden for trucks and trailers to move on some dangerous curves , As the vehicle mass increases , it needs more centripetal force where F α m .. Slowing down in dangerous curves is a must to avoid accidents , As the radius of curve decreases the car needs more centripetal force to turn around , where F α 1/r .. When moving a bucket half filled with water in a vertical circular path at sufficient speed , the water does not spill out from the opening of the bucket because :. The centripetal force acting on water is normal to the direction of its motion , This force changes the direction of velocity without changing its magnitude to keep water inside the bucket rotating in a circular path .. We can make benefit of skidding objects out the circular path when the centripetal force is not sufficient to keep them rotating in the circular path :. 1. Making candy floss .. 2. Rotating barrels in amusement park .. 3. Drying cloths in automatic washing machines : water droplets adhere to clothes by certain forces , when the dryer rotates at great velocity , these adhesive forces will not be sufficient to hold these droplets , They separate from the cloths and move tangential to the circular path .. On using electric sharpener , the glowing metal splints blow in straight lines at tangent velocities .
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# Search by Topic
#### Resources tagged with Creating expressions/formulae similar to Arrange the Digits:
Filter by: Content type:
Stage:
Challenge level:
### There are 85 results
Broad Topics > Algebra > Creating expressions/formulae
### Enriching Experience
##### Stage: 4 Challenge Level:
Find the five distinct digits N, R, I, C and H in the following nomogram
### Seven Up
##### Stage: 3 Challenge Level:
The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?
### Crossed Ends
##### Stage: 3 Challenge Level:
Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Top-heavy Pyramids
##### Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
A car's milometer reads 4631 miles and the trip meter has 173.3 on it. How many more miles must the car travel before the two numbers contain the same digits in the same order?
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Special Sums and Products
##### Stage: 3 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Legs Eleven
##### Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
### Always a Multiple?
##### Stage: 3 Challenge Level:
Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### 2-digit Square
##### Stage: 4 Challenge Level:
A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number?
### Sums of Pairs
##### Stage: 3 and 4 Challenge Level:
Jo has three numbers which she adds together in pairs. When she does this she has three different totals: 11, 17 and 22 What are the three numbers Jo had to start with?”
### Back to Basics
##### Stage: 4 Challenge Level:
Find b where 3723(base 10) = 123(base b).
### Cubes Within Cubes Revisited
##### Stage: 3 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Chocolate Maths
##### Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
### Multiplication Square
##### Stage: 3 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Pair Products
##### Stage: 4 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### Magic W
##### Stage: 4 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
### Around and Back
##### Stage: 4 Challenge Level:
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . .
### Hallway Borders
##### Stage: 3 Challenge Level:
A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.
### Seven Squares
##### Stage: 3 and 4 Challenge Level:
Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?
### Odd Differences
##### Stage: 4 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Plum Tree
##### Stage: 4 and 5 Challenge Level:
Label this plum tree graph to make it totally magic!
### Marbles in a Box
##### Stage: 3 and 4 Challenge Level:
In a three-dimensional version of noughts and crosses, how many winning lines can you make?
### Days and Dates
##### Stage: 4 Challenge Level:
Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
### Steel Cables
##### Stage: 4 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Generating Triples
##### Stage: 4 Challenge Level:
Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?
### One and Three
##### Stage: 4 Challenge Level:
Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . .
### Quick Times
##### Stage: 3 Challenge Level:
32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
### Hike and Hitch
##### Stage: 4 Challenge Level:
Fifteen students had to travel 60 miles. They could use a car, which could only carry 5 students. As the car left with the first 5 (at 40 miles per hour), the remaining 10 commenced hiking along the. . . .
### Christmas Chocolates
##### Stage: 3 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Reasonable Algebra
##### Stage: 4 Challenge Level:
Use algebra to reason why 16 and 32 are impossible to create as the sum of consecutive numbers.
### Three Four Five
##### Stage: 4 Challenge Level:
Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.
### Unit Interval
##### Stage: 4 and 5 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?
### More Number Pyramids
##### Stage: 3 and 4 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Interactive Number Patterns
##### Stage: 4 Challenge Level:
How good are you at finding the formula for a number pattern ?
### Balance Point
##### Stage: 4 Challenge Level:
Attach weights of 1, 2, 4, and 8 units to the four attachment points on the bar. Move the bar from side to side until you find a balance point. Is it possible to predict that position?
### Magic Sums and Products
##### Stage: 3 and 4
How to build your own magic squares.
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Pythagoras Proofs
##### Stage: 4 Challenge Level:
Can you make sense of these three proofs of Pythagoras' Theorem?
### Partially Painted Cube
##### Stage: 4 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Lower Bound
##### Stage: 3 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Screen Shot
##### Stage: 4 Challenge Level:
A moveable screen slides along a mirrored corridor towards a centrally placed light source. A ray of light from that source is directed towards a wall of the corridor, which it strikes at 45 degrees. . . .
##### Stage: 3 Challenge Level:
Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?
##### Stage: 3 Challenge Level:
Think of a number... follow the machine's instructions. I know what your number is! Can you explain how I know?
### ' Tis Whole
##### Stage: 4 and 5 Challenge Level:
Take a few whole numbers away from a triangle number. If you know the mean of the remaining numbers can you find the triangle number and which numbers were removed?
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# Search by Topic. #### Resources tagged with Creating expressions/formulae similar to Arrange the Digits:. Filter by: Content type:. Stage:. Challenge level:. ### There are 85 results. Broad Topics > Algebra > Creating expressions/formulae. ### Enriching Experience. ##### Stage: 4 Challenge Level:. Find the five distinct digits N, R, I, C and H in the following nomogram. ### Seven Up. ##### Stage: 3 Challenge Level:. The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?. ### Crossed Ends. ##### Stage: 3 Challenge Level:. Crosses can be drawn on number grids of various sizes. What do you notice when you add opposite ends?. ### Even So. ##### Stage: 3 Challenge Level:. Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?. ### Number Rules - OK. ##### Stage: 4 Challenge Level:. Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number.... ### Top-heavy Pyramids. ##### Stage: 3 Challenge Level:. Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.. ### Always the Same. ##### Stage: 3 Challenge Level:. Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?. ### How Many Miles to Go?. ##### Stage: 3 Challenge Level:. A car's milometer reads 4631 miles and the trip meter has 173.3 on it. How many more miles must the car travel before the two numbers contain the same digits in the same order?. ### Summing Consecutive Numbers. ##### Stage: 3 Challenge Level:. Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?. ### Special Sums and Products. ##### Stage: 3 Challenge Level:. Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.. ### Legs Eleven. ##### Stage: 3 Challenge Level:. Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?. ### Always a Multiple?. ##### Stage: 3 Challenge Level:. Think of a two digit number, reverse the digits, and add the numbers together. Something special happens.... ### Special Numbers. ##### Stage: 3 Challenge Level:. My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?. ### 2-digit Square. ##### Stage: 4 Challenge Level:. A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number?. ### Sums of Pairs. ##### Stage: 3 and 4 Challenge Level:. Jo has three numbers which she adds together in pairs. When she does this she has three different totals: 11, 17 and 22 What are the three numbers Jo had to start with?”. ### Back to Basics. ##### Stage: 4 Challenge Level:. Find b where 3723(base 10) = 123(base b).. ### Cubes Within Cubes Revisited. ##### Stage: 3 Challenge Level:. Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?. ### Number Pyramids. ##### Stage: 3 Challenge Level:. Try entering different sets of numbers in the number pyramids. How does the total at the top change?. ### Chocolate Maths. ##### Stage: 3 Challenge Level:. Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .. ### Multiplication Square. ##### Stage: 3 Challenge Level:. Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?. ### Pair Products. ##### Stage: 4 Challenge Level:. Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?. ### Magic W. ##### Stage: 4 Challenge Level:. Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.. ### Around and Back. ##### Stage: 4 Challenge Level:.
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A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . .. ### Hallway Borders. ##### Stage: 3 Challenge Level:. A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.. ### Seven Squares. ##### Stage: 3 and 4 Challenge Level:. Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?. ### Odd Differences. ##### Stage: 4 Challenge Level:. The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.. ### Plum Tree. ##### Stage: 4 and 5 Challenge Level:. Label this plum tree graph to make it totally magic!. ### Marbles in a Box. ##### Stage: 3 and 4 Challenge Level:. In a three-dimensional version of noughts and crosses, how many winning lines can you make?. ### Days and Dates. ##### Stage: 4 Challenge Level:. Investigate how you can work out what day of the week your birthday will be on next year, and the year after.... ### Steel Cables. ##### Stage: 4 Challenge Level:. Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?. ### Generating Triples. ##### Stage: 4 Challenge Level:. Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?. ### One and Three. ##### Stage: 4 Challenge Level:. Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . .. ### Quick Times. ##### Stage: 3 Challenge Level:. 32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.. ### Hike and Hitch. ##### Stage: 4 Challenge Level:. Fifteen students had to travel 60 miles. They could use a car, which could only carry 5 students. As the car left with the first 5 (at 40 miles per hour), the remaining 10 commenced hiking along the. . . .. ### Christmas Chocolates. ##### Stage: 3 Challenge Level:. How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?. ### Reasonable Algebra. ##### Stage: 4 Challenge Level:. Use algebra to reason why 16 and 32 are impossible to create as the sum of consecutive numbers.. ### Three Four Five. ##### Stage: 4 Challenge Level:. Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.. ### Unit Interval. ##### Stage: 4 and 5 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?. ### More Number Pyramids. ##### Stage: 3 and 4 Challenge Level:. When number pyramids have a sequence on the bottom layer, some interesting patterns emerge.... ### Interactive Number Patterns. ##### Stage: 4 Challenge Level:. How good are you at finding the formula for a number pattern ?. ### Balance Point. ##### Stage: 4 Challenge Level:. Attach weights of 1, 2, 4, and 8 units to the four attachment points on the bar. Move the bar from side to side until you find a balance point. Is it possible to predict that position?. ### Magic Sums and Products. ##### Stage: 3 and 4. How to build your own magic squares.. ### Painted Cube. ##### Stage: 3 Challenge Level:. Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?. ### Pythagoras Proofs. ##### Stage: 4 Challenge Level:. Can you make sense of these three proofs of Pythagoras' Theorem?. ### Partially Painted Cube. ##### Stage: 4 Challenge Level:. Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?. ### Lower Bound. ##### Stage: 3 Challenge Level:. What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =. ### Screen Shot. ##### Stage: 4 Challenge Level:. A moveable screen slides along a mirrored corridor towards a centrally placed light source. A ray of light from that source is directed towards a wall of the corridor, which it strikes at 45 degrees. . . .. ##### Stage: 3 Challenge Level:. Think of a number and follow my instructions. Tell me your answer, and I'll tell you what you started with! Can you explain how I know?. ##### Stage: 3 Challenge Level:. Think of a number... follow the machine's instructions. I know what your number is! Can you explain how I know?. ### ' Tis Whole. ##### Stage: 4 and 5 Challenge Level:. Take a few whole numbers away from a triangle number. If you know the mean of the remaining numbers can you find the triangle number and which numbers were removed?.
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INDE6620 Optimization and Applications
Question:
A company wants to manufacture five products over a 2-year (24 months) planning horizon. Table below provides the revenue (i.e., selling price) for each product. The table also presents the number of worker-hour to produce one unit of each product.
Product Selling Price (1st yr.) Selling Price (2st yr.) Worker-hour to produce one unit A 1100 1200 11.5 B 1100 1200 12 C 1100 1200 12 D 1100 1200 12.5 E 1100 1200 11.5
The table below provides the maximum demand for each period, which are expected to be similar for all the products: Month
Month 1 2 3 4 5 6 7 8 9 10 11 12 YEAR 1 210 225 240 255 350 350 400 450 450 360 265 180 YEAR 2 215 230 245 260 305 355 405 455 365 270 200 185
The company has a commitment to produce and ship 80% of the maximum demand for each period. The company has 75 workers available. Each worker can work for 160 hours per month. Each worker costs \$33 per hour (including benefits). Overtime is allowed with the cost of 50\$ per hour (including benefits). The manger is also allowed to hire and lay off workers if necessary. The costs of hiring a new worker in the first year and the second year are estimated to be \$2600 and \$2800, respectively. The costs to lay-off a worker in the first year and the second year are estimated to be \$1400 and \$1600, respectively. Moreover, the manager is allowed to keep and use inventory.
The company expects the holding costs to be \$9 for one period (per item) and remains constant over the planning horizon for all the products.
a) Define all the variables and list all your assumptions.
b) Develop the mathematical formulation to maximize the profit.
c) Find the optimal solution. Enough detail and associated computer files should be provided.
d) Suppose that the production manager is allowed to modify the cost elements. What would be your cost-saving suggestion?
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Get Instant Help From 5000+ Experts For. Writing: Get your essay and assignment written from scratch by PhD expert. Rewriting: Paraphrase or rewrite your friend's essay with similar meaning at reduced cost. INDE6620 Optimization and Applications. Question:. A company wants to manufacture five products over a 2-year (24 months) planning horizon. Table below provides the revenue (i.e., selling price) for each product. The table also presents the number of worker-hour to produce one unit of each product.. Product Selling Price (1st yr.) Selling Price (2st yr.) Worker-hour to produce one unit A 1100 1200 11.5 B 1100 1200 12 C 1100 1200 12 D 1100 1200 12.5 E 1100 1200 11.5. The table below provides the maximum demand for each period, which are expected to be similar for all the products: Month. Month 1 2 3 4 5 6 7 8 9 10 11 12 YEAR 1 210 225 240 255 350 350 400 450 450 360 265 180 YEAR 2 215 230 245 260 305 355 405 455 365 270 200 185. The company has a commitment to produce and ship 80% of the maximum demand for each period. The company has 75 workers available.
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Each worker can work for 160 hours per month. Each worker costs \$33 per hour (including benefits). Overtime is allowed with the cost of 50\$ per hour (including benefits). The manger is also allowed to hire and lay off workers if necessary. The costs of hiring a new worker in the first year and the second year are estimated to be \$2600 and \$2800, respectively. The costs to lay-off a worker in the first year and the second year are estimated to be \$1400 and \$1600, respectively. Moreover, the manager is allowed to keep and use inventory.. The company expects the holding costs to be \$9 for one period (per item) and remains constant over the planning horizon for all the products.. a) Define all the variables and list all your assumptions.. b) Develop the mathematical formulation to maximize the profit.. c) Find the optimal solution. Enough detail and associated computer files should be provided.. d) Suppose that the production manager is allowed to modify the cost elements. What would be your cost-saving suggestion?.
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A329864 Number of compositions of n with the same runs-resistance as cuts-resistance. 7
1, 0, 0, 0, 0, 2, 5, 10, 17, 27, 68, 107, 217, 420, 884, 1761, 3679, 7469, 15437, 31396, 64369 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,6 COMMENTS A composition of n is a finite sequence of positive integers summing to n. For the operation of taking the sequence of run-lengths of a finite sequence, runs-resistance is defined to be the number of applications required to reach a singleton. For the operation of shortening all runs by 1, cuts-resistance is defined to be the number of applications required to reach an empty word. LINKS Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003. EXAMPLE The a(5) = 2 through a(8) = 17 compositions: (1112) (1113) (1114) (1115) (2111) (1122) (1222) (1133) (2211) (2221) (3311) (3111) (4111) (5111) (11211) (11122) (11222) (11311) (11411) (21112) (12221) (22111) (21113) (111121) (22211) (121111) (31112) (111131) (111221) (112112) (112211) (122111) (131111) (211211) For example, the runs-resistance of (111221) is 3 because we have: (111221) -> (321) -> (111) -> (3), while the cuts-resistance is also 3 because we have: (111221) -> (112) -> (1) -> (), so (111221) is counted under a(8). MATHEMATICA runsres[q_]:=Length[NestWhileList[Length/@Split[#]&, q, Length[#]>1&]]-1; degdep[q_]:=Length[NestWhileList[Join@@Rest/@Split[#]&, q, Length[#]>0&]]-1; Table[Length[Select[Join@@Permutations/@IntegerPartitions[n], runsres[#]==degdep[#]&]], {n, 0, 10}] CROSSREFS The version for binary expansion is A329865. Compositions counted by runs-resistance are A329744. Compositions counted by cuts-resistance are A329861. Compositions with runs-resistance = cuts-resistance minus 1 are A329869. Cf. A003242, A098504, A114901, A242882, A318928, A319411, A319416, A319420, A319421, A329867, A329868. Sequence in context: A062493 A056871 A246883 * A174910 A301273 A007504 Adjacent sequences: A329861 A329862 A329863 * A329865 A329866 A329867 KEYWORD nonn,more AUTHOR Gus Wiseman, Nov 23 2019 STATUS approved
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Last modified April 15 01:24 EDT 2021. Contains 342974 sequences. (Running on oeis4.)
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!). A329864 Number of compositions of n with the same runs-resistance as cuts-resistance. 7. 1, 0, 0, 0, 0, 2, 5, 10, 17, 27, 68, 107, 217, 420, 884, 1761, 3679, 7469, 15437, 31396, 64369 (list; graph; refs; listen; history; text; internal format). OFFSET 0,6 COMMENTS A composition of n is a finite sequence of positive integers summing to n. For the operation of taking the sequence of run-lengths of a finite sequence, runs-resistance is defined to be the number of applications required to reach a singleton. For the operation of shortening all runs by 1, cuts-resistance is defined to be the number of applications required to reach an empty word. LINKS Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003. EXAMPLE The a(5) = 2 through a(8) = 17 compositions: (1112) (1113) (1114) (1115) (2111) (1122) (1222) (1133) (2211) (2221) (3311) (3111) (4111) (5111) (11211) (11122) (11222) (11311) (11411) (21112) (12221) (22111) (21113) (111121) (22211) (121111) (31112) (111131) (111221) (112112) (112211) (122111) (131111) (211211) For example, the runs-resistance of (111221) is 3 because we have: (111221) -> (321) -> (111) -> (3), while the cuts-resistance is also 3 because we have: (111221) -> (112) -> (1) -> (), so (111221) is counted under a(8). MATHEMATICA runsres[q_]:=Length[NestWhileList[Length/@Split[#]&, q, Length[#]>1&]]-1; degdep[q_]:=Length[NestWhileList[Join@@Rest/@Split[#]&, q, Length[#]>0&]]-1; Table[Length[Select[Join@@Permutations/@IntegerPartitions[n], runsres[#]==degdep[#]&]], {n, 0, 10}] CROSSREFS The version for binary expansion is A329865. Compositions counted by runs-resistance are A329744. Compositions counted by cuts-resistance are A329861.
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Compositions with runs-resistance = cuts-resistance minus 1 are A329869. Cf. A003242, A098504, A114901, A242882, A318928, A319411, A319416, A319420, A319421, A329867, A329868. Sequence in context: A062493 A056871 A246883 * A174910 A301273 A007504 Adjacent sequences: A329861 A329862 A329863 * A329865 A329866 A329867 KEYWORD nonn,more AUTHOR Gus Wiseman, Nov 23 2019 STATUS approved. Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam. Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent. The OEIS Community | Maintained by The OEIS Foundation Inc.. Last modified April 15 01:24 EDT 2021. Contains 342974 sequences. (Running on oeis4.).
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# The distance between the two runners, which is over 50
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The distance between the two runners, which is over 50 [#permalink]
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18 Jun 2012, 21:47
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The distance between the two runners, which is over 50 meters, cannot be made up with only three laps to go in the race.
A)The distance between the two runners, which is over 50 meters
B) The distance between the two runners, who is over 50 meters
C) The distance between the two runners, whom is over 50 meters
D) The distance between the two runners, that is over 50 meters
E) The distance between the two runners, whoever is over 50 meters
B, C, and E are thrown out because the relative pronoun is referring to the distance and not the runners.
My question:
Why is which preferred over that in this situation?
Isn't the information that the distance is over 50 meters necessary? "That" would restrict the distance to "50 meters".
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[Reveal] Spoiler: OA
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Re: Which/That ? detailing measure of distance between runners [#permalink]
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18 Jun 2012, 22:29
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asax wrote:
Yup, i have the same q. Which always modifies the one which is right before it? Don't it?
In this case it is two runners.
Hmm, I'd take that rule with some caution. I do not believe that rule is set in stone. In this situation, the subject is distance and "runners" is the object of a prepositional phrase. Take a look at this link: http://gmatclub.com/blog/2012/04/that-v ... orrection/
It didn't really clarify which versus that in this specific example for me though.
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Re: Which/That ? detailing measure of distance between runners [#permalink]
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18 Jun 2012, 22:11
Yup, i have the same q. Which always modifies the one which is right before it? Don't it?
In this case it is two runners.
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Re: Which/That ? detailing measure of distance between runners [#permalink]
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18 Jun 2012, 23:25
Thanks damham17!
I guess here "which" means, it could be > 50 meter [Non restrictive]?
Like:
The distance between two runners, which is over 50 meters, cannot be made up with only three laps to go in the race.
The distance between two runners that is 50 meters, cannot be made up with only three laps to go in the race.
am i right?
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Re: Which/That ? detailing measure of distance between runners [#permalink]
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You may have overlooked it, but your example with "that" omits the word over and that changes the meaning of the sentence. The answer choices did not omit over.
I tried to poke holes in my answer choice of "that" and this is what I came up with:
The distance between two runners, which is over 50 meters, cannot be made up with only three laps to go in the race.
The distance between two runners, that is over 50 meters cannot be made up with only three laps to go in the race.
Ignore which and that, but instead pay attention to "cannot be made up with only three laps to go in the race." The meaning of the sentence does not change whether or not we include details about the distance. It is stated that it cannot be made up so non restrictive seems appropriate.
Another huge thing that I did not even notice till now is the usage of commas. The comma preceding that is incorrect!!! I cannot believe I missed such an easy tell.
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Re: Which/That ? detailing measure of distance between runners [#permalink]
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19 Jun 2012, 00:30
yup, i notice it just now
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Re: Which/That ? detailing measure of distance between runners [#permalink]
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19 Jun 2012, 00:54
In general, "that" is used when the clause is necessary to the sentence. In addition, "that" is used exlusive of a comma. So we'll go with which here.
As noted, obviously eliminate B, C, and D.
Difficulty of this problem is IMO < 650.
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Re: Which/That ? detailing measure of distance between runners [#permalink]
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19 Jun 2012, 20:25
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asax wrote:
Yup, i have the same q. Which always modifies the one which is right before it? Don't it?
In this case it is two runners.
Which modifies the noun or noun phrase immediately before it.
Because "which is" cannot modify the plural "runners" it modifies the entire phrase "the distance between two runners"
Compare to "The speed of the car, which is 50 miles per hour,..." This sentence likely would not appear in a correct GMAT answer, because the "which is" could refer grammatically to both "the car" and to "the speed of the car."
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Re: The distance between the two runners, which is over 50 [#permalink]
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Re: The distance between the two runners, which is over 50 [#permalink]
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23 Feb 2017, 08:23
damham17 wrote:
The distance between the two runners, which is over 50 meters, cannot be made up with only three laps to go in the race.
A)The distance between the two runners, which is over 50 meters
B) The distance between the two runners, who is over 50 meters
C) The distance between the two runners, whom is over 50 meters
D) The distance between the two runners, that is over 50 meters
E) The distance between the two runners, whoever is over 50 meters
B, C, and E are thrown out because the relative pronoun is referring to the distance and not the runners.
My question:
Why is which preferred over that in this situation?
Isn't the information that the distance is over 50 meters necessary? "That" would restrict the distance to "50 meters".
Source: PowerScore
According to relative pronoun rule Which can modify the nearest thing, But logically in this case which modify the whole thing.There are few problems like this in Og.
Re: The distance between the two runners, which is over 50 [#permalink] 23 Feb 2017, 08:23
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Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack. It is currently 26 May 2017, 03:49. # TODAY:. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. # Events & Promotions. ###### Events & Promotions in June. Open Detailed Calendar. # The distance between the two runners, which is over 50. Author Message. TAGS:. ### Hide Tags. Intern. Joined: 06 Mar 2012. Posts: 29. Followers: 0. Kudos [?]: 24 [2] , given: 31. The distance between the two runners, which is over 50 [#permalink]. ### Show Tags. 18 Jun 2012, 21:47. 2. KUDOS. 2. This post was. BOOKMARKED. 00:00. Difficulty:. 25% (medium). Question Stats:. 51% (01:30) correct 49% (00:23) wrong based on 75 sessions. ### HideShow timer Statistics. The distance between the two runners, which is over 50 meters, cannot be made up with only three laps to go in the race.. A)The distance between the two runners, which is over 50 meters. B) The distance between the two runners, who is over 50 meters. C) The distance between the two runners, whom is over 50 meters. D) The distance between the two runners, that is over 50 meters. E) The distance between the two runners, whoever is over 50 meters. B, C, and E are thrown out because the relative pronoun is referring to the distance and not the runners.. My question:. Why is which preferred over that in this situation?. Isn't the information that the distance is over 50 meters necessary? "That" would restrict the distance to "50 meters".. Source: PowerScore. [Reveal] Spoiler: OA. If you have any questions. New!. Intern. Joined: 06 Mar 2012. Posts: 29. Followers: 0. Kudos [?]: 24 [1] , given: 31. Re: Which/That ? detailing measure of distance between runners [#permalink]. ### Show Tags. 18 Jun 2012, 22:29. 1. KUDOS. asax wrote:. Yup, i have the same q. Which always modifies the one which is right before it? Don't it?. In this case it is two runners.. Hmm, I'd take that rule with some caution. I do not believe that rule is set in stone. In this situation, the subject is distance and "runners" is the object of a prepositional phrase. Take a look at this link: http://gmatclub.com/blog/2012/04/that-v ... orrection/. It didn't really clarify which versus that in this specific example for me though.. Manager. Status: Juggg..Jugggg Go!. Joined: 11 May 2012. Posts: 246. Location: India. GC Meter: A.W.E.S.O.M.E. Concentration: Entrepreneurship, General Management. GMAT 1: 620 Q46 V30. GMAT 2: 720 Q50 V38. Followers: 6. Kudos [?]: 52 [0], given: 239. Re: Which/That ? detailing measure of distance between runners [#permalink]. ### Show Tags. 18 Jun 2012, 22:11. Yup, i have the same q. Which always modifies the one which is right before it? Don't it?. In this case it is two runners.. _________________. You haven't failed, if you haven't given up!. ---. Check out my other posts:. Bschool Deadlines 2013-2014 | Bschool Admission Events 2013 Start your GMAT Prep with Stacey Koprince | Get a head start in MBA finance. Manager. Status: Juggg..Jugggg Go!. Joined: 11 May 2012. Posts: 246. Location: India. GC Meter: A.W.E.S.O.M.E. Concentration: Entrepreneurship, General Management. GMAT 1: 620 Q46 V30. GMAT 2: 720 Q50 V38. Followers: 6. Kudos [?]: 52 [0], given: 239. Re: Which/That ? detailing measure of distance between runners [#permalink]. ### Show Tags. 18 Jun 2012, 23:25. Thanks damham17!. I guess here "which" means, it could be > 50 meter [Non restrictive]?. Like:. The distance between two runners, which is over 50 meters, cannot be made up with only three laps to go in the race.. The distance between two runners that is 50 meters, cannot be made up with only three laps to go in the race.. am i right?. _________________. You haven't failed, if you haven't given up!. ---. Check out my other posts:. Bschool Deadlines 2013-2014 | Bschool Admission Events 2013 Start your GMAT Prep with Stacey Koprince | Get a head start in MBA finance. Intern. Joined: 06 Mar 2012. Posts: 29. Followers: 0. Kudos [?]: 24 [0], given: 31.
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Re: Which/That ? detailing measure of distance between runners [#permalink]. ### Show Tags. 18 Jun 2012, 23:34. 1. This post was. BOOKMARKED. You may have overlooked it, but your example with "that" omits the word over and that changes the meaning of the sentence. The answer choices did not omit over.. I tried to poke holes in my answer choice of "that" and this is what I came up with:. The distance between two runners, which is over 50 meters, cannot be made up with only three laps to go in the race.. The distance between two runners, that is over 50 meters cannot be made up with only three laps to go in the race.. Ignore which and that, but instead pay attention to "cannot be made up with only three laps to go in the race." The meaning of the sentence does not change whether or not we include details about the distance. It is stated that it cannot be made up so non restrictive seems appropriate.. Another huge thing that I did not even notice till now is the usage of commas. The comma preceding that is incorrect!!! I cannot believe I missed such an easy tell.. Manager. Status: Juggg..Jugggg Go!. Joined: 11 May 2012. Posts: 246. Location: India. GC Meter: A.W.E.S.O.M.E. Concentration: Entrepreneurship, General Management. GMAT 1: 620 Q46 V30. GMAT 2: 720 Q50 V38. Followers: 6. Kudos [?]: 52 [0], given: 239. Re: Which/That ? detailing measure of distance between runners [#permalink]. ### Show Tags. 19 Jun 2012, 00:30. yup, i notice it just now. _________________. You haven't failed, if you haven't given up!. ---. Check out my other posts:. Bschool Deadlines 2013-2014 | Bschool Admission Events 2013 Start your GMAT Prep with Stacey Koprince | Get a head start in MBA finance. Senior Manager. Joined: 13 Jan 2012. Posts: 307. Weight: 170lbs. GMAT 1: 740 Q48 V42. GMAT 2: 760 Q50 V42. WE: Analyst (Other). Followers: 17. Kudos [?]: 167 [0], given: 38. Re: Which/That ? detailing measure of distance between runners [#permalink]. ### Show Tags. 19 Jun 2012, 00:54. In general, "that" is used when the clause is necessary to the sentence. In addition, "that" is used exlusive of a comma. So we'll go with which here.. As noted, obviously eliminate B, C, and D.. Difficulty of this problem is IMO < 650.. Kaplan GMAT Instructor. Joined: 25 Aug 2009. Posts: 644. Location: Cambridge, MA. Followers: 84. Kudos [?]: 286 [0], given: 2. Re: Which/That ? detailing measure of distance between runners [#permalink]. ### Show Tags. 19 Jun 2012, 20:25. Expert's post. 1. This post was. BOOKMARKED. asax wrote:. Yup, i have the same q. Which always modifies the one which is right before it? Don't it?. In this case it is two runners.. Which modifies the noun or noun phrase immediately before it.. Because "which is" cannot modify the plural "runners" it modifies the entire phrase "the distance between two runners". Compare to "The speed of the car, which is 50 miles per hour,..." This sentence likely would not appear in a correct GMAT answer, because the "which is" could refer grammatically to both "the car" and to "the speed of the car.". _________________. Eli Meyer. Kaplan Teacher. http://www.kaptest.com/GMAT. Prepare with Kaplan and save \$150 on a course!. Kaplan Reviews. GMAT Club Legend. Joined: 01 Oct 2013. Posts: 10363. Followers: 999. Kudos [?]: 225 [0], given: 0. Re: The distance between the two runners, which is over 50 [#permalink]. ### Show Tags. 22 Feb 2017, 20:00. Hello from the GMAT Club VerbalBot!. Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).. Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.. Manager. Joined: 09 Jan 2016. Posts: 137. Concentration: Finance, Human Resources. WE: General Management (Human Resources). Followers: 2. Kudos [?]: 69 [0], given: 52. Re: The distance between the two runners, which is over 50 [#permalink]. ### Show Tags. 23 Feb 2017, 08:23. damham17 wrote:. The distance between the two runners, which is over 50 meters, cannot be made up with only three laps to go in the race.. A)The distance between the two runners, which is over 50 meters. B) The distance between the two runners, who is over 50 meters. C) The distance between the two runners, whom is over 50 meters. D) The distance between the two runners, that is over 50 meters. E) The distance between the two runners, whoever is over 50 meters. B, C, and E are thrown out because the relative pronoun is referring to the distance and not the runners.. My question:. Why is which preferred over that in this situation?. Isn't the information that the distance is over 50 meters necessary? "That" would restrict the distance to "50 meters".. Source: PowerScore. According to relative pronoun rule Which can modify the nearest thing, But logically in this case which modify the whole thing.There are few problems like this in Og.. Re: The distance between the two runners, which is over 50 [#permalink] 23 Feb 2017, 08:23. Similar topics Replies Last post. Similar. Topics:. Successful mediation between two countries 2 13 Aug 2016, 11:52. 17 The art installation in Central Park, which featured over a 11 24 Dec 2015, 04:14. 24 Taste buds are onion-shaped structures with between 50 and 15 12 Dec 2016, 02:42. 1 confused between two answers 3 18 Oct 2014, 21:00. 38 Taste buds are onion-shaped structures with between 50 and 23 28 Aug 2016, 03:56. Display posts from previous: Sort by.
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# A problem with infinitesimals
1. Jun 25, 2011
### WizardWill
I've been playing around with a free PDF Calculus book lately. But, I have no way to check the logic used to get to a particular answer. I've been trying to find the standard part for:
(1/ɛ)((1/sqrt(4+ɛ))-(1/2))
I've tried every way I could think of to algebraically manipulate this in order to avoid dividing by zero (taking the standard part of 1/ɛ). Just by looking at the problem, I would think it would be undefined...but the odd answers tell me otherwise.
Thanks :)
2. Jun 25, 2011
### micromass
Staff Emeritus
Hi WizardWill!
First, put everything in one fraction:
$$\frac{2-\sqrt{4+\varepsilon}}{\varepsilon 2 \sqrt{4+\varepsilon}}$$
Now, multiplicate numerator and denominator by
$$2+\sqrt{4+\varepsilon}$$
3. Jun 25, 2011
### HallsofIvy
Staff Emeritus
I would be inclined to use a power series for $1/\sqrt{4+ x}$:
$$\frac{1}{\sqrt{4+ x}}= \frac{1}{2}- \frac{1}{16}x+ \frac{3}{128}x^2+ \cdot\cdot\cdot$$
so that
$$\frac{1}{\sqrt{4+ \epsilon}}- \frac{1}{2}= -\frac{1}{16}\epsilon+ \frac{3}{128}\epsilon^2+ \cdot\cdot\cdot$$
4. Jun 25, 2011
### WizardWill
Thanks Micromass and HallsofIvy for your replies :)
I began the problem by distributing the 1/ε term. So, what I had looked a bit messy. I'm surprised I didn't notice to use the conjugate of the numerator...so use to looking in the denominator. Problem solved :)
Thanks again.
5. Jun 26, 2011
### SteveL27
As someone who took calc years ago, I was curious to know if the approach based on nonstandard analysis has become so commonplace these days that it can be freely used.
I can see that the OP is attempting to determine the derivative of f(x) = 1/sqrt(x) at x = 4 directly from the definition, and is doing the equivalent of what would classically be written as
$$f'(4) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} = \lim_{h \rightarrow 0} \frac{\frac{1}{\sqrt{4 + h}} - \frac{1}{2}}{h}$$
Of course the nonstandard approach is mathematically rigorous and has been around for 30 or 40 years now. However I wasn't aware that it had achieved so much "market penetration" that it doesn't need to be remarked on.
I've heard that the nonstandard approach has the drawback that it doesn't work as conveniently in the multi-variable case. I don't know if that's true or not. In any event, someone who learns calculus this way will have trouble reading other calculus texts, or going on to Calc II or real analysis. I don't think they've banished limits in higher math yet!!
Or perhaps the OP already knows calculus and is just learning the nonstandard approach. I wasn't able to tell from the question.
I wonder if someone can put this into the context of modern teaching for me.
Last edited: Jun 26, 2011
6. Jun 26, 2011
### bcrowell
Staff Emeritus
IMO it's the other way around. The Leibniz notation was devised to represent infinitesimals. When infinitesimals were banished ca. 1900, it made it harder for calc students to understand Leibniz notation. A student today who learns calc using infinitesimals will have an easier time understanding the literature, which never stopped using the Leibniz notation.
There was a book by Keisler published back in the 70's, which did freshman calc using infinitesimals. You can find it online for free now. (It may be the book the OP referred to.) AFAIK it did not become a popular way to teach calculus. (One way you can tell that it probably wasn't popular is that the book went out of print and the copyright reverted to Keisler, making him free to put it online.) Education is very conservative, and the textbook market even more so.
One way to check the result is to go to this online calculator I wrote http://www.lightandmatter.com/calc/inf/ and enter the OP's expression as (1/d)*((1/sqrt(4+d))-(1/2)) , where d stands for an infinitesimal. The leading term is -1/16, which agrees with Halls's result.
Last edited: Jun 26, 2011
7. Jun 26, 2011
### SteveL27
Yes you're right, dy/dx doesn't make a lick of sense the way it's taught to calculus students. Perhaps the Leibniz notation should be banned. I should mention that I'm a Newtonian
Yes, that's why I asked the question. I've heard of Robinson's rigorous theory of nonstandard analysis, and I'd also heard about Keisler's book. I wasn't sure if the nonstandard approach had become more widespread since then.
You've written a calculus text that incorporates rigorous infinitesimals! I'd have to yield to your judgment about how to present this material to students. I was just wondering what happens when they get to real analysis? Do infinitesimals make it easer or harder to learn limits?
I also hadn't seen the Levi-Civita field before, that's interesting too. How does that relate to Robinson's system?
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# A problem with infinitesimals. 1. Jun 25, 2011. ### WizardWill. I've been playing around with a free PDF Calculus book lately. But, I have no way to check the logic used to get to a particular answer. I've been trying to find the standard part for:. (1/ɛ)((1/sqrt(4+ɛ))-(1/2)). I've tried every way I could think of to algebraically manipulate this in order to avoid dividing by zero (taking the standard part of 1/ɛ). Just by looking at the problem, I would think it would be undefined...but the odd answers tell me otherwise.. Thanks :). 2. Jun 25, 2011. ### micromass. Staff Emeritus. Hi WizardWill!. First, put everything in one fraction:. $$\frac{2-\sqrt{4+\varepsilon}}{\varepsilon 2 \sqrt{4+\varepsilon}}$$. Now, multiplicate numerator and denominator by. $$2+\sqrt{4+\varepsilon}$$. 3. Jun 25, 2011. ### HallsofIvy. Staff Emeritus. I would be inclined to use a power series for $1/\sqrt{4+ x}$:. $$\frac{1}{\sqrt{4+ x}}= \frac{1}{2}- \frac{1}{16}x+ \frac{3}{128}x^2+ \cdot\cdot\cdot$$. so that. $$\frac{1}{\sqrt{4+ \epsilon}}- \frac{1}{2}= -\frac{1}{16}\epsilon+ \frac{3}{128}\epsilon^2+ \cdot\cdot\cdot$$. 4. Jun 25, 2011. ### WizardWill. Thanks Micromass and HallsofIvy for your replies :). I began the problem by distributing the 1/ε term. So, what I had looked a bit messy. I'm surprised I didn't notice to use the conjugate of the numerator...so use to looking in the denominator. Problem solved :). Thanks again.. 5. Jun 26, 2011. ### SteveL27. As someone who took calc years ago, I was curious to know if the approach based on nonstandard analysis has become so commonplace these days that it can be freely used.. I can see that the OP is attempting to determine the derivative of f(x) = 1/sqrt(x) at x = 4 directly from the definition, and is doing the equivalent of what would classically be written as.
|
$$f'(4) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} = \lim_{h \rightarrow 0} \frac{\frac{1}{\sqrt{4 + h}} - \frac{1}{2}}{h}$$. Of course the nonstandard approach is mathematically rigorous and has been around for 30 or 40 years now. However I wasn't aware that it had achieved so much "market penetration" that it doesn't need to be remarked on.. I've heard that the nonstandard approach has the drawback that it doesn't work as conveniently in the multi-variable case. I don't know if that's true or not. In any event, someone who learns calculus this way will have trouble reading other calculus texts, or going on to Calc II or real analysis. I don't think they've banished limits in higher math yet!!. Or perhaps the OP already knows calculus and is just learning the nonstandard approach. I wasn't able to tell from the question.. I wonder if someone can put this into the context of modern teaching for me.. Last edited: Jun 26, 2011. 6. Jun 26, 2011. ### bcrowell. Staff Emeritus. IMO it's the other way around. The Leibniz notation was devised to represent infinitesimals. When infinitesimals were banished ca. 1900, it made it harder for calc students to understand Leibniz notation. A student today who learns calc using infinitesimals will have an easier time understanding the literature, which never stopped using the Leibniz notation.. There was a book by Keisler published back in the 70's, which did freshman calc using infinitesimals. You can find it online for free now. (It may be the book the OP referred to.) AFAIK it did not become a popular way to teach calculus. (One way you can tell that it probably wasn't popular is that the book went out of print and the copyright reverted to Keisler, making him free to put it online.) Education is very conservative, and the textbook market even more so.. One way to check the result is to go to this online calculator I wrote http://www.lightandmatter.com/calc/inf/ and enter the OP's expression as (1/d)*((1/sqrt(4+d))-(1/2)) , where d stands for an infinitesimal. The leading term is -1/16, which agrees with Halls's result.. Last edited: Jun 26, 2011. 7. Jun 26, 2011. ### SteveL27. Yes you're right, dy/dx doesn't make a lick of sense the way it's taught to calculus students. Perhaps the Leibniz notation should be banned. I should mention that I'm a Newtonian. Yes, that's why I asked the question. I've heard of Robinson's rigorous theory of nonstandard analysis, and I'd also heard about Keisler's book. I wasn't sure if the nonstandard approach had become more widespread since then.. You've written a calculus text that incorporates rigorous infinitesimals! I'd have to yield to your judgment about how to present this material to students. I was just wondering what happens when they get to real analysis? Do infinitesimals make it easer or harder to learn limits?. I also hadn't seen the Levi-Civita field before, that's interesting too. How does that relate to Robinson's system?.
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# What is Instantaneous velocity: Definition and 113 Discussions
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north). Velocity is a fundamental concept in kinematics, the branch of classical mechanics that describes the motion of bodies.
Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s or m⋅s−1). For example, "5 metres per second" is a scalar, whereas "5 metres per second east" is a vector. If there is a change in speed, direction or both, then the object has a changing velocity and is said to be undergoing an acceleration.
View More On Wikipedia.org
1. ### I Is an accelerating frame the same as an inertial frame at a point?
If we have an observer that is accelerating in one direction (perhaps a rocket ship accelerating towards the sun), would its reference frame be identical to an observer at the same point that is not accelerating, but has the same instantaneous velocity? In other words, is an accelerating...
2. ### I When does the instantaneous velocity exist?
The instantaneous velocity at time a is defined as derivative of motion function f(t). It is not similar to average velocity in an interval of time. From the Newton law. If an object is at rest, we must exert a force to make it move, assume that there is no friction. Depend on the weight of...
4. ### MHB Advanced Functions Average vs. Instantaneous velocity
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5. ### Can instantaneous velocity be different from average velocity?
Homework Statement Is it possible for the instantaneous velocity of an object at some time, t1, to not be parallel to the average velocity over a short time interval, Δt=t2-t1? If it is not possible, explain why not. If it is possible, explain what this situation implies about the motion of...
6. ### Midway Instantaneous Velocity
Homework Statement The position of a particle moving along the x axis is given in centimeters by x = 9.35 + 1.03 t3, where t is in seconds. Calculate the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s. Homework EquationsThe Attempt at a...
7. ### How to Find Instantaneous Velocity Using Force
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8. ### Instantaneous velocity calculation from position data
Hello Forum, Given position and related time data, what is the most correct way to calculate the instantaneous speed? For example, given the data (x1, t1) , (x2, t2), (x3, t3) , (x4, t4), (x5, t5), is the instantaneous speed at time t3 given more correctly by v(t_3) = \frac{(x_4-x_2)}...
9. ### Calculate Instantaneous Velocity at t=2s
Homework Statement Homework EquationsThe Attempt at a Solution I tried to find the slope of the tangent line, but this gave me 3.66 and the answer is 3.8 how do I find this?
10. ### Difference between instantaneous velocity and acceleration
Hi, I'm just beginning to learn physics on my own. It seems that instantaneous velocity and acceleration are the same thing. Is it reasonable for me to be confused about the two? If acceleration = final velocity - initial velocity / change in time ... then how this any different from...
11. ### Estimate Instantaneous Velocity
Homework Statement If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following. (i) 0.5 seconds (ii) 0.1 seconds...
12. ### Instantaneous Velocity: Can We Find it at a Point?
1.one question i have just read n my book that find inst. velocity at a point ? i think it is a wrong quotation since we need atleast two points for measuring dx length so my teacher taught me that assume point p where we have to find inst. velocity thank take one point before and after p that...
13. ### Quick question about instantaneous velocity and acceleration
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14. ### Instantaneous Speed = Instantaneous Velocity?
Hi everyone, I am taking my first physics class (algebra based), and yesterday we were covering 1-Dimensional motion. I'm enrolled in Cal. 2, so I have some understanding of instantaneous velocity, etc. However, my professor said that instantaneous speed is always equal too instantaneous...
15. ### Instantaneous Velocity of a electron with a provided formula
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16. ### Instantaneous Velocity from Strobe Diagram
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Hello. I've seen on Wikipedia (http://en.wikipedia.org/wiki/Airsoft_pellets#pellet_ballistics) a formula to find the aerodynamic drag (velocity dependent negative acceleration) for a spherical body. Knowing body's mass, diameter and air density I simplified it to the form: A = vt2 * 0.04 where A...
21. ### Using v= omega cross r to find instantaneous velocity
Homework Statement An object is rotating at 4 rad/s about an axis in direction of (2 \hat{i} - 4 \hat{j} + 3\hat{k} ) which then passes through a point (1,2,0)m . Calculate the instantaneous velocity at the point (2,0,3)m (Hint: use \vec{v} = \vec{\omega} \times \vec{r} )Homework Equations...
22. ### Determine the Magnitude of the instantaneous Velocity
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23. ### What is the Instantaneous Velocity Question and How to Solve It?
Homework Statement I am having trouble with the last question my teacher gave me. I have attached a picture. It is about instantaneous velocity. I did the same thing for all the other questions?[/B] Homework Equations average velocity = delta x / delta tThe Attempt at a Solution I did...
24. ### Instantaneous velocity and a catapult
Homework Statement A stone of mass 5 g is projected with a rubber catapult. If the catapult is streched through a distance of 7 cm by an average force of 70 N, calculate the instantaneous velocity of the stone when released. Homework Equations work done in elastic material w = 1/2 fe =...
25. ### Finding the instantaneous velocity of a scalar function?
Homework Statement Find the instantaneous velocity of f(x, y, z) = xyz, at (1, 2, 1) Homework Equations The Attempt at a Solution I think this problem our proffesor gave us wasn't formulated correctly. The only time when we calculated instantaneous velocity was when we had a...
26. ### Kinematics motion-setting up instantaneous velocity function
Homework Statement An object starts moving in a straight line from position xi, at time t =0, with velocity vi. Its acceleration is given by a = ai + bt, where ai and b are constant. Find expressions for a)instantaneous velocity b) position as functions of time. Homework Equations...
27. ### MHB What is the meaning of instantaneous velocity and how is it defined?
I'm not really sure how this concept makes sense. We give it meaning by way of a limit but, to my mind, at any instant the car is motionless so how can it have a velocity? What, in essence, does it mean to say that at some point in time a car is going 2 mph?
28. ### When can instantaneous velocity equal average velocity during a 10-second trip?
The position of a particle moving in a straight line during a 10–second trip is s(t) = 3t2 − 3t + 5 cm.Find a time t at which the instantaneous velocity is equal to the average velocity for the entire trip
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30. ### Trying to find instantaneous velocity at a point
Homework Statement Given the function s(t) = -16t2 + 100t which represents the velocity of an object in meters, what is the instantaneous velocity at t = 3 seconds? Homework Equations I think these are correct: Average velocity equals the slope of the secant line connecting any two points...
31. ### What is the Instantaneous Velocity at t = 4.0s?
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33. ### Terminal Velocity Given Acceleration at Instantaneous Velocity
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34. ### Find the instantaneous velocity
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35. ### Average and instantaneous velocity
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36. ### Determine the Instantaneous Velocity of two coordinates on a graph
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37. ### Derivation of Instantaneous Velocity
Find the instantaneous velocity where r is the position vector as a function of time: r(t)=(3.0m/s^2)t\hat{x}+(4.0m/s)t\hat{y} I attempted to find the derivative of this to find instantaneous velocity, but the book's solution was different. I think the author of the book may have made a...
38. ### Instantaneous velocity in one dimension
I started recently to study physics for motions, it is interesting so far but sadly I need some basic knowledge in calculus, I can get the Instantaneous velocity if the question gave me a function of time, but I don't know how to get it from a graph. I used the equation of calculating average...
39. ### Representing instantaneous velocity vs time on a graph
Homework Statement I need to draw an instantaneous velocity vs time graph of an object moving at a constant velocity, provide an equation, and explain what it all means. I know that instantaneous velocity is typically shown as dx/dt (the derivative of change in position with respect to time...
40. ### Find the Instantaneous Velocity of D(x) = x + 2cos(x) in (0,2π)
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41. ### Instantaneous velocity, help me please
Homework Statement I need to find the instantaneous velocity directly before a car collides with another at an intersection. It travels 24 metres with an acceleration of -5.39 m s2. The only time factor given is the light sequence: Green for 45 seconds, amber for 3 seconds and red for 5...
42. ### Instantaneous velocity of an object with varying mass at the bottom of a slope
I conducted an experiment to investigate whether the mass of an object will affect the object speed at the bottom of a slope with a constant gradient. The experiment showed that as the mass of the object (car) increases, the speed of the car at the bottom of the slope increases. I still do not...
43. ### Instantaneous velocity from avg velocity with constant accelartion
Homework Statement With constant acceleration prove that the average velocity from t1 to t2 =t1 + Δt is equal to the instanous velocity in the middle of the time interval between t1 and t2. Homework Equations What I am looking for is a general equation that does not involve accelaration. All I...
44. ### Instantaneous Velocity from the graph
Homework Statement The problem is finding the instantaneous velocity during certain time intervals from the graph. Examples of time intervals: 1.1s, 7.4s, 2.7s Homework Equations I know this is the formula for finding the instantaneous velocity lim Δv=(Δx/Δt) t->0 The...
45. ### Instantaneous Velocity: drag racing problem
The figure shows a graph of actual position-versus-time data for a particular type of drag racer known as a "funny car." Estimate the car's velocity at 2.0s Equation: v = m/s I thought I knew how to solve this, and I made a tangent line at the curve and 2s, and when I do the slope at...
46. ### Help finding instantaneous velocity graphically
A graph of position versus time for a certain particle moving along the x-axis is shown in the figure below. Find the instantaneous velocity at the following instants. (a) t = 1.00 s (b) t = 3.00 s (c) t = 4.50 s (d) t = 7.50 s I know that a) is 5 m/s and c) is 0 m/s, but I need...
47. ### Effect of Instantaneous Velocity Change on Orbit
Homework Statement A small particle of mass m is on a circular orbit of radius R around a much larger mass M. Suppose we suddenly increase the speed at which the mass m is moving by a factor (that is, v_{final} = α * v_{initial}, with α > 1). Compute the major axis, minor axis, pericentre...
48. ### What is the problem with this physics question?
Homework Statement What is the instantaneous velocity of a freely falling object 19 s after it is released from a position of rest? The acceleration of gravity is 9.8 m/s^2 Answer in units of m/s Homework Equations Δv=aΔt The Attempt at a Solution 186.2
49. ### Is acceleration correlated with an instantaneous velocity?
Is acceleration correlated with an instantaneous velocity? Velocity = Instantaneous Velocity Acceleration = Instantaneous Acceleration Change in displacement/time elapsed = Δs/Δt Change in velocity/time elapsed = Δv/Δt thanks
50. ### Is acceleration correlated with an instantaneous velocity?
Is acceleration correlated with an instantaneous velocity? tia
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# What is Instantaneous velocity: Definition and 113 Discussions. The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north). Velocity is a fundamental concept in kinematics, the branch of classical mechanics that describes the motion of bodies.. Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s or m⋅s−1). For example, "5 metres per second" is a scalar, whereas "5 metres per second east" is a vector. If there is a change in speed, direction or both, then the object has a changing velocity and is said to be undergoing an acceleration.. View More On Wikipedia.org. 1. ### I Is an accelerating frame the same as an inertial frame at a point?. If we have an observer that is accelerating in one direction (perhaps a rocket ship accelerating towards the sun), would its reference frame be identical to an observer at the same point that is not accelerating, but has the same instantaneous velocity? In other words, is an accelerating.... 2. ### I When does the instantaneous velocity exist?. The instantaneous velocity at time a is defined as derivative of motion function f(t). It is not similar to average velocity in an interval of time. From the Newton law. If an object is at rest, we must exert a force to make it move, assume that there is no friction. Depend on the weight of.... 4. ### MHB Advanced Functions Average vs. Instantaneous velocity. What do the average velocities on the very short time intervals [2,2.01] and [1.99,2] approximate? What relationship does this suggest exist between a velocity on an interval [a,b] and a velocity near t=a+b/2 for this type of polynomial?. 5. ### Can instantaneous velocity be different from average velocity?. Homework Statement Is it possible for the instantaneous velocity of an object at some time, t1, to not be parallel to the average velocity over a short time interval, Δt=t2-t1? If it is not possible, explain why not. If it is possible, explain what this situation implies about the motion of.... 6. ### Midway Instantaneous Velocity. Homework Statement The position of a particle moving along the x axis is given in centimeters by x = 9.35 + 1.03 t3, where t is in seconds. Calculate the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s. Homework EquationsThe Attempt at a.... 7. ### How to Find Instantaneous Velocity Using Force. Homework Statement Let's suppose we have two objects on ##x## axis.They have mass ##m## and ##M##, They are exerting a gravitational force on each other and its magnitude can be written as ##F=\frac {mMG} {R^2}##. So my question is can we find Instantaneous velocity of ##m##, ##{V_m}## .... 8. ### Instantaneous velocity calculation from position data. Hello Forum, Given position and related time data, what is the most correct way to calculate the instantaneous speed? For example, given the data (x1, t1) , (x2, t2), (x3, t3) , (x4, t4), (x5, t5), is the instantaneous speed at time t3 given more correctly by v(t_3) = \frac{(x_4-x_2)}.... 9. ### Calculate Instantaneous Velocity at t=2s. Homework Statement Homework EquationsThe Attempt at a Solution I tried to find the slope of the tangent line, but this gave me 3.66 and the answer is 3.8 how do I find this?. 10. ### Difference between instantaneous velocity and acceleration. Hi, I'm just beginning to learn physics on my own. It seems that instantaneous velocity and acceleration are the same thing. Is it reasonable for me to be confused about the two? If acceleration = final velocity - initial velocity / change in time ... then how this any different from.... 11. ### Estimate Instantaneous Velocity. Homework Statement If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2. (a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following. (i) 0.5 seconds (ii) 0.1 seconds.... 12. ### Instantaneous Velocity: Can We Find it at a Point?. 1.one question i have just read n my book that find inst. velocity at a point ? i think it is a wrong quotation since we need atleast two points for measuring dx length so my teacher taught me that assume point p where we have to find inst. velocity thank take one point before and after p that.... 13. ### Quick question about instantaneous velocity and acceleration. Homework Statement [/B] Homework Equations The Attempt at a Solution [/B] Hey, I did part a and b. Although I need a little help with part c. I know you're supposed to solve for t^2 but I don't know what value to use for acceleration. I'd use the quadratic formula to solve for t^2, but.... 14. ### Instantaneous Speed = Instantaneous Velocity?. Hi everyone, I am taking my first physics class (algebra based), and yesterday we were covering 1-Dimensional motion. I'm enrolled in Cal. 2, so I have some understanding of instantaneous velocity, etc. However, my professor said that instantaneous speed is always equal too instantaneous.... 15. ### Instantaneous Velocity of a electron with a provided formula. Homework Statement The motion of an electron is given by x(t)=pt^3 +qt^2 +r, with p = -1.9 m/s^3 , q = +1.3 m/s^2 , and r = +9.0 m. What is the velocity at: a) t=0s b) t=1s c) t=2s d t=3s Homework Equations v=x/t The Attempt at a Solution I have tried plugging in the time to equal to t, in.... 16. ### Instantaneous Velocity from Strobe Diagram. Hi guys! My first post here. I've been frequenting Physics Forums and have found a wealth of information, and I'm hoping I can get some specific help about this concept. Thanks! Homework Statement "Cart A and B move along a horizontal track. The top-view strobe diagram below shows the.... 17. ### Average Velocity and Final Instantaneous Velocity. Suppose a body moving in a curved path at a constant speed would its average velocity for a specific time period equal its final instantaneous velocity at the end of this period ?. 18. ### How Far Ahead Should You Monitor for Obstacles When Controlling a Mars Rover?. You sit at NASA to control the Mars rover across the Martian surface 2.0 x 10^8 km away. The communication travels at the speed of light between Earth and Mars, and the rover's top speed is 2.0 m/min. How far ahead in the rover's field of view you have to watch out for a Martian cliff? My first.... 19. ### Instantaneous Velocity and Acceleration. Homework Statement A car is stopped at a traffic light. It then travels along a straight road so that its velocity is given by the equation v(t)=(7.20m/s^2)t-(0.720m/s^3)t^2 Determine the following a. The average velocity of the car from t=0s to t=10s b. The instantaneous velocity of the car at.... 20. ### Instantaneous velocity when non-constant acceleration. Hello. I've seen on Wikipedia (http://en.wikipedia.org/wiki/Airsoft_pellets#pellet_ballistics) a formula to find the aerodynamic drag (velocity dependent negative acceleration) for a spherical body. Knowing body's mass, diameter and air density I simplified it to the form: A = vt2 * 0.04 where A.... 21. ### Using v= omega cross r to find instantaneous velocity. Homework Statement An object is rotating at 4 rad/s about an axis in direction of (2 \hat{i} - 4 \hat{j} + 3\hat{k} ) which then passes through a point (1,2,0)m . Calculate the instantaneous velocity at the point (2,0,3)m (Hint: use \vec{v} = \vec{\omega} \times \vec{r} )Homework Equations.... 22. ### Determine the Magnitude of the instantaneous Velocity. Homework Statement Determine the Magnitude of the instantaneous velocity at T=5s Function is x=AT^2 + B A=2.1m/s^2 B=2.8 M Homework Equations The previous questions answered on this problem are: AVG Velocity is 16.8 Displacement is 33.6 from 3s to 5s The Attempt at a Solution I'm not sure how.... 23. ### What is the Instantaneous Velocity Question and How to Solve It?. Homework Statement I am having trouble with the last question my teacher gave me. I have attached a picture. It is about instantaneous velocity. I did the same thing for all the other questions?[/B] Homework Equations average velocity = delta x / delta tThe Attempt at a Solution I did.... 24.
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### Instantaneous velocity and a catapult. Homework Statement A stone of mass 5 g is projected with a rubber catapult. If the catapult is streched through a distance of 7 cm by an average force of 70 N, calculate the instantaneous velocity of the stone when released. Homework Equations work done in elastic material w = 1/2 fe =.... 25. ### Finding the instantaneous velocity of a scalar function?. Homework Statement Find the instantaneous velocity of f(x, y, z) = xyz, at (1, 2, 1) Homework Equations The Attempt at a Solution I think this problem our proffesor gave us wasn't formulated correctly. The only time when we calculated instantaneous velocity was when we had a.... 26. ### Kinematics motion-setting up instantaneous velocity function. Homework Statement An object starts moving in a straight line from position xi, at time t =0, with velocity vi. Its acceleration is given by a = ai + bt, where ai and b are constant. Find expressions for a)instantaneous velocity b) position as functions of time. Homework Equations.... 27. ### MHB What is the meaning of instantaneous velocity and how is it defined?. I'm not really sure how this concept makes sense. We give it meaning by way of a limit but, to my mind, at any instant the car is motionless so how can it have a velocity? What, in essence, does it mean to say that at some point in time a car is going 2 mph?. 28. ### When can instantaneous velocity equal average velocity during a 10-second trip?. The position of a particle moving in a straight line during a 10–second trip is s(t) = 3t2 − 3t + 5 cm.Find a time t at which the instantaneous velocity is equal to the average velocity for the entire trip. Here is the question: I have posted a link there to this topic so the OP can see my work.. 30. ### Trying to find instantaneous velocity at a point. Homework Statement Given the function s(t) = -16t2 + 100t which represents the velocity of an object in meters, what is the instantaneous velocity at t = 3 seconds? Homework Equations I think these are correct: Average velocity equals the slope of the secant line connecting any two points.... 31. ### What is the Instantaneous Velocity at t = 4.0s?. Homework Statement Determine the instantaneous velocity at time t = 4, 0 s Homework Equations The Attempt at a Solution I believe it has something to do with the derivative of the point at t = 4.0s but I don't know how to solve this. I need some guidance.. 32. ### Confusion regarding notation of instantaneous velocity wrt something.. I know that v(t)=dx/dt Then what is v(x) and how? Is it also dx/dt or something else?. 33. ### Terminal Velocity Given Acceleration at Instantaneous Velocity. Homework Statement This is a repost of a homework like question. The previous thread I did not understand. In the assumption that drag is proportional to velocity, and when v = 20 m/s, a = 7.35 m/s^2, find the terminal velocity. Homework Equations The thread stated that the equation most.... 34. ### Find the instantaneous velocity. Homework Statement a ball is thrown in the air. its height from the ground in metres after t seconds is modeled by h(t) = -5t^2 + 20t + 1. What is the instantaneous velocity of the ball at t=2 seconds? Homework Equations (f(a+h)-f(a))/h The Attempt at a Solution h(2) = -5(2)2 +.... 35. ### Average and instantaneous velocity. Homework Statement Given the following formula for distance, find the average velocity on the interval [1,3] S(t) = t/(1+t^2) Homework Equations Vavg = (S(t0) - S(t1)) / (t0 - t1) or Vavg = (V0 + V1)/2The Attempt at a Solution I get two different answers and I need help understand why.... 36. ### Determine the Instantaneous Velocity of two coordinates on a graph. I can't provide all the information, because I'm on my mobile phone. Here is a shot of the problem: The tangent of each curve is the slope, right? So how do we use that to find vf (instantaneous velocity) at 4s and 8s. 37. ### Derivation of Instantaneous Velocity. Find the instantaneous velocity where r is the position vector as a function of time: r(t)=(3.0m/s^2)t\hat{x}+(4.0m/s)t\hat{y} I attempted to find the derivative of this to find instantaneous velocity, but the book's solution was different. I think the author of the book may have made a.... 38. ### Instantaneous velocity in one dimension. I started recently to study physics for motions, it is interesting so far but sadly I need some basic knowledge in calculus, I can get the Instantaneous velocity if the question gave me a function of time, but I don't know how to get it from a graph. I used the equation of calculating average.... 39. ### Representing instantaneous velocity vs time on a graph. Homework Statement I need to draw an instantaneous velocity vs time graph of an object moving at a constant velocity, provide an equation, and explain what it all means. I know that instantaneous velocity is typically shown as dx/dt (the derivative of change in position with respect to time.... 40. ### Find the Instantaneous Velocity of D(x) = x + 2cos(x) in (0,2π). Homework Statement if d(x)=x+2cosx represents the distance traveled by a particle from x=0 to x= 2pi.Find an x value in (0,2pi) for which the instantaneous velocity equals the average velocity. The Attempt at a Solution This question is a direct application of the mean value.... 41. ### Instantaneous velocity, help me please. Homework Statement I need to find the instantaneous velocity directly before a car collides with another at an intersection. It travels 24 metres with an acceleration of -5.39 m s2. The only time factor given is the light sequence: Green for 45 seconds, amber for 3 seconds and red for 5.... 42. ### Instantaneous velocity of an object with varying mass at the bottom of a slope. I conducted an experiment to investigate whether the mass of an object will affect the object speed at the bottom of a slope with a constant gradient. The experiment showed that as the mass of the object (car) increases, the speed of the car at the bottom of the slope increases. I still do not.... 43. ### Instantaneous velocity from avg velocity with constant accelartion. Homework Statement With constant acceleration prove that the average velocity from t1 to t2 =t1 + Δt is equal to the instanous velocity in the middle of the time interval between t1 and t2. Homework Equations What I am looking for is a general equation that does not involve accelaration. All I.... 44. ### Instantaneous Velocity from the graph. Homework Statement The problem is finding the instantaneous velocity during certain time intervals from the graph. Examples of time intervals: 1.1s, 7.4s, 2.7s Homework Equations I know this is the formula for finding the instantaneous velocity lim Δv=(Δx/Δt) t->0 The.... 45. ### Instantaneous Velocity: drag racing problem. The figure shows a graph of actual position-versus-time data for a particular type of drag racer known as a "funny car." Estimate the car's velocity at 2.0s Equation: v = m/s I thought I knew how to solve this, and I made a tangent line at the curve and 2s, and when I do the slope at.... 46. ### Help finding instantaneous velocity graphically. A graph of position versus time for a certain particle moving along the x-axis is shown in the figure below. Find the instantaneous velocity at the following instants. (a) t = 1.00 s (b) t = 3.00 s (c) t = 4.50 s (d) t = 7.50 s I know that a) is 5 m/s and c) is 0 m/s, but I need.... 47. ### Effect of Instantaneous Velocity Change on Orbit. Homework Statement A small particle of mass m is on a circular orbit of radius R around a much larger mass M. Suppose we suddenly increase the speed at which the mass m is moving by a factor (that is, v_{final} = α * v_{initial}, with α > 1). Compute the major axis, minor axis, pericentre.... 48. ### What is the problem with this physics question?. Homework Statement What is the instantaneous velocity of a freely falling object 19 s after it is released from a position of rest? The acceleration of gravity is 9.8 m/s^2 Answer in units of m/s Homework Equations Δv=aΔt The Attempt at a Solution 186.2. 49. ### Is acceleration correlated with an instantaneous velocity?. Is acceleration correlated with an instantaneous velocity? Velocity = Instantaneous Velocity Acceleration = Instantaneous Acceleration Change in displacement/time elapsed = Δs/Δt Change in velocity/time elapsed = Δv/Δt thanks. 50. ### Is acceleration correlated with an instantaneous velocity?. Is acceleration correlated with an instantaneous velocity? tia.
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# Power; Rotational Energy - PowerPoint PPT Presentation
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Power; Rotational Energy
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Power; Rotational Energy
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## Power; Rotational Energy
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1. Power; Rotational Energy • Power • Rotational work, power, and kinetic energy. Serway & Jewett 7.5, 10.4, 10.8 Physics 1D03 - Lecture 21
2. Recall: Physics 1D03 - Lecture 21
3. Power Power is the rate at which work is done: Average power = Work/time units: 1 J/s =1 watt (W) Instantaneous power: Average over an infinitesimal time dt, displacement ds; the work is dW = F • ds, and power is Physics 1D03 - Lecture 21
4. f F r ds = rdq d Rotational Work A bit of work, dW, is done in turning a nut through a tiny angle d : So, and so for a constant torque, Physics 1D03 - Lecture 21
5. Power: So, (again, angular velocity must be expressed in radians/second). Physics 1D03 - Lecture 21
6. Quiz A power screwdriver is intended to provide a torque of 0.5 N·m while turning at 120 revolutions per minute. The minimum power needed from the motor will be about • 60W • 6 W • 1 W Physics 1D03 - Lecture 21
7. vi w but so K = ½ Iw 2 Kinetic energy of a rotating rigid body: Add up the kinetic energies of the particles: Physics 1D03 - Lecture 21
8. Quiz A wheel is spun up to speed by a motor that produces a constant power. It takes time t to reach an angular velocity . Assuming negligible friction at the axle, how long does it take to reach twice this angular velocity? Physics 1D03 - Lecture 21
9. Example A computer hard drive has four 100-gram platters (disks), 10 cm in diameter. (Uniform thin disk: I= ½ M R2) • How much kinetic energy do they have at 7200 rpm? • How long does a 7-watt motor take to get the drive up to speed? Physics 1D03 - Lecture 21
10. Example: Big Ben, a tower clock in London has an hour hand 2.7m long with a mass of 60kg and a minute hand 4.5m long with a mass of 100kg. Calculate the rotational kinetic energy of the two hands. (I=1/3 ML2) Big Ben Physics 1D03 - Lecture 21
11. Quiz A cone-shaped top is launched by winding a string of length L around the top, and pulling with a constant force F. How should the string be wound to do the greatest amount of work on the top? • wind it around the thick end • wind it around the thin end • it doesn’t matter how it is wound • not enough information F Physics 1D03 - Lecture 21
12. Summary • Power: P=dW/dt = F • v • Rotation: dW = t dq, P = t w, K = ½ Iw 2 • Suggested Problems:Chapter 7, problems 35, 40a (5910W) • Chapter 10, problems 21.(5th ed): • Chapter 7, problems 37, 47a • Chapter 10, problems 23. Physics 1D03 - Lecture 21
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# Power; Rotational Energy - PowerPoint PPT Presentation. Download Presentation. Power; Rotational Energy. 1 / 12. Power; Rotational Energy. Download Presentation. ## Power; Rotational Energy. - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -. ##### Presentation Transcript. 1. Power; Rotational Energy • Power • Rotational work, power, and kinetic energy. Serway & Jewett 7.5, 10.4, 10.8 Physics 1D03 - Lecture 21. 2. Recall: Physics 1D03 - Lecture 21. 3. Power Power is the rate at which work is done: Average power = Work/time units: 1 J/s =1 watt (W) Instantaneous power: Average over an infinitesimal time dt, displacement ds; the work is dW = F • ds, and power is Physics 1D03 - Lecture 21. 4. f F r ds = rdq d Rotational Work A bit of work, dW, is done in turning a nut through a tiny angle d : So, and so for a constant torque, Physics 1D03 - Lecture 21. 5. Power: So, (again, angular velocity must be expressed in radians/second). Physics 1D03 - Lecture 21. 6.
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Quiz A power screwdriver is intended to provide a torque of 0.5 N·m while turning at 120 revolutions per minute. The minimum power needed from the motor will be about • 60W • 6 W • 1 W Physics 1D03 - Lecture 21. 7. vi w but so K = ½ Iw 2 Kinetic energy of a rotating rigid body: Add up the kinetic energies of the particles: Physics 1D03 - Lecture 21. 8. Quiz A wheel is spun up to speed by a motor that produces a constant power. It takes time t to reach an angular velocity . Assuming negligible friction at the axle, how long does it take to reach twice this angular velocity? Physics 1D03 - Lecture 21. 9. Example A computer hard drive has four 100-gram platters (disks), 10 cm in diameter. (Uniform thin disk: I= ½ M R2) • How much kinetic energy do they have at 7200 rpm? • How long does a 7-watt motor take to get the drive up to speed? Physics 1D03 - Lecture 21. 10. Example: Big Ben, a tower clock in London has an hour hand 2.7m long with a mass of 60kg and a minute hand 4.5m long with a mass of 100kg. Calculate the rotational kinetic energy of the two hands. (I=1/3 ML2) Big Ben Physics 1D03 - Lecture 21. 11. Quiz A cone-shaped top is launched by winding a string of length L around the top, and pulling with a constant force F. How should the string be wound to do the greatest amount of work on the top? • wind it around the thick end • wind it around the thin end • it doesn’t matter how it is wound • not enough information F Physics 1D03 - Lecture 21. 12. Summary • Power: P=dW/dt = F • v • Rotation: dW = t dq, P = t w, K = ½ Iw 2 • Suggested Problems:Chapter 7, problems 35, 40a (5910W) • Chapter 10, problems 21.(5th ed): • Chapter 7, problems 37, 47a • Chapter 10, problems 23. Physics 1D03 - Lecture 21.
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# Homework Help: Differential equations - exact equations
1. Feb 24, 2007
Find the solution to the initial value problem.
2ty^3 + 3t^2y^2 dy/dt = 0 , y(1) = 1
I found out whether the equation was exact and it was, and i continued as follows. NOTE: the answer is y(t) = t^(-4/3) i just don't know how to get it. or maybe my solution is just in another form.
im going to use f instead of that greek symbol i do not how to type out here.
M(t,y) = 2ty^3
N(t,y) = 3t^2y^2
don't know why, just copied from textbook.
M(t,y) = df(t,y)/dt = 2ty^3
N(t,y) = df(t,y)/dy = 3t^2y^2
f(t,y) = integ(M(t,y)dt) + h(y)
f(t,y) = integ(2ty^3 dt) + h(y)
f(t,y) = 2y^3(1/2t^2) + h(y)
f(t,y) = y^3t^2 + h(y)
df(t,y)/dy = t^2(3y^2) + dh(y)/dy
df(t,y)/dy = 3t^2y^2 + dh(y)/dy
3t^2y^2 = 3t^2y^2 + dh(y)/dy
dh(y)/dy = 1
h(y) = integ(1 dy) + c
h(y) = y + c
f(t,y) = y^3t^2 + y + c
y^3t^2 + y = C
sub y(1) = 1
(1)^3(1)^2 + 1 = C
C= 2
.'. y(t)^3t^2 + y(t) = 2
and i can't isolate y(t) so i left that as final solution. any thoughts on whether this is right or wrong.
2. Feb 24, 2007
### HallsofIvy
You've made one silly little error- and it isn't "differential equations", it's basic algebra- maybe an arithmetic error!
Surely your textbook defines M and N?
Your differential equation is $2ty^3+ 3t^2y^2dy/dt= 0$ which you can write in "differential form" as $(2ty^3)dt+ (3t^3y^2)dy= 0$.
You want to determine whether the left side is an "exact differential"- that is, if it can be written as df for some function f(t,y). By the chain rule, $$df= \frac{\partial f}{\partial t}dt+ \frac{\partial f}{\partial y}dy$$
So you want to know if there exist f such that
$$\frac{\partial f}{\partial t}= 2ty^3$$
and
$$\frac{\partial f}{\partial t}= 3t^2y^2$$
One way of determining whether such an f exists without finding it is to remember that the "mixed" derivatives are equal (as long as they are continuous). Here, if such an f exists,
$$\frac{\partial^2 f}{\partial y\partial t}= \frac{\partial (2ty^3}{\partial y}= 6ty^2$$
and
$$\frac{\partial^2 f}{\partial t\partial y}= \frac{\partial (3t^2y^2}{\partial t}= 6ty^2$$
Yes, those are the same so this equation is exact and such a function f(t,y) exists!
Good. In "reversing" partial integration with respect to t, in which you treat y as a constant, the "constant of integration might be a function of y- that's your "h(y)".
Strictly speaking, that should be $\frac{\partial f}{\partial y}$ but that's fine. The dh/dy is an "ordinary" derivative since h depends only on y.
Yes, that partial derivative must be equal to the "N(t,y)" from the equation.
What? Is that a typo? Surely, subtracting $3t^2y^2$ from both sides, dh/dy= 0!
Since dh/dy= 0, h(y)= c, a constant. (And since h depends only on y, it really is a constant.)
First, it should be $f(t,y)= y^3t^2+ c$. Second you don't really need the "c" here but it doesn't hurt. Since the equation is basically df= 0, f(t,y)= C and you can combine c and C- exactly as you do next.
Correction: $y^3t^2= C$
With the correction $(1^3)(1^2)= C$ so C= 1.
$y^3t^2= 1$
Well, with the correction, you can write $y= ^3\sqrt{1/t^2}= t^{-\frac{2}{3}}$ but you should be aware that, with first order differential equations you quite often can't solve for one variable as a function of the other. You could, for example, use "implicit differentiation" to see if the solution satisfies the differential equation: differentiating $y^3t^2= 1$ with respect to t, we get $3y^2t^2 y'+ 2y^3t= 0$. Yes, that's exactly the original equation.
If we check you erroneous solution, $y(t)^3t^2 + y(t) = 2$ we get $3y^2t^2y'+ 2y^3t+ y'= 0$ or $2y^3t+ (3y^2t^2+ 1)y'= 0$ which is NOT the original equation.
By the way, in addition to being exact, this is also a "separable" equation: we can write $2ty^3 + 3t^2y^2 dy/dt = 0$ as $3t^2y^2 dy/dt= -2ty^3$ and then, dividing both sides by $y^3t^2$, get $3dy/y= -2dt/t$. Integrating both sides, 3 ln(y)= -2ln(t)+ C so
$ln(y^3)= ln(t^{-2})+ C$, $y^3= Ct^{-2}$ and, finally, $t^2y^3= C$ as before.
3. Feb 24, 2007
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# Homework Help: Differential equations - exact equations. 1. Feb 24, 2007. Find the solution to the initial value problem.. 2ty^3 + 3t^2y^2 dy/dt = 0 , y(1) = 1. I found out whether the equation was exact and it was, and i continued as follows. NOTE: the answer is y(t) = t^(-4/3) i just don't know how to get it. or maybe my solution is just in another form.. im going to use f instead of that greek symbol i do not how to type out here.. M(t,y) = 2ty^3. N(t,y) = 3t^2y^2. don't know why, just copied from textbook.. M(t,y) = df(t,y)/dt = 2ty^3. N(t,y) = df(t,y)/dy = 3t^2y^2. f(t,y) = integ(M(t,y)dt) + h(y). f(t,y) = integ(2ty^3 dt) + h(y). f(t,y) = 2y^3(1/2t^2) + h(y). f(t,y) = y^3t^2 + h(y). df(t,y)/dy = t^2(3y^2) + dh(y)/dy. df(t,y)/dy = 3t^2y^2 + dh(y)/dy. 3t^2y^2 = 3t^2y^2 + dh(y)/dy. dh(y)/dy = 1. h(y) = integ(1 dy) + c. h(y) = y + c. f(t,y) = y^3t^2 + y + c. y^3t^2 + y = C. sub y(1) = 1. (1)^3(1)^2 + 1 = C. C= 2. .'. y(t)^3t^2 + y(t) = 2. and i can't isolate y(t) so i left that as final solution. any thoughts on whether this is right or wrong.. 2. Feb 24, 2007. ### HallsofIvy.
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You've made one silly little error- and it isn't "differential equations", it's basic algebra- maybe an arithmetic error!. Surely your textbook defines M and N?. Your differential equation is $2ty^3+ 3t^2y^2dy/dt= 0$ which you can write in "differential form" as $(2ty^3)dt+ (3t^3y^2)dy= 0$.. You want to determine whether the left side is an "exact differential"- that is, if it can be written as df for some function f(t,y). By the chain rule, $$df= \frac{\partial f}{\partial t}dt+ \frac{\partial f}{\partial y}dy$$. So you want to know if there exist f such that. $$\frac{\partial f}{\partial t}= 2ty^3$$. and. $$\frac{\partial f}{\partial t}= 3t^2y^2$$. One way of determining whether such an f exists without finding it is to remember that the "mixed" derivatives are equal (as long as they are continuous). Here, if such an f exists,. $$\frac{\partial^2 f}{\partial y\partial t}= \frac{\partial (2ty^3}{\partial y}= 6ty^2$$. and. $$\frac{\partial^2 f}{\partial t\partial y}= \frac{\partial (3t^2y^2}{\partial t}= 6ty^2$$. Yes, those are the same so this equation is exact and such a function f(t,y) exists!. Good. In "reversing" partial integration with respect to t, in which you treat y as a constant, the "constant of integration might be a function of y- that's your "h(y)".. Strictly speaking, that should be $\frac{\partial f}{\partial y}$ but that's fine. The dh/dy is an "ordinary" derivative since h depends only on y.. Yes, that partial derivative must be equal to the "N(t,y)" from the equation.. What? Is that a typo? Surely, subtracting $3t^2y^2$ from both sides, dh/dy= 0!. Since dh/dy= 0, h(y)= c, a constant. (And since h depends only on y, it really is a constant.). First, it should be $f(t,y)= y^3t^2+ c$. Second you don't really need the "c" here but it doesn't hurt. Since the equation is basically df= 0, f(t,y)= C and you can combine c and C- exactly as you do next.. Correction: $y^3t^2= C$. With the correction $(1^3)(1^2)= C$ so C= 1.. $y^3t^2= 1$. Well, with the correction, you can write $y= ^3\sqrt{1/t^2}= t^{-\frac{2}{3}}$ but you should be aware that, with first order differential equations you quite often can't solve for one variable as a function of the other. You could, for example, use "implicit differentiation" to see if the solution satisfies the differential equation: differentiating $y^3t^2= 1$ with respect to t, we get $3y^2t^2 y'+ 2y^3t= 0$. Yes, that's exactly the original equation.. If we check you erroneous solution, $y(t)^3t^2 + y(t) = 2$ we get $3y^2t^2y'+ 2y^3t+ y'= 0$ or $2y^3t+ (3y^2t^2+ 1)y'= 0$ which is NOT the original equation.. By the way, in addition to being exact, this is also a "separable" equation: we can write $2ty^3 + 3t^2y^2 dy/dt = 0$ as $3t^2y^2 dy/dt= -2ty^3$ and then, dividing both sides by $y^3t^2$, get $3dy/y= -2dt/t$. Integrating both sides, 3 ln(y)= -2ln(t)+ C so. $ln(y^3)= ln(t^{-2})+ C$, $y^3= Ct^{-2}$ and, finally, $t^2y^3= C$ as before.. 3. Feb 24, 2007.
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# Three nearby numbers in three arrays
Given three sorted floating-point arrays a[], b[], and c[], design a linearithmic algorithm to find three integers i, j, and k such that |a[i] - b[j]| + |b[j] - c[k]| + |c[k] - a[i]| is minimum.
I do have a solution in mind but I don't think that is linearithmic. This is what I have right now:
`````` assume minDiff = // some huge value
for each entry in 'a'
find an entry closest to it in 'b' and call it 'closestToA'
find an entry closest to 'closestToA' in 'c' and call it 'closestToB'
compute the diff:
int currDiff = Math.abs(a[i] - closestToA) + Math.abs(closestToA - closestToB) + Math.abs(closestToB - a[i]);
Replace minDiff with currDiff, if currDiff < minDiff
``````
First of all, I'd like to know if there is any better solution? If not, then am I right in thinking that this solution doesn't have linearithmic complexity? The closest number can be found using binary search.
The question is from "Algorithms - 4th Ed." by Robert Sedgewick and Kevin Wayne and I'm preparing for an upcoming interview.
Somewhat Similar question: Match Three or More Nearest Numbers from arrays
-
I haven't read the whole thing through yet but finding the closest element in a sorted array should be done in `logn` time by binary search. – biziclop Jul 19 '14 at 12:29
@biziclop yes, you're right so that fixes one part of the approach. But I still don't know if the approach is correct in the first place. – user1071840 Jul 19 '14 at 12:44
Let us look at some potential ordering of the elements:
``````a[i] < b[j] < c[k]
``````
Then we can see the following claim holds:
``````Target = |a[i] - b[j]| + |b[j] - c[k]| + |c[k] - a[i]|
= b[j] - a[i] + c[k] - b[j] + c[k] - a[i]
= 2 * (c[k] - a[i])
``````
So, for any possible ordering, this is the minimization of the difference between two elements in two different arrays. So, simply minimize for every possible combination (`a` and `b`, `b` and `c`, `c` and `a`) as shown in the question you gave a reference to (can be done in linearithmic time for each pair).
Once you found the minimization for a pair, finding the matching element from the third array should be quite easy - simply go over that array and check each element.
-
linearithmic time for each pair is not linearitmic – Silnik Jul 19 '14 at 13:14
@Silnik: You can find the pair that minimizes between `a` and `b` in linearithmic time. Do the same for `b` and `c`, and also for `c` and `a` – LightningIsMyName Jul 19 '14 at 13:15
What if such an element (b [j]) does not exist in the third array? – user1990169 Jul 19 '14 at 13:17
@LightningIsMyName: How will your algorithm work for arrays: 1 4 , 2 5 , 1 3 6 ? – Silnik Jul 19 '14 at 13:24
@LightningIsMyName I think you are right because there is a similar hint in the book but for a diff question. I've not read your solution because I've to ask a silly question first. How did you remove the modulus sign before arriving at 2 * (c[k] - a[i])? – user1071840 Jul 19 '14 at 13:31
The following algorithm is almost like merging three sorted arrays into one sorted one.
Keep one pointer for each array (i,j,k for A, B and C respectively). Initialize them to 0.
Then compute the difference between A[i], B[j], C[k] and update the lowest value achieved till now if necessary.
Increment the index in the array for which
``````array[index] = min(A[i], B[j] and C[k])
``````
if it has not reached the end.
That is:
``````If ( A[i] is the least ), then increment i.
else If ( B[j] is the least ), then increment j.
else increment k.
``````
Keep doing the above till any one index runs past the end or you find a situation where all three A[i], B[j] and C[k] are equal.
EDIT:
If there are two duplicate candidates (say A[i] == B[j]), then increment both i and j. See for yourself why.
Also, if A[i+1] == A[i], then simply increment i again.
End Edit:
The above algorithm has O(N) time complexity.
Proof of correctness:
As shown by other answer, the difference depends only on two extremes of A[i], B[j], C[k].
So if A[i] < B[j] < C[k], then difference = 2*(C[k] - A[i]). Hence if we increment either j or k, then the difference can only increase. Hence we increment i.
-
Can you add a sample run-though of this algorithm? – user1071840 Jul 19 '14 at 13:51
@user1071840 Sorry I can't. If there is an error, or if there is anything you don't understand, please point it out. – user1990169 Jul 19 '14 at 13:53
@Downvoter, please tell me where I am wrong so that I can correct my error or delete this answer. – user1990169 Jul 19 '14 at 13:57
What happens in case of duplicates? – user1071840 Jul 19 '14 at 14:01
@user1071840 No, if the lowest valued index runs past the end, or if you reach the instance where all three values A[i] = B[j] = C[k], then there is no way you can decrease the difference further. Also, sorry I have to leave now. – user1990169 Jul 19 '14 at 14:18
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# Three nearby numbers in three arrays. Given three sorted floating-point arrays a[], b[], and c[], design a linearithmic algorithm to find three integers i, j, and k such that |a[i] - b[j]| + |b[j] - c[k]| + |c[k] - a[i]| is minimum.. I do have a solution in mind but I don't think that is linearithmic. This is what I have right now:. `````` assume minDiff = // some huge value. for each entry in 'a'. find an entry closest to it in 'b' and call it 'closestToA'. find an entry closest to 'closestToA' in 'c' and call it 'closestToB'. compute the diff:. int currDiff = Math.abs(a[i] - closestToA) + Math.abs(closestToA - closestToB) + Math.abs(closestToB - a[i]);. Replace minDiff with currDiff, if currDiff < minDiff. ``````. First of all, I'd like to know if there is any better solution? If not, then am I right in thinking that this solution doesn't have linearithmic complexity? The closest number can be found using binary search.. The question is from "Algorithms - 4th Ed." by Robert Sedgewick and Kevin Wayne and I'm preparing for an upcoming interview.. Somewhat Similar question: Match Three or More Nearest Numbers from arrays. -. I haven't read the whole thing through yet but finding the closest element in a sorted array should be done in `logn` time by binary search. – biziclop Jul 19 '14 at 12:29. @biziclop yes, you're right so that fixes one part of the approach. But I still don't know if the approach is correct in the first place. – user1071840 Jul 19 '14 at 12:44. Let us look at some potential ordering of the elements:. ``````a[i] < b[j] < c[k]. ``````. Then we can see the following claim holds:. ``````Target = |a[i] - b[j]| + |b[j] - c[k]| + |c[k] - a[i]|. = b[j] - a[i] + c[k] - b[j] + c[k] - a[i]. = 2 * (c[k] - a[i]). ``````. So, for any possible ordering, this is the minimization of the difference between two elements in two different arrays. So, simply minimize for every possible combination (`a` and `b`, `b` and `c`, `c` and `a`) as shown in the question you gave a reference to (can be done in linearithmic time for each pair).. Once you found the minimization for a pair, finding the matching element from the third array should be quite easy - simply go over that array and check each element.. -. linearithmic time for each pair is not linearitmic – Silnik Jul 19 '14 at 13:14. @Silnik: You can find the pair that minimizes between `a` and `b` in linearithmic time. Do the same for `b` and `c`, and also for `c` and `a` – LightningIsMyName Jul 19 '14 at 13:15. What if such an element (b [j]) does not exist in the third array? – user1990169 Jul 19 '14 at 13:17. @LightningIsMyName: How will your algorithm work for arrays: 1 4 , 2 5 , 1 3 6 ? – Silnik Jul 19 '14 at 13:24.
|
@LightningIsMyName I think you are right because there is a similar hint in the book but for a diff question. I've not read your solution because I've to ask a silly question first. How did you remove the modulus sign before arriving at 2 * (c[k] - a[i])? – user1071840 Jul 19 '14 at 13:31. The following algorithm is almost like merging three sorted arrays into one sorted one.. Keep one pointer for each array (i,j,k for A, B and C respectively). Initialize them to 0.. Then compute the difference between A[i], B[j], C[k] and update the lowest value achieved till now if necessary.. Increment the index in the array for which. ``````array[index] = min(A[i], B[j] and C[k]). ``````. if it has not reached the end.. That is:. ``````If ( A[i] is the least ), then increment i.. else If ( B[j] is the least ), then increment j.. else increment k.. ``````. Keep doing the above till any one index runs past the end or you find a situation where all three A[i], B[j] and C[k] are equal.. EDIT:. If there are two duplicate candidates (say A[i] == B[j]), then increment both i and j. See for yourself why.. Also, if A[i+1] == A[i], then simply increment i again.. End Edit:. The above algorithm has O(N) time complexity.. Proof of correctness:. As shown by other answer, the difference depends only on two extremes of A[i], B[j], C[k].. So if A[i] < B[j] < C[k], then difference = 2*(C[k] - A[i]). Hence if we increment either j or k, then the difference can only increase. Hence we increment i.. -. Can you add a sample run-though of this algorithm? – user1071840 Jul 19 '14 at 13:51. @user1071840 Sorry I can't. If there is an error, or if there is anything you don't understand, please point it out. – user1990169 Jul 19 '14 at 13:53. @Downvoter, please tell me where I am wrong so that I can correct my error or delete this answer. – user1990169 Jul 19 '14 at 13:57. What happens in case of duplicates? – user1071840 Jul 19 '14 at 14:01. @user1071840 No, if the lowest valued index runs past the end, or if you reach the instance where all three values A[i] = B[j] = C[k], then there is no way you can decrease the difference further. Also, sorry I have to leave now. – user1990169 Jul 19 '14 at 14:18.
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What is the rectangular coordinate form of the polar coordinates (r, pi/6) ?
$\left(r , \frac{\pi}{6}\right)$ in polar form is $\left(\frac{\sqrt{3}}{2} r , \frac{r}{2}\right)$ in rectangular form
$\theta = \frac{\pi}{6}$ only gives you the angle.
You need a radius too, but $\left(r , \frac{\pi}{6}\right)$ in polar form is $\left(\frac{\sqrt{3}}{2} r , \frac{r}{2}\right)$ in rectangular form.
$\frac{\pi}{6}$ is one half of an internal angle of an equilateral triangle.
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What is the rectangular coordinate form of the polar coordinates (r, pi/6) ?. $\left(r , \frac{\pi}{6}\right)$ in polar form is $\left(\frac{\sqrt{3}}{2} r , \frac{r}{2}\right)$ in rectangular form. $\theta = \frac{\pi}{6}$ only gives you the angle.
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You need a radius too, but $\left(r , \frac{\pi}{6}\right)$ in polar form is $\left(\frac{\sqrt{3}}{2} r , \frac{r}{2}\right)$ in rectangular form.. $\frac{\pi}{6}$ is one half of an internal angle of an equilateral triangle.
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# Heat generated in a circuit is given by $H=I^2Rt$, $I$ is current $R$ is resistance, $t$ is time. If percentage error in measuring $I$, $R$ and $t$ are $2 \%,1\%$ and $1\%$ respectively. what is the maximum error in measuring heat
$(a)\;2 \%\quad (b)\;3\% \quad (c)\;4 \%\quad (d)\;6 \%$
$\large\frac{\Delta H}{H}$$=2 \large\frac{\Delta I}{I}+\frac{\Delta R}{R}+\frac{\Delta t}{t} \large\frac{\Delta H}{H}$$\times 100= 2 \times 2 \%+1 \%+1 \%$
$\qquad\qquad=6 \%$
Hence the correct option is $d$
edited Jun 20, 2013 by meena.p
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# Heat generated in a circuit is given by $H=I^2Rt$, $I$ is current $R$ is resistance, $t$ is time. If percentage error in measuring $I$, $R$ and $t$ are $2 \%,1\%$ and $1\%$ respectively. what is the maximum error in measuring heat. $(a)\;2 \%\quad (b)\;3\% \quad (c)\;4 \%\quad (d)\;6 \%$. $\large\frac{\Delta H}{H}$$=2 \large\frac{\Delta I}{I}+\frac{\Delta R}{R}+\frac{\Delta t}{t} \large\frac{\Delta H}{H}$$\times 100= 2 \times 2 \%+1 \%+1 \%$.
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$\qquad\qquad=6 \%$. Hence the correct option is $d$. edited Jun 20, 2013 by meena.p.
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https://www.esaral.com/q/if-two-positive-integers-m-and-n-are-expressible-in-the-form-m-88383
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# If two positive integers m and n are expressible in the form m
Question:
If two positive integers $m$ and $n$ are expressible in the form $m=p q^{3}$ and $n=p^{3} q^{2}$, where $p, q$ are prime numbers, then $\operatorname{HCF}(m, n)=$
(a) $p q$
(b) pq2
(c) $p^{3} q^{2}$
(d) $p^{2} q^{2}$
Solution:
Two positive integers are expressed as follows:
$m=p q^{3}$
$n=p^{3} q^{2}$
p and q are prime numbers.
Then, taking the smallest powers of and q in the values for m and n we get
$\operatorname{HCF}(m, n)=p q^{2}$
Hence the correct choice is (b).
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# If two positive integers m and n are expressible in the form m. Question:. If two positive integers $m$ and $n$ are expressible in the form $m=p q^{3}$ and $n=p^{3} q^{2}$, where $p, q$ are prime numbers, then $\operatorname{HCF}(m, n)=$. (a) $p q$. (b) pq2. (c) $p^{3} q^{2}$. (d) $p^{2} q^{2}$. Solution:.
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Two positive integers are expressed as follows:. $m=p q^{3}$. $n=p^{3} q^{2}$. p and q are prime numbers.. Then, taking the smallest powers of and q in the values for m and n we get. $\operatorname{HCF}(m, n)=p q^{2}$. Hence the correct choice is (b).
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## 5.6 Forecasting using transformations
Some common transformations which can be used when modelling were discussed in Section 3.1. When forecasting from a model with transformations, we first produce forecasts of the transformed data. Then, we need to reverse the transformation (or back-transform) to obtain forecasts on the original scale. The reverse Box-Cox transformation is given by $$$\tag{5.1} y_{t} = \begin{cases} \exp(w_{t}) & \text{if \lambda=0};\\ (\lambda w_t+1)^{1/\lambda} & \text{otherwise}. \end{cases}$$$
The fable package will automatically back-transform the forecasts whenever a transformation has been used in the model definition. The back-transformed forecast distribution is then a “transformed Normal” distribution.
### Prediction intervals with transformations
If a transformation has been used, then the prediction interval is first computed on the transformed scale, then the end points are back-transformed to give a prediction interval on the original scale. This approach preserves the probability coverage of the prediction interval, although it will no longer be symmetric around the point forecast.
The back-transformation of prediction intervals is done automatically for fable models, provided that you have used a transformation in the model formula.
Transformations sometimes make little difference to the point forecasts but have a large effect on prediction intervals.
### Forecasting with constraints
One common use of transformations is to ensure the forecasts remain on the appropriate scale. For example, log transformations constrain the forecasts to stay positive.
Another useful transformation is the scaled logit, which can be used to ensure that the forecasts are kept within a specific interval. A scaled logit that ensures the forecasted values are between $$a$$ and $$b$$ (where $$a<b$$) is given by: $f(x) = \log\left(\dfrac{x-a}{b-x}\right).$ Inverting this transformation gives the appropriate back-transformation of: $f^{-1}(x) = \dfrac{a + be^x}{1 + e^x} = \dfrac{(b-a)e^x}{1 + e^x} + a.$
To use this transformation when modelling, we can create a new transformation with the new_transformation() function. This allows us to define two functions that accept the same parameters, where the observations are provided as the first argument. The first function is used to transform the data, the second is used to back-transform forecasts.
scaled_logit <- new_transformation(
transformation = function(x, lower=0, upper=1){
log((x-lower)/(upper-x))
},
inverse = function(x, lower=0, upper=1){
(upper-lower)*exp(x)/(1+exp(x)) + lower
}
)
With this new transformation function defined, it is now possible to restrict forecasts to be within a specified interval. For example, to restrict the forecasts to be between 0 and 100 you could use scaled_logit(y, 0, 100) as the model’s left hand side formula.
One issue with using mathematical transformations such as Box-Cox transformations is that the back-transformed point forecast will not be the mean of the forecast distribution. In fact, it will usually be the median of the forecast distribution (assuming that the distribution on the transformed space is symmetric). For many purposes, this is acceptable, but occasionally the mean forecast is required. For example, you may wish to add up sales forecasts from various regions to form a forecast for the whole country. But medians do not add up, whereas means do.
For a Box-Cox transformation, the back-transformed mean is given by $$$\tag{5.2} y_t = \begin{cases} \exp(w_t)\left[1 + \frac{\sigma_h^2}{2}\right] & \text{if \lambda=0;}\\ (\lambda w_t+1)^{1/\lambda}\left[1 + \frac{\sigma_h^2(1-\lambda)}{2(\lambda w_t+1)^{2}}\right] & \text{otherwise;} \end{cases}$$$ where $$\sigma_h^2$$ is the $$h$$-step forecast variance on the transformed scale. The larger the forecast variance, the bigger the difference between the mean and the median.
The difference between the simple back-transformed forecast given by (5.1) and the mean given by (5.2) is called the bias. When we use the mean, rather than the median, we say the point forecasts have been bias-adjusted.
To see how much difference this bias-adjustment makes, consider the following example, where we forecast average annual price of eggs using the drift method with a log transformation $$(\lambda=0)$$. The log transformation is useful in this case to ensure the forecasts and the prediction intervals stay positive.
eggs <- as_tsibble(fma::eggs)
eggs %>%
model(RW(log(value) ~ drift())) %>%
forecast(h=50) %>%
autoplot(eggs, level = 80, point_forecast = lst(mean, median))
#> Warning: Ignoring unknown aesthetics: linetype
The dashed line in Figure 5.11 shows the forecast medians while the solid line shows the forecast means. Notice how the skewed forecast distribution pulls up the forecast distribution’s mean, this is a result of the added term from the bias adjustment.
Bias adjusted forecast means are automatically computed in the fable package when using mean() on a distribution.. The forecast median (point forecast prior to bias adjustment) can be obtained using the median() function on the distribution.
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## 5.6 Forecasting using transformations. Some common transformations which can be used when modelling were discussed in Section 3.1. When forecasting from a model with transformations, we first produce forecasts of the transformed data. Then, we need to reverse the transformation (or back-transform) to obtain forecasts on the original scale. The reverse Box-Cox transformation is given by $$$\tag{5.1} y_{t} = \begin{cases} \exp(w_{t}) & \text{if \lambda=0};\\ (\lambda w_t+1)^{1/\lambda} & \text{otherwise}. \end{cases}$$$. The fable package will automatically back-transform the forecasts whenever a transformation has been used in the model definition. The back-transformed forecast distribution is then a “transformed Normal” distribution.. ### Prediction intervals with transformations. If a transformation has been used, then the prediction interval is first computed on the transformed scale, then the end points are back-transformed to give a prediction interval on the original scale. This approach preserves the probability coverage of the prediction interval, although it will no longer be symmetric around the point forecast.. The back-transformation of prediction intervals is done automatically for fable models, provided that you have used a transformation in the model formula.. Transformations sometimes make little difference to the point forecasts but have a large effect on prediction intervals.. ### Forecasting with constraints. One common use of transformations is to ensure the forecasts remain on the appropriate scale. For example, log transformations constrain the forecasts to stay positive.. Another useful transformation is the scaled logit, which can be used to ensure that the forecasts are kept within a specific interval. A scaled logit that ensures the forecasted values are between $$a$$ and $$b$$ (where $$a<b$$) is given by: $f(x) = \log\left(\dfrac{x-a}{b-x}\right).$ Inverting this transformation gives the appropriate back-transformation of: $f^{-1}(x) = \dfrac{a + be^x}{1 + e^x} = \dfrac{(b-a)e^x}{1 + e^x} + a.$. To use this transformation when modelling, we can create a new transformation with the new_transformation() function. This allows us to define two functions that accept the same parameters, where the observations are provided as the first argument. The first function is used to transform the data, the second is used to back-transform forecasts.. scaled_logit <- new_transformation(. transformation = function(x, lower=0, upper=1){. log((x-lower)/(upper-x)). },. inverse = function(x, lower=0, upper=1){.
|
(upper-lower)*exp(x)/(1+exp(x)) + lower. }. ). With this new transformation function defined, it is now possible to restrict forecasts to be within a specified interval. For example, to restrict the forecasts to be between 0 and 100 you could use scaled_logit(y, 0, 100) as the model’s left hand side formula.. One issue with using mathematical transformations such as Box-Cox transformations is that the back-transformed point forecast will not be the mean of the forecast distribution. In fact, it will usually be the median of the forecast distribution (assuming that the distribution on the transformed space is symmetric). For many purposes, this is acceptable, but occasionally the mean forecast is required. For example, you may wish to add up sales forecasts from various regions to form a forecast for the whole country. But medians do not add up, whereas means do.. For a Box-Cox transformation, the back-transformed mean is given by $$$\tag{5.2} y_t = \begin{cases} \exp(w_t)\left[1 + \frac{\sigma_h^2}{2}\right] & \text{if \lambda=0;}\\ (\lambda w_t+1)^{1/\lambda}\left[1 + \frac{\sigma_h^2(1-\lambda)}{2(\lambda w_t+1)^{2}}\right] & \text{otherwise;} \end{cases}$$$ where $$\sigma_h^2$$ is the $$h$$-step forecast variance on the transformed scale. The larger the forecast variance, the bigger the difference between the mean and the median.. The difference between the simple back-transformed forecast given by (5.1) and the mean given by (5.2) is called the bias. When we use the mean, rather than the median, we say the point forecasts have been bias-adjusted.. To see how much difference this bias-adjustment makes, consider the following example, where we forecast average annual price of eggs using the drift method with a log transformation $$(\lambda=0)$$. The log transformation is useful in this case to ensure the forecasts and the prediction intervals stay positive.. eggs <- as_tsibble(fma::eggs). eggs %>%. model(RW(log(value) ~ drift())) %>%. forecast(h=50) %>%. autoplot(eggs, level = 80, point_forecast = lst(mean, median)). #> Warning: Ignoring unknown aesthetics: linetype. The dashed line in Figure 5.11 shows the forecast medians while the solid line shows the forecast means. Notice how the skewed forecast distribution pulls up the forecast distribution’s mean, this is a result of the added term from the bias adjustment.. Bias adjusted forecast means are automatically computed in the fable package when using mean() on a distribution.. The forecast median (point forecast prior to bias adjustment) can be obtained using the median() function on the distribution.
|
https://www.topperlearning.com/frank-solutions/icse-class-10-physics/frank-modern-certificate-physics-part-ii/current-electricity-exercises-and-mcq
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# Class 10 FRANK Solutions Physics Chapter 4 - Current Electricity - Exercises and MCQ
Study from Frank Solutions for ICSE Class 10 Physics Chapter 4 Current Electricity Exercises and MCQ to enhance your Physics capabilities. Learn to understand and work with circuit diagrams with our step-by-step explanations. Also, revise chapter definitions such as semiconductors, ohm, electromotive force etc. with our Frank textbook solutions.
Understand the construction and functioning of a filament bulb with TopperLearning’s ICSE Class 10 Frank Solutions and Selina Solutions. For chapter-specific self-assessments, look at our online practice tests which can be used as many times as you need.
## Current Electricity - Exercises and MCQ Exercise 209
### Solution 1
(a)
(i) 2/3 A
(b)
(ii) double in value
(c)
(iv) ohmic conductors
(d)
(iv) 6 ?
(e)
(ii) voltage
(f)
(i) protect the user from electric shock by short circuiting and consequently breaking the circuit.
(g)
(iii) A 13 ampere fuse is the most suitable rating to use
### Solution 2
(a) Potential difference: The potential difference between two points may be defined as the work done in moving a unit positive charge from one point to the other.
(b) (i) Coulomb: It is the unit of charge.
(ii) Ohm: It is the unit of resistance. The resistance of a conductor is said to be 1 ohm, if 1 ampere current flows through it, when the potential difference across its ends is 1 volt.
(c) Electromotive force: When no current is drawn from a cell, when the cell is in open circuit, the potential difference between the terminals of the cell is called its electromotive force (or e.m.f.).
(d) Semiconductors: Substances whose resistance decreases with the increase in temperature are named as semiconductors. E.g. manganin, constantan etc.
(e) Superconductors: Substance whose resistance decreases tremendously with the decrease in temperature and reaches nearly zero around absolute zero temperature are named as superconductors; e.g. lead, tin etc.
### Solution 3
According to Ohm's law, the current flowing in a conductor is directly proportional to the potential difference across its ends provided the physical conditions and temperature of conductor remains constant.
Limitations of Ohm's law:
1. Ohm's law does not apply to conductors such as diode, radio valves, metal rectifiers, where electricity passes through gases.
2. Ohm's law is applicable only when the physical conditions remain constant.
3. Ohm's law is applicable only when the temperature of the conductor is constant.
### Solution 4
Factors on which the resistance of a conductor depends are:
(i) Nature of conductor: different materials have different concentration of free electrons and therefore resistance of a conductor depends on its material.
(ii) Length of conductor: Resistance of a conductor is directly proportional to the length of a conductor.
(iii) Area of cross-section of a conductor: Resistance of a conductor is inversely proportional to the area of cross-section of the uniform wire.
(iv) Temperature of conductor: In general for metallic conductors, higher the temperature larger is the resistance.
Materials which allow electric charges to flow through them easily are known as conductors. E.g. metals and materials which do not allow the electric charge to flow through them are known as insulators. E.g. rubber, dry wood etc.
## Current Electricity - Exercises and MCQ Exercise 210
### Solution 5
Functions:
(A) Cell- It provides the potential difference in the circuit.
(B) Key- It serves as a switch in the circuit. It supplies or cuts off current as required.
(C) Ammeter- It measures the current in the circuit.
(D) Rheostat- It helps to change the resistance of the circuit without changing its voltage.
(E) Resistor- It provides a constant resistance in the circuit.
(F) Voltmeter- It measure the potential drop across the resistor.
## Current Electricity - Exercises and MCQ Exercise 212
### Solution 23
(a) Brown wire or live wire should be connected to terminal C.
Blue wire or neutral wire should be connected to terminal B.
Green wire or earth wire should be connected to terminal A.
(b) No, current passes through the earth terminal i.e. terminal A in normal circumstances.
(c) The metal case of an electrical appliance is earthed so that in any case of accidental contact of live wire with the metallic body of the appliance, the earth wire would provide a safe and easy path for the electric charges to flow down to the earth which acts as very large sink. Thus, user is thereby protected from any fatal electric shock.
### Solution 24
'kilowatt-hour' is the commercial unit of electricity. One kilo-watt hour is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for 1 hour.
## Current Electricity - Exercises and MCQ Exercise 213
### Solution 25
Construction and working of filament bulb
List of materials used:
Light bulbs have two metal contacts, which connect to the ends of an electrical circuit. The metal contacts are attached to two stiff wires, which are attached to a thin metal filament. The filament sits in the middle of the bulb, held up by a glass mount. The wires and the filament are housed in a glass bulb, which is filled with an inert gas, such as argon.
When the bulb is connected to a power supply, an electric current flows from one contact to the other, through the wires and the filament.
As the electrons zip along through the filament, they are constantly bumping into the atoms that make up the filament. The energy of each impact vibrates an atom -- in other words, the current heats the atoms up.
Metal atoms release mostly infrared light photons, which are invisible to the human eye. But if they are heated to a high enough level -- around 4,000 degrees Fahrenheit in the case of a light bulb -- they will emit a good deal of visible light.
Tungsten is used in nearly all incandescent light bulbs because it is an ideal filament material.
In a modern light bulb, inert gases, typically argon, greatly reduce this loss of tungsten. At extreme temperatures, the occasional tungsten atom vibrates enough to detach from the atoms around it and flies into the air resulting in its evaporation. In the rpesence of argon gas around it, the chances are that it will collide with an argon atom and bounce right back toward the filament, where it will rejoin the solid structure. Also since inert gases normally don't react with other elements, there is no chance of the elements combining in a combustion reaction.
### Solution 27
(a) Connecting wires: Materials having low resistance low resistivity and high melting point e.g. copper, aluminium.
(b) Fuse wire: Materials having high resistance and low melting point e.g. solder an alloy of lead and tin.
(c) Heating element: Materials having high resistivity and high melting point e.g. tungsten.
(d) Connecting wire of a power line: Materials having low resistance and non-corosive properties e.g. high tension wires.
(e) Earthing elements: Materials which are good conductors of electricity. Earthing elements are copper wire, copper plate, salt.
### Solution 31
(a) The ring system consists of a ring of three wires namely live wire, earth wire and neutral wire, which originate from the main fuse box and after running around the rooms in the house comes back to the main fuse box, thus, completing a ring. In ring system, a separate connection is taken from the live wire of the ring for each appliance. In the ring circuit, all appliances are connected in parallel.
(b) Advantages of a parallel connection are:
(i) In parallel arrangement, each appliance works at the same voltage. For example, if several bulbs are connected in parallel, each bulb glows at the same voltage. Therefore, the glow of a bulb is unaffected if another bulb is switched on or off.
(ii) In parallel arrangement, if one bulb (or appliance
### Solution 33
When no current is drawn from the cell i.e. when the cell is in open circuit, the potential difference between the terminals of the cell is called its electromotive force (or e.m.f).
### Solution 37
Incorrect statement:
(a) A 13A fuse is the most appropriate value to use
### Solution 38
(i) A transformer works on the principle of electromagnetic induction.
(ii) Function of a step-up transformer is to increase the a.c. voltage and decrease the current.
(iii) No, a transformer cannot work on a d.c. source. With a d.c. source, there will be no change in magnetic flux linked with the secondary coil.
### Solution 39
(i) Current at the end B of the coil X is anticlockwise therefore at this end there is north pole.
(ii) While closing the key, polarity at the end C of the coil Y will be north. There will be no polarity at the end C of the coil Y when the current becomes steady in the coil X.
(iii) (a) While the coil Y is moved towards the coil X, the polarity at the end C of the coil Y is north.
(b) While the coil Y is moved away the coil X, the polarity at the end C of the coil Y is south.
## Current Electricity - Exercises and MCQ Exercise 214
### Solution 40
(a) (i) When the magnet is moved rapidly in the direction of arrow, the magnetic flux linked with the coil changes and there is a deflection in the galvanometer, indicating a flow of current through the coil.
(ii) On keeping the magnet still, the magnetic flux linked with the coil does not change and there is no deflection in the galvanometer, indicating that no current is flowing through the coil.
(iii) When the magnet is rapidly pulled out, there is again change in the magnetic flux linked with the coil and the galvanometer shows a deflection but this time in opposite direction, indicating that a current is flowing in opposite direction in the coil.
(b) If a more powerful magnet is used, deflection in the galvanometer will be large, indicating a greater amount of current.
### Solution 41
(a) (v) left left left
(b) (ii) repulsion and attraction respectively
(c) (i) connecting a large resistor in series
(d) (iv) upwards and perpendicular to XY
(e) (i) 250 V
### Solution 42
(a) Plastic being non-magnetic cannot be used as a material for core. It shall not intensify the formed magnetic field.
(b) Steel has high retentivity. Hence, after prolonged use even when the switch is off, it may retain some magnetic property and attract the armature.
(c) Using copper as a material for core will introduce eddy currents in the core and thus, interfere with the working of the bell.
## Current Electricity - Exercises and MCQ Exercise 215
### Solution 45
(i) This is due to change in magnetic flux in the coil. Due to change in magnetic flux an induced emf is produced in the coil. Hence, a current flows through the galvanometer.
(ii) The current appears anticlockwise when viewed from end A because end A will form north-pole.
(iii) The galvanometer now deflects towards left.
(iv) No deflection is observed as there is no relative motion between the magnet and the coil.
### Solution 47
A moving coil galvanometer can be converted into an ammeter by connecting a low resistance (called a shunt) in parallel to the galvanometer.
A moving coil galvanometer can be converted into a voltmeter by connecting a high resistance in series with the galvanometer.
### Solution 48
An ammeter is a low resistance device; hence it is connected in series.
A voltmeter is a high resistance device; hence it is connected in parallel.
## Current Electricity - Exercises and MCQ Exercise 216
### Solution 50
(a) We will observe that the needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in coil connected to galvanometer.
(b) Introducing an iron bar in the tube, will increase the amount of induced current and the galvanometer will show a greater deflection.
### Solution 53
It is a diagram of step-down transformer.
(a) Brightness of bulb will increase because increasing the number of turns in the secondary will increase the change in magnetic flux linked with the coil.
(b) Part X is the core of the transformer is removed, it shall become an open core and there shall be magnetic flux link loss; i.e. the entire magnetic field lines produced by the primary shall not be linked with the secondary.
(c) If the core of the transformer is made of copper due to the formation of eddy currents a lot of energy shall be lost.
(d) A transformer cannot be used with direct current (d.c.) since its working is based on the principle that when there is a change of magnetic field lines due to varying current of same in one coil, an induced varying current of same frequency flows in the other coil. If the current in one coil is constant (i.e. d.c.), no induced current will flow in the other coil since there will be no change in the magnetic field lines linked with the coil.
## Current Electricity - Exercises and MCQ Exercise 217
### Solution 56
(a) magnet - soft iron
(b) core- soft iron
(c) core- soft iron, magnet - steel
(d) core- soft iron, magnet - steel
(e) core-soft iron
### Solution 55
Features which provide greater efficiency to a transformer are:
(i) The core of the transformer is laminated which prevents the formation of eddy currents.
(ii) A closed soft-iron core is used which reduces the magnetic field link loss and hysteresis loss.
### Solution 60
The ratio of number of turns NS in secondary coil to the number of turns NP in the primary coil (NS / NP) is called the turns ratio.
A transformer cannot be used with direct current (d.c.) since its working is based on the principle that when there is a change of magnetic field lines due to varying current of same in one coil, an induced varying current of same frequency flows in the other coil. If the current in one coil is constant (i.e. d.c.), no induced current will flow in the other coil since there will be no change in the magnetic field lines linked with the coil.
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Request a call back. # Class 10 FRANK Solutions Physics Chapter 4 - Current Electricity - Exercises and MCQ. Study from Frank Solutions for ICSE Class 10 Physics Chapter 4 Current Electricity Exercises and MCQ to enhance your Physics capabilities. Learn to understand and work with circuit diagrams with our step-by-step explanations. Also, revise chapter definitions such as semiconductors, ohm, electromotive force etc. with our Frank textbook solutions.. Understand the construction and functioning of a filament bulb with TopperLearning’s ICSE Class 10 Frank Solutions and Selina Solutions. For chapter-specific self-assessments, look at our online practice tests which can be used as many times as you need.. ## Current Electricity - Exercises and MCQ Exercise 209. ### Solution 1. (a). (i) 2/3 A. (b). (ii) double in value. (c). (iv) ohmic conductors. (d). (iv) 6 ?. (e). (ii) voltage. (f). (i) protect the user from electric shock by short circuiting and consequently breaking the circuit.. (g). (iii) A 13 ampere fuse is the most suitable rating to use. ### Solution 2. (a) Potential difference: The potential difference between two points may be defined as the work done in moving a unit positive charge from one point to the other.. (b) (i) Coulomb: It is the unit of charge.. (ii) Ohm: It is the unit of resistance. The resistance of a conductor is said to be 1 ohm, if 1 ampere current flows through it, when the potential difference across its ends is 1 volt.. (c) Electromotive force: When no current is drawn from a cell, when the cell is in open circuit, the potential difference between the terminals of the cell is called its electromotive force (or e.m.f.).. (d) Semiconductors: Substances whose resistance decreases with the increase in temperature are named as semiconductors. E.g. manganin, constantan etc.. (e) Superconductors: Substance whose resistance decreases tremendously with the decrease in temperature and reaches nearly zero around absolute zero temperature are named as superconductors; e.g. lead, tin etc.. ### Solution 3. According to Ohm's law, the current flowing in a conductor is directly proportional to the potential difference across its ends provided the physical conditions and temperature of conductor remains constant.. Limitations of Ohm's law:. 1. Ohm's law does not apply to conductors such as diode, radio valves, metal rectifiers, where electricity passes through gases.. 2. Ohm's law is applicable only when the physical conditions remain constant.. 3. Ohm's law is applicable only when the temperature of the conductor is constant.. ### Solution 4. Factors on which the resistance of a conductor depends are:. (i) Nature of conductor: different materials have different concentration of free electrons and therefore resistance of a conductor depends on its material.. (ii) Length of conductor: Resistance of a conductor is directly proportional to the length of a conductor.. (iii) Area of cross-section of a conductor: Resistance of a conductor is inversely proportional to the area of cross-section of the uniform wire.. (iv) Temperature of conductor: In general for metallic conductors, higher the temperature larger is the resistance.. Materials which allow electric charges to flow through them easily are known as conductors. E.g. metals and materials which do not allow the electric charge to flow through them are known as insulators. E.g. rubber, dry wood etc.. ## Current Electricity - Exercises and MCQ Exercise 210. ### Solution 5. Functions:. (A) Cell- It provides the potential difference in the circuit.. (B) Key- It serves as a switch in the circuit. It supplies or cuts off current as required.. (C) Ammeter- It measures the current in the circuit.. (D) Rheostat- It helps to change the resistance of the circuit without changing its voltage.. (E) Resistor- It provides a constant resistance in the circuit.. (F) Voltmeter- It measure the potential drop across the resistor.. ## Current Electricity - Exercises and MCQ Exercise 212. ### Solution 23. (a) Brown wire or live wire should be connected to terminal C.. Blue wire or neutral wire should be connected to terminal B.. Green wire or earth wire should be connected to terminal A.. (b) No, current passes through the earth terminal i.e. terminal A in normal circumstances.. (c) The metal case of an electrical appliance is earthed so that in any case of accidental contact of live wire with the metallic body of the appliance, the earth wire would provide a safe and easy path for the electric charges to flow down to the earth which acts as very large sink. Thus, user is thereby protected from any fatal electric shock.. ### Solution 24. 'kilowatt-hour' is the commercial unit of electricity. One kilo-watt hour is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for 1 hour.. ## Current Electricity - Exercises and MCQ Exercise 213. ### Solution 25. Construction and working of filament bulb. List of materials used:. Light bulbs have two metal contacts, which connect to the ends of an electrical circuit. The metal contacts are attached to two stiff wires, which are attached to a thin metal filament. The filament sits in the middle of the bulb, held up by a glass mount. The wires and the filament are housed in a glass bulb, which is filled with an inert gas, such as argon.. When the bulb is connected to a power supply, an electric current flows from one contact to the other, through the wires and the filament.. As the electrons zip along through the filament, they are constantly bumping into the atoms that make up the filament. The energy of each impact vibrates an atom -- in other words, the current heats the atoms up.. Metal atoms release mostly infrared light photons, which are invisible to the human eye. But if they are heated to a high enough level -- around 4,000 degrees Fahrenheit in the case of a light bulb -- they will emit a good deal of visible light.. Tungsten is used in nearly all incandescent light bulbs because it is an ideal filament material.. In a modern light bulb, inert gases, typically argon, greatly reduce this loss of tungsten. At extreme temperatures, the occasional tungsten atom vibrates enough to detach from the atoms around it and flies into the air resulting in its evaporation. In the rpesence of argon gas around it, the chances are that it will collide with an argon atom and bounce right back toward the filament, where it will rejoin the solid structure. Also since inert gases normally don't react with other elements, there is no chance of the elements combining in a combustion reaction.. ### Solution 27. (a) Connecting wires: Materials having low resistance low resistivity and high melting point e.g. copper, aluminium.. (b) Fuse wire: Materials having high resistance and low melting point e.g.
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solder an alloy of lead and tin.. (c) Heating element: Materials having high resistivity and high melting point e.g. tungsten.. (d) Connecting wire of a power line: Materials having low resistance and non-corosive properties e.g. high tension wires.. (e) Earthing elements: Materials which are good conductors of electricity. Earthing elements are copper wire, copper plate, salt.. ### Solution 31. (a) The ring system consists of a ring of three wires namely live wire, earth wire and neutral wire, which originate from the main fuse box and after running around the rooms in the house comes back to the main fuse box, thus, completing a ring. In ring system, a separate connection is taken from the live wire of the ring for each appliance. In the ring circuit, all appliances are connected in parallel.. (b) Advantages of a parallel connection are:. (i) In parallel arrangement, each appliance works at the same voltage. For example, if several bulbs are connected in parallel, each bulb glows at the same voltage. Therefore, the glow of a bulb is unaffected if another bulb is switched on or off.. (ii) In parallel arrangement, if one bulb (or appliance. ### Solution 33. When no current is drawn from the cell i.e. when the cell is in open circuit, the potential difference between the terminals of the cell is called its electromotive force (or e.m.f).. ### Solution 37. Incorrect statement:. (a) A 13A fuse is the most appropriate value to use. ### Solution 38. (i) A transformer works on the principle of electromagnetic induction.. (ii) Function of a step-up transformer is to increase the a.c. voltage and decrease the current.. (iii) No, a transformer cannot work on a d.c. source. With a d.c. source, there will be no change in magnetic flux linked with the secondary coil.. ### Solution 39. (i) Current at the end B of the coil X is anticlockwise therefore at this end there is north pole.. (ii) While closing the key, polarity at the end C of the coil Y will be north. There will be no polarity at the end C of the coil Y when the current becomes steady in the coil X.. (iii) (a) While the coil Y is moved towards the coil X, the polarity at the end C of the coil Y is north.. (b) While the coil Y is moved away the coil X, the polarity at the end C of the coil Y is south.. ## Current Electricity - Exercises and MCQ Exercise 214. ### Solution 40. (a) (i) When the magnet is moved rapidly in the direction of arrow, the magnetic flux linked with the coil changes and there is a deflection in the galvanometer, indicating a flow of current through the coil.. (ii) On keeping the magnet still, the magnetic flux linked with the coil does not change and there is no deflection in the galvanometer, indicating that no current is flowing through the coil.. (iii) When the magnet is rapidly pulled out, there is again change in the magnetic flux linked with the coil and the galvanometer shows a deflection but this time in opposite direction, indicating that a current is flowing in opposite direction in the coil.. (b) If a more powerful magnet is used, deflection in the galvanometer will be large, indicating a greater amount of current.. ### Solution 41. (a) (v) left left left. (b) (ii) repulsion and attraction respectively. (c) (i) connecting a large resistor in series. (d) (iv) upwards and perpendicular to XY. (e) (i) 250 V. ### Solution 42. (a) Plastic being non-magnetic cannot be used as a material for core. It shall not intensify the formed magnetic field.. (b) Steel has high retentivity. Hence, after prolonged use even when the switch is off, it may retain some magnetic property and attract the armature.. (c) Using copper as a material for core will introduce eddy currents in the core and thus, interfere with the working of the bell.. ## Current Electricity - Exercises and MCQ Exercise 215. ### Solution 45. (i) This is due to change in magnetic flux in the coil. Due to change in magnetic flux an induced emf is produced in the coil. Hence, a current flows through the galvanometer.. (ii) The current appears anticlockwise when viewed from end A because end A will form north-pole.. (iii) The galvanometer now deflects towards left.. (iv) No deflection is observed as there is no relative motion between the magnet and the coil.. ### Solution 47. A moving coil galvanometer can be converted into an ammeter by connecting a low resistance (called a shunt) in parallel to the galvanometer.. A moving coil galvanometer can be converted into a voltmeter by connecting a high resistance in series with the galvanometer.. ### Solution 48. An ammeter is a low resistance device; hence it is connected in series.. A voltmeter is a high resistance device; hence it is connected in parallel.. ## Current Electricity - Exercises and MCQ Exercise 216. ### Solution 50. (a) We will observe that the needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in coil connected to galvanometer.. (b) Introducing an iron bar in the tube, will increase the amount of induced current and the galvanometer will show a greater deflection.. ### Solution 53. It is a diagram of step-down transformer.. (a) Brightness of bulb will increase because increasing the number of turns in the secondary will increase the change in magnetic flux linked with the coil.. (b) Part X is the core of the transformer is removed, it shall become an open core and there shall be magnetic flux link loss; i.e. the entire magnetic field lines produced by the primary shall not be linked with the secondary.. (c) If the core of the transformer is made of copper due to the formation of eddy currents a lot of energy shall be lost.. (d) A transformer cannot be used with direct current (d.c.) since its working is based on the principle that when there is a change of magnetic field lines due to varying current of same in one coil, an induced varying current of same frequency flows in the other coil. If the current in one coil is constant (i.e. d.c.), no induced current will flow in the other coil since there will be no change in the magnetic field lines linked with the coil.. ## Current Electricity - Exercises and MCQ Exercise 217. ### Solution 56. (a) magnet - soft iron. (b) core- soft iron. (c) core- soft iron, magnet - steel. (d) core- soft iron, magnet - steel. (e) core-soft iron. ### Solution 55. Features which provide greater efficiency to a transformer are:. (i) The core of the transformer is laminated which prevents the formation of eddy currents.. (ii) A closed soft-iron core is used which reduces the magnetic field link loss and hysteresis loss.. ### Solution 60. The ratio of number of turns NS in secondary coil to the number of turns NP in the primary coil (NS / NP) is called the turns ratio.. A transformer cannot be used with direct current (d.c.) since its working is based on the principle that when there is a change of magnetic field lines due to varying current of same in one coil, an induced varying current of same frequency flows in the other coil. If the current in one coil is constant (i.e. d.c.), no induced current will flow in the other coil since there will be no change in the magnetic field lines linked with the coil.
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# Matrix in vector notation?
Let, $$\mathbf{A} \in \mathbb{R}^{n \times n}$$, $$x \in \mathbb{R}^{n}$$, and $$\mathbf{I}$$ be an $$n$$ by $$n$$ identity matrix. What does,
$$\begin{bmatrix}\mathbf{A} \\ \mathbf{I} \end{bmatrix}x,$$
mean? I see this notation often used in books and no idea what it implies. Thanks!
• See block matrix Jul 5, 2021 at 8:11
It denotes a $$2n\times 1$$ vector with coordinates $$[A_1x, \ldots, A_nx, x_1,\ldots,x_n]^{T}$$ Where $$A_i$$ denotes the $$i$$-th row of matrix $$A$$.
I understand it as "expanding" the matrices $$\mathbf{A}$$ and the identity inside the brackets, and then, performing the matrix multiplication:
$$\begin{bmatrix}\mathbf{A} \\ \mathbf{I} \end{bmatrix}x \equiv \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \dots \\ a_{n1} & a_{n2} & \dots & a_{nn} \\ 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \dots\\ 0 & 0 & \dots & 1 \end{bmatrix}\cdot\begin{bmatrix} x_1 \\ x_2 \\ \dots \\ x_n \end{bmatrix}$$
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# Matrix in vector notation?. Let, $$\mathbf{A} \in \mathbb{R}^{n \times n}$$, $$x \in \mathbb{R}^{n}$$, and $$\mathbf{I}$$ be an $$n$$ by $$n$$ identity matrix. What does,. $$\begin{bmatrix}\mathbf{A} \\ \mathbf{I} \end{bmatrix}x,$$. mean? I see this notation often used in books and no idea what it implies. Thanks!.
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• See block matrix Jul 5, 2021 at 8:11. It denotes a $$2n\times 1$$ vector with coordinates $$[A_1x, \ldots, A_nx, x_1,\ldots,x_n]^{T}$$ Where $$A_i$$ denotes the $$i$$-th row of matrix $$A$$.. I understand it as "expanding" the matrices $$\mathbf{A}$$ and the identity inside the brackets, and then, performing the matrix multiplication:. $$\begin{bmatrix}\mathbf{A} \\ \mathbf{I} \end{bmatrix}x \equiv \begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \dots \\ a_{n1} & a_{n2} & \dots & a_{nn} \\ 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \dots\\ 0 & 0 & \dots & 1 \end{bmatrix}\cdot\begin{bmatrix} x_1 \\ x_2 \\ \dots \\ x_n \end{bmatrix}$$.
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# Thread: Vector space and maximal proper subspace
1. ## Vector space and maximal proper subspace
Suppose that V is a vector space over F and $\displaystyle \vec{v} \in V$ is any nonzero vector.
Show that there is a subspace W of V maximal subject to the condition
that $\displaystyle \vec{v} \notin W$ ; show that any such W is a maximal proper subspace of V .
(Do NOT assume that V is finite-dimensional.)
For the first part I thought that I could say :
1. Prove that we can exchange one of the vector in basis of V by $\displaystyle \vec{v}$.
2. Prove that the basis (B) is still one.
3.Let $\displaystyle W = B \backslash \vec{v}$.
4. $\displaystyle B \backslash \vec{v}$ is a basis and thus a maximal independant subset that doesn't contain $\displaystyle \vec{v}$.
5. W is a subspace of V.
Is it correct?
For the second part I don't have a good idea of what to do.
Any hint would be great. Thanks!
2. Originally Posted by vincisonfire
Suppose that V is a vector space over F and $\displaystyle \vec{v} \in V$ is any nonzero vector.
Show that there is a subspace W of V maximal subject to the condition
that $\displaystyle \vec{v} \notin W$ ; show that any such W is a maximal proper subspace of V .
(Do NOT assume that V is finite-dimensional.)
the first part is a quick result of Zorn's lemma: let $\displaystyle C$ be the set of all subspaces W of V such that $\displaystyle v \notin W.$ since $\displaystyle v \neq 0,$ the set $\displaystyle C$ contains the 0 subspace and hence it's not empty.
choose any chain (totally ordered collection) of elements of $\displaystyle (C, \subseteq).$ then the union of those elements is still in $\displaystyle C.$ thus $\displaystyle C$ has a maximal element by Zorn's lemma.
for the second part, suppose $\displaystyle W$ is a maximal element of $\displaystyle C$ and $\displaystyle W \subsetneq W_1,$ for some subspace $\displaystyle W_1$ of $\displaystyle V.$ the claim is that $\displaystyle W_1=V$: by maximality of $\displaystyle W$ in $\displaystyle C,$ we must have $\displaystyle v \in W_1.$
thus $\displaystyle W+<v> \subseteq W_1.$ so we only need to show that $\displaystyle W+<v>=V.$ suppose $\displaystyle W+<v> \neq V,$ and let $\displaystyle v_1 \notin W + <v>.$ then $\displaystyle v \notin W + <v_1>$ and thus $\displaystyle W + <v_1> \in C,$ which
contradicts maximality of $\displaystyle W$ in $\displaystyle C. \ \Box$
3. The version of Zorn's lemma I know says that $\displaystyle C$ needs to be closed under union of chains.
Here it is
Let U be any set, and B any nonempty collection of subsets of U. Suppose tnat B is closed under unions of chains, then B has a maximal element.
How is define the union of these chains?
4. Originally Posted by vincisonfire
How is define the union of these chains?
if $\displaystyle \{W_{\alpha}: \ \alpha \in I \}$ is a chain of subspaces (under inclusion), then $\displaystyle \bigcup_{\alpha \in I} W_{\alpha}$ is again a subspace. if $\displaystyle \forall \alpha \in I: \ W_{\alpha} \in C,$ as defined in my solution, then $\displaystyle \bigcup_{\alpha \in I} W_{\alpha} \in C.$
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# definemaximak proper subspace
Click on a term to search for related topics.
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# Thread: Vector space and maximal proper subspace. 1. ## Vector space and maximal proper subspace. Suppose that V is a vector space over F and $\displaystyle \vec{v} \in V$ is any nonzero vector.. Show that there is a subspace W of V maximal subject to the condition. that $\displaystyle \vec{v} \notin W$ ; show that any such W is a maximal proper subspace of V .. (Do NOT assume that V is finite-dimensional.). For the first part I thought that I could say :. 1. Prove that we can exchange one of the vector in basis of V by $\displaystyle \vec{v}$.. 2. Prove that the basis (B) is still one.. 3.Let $\displaystyle W = B \backslash \vec{v}$.. 4. $\displaystyle B \backslash \vec{v}$ is a basis and thus a maximal independant subset that doesn't contain $\displaystyle \vec{v}$.. 5. W is a subspace of V.. Is it correct?. For the second part I don't have a good idea of what to do.. Any hint would be great. Thanks!. 2. Originally Posted by vincisonfire. Suppose that V is a vector space over F and $\displaystyle \vec{v} \in V$ is any nonzero vector.. Show that there is a subspace W of V maximal subject to the condition. that $\displaystyle \vec{v} \notin W$ ; show that any such W is a maximal proper subspace of V .. (Do NOT assume that V is finite-dimensional.).
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the first part is a quick result of Zorn's lemma: let $\displaystyle C$ be the set of all subspaces W of V such that $\displaystyle v \notin W.$ since $\displaystyle v \neq 0,$ the set $\displaystyle C$ contains the 0 subspace and hence it's not empty.. choose any chain (totally ordered collection) of elements of $\displaystyle (C, \subseteq).$ then the union of those elements is still in $\displaystyle C.$ thus $\displaystyle C$ has a maximal element by Zorn's lemma.. for the second part, suppose $\displaystyle W$ is a maximal element of $\displaystyle C$ and $\displaystyle W \subsetneq W_1,$ for some subspace $\displaystyle W_1$ of $\displaystyle V.$ the claim is that $\displaystyle W_1=V$: by maximality of $\displaystyle W$ in $\displaystyle C,$ we must have $\displaystyle v \in W_1.$. thus $\displaystyle W+<v> \subseteq W_1.$ so we only need to show that $\displaystyle W+<v>=V.$ suppose $\displaystyle W+<v> \neq V,$ and let $\displaystyle v_1 \notin W + <v>.$ then $\displaystyle v \notin W + <v_1>$ and thus $\displaystyle W + <v_1> \in C,$ which. contradicts maximality of $\displaystyle W$ in $\displaystyle C. \ \Box$. 3. The version of Zorn's lemma I know says that $\displaystyle C$ needs to be closed under union of chains.. Here it is. Let U be any set, and B any nonempty collection of subsets of U. Suppose tnat B is closed under unions of chains, then B has a maximal element.. How is define the union of these chains?. 4. Originally Posted by vincisonfire. How is define the union of these chains?. if $\displaystyle \{W_{\alpha}: \ \alpha \in I \}$ is a chain of subspaces (under inclusion), then $\displaystyle \bigcup_{\alpha \in I} W_{\alpha}$ is again a subspace. if $\displaystyle \forall \alpha \in I: \ W_{\alpha} \in C,$ as defined in my solution, then $\displaystyle \bigcup_{\alpha \in I} W_{\alpha} \in C.$. ,. ,. ,. ,. ,. ,. # definemaximak proper subspace. Click on a term to search for related topics.
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Test Series-06
## Instructions
• Your test contains 10 multiple choice questions with only one answer
• This is 30 min test.Please make sure you complete it in stipulated time
• You can Finish this test any time using 'Submit' button.
• Once finished you get a chance to review all question with correct answers and their solutions.
1.Which of the following is terminating decimal expression?
• 11/17
• 15/1600
• 29/343
• 11/9
2) Which of the following is not a irrational number
• √80
• (√3-√7)(√3+√7)
• √110
• (√3 -√5)(√5-√3)
3. LCM(a,b)=48 and HCF( a,b) =4 then a and b can be ?
• 12 and 16
• 24 and 8
• 32 and 6
• None of these
4. If x and y are prime number then?
• HCF (x,y) =y/x
• LCM( x,y)= x/y
• HCF(x,y) =1
• None of these
5. Which one of these is false?
• √110 is a real number
• There are few rational number between √3 and 2
• Irrational number are non terminating and non repeating
• There are infinite irrational number between two rational number
6. if a =23X 5X 32 and b =22X 7X 33
• LCM(a,b) =2520
• HCF(a,b)=1260
• LCM(a,b) =1260
• a/b=5/7
7. Which of these are true
• Sum of rational number and irrational number is always a irrational number
• Sum of two irrational number is always a irrational number
• Product of two irrational number is always a irrational number
• Substraction of two irrational number is always a irrational number
8. Let a and b are two number and a > b. Then which of the following is true?
• a=bq+r where 0 < r < b
• a=bq+r where 0 < q < b
• a=bq+r where 0 < b < r
• a=bq+r where a < r < b
9. Which of these is False
• 6n can end in zero for any n where n belongs to natural number
• Every positive even number is of the form 2n and every odd number is of the form 2n+1 where n belongs to integer
• Square of any positive integer can be written in form 3n or 3n+1 where n belongs to integer
• None of these
10.two statement Statement A: Every composite number can be represent as product of primes Statement B: There is only 1 even prime number
• A is correct only
• B is correct only
• A and B both are correct
• A and B both are incorrect
• Note to our visitors :-
Thanks for visiting our website. From feedback of our visitors we came to know that sometimes you are not able to see the answers given under "Answers" tab below questions. This might happen sometimes as we use javascript there. So you can view answers where they are available by reloding the page and letting it reload properly by waiting few more seconds before clicking the button.
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Test Series-06. ## Instructions. • Your test contains 10 multiple choice questions with only one answer. • This is 30 min test.Please make sure you complete it in stipulated time. • You can Finish this test any time using 'Submit' button.. • Once finished you get a chance to review all question with correct answers and their solutions.. 1.Which of the following is terminating decimal expression?. • 11/17. • 15/1600. • 29/343. • 11/9. 2) Which of the following is not a irrational number. • √80. • (√3-√7)(√3+√7). • √110. • (√3 -√5)(√5-√3). 3. LCM(a,b)=48 and HCF( a,b) =4 then a and b can be ?. • 12 and 16. • 24 and 8. • 32 and 6. • None of these. 4. If x and y are prime number then?. • HCF (x,y) =y/x. • LCM( x,y)= x/y. • HCF(x,y) =1. • None of these. 5. Which one of these is false?. • √110 is a real number. • There are few rational number between √3 and 2. • Irrational number are non terminating and non repeating. • There are infinite irrational number between two rational number. 6. if a =23X 5X 32 and b =22X 7X 33. • LCM(a,b) =2520.
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• HCF(a,b)=1260. • LCM(a,b) =1260. • a/b=5/7. 7. Which of these are true. • Sum of rational number and irrational number is always a irrational number. • Sum of two irrational number is always a irrational number. • Product of two irrational number is always a irrational number. • Substraction of two irrational number is always a irrational number. 8. Let a and b are two number and a > b. Then which of the following is true?. • a=bq+r where 0 < r < b. • a=bq+r where 0 < q < b. • a=bq+r where 0 < b < r. • a=bq+r where a < r < b. 9. Which of these is False. • 6n can end in zero for any n where n belongs to natural number. • Every positive even number is of the form 2n and every odd number is of the form 2n+1 where n belongs to integer. • Square of any positive integer can be written in form 3n or 3n+1 where n belongs to integer. • None of these. 10.two statement Statement A: Every composite number can be represent as product of primes Statement B: There is only 1 even prime number. • A is correct only. • B is correct only. • A and B both are correct. • A and B both are incorrect. • Note to our visitors :-. Thanks for visiting our website. From feedback of our visitors we came to know that sometimes you are not able to see the answers given under "Answers" tab below questions. This might happen sometimes as we use javascript there. So you can view answers where they are available by reloding the page and letting it reload properly by waiting few more seconds before clicking the button.. We really do hope that this resolve the issue. If you still hare facing problems then feel free to contact us using feedback button or contact us directly by sending is an email at [email protected]. We are aware that our users want answers to all the questions in the website. Since ours is more or less a one man army we are working towards providing answers to questions available at our website.
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# Water drop problem
Consider a pipe of length L. The pipe has N water droplets at N different positions within it. Each water droplet is moving towards the end of the pipe(x=L) at different rates. When a water droplet mixes with another water droplet, it assumes the speed of the water droplet it is mixing with. Determine the no of droplets that come out of the end of the pipe.
Refer to the figure below:
The numbers on circles indicates speed of water droplets
Examples:
```Input: length = 12, position = [10, 8, 0, 5, 3],
speed = [2, 4, 1, 1, 3]
Output: 3
Explanation:
Droplets starting at x=10 and x=8 become a droplet,
meeting each other at x=12 at time =1 sec.
The droplet starting at 0 doesn't mix with any
other droplet, so it is a drop by itself.
Droplets starting at x=5 and x=3 become a single
drop, mixing with each other at x=6 at time = 1 sec.
Note that no other droplets meet these drops before
the end of the pipe, so the answer is 3.
Refer to the figure below
Numbers on circles indicates speed of water droplets.```
Approach:
This problem uses greedy technique.
A drop will mix with another drop if two conditions are met:
1. 1. If the drop is faster than the drop it is mixing with
2. If the position of the faster drop is behind the slower drop.
We use an array of pairs to store the position and the time that ith drop would take to reach the end of the pipe. Then we sort the array according to the position of the drops. Now we have a fair idea of which drops lie behind which drops and their respective time taken to reach the end. More time means less speed and less time means more speed. Now all the drops before a slower drop will mix with it. And all the drops after the slower drop with mix with the next slower drop and so on.
For example, if the times to reach the end are: 12, 3, 7, 8, 1 (sorted according to positions)
0th drop is slowest, it won’t mix with the next drop
1st drop is faster than the 2nd drop, So they will mix and 2nd drop is faster than the third drop so all three will mix together. They cannot mix with the 4th drop because that is faster.
No of local maximal + residue(drops after last local maxima) = Total number of drops.
Below is the implementation of the above approach:
## C++
`#include ``using` `namespace` `std;` `// Function to find the number``// of the drops that come out of the``// pipe``int` `drops(``int` `length, ``int` `position[],`` ``int` `speed[], ``int` `n)``{ `` ``// stores position and time`` ``// taken by a single`` ``// drop to reach the end as a pair`` ``vector > m(n);` ` ``int` `i;`` ``for` `(i = 0; i < n; i++) {` ` ``// calculates distance needs to be`` ``// covered by the ith drop`` ``int` `p = length - position[i];` ` ``// inserts initial position of the`` ``// ith drop to the pair `` ``m[i].first = position[i];` ` ``// inserts time taken by ith`` ``// drop to reach`` ``// the end to the pair`` ``m[i].second = p * 1.0 / speed[i];`` ``}` ` ``// sorts the pair according to increasing`` ``// order of their positions`` ``sort(m.begin(), m.end());`` ``int` `k = 0; ``// counter for no of final drops` ` ``int` `curr_max = m[n-1].second;`` ``// we traverse the array demo`` ``// right to left`` ``// to determine the slower drop`` ``for` `(i = n - 2; i >= 0; i--)`` ``{`` ``// checks for next slower drop`` ``if` `(m[i].second > curr_max)`` ``{`` ``k++;`` ``curr_max=m[i].second;`` ``}`` ``}` ` ``// calculating residual`` ``// drops in the pipe`` ``k++;`` ``return` `k;``}` `// Driver Code``int` `main()``{`` ``// length of pipe`` ``int` `length = 12;`` ` ` ``// position of droplets`` ``int` `position[] = { 10, 8, 0, 5, 3 };`` ` ` ``// speed of each droplets`` ``int` `speed[] = { 2, 4, 1, 1, 3 };`` ``int` `n = ``sizeof``(speed)/``sizeof``(speed[0]);`` ``cout << drops(length, position, speed, n);`` ``return` `0;``}`
## Java
`import` `java.util.*;` `// User defined Pair class``class` `Pair {`` ``int` `x;`` ``int` `y;`` ` ` ``// Constructor`` ``public` `Pair(``int` `x, ``int` `y)`` ``{`` ``this``.x = x;`` ``this``.y = y;`` ``}``}` `// class to define user defined conparator``class` `Compare {`` ` ` ``static` `void` `compare(Pair arr[], ``int` `n)`` ``{`` ``// Comparator to sort the pair according to second element`` ``Arrays.sort(arr, ``new` `Comparator() {`` ``@Override` `public` `int` `compare(Pair p1, Pair p2)`` ``{`` ``return` `p1.x - p2.x;`` ``}`` ``});`` ``}``}` `public` `class` `Main``{`` ``// Function to find the number`` ``// of the drops that come out of the`` ``// pipe`` ``static` `int` `drops(``int` `length, ``int``[] position, ``int``[] speed, ``int` `n)`` ``{`` ``// stores position and time`` ``// taken by a single`` ``// drop to reach the end as a pair`` ``Pair m[] = ``new` `Pair[n];`` ``int` `i;`` ``for` `(i = ``0``; i < n; i++)`` ``{`` ` ` ``// calculates distance needs to be`` ``// covered by the ith drop`` ``int` `p = length - position[i];`` ` ` ``// inserts initial position of the`` ``// ith drop to the pair`` ``// inserts time taken by ith`` ``// drop to reach`` ``// the end to the pair`` ``m[i] = ``new` `Pair(position[i], p / speed[i]);`` ``}`` ` ` ``// sorts the pair according to increasing`` ``// order of their positions`` ``Compare obj = ``new` `Compare();`` ``obj.compare(m, n);`` ``int` `k = ``0``; ``// counter for no of final drops`` ``int` `curr_max = (``int``)(m[n - ``1``].y);`` ` ` ``// we traverse the array demo`` ``// right to left`` ``// to determine the slower drop`` ``for` `(i = n - ``2``; i >= ``0``; i--)`` ``{`` ` ` ``// checks for next slower drop`` ``if` `(m[i].y > curr_max)`` ``{`` ``k++;`` ``curr_max = (``int``)(m[i].y);`` ``}`` ``}`` ` ` ``// calculating residual`` ``// drops in the pipe`` ``k++;`` ``return` `k;`` ``}`` ` ` ``public` `static` `void` `main(String[] args) {`` ``// length of pipe`` ``int` `length = ``12``;`` ` ` ``// position of droplets`` ``int``[] position = { ``10``, ``8``, ``0``, ``5``, ``3` `};`` ` ` ``// speed of each droplets`` ``int``[] speed = { ``2``, ``4``, ``1``, ``1``, ``3` `};`` ``int` `n = speed.length;`` ``System.out.println(drops(length, position, speed, n));`` ``}``}` `// This code is contributed by decode2207.`
## Python3
`# Function to find the number``# of the drops that come out of the``# pipe``def` `drops(length, position, speed, n):`` ` ` ``# Stores position and time`` ``# taken by a single drop to`` ``# reach the end as a pair`` ``m ``=` `[]` ` ``for` `i ``in` `range``(n):`` ` ` ``# Calculates distance needs to be`` ``# covered by the ith drop`` ``p ``=` `length ``-` `position[i]` ` ``# Inserts initial position of the`` ``# ith drop to the pair`` ``# inserts time taken by ith`` ``# drop to reach`` ``# the end to the pair`` ``m.append([position[i], (p ``*` `1.0``) ``/` `speed[i]])` ` ``# Sorts the pair according to increasing`` ``# order of their positions`` ``m.sort()`` ` ` ``# Counter for no of final drops`` ``k ``=` `0`` ` ` ``curr_max ``=` `m[n ``-` `1``][``1``]`` ` ` ``# We traverse the array demo`` ``# right to left`` ``# to determine the slower drop`` ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``):`` ` ` ``# Checks for next slower drop`` ``if` `(m[i][``1``] > curr_max):`` ``k ``+``=` `1`` ``curr_max ``=` `m[i][``1``]`` ` ` ``# Calculating residual`` ``# drops in the pipe`` ``k ``+``=` `1`` ``return` `k`` ` `# Driver Code` `# Length of pipe``length ``=` `12` `# Position of droplets``position ``=` `[ ``10``, ``8``, ``0``, ``5``, ``3` `]` `# Speed of each droplets``speed ``=` `[ ``2``, ``4``, ``1``, ``1``, ``3` `]``n ``=` `len``(speed)` `print``(drops(length, position, speed, n))` `# This code is contributed by divyeshrabadiya07`
## C#
`using` `System;``using` `System.Collections.Generic;``class` `GFG``{` ` ``// Function to find the number`` ``// of the drops that come out of the`` ``// pipe`` ``static` `int` `drops(``int` `length, ``int``[] position,`` ``int``[] speed, ``int` `n)`` ``{` ` ``// stores position and time`` ``// taken by a single`` ``// drop to reach the end as a pair`` ``List> m = ``new` `List>();`` ``int` `i;`` ``for` `(i = 0; i < n; i++)`` ``{` ` ``// calculates distance needs to be`` ``// covered by the ith drop`` ``int` `p = length - position[i];` ` ``// inserts initial position of the`` ``// ith drop to the pair`` ``// inserts time taken by ith`` ``// drop to reach`` ``// the end to the pair`` ``m.Add(``new` `Tuple<``int``,``double``>(position[i], p * 1.0 / speed[i]));`` ``}` ` ``// sorts the pair according to increasing`` ``// order of their positions`` ``m.Sort();`` ``int` `k = 0; ``// counter for no of final drops`` ``int` `curr_max = (``int``)m[n - 1].Item2;` ` ``// we traverse the array demo`` ``// right to left`` ``// to determine the slower drop`` ``for` `(i = n - 2; i >= 0; i--)`` ``{` ` ``// checks for next slower drop`` ``if` `(m[i].Item2 > curr_max)`` ``{`` ``k++;`` ``curr_max = (``int``)m[i].Item2;`` ``}`` ``}` ` ``// calculating residual`` ``// drops in the pipe`` ``k++;`` ``return` `k;`` ``}` ` ``// Driver code`` ``static` `void` `Main()`` ``{` ` ``// length of pipe`` ``int` `length = 12;` ` ``// position of droplets`` ``int``[] position = { 10, 8, 0, 5, 3 };` ` ``// speed of each droplets`` ``int``[] speed = { 2, 4, 1, 1, 3 };`` ``int` `n = speed.Length;`` ``Console.WriteLine(drops(length, position, speed, n));`` ``}``}` `// This code is contributed by divyesh072019`
## Javascript
``
Output
`3`
Complexity Analysis:
• Time Complexity: O(nlog(n))
• Auxiliary Space: O(n)
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Get the best out of our app. GeeksforGeeks App. Open App. Browser. Continue. # Water drop problem. Consider a pipe of length L. The pipe has N water droplets at N different positions within it. Each water droplet is moving towards the end of the pipe(x=L) at different rates. When a water droplet mixes with another water droplet, it assumes the speed of the water droplet it is mixing with. Determine the no of droplets that come out of the end of the pipe.. Refer to the figure below:. The numbers on circles indicates speed of water droplets. Examples:. ```Input: length = 12, position = [10, 8, 0, 5, 3],. speed = [2, 4, 1, 1, 3]. Output: 3. Explanation:. Droplets starting at x=10 and x=8 become a droplet,. meeting each other at x=12 at time =1 sec.. The droplet starting at 0 doesn't mix with any. other droplet, so it is a drop by itself.. Droplets starting at x=5 and x=3 become a single. drop, mixing with each other at x=6 at time = 1 sec.. Note that no other droplets meet these drops before. the end of the pipe, so the answer is 3.. Refer to the figure below. Numbers on circles indicates speed of water droplets.```. Approach:. This problem uses greedy technique.. A drop will mix with another drop if two conditions are met:. 1. 1.
|
If the drop is faster than the drop it is mixing with. 2. If the position of the faster drop is behind the slower drop.. We use an array of pairs to store the position and the time that ith drop would take to reach the end of the pipe. Then we sort the array according to the position of the drops. Now we have a fair idea of which drops lie behind which drops and their respective time taken to reach the end. More time means less speed and less time means more speed. Now all the drops before a slower drop will mix with it. And all the drops after the slower drop with mix with the next slower drop and so on.. For example, if the times to reach the end are: 12, 3, 7, 8, 1 (sorted according to positions). 0th drop is slowest, it won’t mix with the next drop. 1st drop is faster than the 2nd drop, So they will mix and 2nd drop is faster than the third drop so all three will mix together. They cannot mix with the 4th drop because that is faster.. No of local maximal + residue(drops after last local maxima) = Total number of drops.. Below is the implementation of the above approach:. ## C++. `#include ``using` `namespace` `std;` `// Function to find the number``// of the drops that come out of the``// pipe``int` `drops(``int` `length, ``int` `position[],`` ``int` `speed[], ``int` `n)``{ `` ``// stores position and time`` ``// taken by a single`` ``// drop to reach the end as a pair`` ``vector > m(n);` ` ``int` `i;`` ``for` `(i = 0; i < n; i++) {` ` ``// calculates distance needs to be`` ``// covered by the ith drop`` ``int` `p = length - position[i];` ` ``// inserts initial position of the`` ``// ith drop to the pair `` ``m[i].first = position[i];` ` ``// inserts time taken by ith`` ``// drop to reach`` ``// the end to the pair`` ``m[i].second = p * 1.0 / speed[i];`` ``}` ` ``// sorts the pair according to increasing`` ``// order of their positions`` ``sort(m.begin(), m.end());`` ``int` `k = 0; ``// counter for no of final drops` ` ``int` `curr_max = m[n-1].second;`` ``// we traverse the array demo`` ``// right to left`` ``// to determine the slower drop`` ``for` `(i = n - 2; i >= 0; i--)`` ``{`` ``// checks for next slower drop`` ``if` `(m[i].second > curr_max)`` ``{`` ``k++;`` ``curr_max=m[i].second;`` ``}`` ``}` ` ``// calculating residual`` ``// drops in the pipe`` ``k++;`` ``return` `k;``}` `// Driver Code``int` `main()``{`` ``// length of pipe`` ``int` `length = 12;`` ` ` ``// position of droplets`` ``int` `position[] = { 10, 8, 0, 5, 3 };`` ` ` ``// speed of each droplets`` ``int` `speed[] = { 2, 4, 1, 1, 3 };`` ``int` `n = ``sizeof``(speed)/``sizeof``(speed[0]);`` ``cout << drops(length, position, speed, n);`` ``return` `0;``}`. ## Java. `import` `java.util.*;` `// User defined Pair class``class` `Pair {`` ``int` `x;`` ``int` `y;`` ` ` ``// Constructor`` ``public` `Pair(``int` `x, ``int` `y)`` ``{`` ``this``.x = x;`` ``this``.y = y;`` ``}``}` `// class to define user defined conparator``class` `Compare {`` ` ` ``static` `void` `compare(Pair arr[], ``int` `n)`` ``{`` ``// Comparator to sort the pair according to second element`` ``Arrays.sort(arr, ``new` `Comparator() {`` ``@Override` `public` `int` `compare(Pair p1, Pair p2)`` ``{`` ``return` `p1.x - p2.x;`` ``}`` ``});`` ``}``}` `public` `class` `Main``{`` ``// Function to find the number`` ``// of the drops that come out of the`` ``// pipe`` ``static` `int` `drops(``int` `length, ``int``[] position, ``int``[] speed, ``int` `n)`` ``{`` ``// stores position and time`` ``// taken by a single`` ``// drop to reach the end as a pair`` ``Pair m[] = ``new` `Pair[n];`` ``int` `i;`` ``for` `(i = ``0``; i < n; i++)`` ``{`` ` ` ``// calculates distance needs to be`` ``// covered by the ith drop`` ``int` `p = length - position[i];`` ` ` ``// inserts initial position of the`` ``// ith drop to the pair`` ``// inserts time taken by ith`` ``// drop to reach`` ``// the end to the pair`` ``m[i] = ``new` `Pair(position[i], p / speed[i]);`` ``}`` ` ` ``// sorts the pair according to increasing`` ``// order of their positions`` ``Compare obj = ``new` `Compare();`` ``obj.compare(m, n);`` ``int` `k = ``0``; ``// counter for no of final drops`` ``int` `curr_max = (``int``)(m[n - ``1``].y);`` ` ` ``// we traverse the array demo`` ``// right to left`` ``// to determine the slower drop`` ``for` `(i = n - ``2``; i >= ``0``; i--)`` ``{`` ` ` ``// checks for next slower drop`` ``if` `(m[i].y > curr_max)`` ``{`` ``k++;`` ``curr_max = (``int``)(m[i].y);`` ``}`` ``}`` ` ` ``// calculating residual`` ``// drops in the pipe`` ``k++;`` ``return` `k;`` ``}`` ` ` ``public` `static` `void` `main(String[] args) {`` ``// length of pipe`` ``int` `length = ``12``;`` ` ` ``// position of droplets`` ``int``[] position = { ``10``, ``8``, ``0``, ``5``, ``3` `};`` ` ` ``// speed of each droplets`` ``int``[] speed = { ``2``, ``4``, ``1``, ``1``, ``3` `};`` ``int` `n = speed.length;`` ``System.out.println(drops(length, position, speed, n));`` ``}``}` `// This code is contributed by decode2207.`. ## Python3. `# Function to find the number``# of the drops that come out of the``# pipe``def` `drops(length, position, speed, n):`` ` ` ``# Stores position and time`` ``# taken by a single drop to`` ``# reach the end as a pair`` ``m ``=` `[]` ` ``for` `i ``in` `range``(n):`` ` ` ``# Calculates distance needs to be`` ``# covered by the ith drop`` ``p ``=` `length ``-` `position[i]` ` ``# Inserts initial position of the`` ``# ith drop to the pair`` ``# inserts time taken by ith`` ``# drop to reach`` ``# the end to the pair`` ``m.append([position[i], (p ``*` `1.0``) ``/` `speed[i]])` ` ``# Sorts the pair according to increasing`` ``# order of their positions`` ``m.sort()`` ` ` ``# Counter for no of final drops`` ``k ``=` `0`` ` ` ``curr_max ``=` `m[n ``-` `1``][``1``]`` ` ` ``# We traverse the array demo`` ``# right to left`` ``# to determine the slower drop`` ``for` `i ``in` `range``(n ``-` `2``, ``-``1``, ``-``1``):`` ` ` ``# Checks for next slower drop`` ``if` `(m[i][``1``] > curr_max):`` ``k ``+``=` `1`` ``curr_max ``=` `m[i][``1``]`` ` ` ``# Calculating residual`` ``# drops in the pipe`` ``k ``+``=` `1`` ``return` `k`` ` `# Driver Code` `# Length of pipe``length ``=` `12` `# Position of droplets``position ``=` `[ ``10``, ``8``, ``0``, ``5``, ``3` `]` `# Speed of each droplets``speed ``=` `[ ``2``, ``4``, ``1``, ``1``, ``3` `]``n ``=` `len``(speed)` `print``(drops(length, position, speed, n))` `# This code is contributed by divyeshrabadiya07`. ## C#. `using` `System;``using` `System.Collections.Generic;``class` `GFG``{` ` ``// Function to find the number`` ``// of the drops that come out of the`` ``// pipe`` ``static` `int` `drops(``int` `length, ``int``[] position,`` ``int``[] speed, ``int` `n)`` ``{` ` ``// stores position and time`` ``// taken by a single`` ``// drop to reach the end as a pair`` ``List> m = ``new` `List>();`` ``int` `i;`` ``for` `(i = 0; i < n; i++)`` ``{` ` ``// calculates distance needs to be`` ``// covered by the ith drop`` ``int` `p = length - position[i];` ` ``// inserts initial position of the`` ``// ith drop to the pair`` ``// inserts time taken by ith`` ``// drop to reach`` ``// the end to the pair`` ``m.Add(``new` `Tuple<``int``,``double``>(position[i], p * 1.0 / speed[i]));`` ``}` ` ``// sorts the pair according to increasing`` ``// order of their positions`` ``m.Sort();`` ``int` `k = 0; ``// counter for no of final drops`` ``int` `curr_max = (``int``)m[n - 1].Item2;` ` ``// we traverse the array demo`` ``// right to left`` ``// to determine the slower drop`` ``for` `(i = n - 2; i >= 0; i--)`` ``{` ` ``// checks for next slower drop`` ``if` `(m[i].Item2 > curr_max)`` ``{`` ``k++;`` ``curr_max = (``int``)m[i].Item2;`` ``}`` ``}` ` ``// calculating residual`` ``// drops in the pipe`` ``k++;`` ``return` `k;`` ``}` ` ``// Driver code`` ``static` `void` `Main()`` ``{` ` ``// length of pipe`` ``int` `length = 12;` ` ``// position of droplets`` ``int``[] position = { 10, 8, 0, 5, 3 };` ` ``// speed of each droplets`` ``int``[] speed = { 2, 4, 1, 1, 3 };`` ``int` `n = speed.Length;`` ``Console.WriteLine(drops(length, position, speed, n));`` ``}``}` `// This code is contributed by divyesh072019`. ## Javascript. ``. Output. `3`. Complexity Analysis:. • Time Complexity: O(nlog(n)). • Auxiliary Space: O(n). My Personal Notes arrow_drop_up.
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# Transmission Line Standing Wave Ratio (SWR) Formula & Calculator
## Load to Characteristic Impedance Ratio Calculation
p - Reflection coefficient
standing wave ratio (SWR) calculator - step by step calculation, formula & solved example problem to find the ratio of load impedance matching to the transmission line or wave guide characteristic impedance (Z0). Zterm - load or terminal impedance & Z0 - characteristic impedance are the key terms of this calculation. Higher & lower SWR ratio values indicate that the higher & lower standing wave along the transmission line.
## Formula
The below mathematical formula is used in this calculator to find the standing wave ratio (SWR) of the transmission line
### Solved Example
The below step by step solved example problem may helpful for users to understand how the input values are being used in such calculations to find the ratio of load impedance matching to the transmission line or wave guide characteristic impedance (Z0).
Example Problem
Find the SWR of the transmission line or wave guide whose load or terminal impedance Zterm = 7.5 kilo ohms & characteristic impedance Z0 = 6 kilo ohms.
Solution
The given data
terminal impedance Zterm = 7.5 kilo Ω
characteristic impedance Z0 = 6 kilo Ω
Step by step calculation
Formula to find ρ = (Zterm - Z0)/(Zterm + Z0)
substitute the values in the above formula
= (7500 - 6000)/(7500 + 6000)
ρ = 0.11
Step by step calculation
Formula to find SWR = (1 + |ρ|)/(1 - |ρ|)
substitute the values in the above formula
= (1 + 0.11)/(1 - 0.11)
SWR = 1.24
In the field of electrical engineering, it's important to calculate the standing wave ratio of the medium to analyse the transmission lines characteristics. The above formula, step by step calculation & solved example problem may be useful for users to understand how the values are being used in the formula to find ratio of load impedance (Zterm) matching to the transmission line or wave guide characteristic impedance (Z0), however, when it comes to online for quick calculations, this calculator helps the user to perform & verify such calculations as quick as possible.
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# Transmission Line Standing Wave Ratio (SWR) Formula & Calculator. ## Load to Characteristic Impedance Ratio Calculation. p - Reflection coefficient. standing wave ratio (SWR) calculator - step by step calculation, formula & solved example problem to find the ratio of load impedance matching to the transmission line or wave guide characteristic impedance (Z0). Zterm - load or terminal impedance & Z0 - characteristic impedance are the key terms of this calculation. Higher & lower SWR ratio values indicate that the higher & lower standing wave along the transmission line.. ## Formula. The below mathematical formula is used in this calculator to find the standing wave ratio (SWR) of the transmission line. ### Solved Example. The below step by step solved example problem may helpful for users to understand how the input values are being used in such calculations to find the ratio of load impedance matching to the transmission line or wave guide characteristic impedance (Z0).. Example Problem. Find the SWR of the transmission line or wave guide whose load or terminal impedance Zterm = 7.5 kilo ohms & characteristic impedance Z0 = 6 kilo ohms.. Solution. The given data. terminal impedance Zterm = 7.5 kilo Ω. characteristic impedance Z0 = 6 kilo Ω. Step by step calculation.
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Formula to find ρ = (Zterm - Z0)/(Zterm + Z0). substitute the values in the above formula. = (7500 - 6000)/(7500 + 6000). ρ = 0.11. Step by step calculation. Formula to find SWR = (1 + |ρ|)/(1 - |ρ|). substitute the values in the above formula. = (1 + 0.11)/(1 - 0.11). SWR = 1.24. In the field of electrical engineering, it's important to calculate the standing wave ratio of the medium to analyse the transmission lines characteristics. The above formula, step by step calculation & solved example problem may be useful for users to understand how the values are being used in the formula to find ratio of load impedance (Zterm) matching to the transmission line or wave guide characteristic impedance (Z0), however, when it comes to online for quick calculations, this calculator helps the user to perform & verify such calculations as quick as possible.. Related Calculators. KVA Calculator Solar Panel Calculator Horsepower to Torque Calculator Speed to Horsepower Converter Torque to Horsepower Converter. Related Calculators. KVA Conversion Propagation Constant Calculator Characteristic Impedance Calculator Reflection Coefficient Calculator Terminal Voltage Magnitude Calculator Wavelength Calculator.
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# Thread: system of differential equations
1. ## system of differential equations
Consider the system of first-order differential equations:
Identify the two nullclines of this system, and find the two equilibrium solution coordinates.
I found the vertical nullcline to be $0=x-y^2$ and the horizontal to be $0=x-y^2$
And I know equilibrium solutions are when both vertical and horizontal null clines are zero...but how exactly do I solve for that? Do I just set the two null clines equal to each other...? Any help is greatly appreciated!
2. Originally Posted by cdlegendary
Consider the system of first-order differential equations:
Identify the two nullclines of this system, and find the two equilibrium solution coordinates.
I found the vertical nullcline to be $0=x-y^2$ and the horizontal to be $0=x-y^2$
??? NO! A vertical nullcline is when dx/dt= 0 which means 1- x- y= 0. Was that a typo?
And I know equilibrium solutions are when both vertical and horizontal null clines are zero...but how exactly do I solve for that? Do I just set the two null clines equal to each other...? Any help is greatly appreciated!
No, don't set them equal to each other, set them equal to 0! What values of x and y satisfy both 1- x- y= 0 and $x- y^2= 0$?
3. Well in this differential equation the method of 3rd rule chain method can be applied. Also you can use method in the form du/dv
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# Thread: system of differential equations. 1. ## system of differential equations. Consider the system of first-order differential equations:. Identify the two nullclines of this system, and find the two equilibrium solution coordinates.. I found the vertical nullcline to be $0=x-y^2$ and the horizontal to be $0=x-y^2$. And I know equilibrium solutions are when both vertical and horizontal null clines are zero...but how exactly do I solve for that? Do I just set the two null clines equal to each other...? Any help is greatly appreciated!. 2. Originally Posted by cdlegendary. Consider the system of first-order differential equations:.
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Identify the two nullclines of this system, and find the two equilibrium solution coordinates.. I found the vertical nullcline to be $0=x-y^2$ and the horizontal to be $0=x-y^2$. ??? NO! A vertical nullcline is when dx/dt= 0 which means 1- x- y= 0. Was that a typo?. And I know equilibrium solutions are when both vertical and horizontal null clines are zero...but how exactly do I solve for that? Do I just set the two null clines equal to each other...? Any help is greatly appreciated!. No, don't set them equal to each other, set them equal to 0! What values of x and y satisfy both 1- x- y= 0 and $x- y^2= 0$?. 3. Well in this differential equation the method of 3rd rule chain method can be applied. Also you can use method in the form du/dv.
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http://www.puzzlevilla.com/puzzles/puzzle/74
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# PUZZLEVILLA
or
Remember me
### Register
OR
2+3=10, 7+2=63, 6+5=66, 8+4=96 Puzzle
Difficulty Level
If 2+3=10, 7+2=63, 6+5=66, 8+4=96
Then find 9+7 = ?
2+3=10
7+2=63
6+5=66
8+4=96
9+7=144
The operation "+" here is apparently just to add the two integers and then multiply the answer by the first.
2 + 3 = 5 (this answer x first number)
5 x 2 = 10
7 + 2 = 9
9 x 7 = 63
6 + 5 = 11
11 x 6 = 66
8 + 4 = 12
12 x 8 = 96
Then, 9 + 7= 16
16 x 9 = 144
There's one other way to derive the answer, These type of questions are more tricky.
It's (a)2 + (a*b) = x
4+6=10
49+14=63
36+30=66
64+32=96
81+63=144
Guest Said:Posted On 2017-03-25
144
Guest Said:Posted On 2017-01-20
another way : 9+7+(9 square)
Guest Said:Posted On 2017-01-20
another way is : 9+7=(9+7)*9square
Guest Said:Posted On 2016-11-04
Nice
Guest Said:Posted On 2016-03-09
144
Submit
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# PUZZLEVILLA. or. Remember me. ### Register. OR. 2+3=10, 7+2=63, 6+5=66, 8+4=96 Puzzle. Difficulty Level. If 2+3=10, 7+2=63, 6+5=66, 8+4=96. Then find 9+7 = ?. 2+3=10. 7+2=63. 6+5=66. 8+4=96. 9+7=144. The operation "+" here is apparently just to add the two integers and then multiply the answer by the first.. 2 + 3 = 5 (this answer x first number). 5 x 2 = 10. 7 + 2 = 9. 9 x 7 = 63. 6 + 5 = 11. 11 x 6 = 66. 8 + 4 = 12.
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12 x 8 = 96. Then, 9 + 7= 16. 16 x 9 = 144. There's one other way to derive the answer, These type of questions are more tricky.. It's (a)2 + (a*b) = x. 4+6=10. 49+14=63. 36+30=66. 64+32=96. 81+63=144. Guest Said:Posted On 2017-03-25. 144. Guest Said:Posted On 2017-01-20. another way : 9+7+(9 square). Guest Said:Posted On 2017-01-20. another way is : 9+7=(9+7)*9square. Guest Said:Posted On 2016-11-04. Nice. Guest Said:Posted On 2016-03-09. 144. Submit.
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https://www.jiskha.com/questions/1471807/a-passenger-on-a-ride-at-the-firemens-field-days-sits-in-a-chair-and-is-swung-around-at
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# Physics
A passenger on a ride at the firemen's field days sits in a chair and is swung around at the end of a cable connected to a central tower. At full speed, the cable makes an angle of 56o with the vertical, and the chair is 46 m from the pole.
1)What is the speed of the passenger?
Please help ASAP!!
1. 👍
2. 👎
3. 👁
1. I still need help!!
1. 👍
2. 👎
## Similar Questions
1. ### physics
A chair of weight 150 lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of = 38.0 directed at an angle of 41.0 below the horizontal and the chair slides along the floor.Using Newton's
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4. ### Math Algebra 2
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# Physics. A passenger on a ride at the firemen's field days sits in a chair and is swung around at the end of a cable connected to a central tower. At full speed, the cable makes an angle of 56o with the vertical, and the chair is 46 m from the pole.. 1)What is the speed of the passenger?. Please help ASAP!!. 1. 👍. 2. 👎. 3. 👁. 1. I still need help!!. 1. 👍. 2. 👎. ## Similar Questions. 1. ### physics. A chair of weight 150 lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of = 38.0 directed at an angle of 41.0 below the horizontal and the chair slides along the floor.Using Newton's. 2. ### physics. The Space Shuttle astronauts use a massing chair to measure their mass. The chair is attached to a spring and is free to oscillate back and forth. The frequency of the oscillation is measured and that is used to calculate the. 3. ### Algebra 2 Sine Function. A power outage occurs 6 min after the ride started passengers must wait for their cage to be manually cranked into the lowest position in order to exit the ride. Sine function model: 82.5 sin 3 pi (t+0.5)+97.5 where h is the. 4. ### science. A chair of mass 15.5 is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force = 38.0 that is directed at an angle of 42.0 below the horizontal and the chair slides along the floor.Use.
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1. ### math. Determine the number of ways three trumpet players out of 6 are chosen for 1st chair, 2nd chair, and 3rd chair.. 2. ### physics. A chair of weight 100N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 43.0N directed at an angle of 35.0^\circ below the horizontal and the chair slides along the floor.. 3. ### Physics. A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 44.0 N directed at an angle of 42.0 degrees below the horizontal and the chair slides along the floor.. 4. ### Physics. A 60kg front seat passenger in a car moving initially with a speed of 18m/s (40 MPH) is brought to rest by an air bag in a time of 0.4s a. What is the impulse acting on the passenger? b.What is the average force acting on the. 1. ### math. The London Eye is a large ferris wheel. Each sealed and air-conditioned passenger capsule holds about 25 passengers. The diameter of the wheel is 135 m, and the wheel takes about half an hour to complete one revolution. a). 2. ### math. Anthony plans to ride 175 km on his bike and Samanta plans to ride 250km. They both plan to complete their journeys in a whole number of days. How many kilometres could each person travel if they both ride the same number of. 3. ### Geometry. A Ferris wheel has a 50 meter radius. How many kilometers will a passenger travel during a ride if the wheel makes 10 revolutions? Round your answer to the nearest tenth of a kilometer.. 4. ### Math Algebra 2. whobatman Please Help?! It might be a little hard but I will upvote if it's right!! The Colossus Ferris wheel debuted at the 1984 New Orleans World’s Fair. The ride is 180 ft tall, and passengers board the ride at an initial.
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# Algebra of random variables
The algebra of random variables in statistics, provides rules for the symbolic manipulation of random variables, while avoiding delving too deeply into the mathematically sophisticated ideas of probability theory. Its symbolism allows the treatment of sums, products, ratios and general functions of random variables, as well as dealing with operations such as finding the probability distributions and the expectations (or expected values), variances and covariances of such combinations.
In principle, the elementary algebra of random variables is equivalent to that of conventional non-random (or deterministic) variables. However, the changes occurring on the probability distribution of a random variable obtained after performing algebraic operations are not straightforward. Therefore, the behavior of the different operators of the probability distribution, such as expected values, variances, covariances, and moments, may be different from that observed for the random variable using symbolic algebra. It is possible to identify some key rules for each of those operators, resulting in different types of algebra for random variables, apart from the elementary symbolic algebra: Expectation algebra, Variance algebra, Covariance algebra, Moment algebra, etc.
## Elementary symbolic algebra of random variables
Considering two random variables ${\displaystyle X}$ and ${\displaystyle Y}$, the following algebraic operations are possible:
• Addition: ${\displaystyle Z=X+Y=Y+X}$
• Subtraction: ${\displaystyle Z=X-Y=-Y+X}$
• Multiplication: ${\displaystyle Z=XY=YX}$
• Division: ${\displaystyle Z=X/Y=X\cdot (1/Y)=(1/Y)\cdot X}$
• Exponentiation: ${\displaystyle Z=X^{Y}=e^{Y\ln(X)}}$
In all cases, the variable ${\displaystyle Z}$ resulting from each operation is also a random variable. All commutative and associative properties of conventional algebraic operations are also valid for random variables. If any of the random variables is replaced by a deterministic variable or by a constant value, all the previous properties remain valid.
## Expectation algebra for random variables
The expected value ${\displaystyle E}$ of the random variable ${\displaystyle Z}$ resulting from an algebraic operation between two random variables can be calculated using the following set of rules:
• Addition: ${\displaystyle E[Z]=E[X+Y]=E[X]+E[Y]=E[Y]+E[X]}$
• Subtraction: ${\displaystyle E[Z]=E[X-Y]=E[X]-E[Y]=-E[Y]+E[X]}$
• Multiplication: ${\displaystyle E[Z]=E[XY]=E[YX]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle E[XY]=E[X]\cdot E[Y]=E[Y]\cdot E[X]}$.
• Division: ${\displaystyle E[Z]=E[X/Y]=E[X\cdot (1/Y)]=E[(1/Y)\cdot X]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle E[X/Y]=E[X]\cdot E[1/Y]=E[1/Y]\cdot E[X]}$.
• Exponentiation: ${\displaystyle E[Z]=E[X^{Y}]=E[e^{Y\ln(X)}]}$
If any of the random variables is replaced by a deterministic variable or by a constant value (${\displaystyle k}$), the previous properties remain valid considering that ${\displaystyle P[X=k]=1}$ and, therefore, ${\displaystyle E[X]=k}$.
If ${\displaystyle Z}$ is defined as a general non-linear algebraic function ${\displaystyle f}$ of a random variable ${\displaystyle X}$, then:
${\displaystyle E[Z]=E[f(X)]\neq f(E[X])}$
Some examples of this property include:
• ${\displaystyle E[X^{2}]\neq E[X]^{2}}$
• ${\displaystyle E[1/X]\neq 1/E[X]}$
• ${\displaystyle E[e^{X}]\neq e^{E[X]}}$
• ${\displaystyle E[\ln(X)]\neq \ln(E[X])}$
The exact value of the expectation of the non-linear function will depend on the particular probability distribution of the random variable ${\displaystyle X}$.
## Variance algebra for random variables
The variance ${\displaystyle \mathrm {Var} }$ of the random variable ${\displaystyle Z}$ resulting from an algebraic operation between random variables can be calculated using the following set of rules:
• Addition: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [X+Y]=\mathrm {Var} [X]+2\mathrm {Cov} [X,Y]+\mathrm {Var} [Y]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle \mathrm {Var} [X+Y]=\mathrm {Var} [X]+\mathrm {Var} [Y]}$.
• Subtraction: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [X-Y]=\mathrm {Var} [X]-2\mathrm {Cov} [X,Y]+\mathrm {Var} [Y]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle \mathrm {Var} [X-Y]=\mathrm {Var} [X]+\mathrm {Var} [Y]}$. That is, for independent random variables the variance is the same for additions and subtractions: ${\displaystyle \mathrm {Var} [X+Y]=\mathrm {Var} [X-Y]=\mathrm {Var} [Y-X]=\mathrm {Var} [-X-Y]}$
• Multiplication: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [XY]=\mathrm {Var} [YX]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle \mathrm {Var} [XY]=E[X^{2}]\cdot E[Y^{2}]-(E[X]\cdot E[Y])^{2}=\mathrm {Var} [X]\cdot \mathrm {Var} [Y]+\mathrm {Var} [X]\cdot (E[Y])^{2}+\mathrm {Var} [Y]\cdot (E[X])^{2}}$.
• Division: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [X/Y]=\mathrm {Var} [X\cdot (1/Y)]=\mathrm {Var} [(1/Y)\cdot X]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle \mathrm {Var} [X/Y]=E[X^{2}]\cdot E[1/Y^{2}]-(E[X]\cdot E[1/Y])^{2}=\mathrm {Var} [X]\cdot \mathrm {Var} [1/Y]+\mathrm {Var} [X]\cdot (E[1/Y])^{2}+\mathrm {Var} [1/Y]\cdot (E[X])^{2}}$.
• Exponentiation: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [X^{Y}]=\mathrm {Var} [e^{Y\ln(X)}]}$
where ${\displaystyle \mathrm {Cov} [X,Y]=\mathrm {Cov} [Y,X]}$ represents the covariance operator between random variables ${\displaystyle X}$ and ${\displaystyle Y}$.
The variance of a random variable can also be expressed directly in terms of the covariance or in terms of the expected value:
${\displaystyle \mathrm {Var} [X]=\mathrm {Cov} (X,X)=E[X^{2}]-E[X]^{2}}$
If any of the random variables is replaced by a deterministic variable or by a constant value (${\displaystyle k}$), the previous properties remain valid considering that ${\displaystyle P[X=k]=1}$ and ${\displaystyle E[X]=k}$, ${\displaystyle \mathrm {Var} [X]=0}$ and ${\displaystyle \mathrm {Cov} [Y,k]=0}$. Special cases are the addition and multiplication of a random variable with a deterministic variable or a constant, where:
• ${\displaystyle \mathrm {Var} [k+Y]=\mathrm {Var} [Y]}$
• ${\displaystyle \mathrm {Var} [kY]=k^{2}\mathrm {Var} [Y]}$
If ${\displaystyle Z}$ is defined as a general non-linear algebraic function ${\displaystyle f}$ of a random variable ${\displaystyle X}$, then:
${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [f(X)]\neq f(\mathrm {Var} [X])}$
The exact value of the variance of the non-linear function will depend on the particular probability distribution of the random variable ${\displaystyle X}$.
## Covariance algebra for random variables
The covariance (${\displaystyle \mathrm {Cov} }$) between the random variable ${\displaystyle Z}$ resulting from an algebraic operation and the random variable ${\displaystyle X}$ can be calculated using the following set of rules:
• Addition: ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [X+Y,X]=\mathrm {Var} [X]+\mathrm {Cov} [X,Y]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [X+Y,X]=\mathrm {Var} [X]}$.
• Subtraction: ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [X-Y,X]=\mathrm {Var} [X]-\mathrm {Cov} [X,Y]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [X-Y,X]=\mathrm {Var} [X]}$.
• Multiplication: ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [XY,X]=E[X^{2}Y]-E[XY]E[X]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [XY,X]=\mathrm {Var} [X]\cdot E[Y]}$.
• Division (covariance with respect to the numerator): ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [X/Y,X]=E[X^{2}/Y]-E[X/Y]E[X]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [X/Y,X]=\mathrm {Var} [X]\cdot E[1/Y]}$.
• Division (covariance with respect to the denominator): ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [Y/X,X]=E[Y]-E[Y/X]E[X]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [Y/X,X]=E[Y]\cdot (1-E[X]\cdot E[1/X])}$.
• Exponentiation (covariance with respect to the base): ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [X^{Y},X]=E[X^{Y+1}]-E[X^{Y}]E[X]}$.
• Exponentiation (covariance with respect to the power): ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [Y^{X},X]=E[XY^{X}]-E[Y^{X}]E[X]}$.
The covariance of a random variable can also be expressed directly in terms of the expected value:
${\displaystyle \mathrm {Cov} (X,Y)=E[XY]-E[X]E[Y]}$
If any of the random variables is replaced by a deterministic variable or by a constant value ( ${\displaystyle k}$), the previous properties remain valid considering that ${\displaystyle E[k]=k}$, ${\displaystyle \mathrm {Var} [k]=0}$ and ${\displaystyle \mathrm {Cov} [X,k]=0}$.
If ${\displaystyle Z}$ is defined as a general non-linear algebraic function ${\displaystyle f}$of a random variable ${\displaystyle X}$, then:
${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [f(X),X]=E[Xf(X)]-E[f(X)]E[X]}$
The exact value of the variance of the non-linear function will depend on the particular probability distribution of the random variable ${\displaystyle X}$.
## Approximations by Taylor series expansions of moments
If the moments of a certain random variable ${\displaystyle X}$are known (or can be determined by integration if the probability density function is known), then it is possible to approximate the expected value of any general non-linear function ${\displaystyle f(X)}$as a Taylor series expansion of the moments, as follows:
${\displaystyle f(X)=\displaystyle \sum _{n=0}^{\infty }\displaystyle {\frac {1}{n!}}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }(X-\mu )^{n}}$, where ${\displaystyle \mu =E[X]}$is the mean value of ${\displaystyle X}$.
${\displaystyle E[f(X)]=E{\biggl (}\textstyle \sum _{n=0}^{\infty }\displaystyle {1 \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }(X-\mu )^{n}{\biggr )}=\displaystyle \sum _{n=0}^{\infty }\displaystyle {1 \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }E[(X-\mu )^{n}]=\textstyle \sum _{n=0}^{\infty }\displaystyle {\frac {1}{n!}}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }\mu _{n}(X)}$, where ${\displaystyle \mu _{n}(X)=E[(X-\mu )^{n}]}$is the n-th moment of ${\displaystyle X}$ about its mean. Note that by their definition, ${\displaystyle \mu _{0}(X)=1}$ and ${\displaystyle \mu _{1}(X)=0}$. The first order term always vanishes but was kept to obtain a closed form expression.
Then,
${\displaystyle E[f(X)]\approx \textstyle \sum _{n=0}^{n_{max}}\displaystyle {1 \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }\mu _{n}(X)}$, where the Taylor expansion is truncated after the ${\displaystyle n_{max}}$-th moment.
Particularly for functions of normal random variables, it is possible to obtain a Taylor expansion in terms of the standard normal distribution:[1]
${\displaystyle f(X)=\textstyle \sum _{n=0}^{\infty }\displaystyle {\sigma ^{n} \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }\mu _{n}(Z)}$, where ${\displaystyle X\sim N(\mu ,\sigma ^{2})}$is a normal random variable, and ${\displaystyle Z\sim N(0,1)}$is the standard normal distribution. Thus,
${\displaystyle E[f(X)]\approx \textstyle \sum _{n=0}^{n_{max}}\displaystyle {\sigma ^{n} \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }\mu _{n}(Z)}$, where the moments of the standard normal distribution are given by:
${\displaystyle \mu _{n}(Z)={\begin{cases}\prod _{i=1}^{n/2}(2i-1),&{\text{if }}n{\text{ is even}}\\0,&{\text{if }}n{\text{ is odd}}\end{cases}}}$
Similarly for normal random variables, it is also possible to approximate the variance of the non-linear function as a Taylor series expansion as:
${\displaystyle Var[f(X)]\approx \textstyle \sum _{n=1}^{n_{max}}\displaystyle {\biggl (}{\sigma ^{n} \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }{\biggr )}^{2}Var[Z^{n}]+\textstyle \sum _{n=1}^{n_{max}}\displaystyle \textstyle \sum _{m\neq n}\displaystyle {\sigma ^{n+m} \over {n!m!}}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }{\biggl (}{d^{m}f \over dX^{m}}{\biggr )}_{X=\mu }Cov[Z^{n},Z^{m}]}$, where
${\displaystyle Var[Z^{n}]={\begin{cases}\prod _{i=1}^{n}(2i-1)-\prod _{i=1}^{n/2}(2i-1)^{2},&{\text{if }}n{\text{ is even}}\\\prod _{i=1}^{n}(2i-1),&{\text{if }}n{\text{ is odd}}\end{cases}}}$, and
${\displaystyle Cov[Z^{n},Z^{m}]={\begin{cases}\prod _{i=1}^{(n+m)/2}(2i-1)-\prod _{i=1}^{n/2}(2i-1)\prod _{j=1}^{m/2}(2j-1),&{\text{if }}n{\text{ and }}m{\text{ are even}}\\\prod _{i=1}^{(n+m)/2}(2i-1),&{\text{if }}n{\text{ and }}m{\text{ are odd}}\\0,&{\text{otherwise}}\end{cases}}}$
## Algebra of complex random variables
In the algebraic axiomatization of probability theory, the primary concept is not that of probability of an event, but rather that of a random variable. Probability distributions are determined by assigning an expectation to each random variable. The measurable space and the probability measure arise from the random variables and expectations by means of well-known representation theorems of analysis. One of the important features of the algebraic approach is that apparently infinite-dimensional probability distributions are not harder to formalize than finite-dimensional ones.
Random variables are assumed to have the following properties:
1. complex constants are possible realizations of a random variable;
2. the sum of two random variables is a random variable;
3. the product of two random variables is a random variable;
4. addition and multiplication of random variables are both commutative; and
5. there is a notion of conjugation of random variables, satisfying (XY)* = Y*X* and X** = X for all random variables X,Y and coinciding with complex conjugation if X is a constant.
This means that random variables form complex commutative *-algebras. If X = X* then the random variable X is called "real".
An expectation E on an algebra A of random variables is a normalized, positive linear functional. What this means is that
1. E[k] = k where k is a constant;
2. E[X*X] ≥ 0 for all random variables X;
3. E[X + Y] = E[X] + E[Y] for all random variables X and Y; and
4. E[kX] = kE[X] if k is a constant.
One may generalize this setup, allowing the algebra to be noncommutative. This leads to other areas of noncommutative probability such as quantum probability, random matrix theory, and free probability.
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# Algebra of random variables. The algebra of random variables in statistics, provides rules for the symbolic manipulation of random variables, while avoiding delving too deeply into the mathematically sophisticated ideas of probability theory. Its symbolism allows the treatment of sums, products, ratios and general functions of random variables, as well as dealing with operations such as finding the probability distributions and the expectations (or expected values), variances and covariances of such combinations.. In principle, the elementary algebra of random variables is equivalent to that of conventional non-random (or deterministic) variables. However, the changes occurring on the probability distribution of a random variable obtained after performing algebraic operations are not straightforward. Therefore, the behavior of the different operators of the probability distribution, such as expected values, variances, covariances, and moments, may be different from that observed for the random variable using symbolic algebra. It is possible to identify some key rules for each of those operators, resulting in different types of algebra for random variables, apart from the elementary symbolic algebra: Expectation algebra, Variance algebra, Covariance algebra, Moment algebra, etc.. ## Elementary symbolic algebra of random variables. Considering two random variables ${\displaystyle X}$ and ${\displaystyle Y}$, the following algebraic operations are possible:. • Addition: ${\displaystyle Z=X+Y=Y+X}$. • Subtraction: ${\displaystyle Z=X-Y=-Y+X}$. • Multiplication: ${\displaystyle Z=XY=YX}$. • Division: ${\displaystyle Z=X/Y=X\cdot (1/Y)=(1/Y)\cdot X}$. • Exponentiation: ${\displaystyle Z=X^{Y}=e^{Y\ln(X)}}$. In all cases, the variable ${\displaystyle Z}$ resulting from each operation is also a random variable. All commutative and associative properties of conventional algebraic operations are also valid for random variables. If any of the random variables is replaced by a deterministic variable or by a constant value, all the previous properties remain valid.. ## Expectation algebra for random variables. The expected value ${\displaystyle E}$ of the random variable ${\displaystyle Z}$ resulting from an algebraic operation between two random variables can be calculated using the following set of rules:. • Addition: ${\displaystyle E[Z]=E[X+Y]=E[X]+E[Y]=E[Y]+E[X]}$. • Subtraction: ${\displaystyle E[Z]=E[X-Y]=E[X]-E[Y]=-E[Y]+E[X]}$. • Multiplication: ${\displaystyle E[Z]=E[XY]=E[YX]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle E[XY]=E[X]\cdot E[Y]=E[Y]\cdot E[X]}$.. • Division: ${\displaystyle E[Z]=E[X/Y]=E[X\cdot (1/Y)]=E[(1/Y)\cdot X]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle E[X/Y]=E[X]\cdot E[1/Y]=E[1/Y]\cdot E[X]}$.. • Exponentiation: ${\displaystyle E[Z]=E[X^{Y}]=E[e^{Y\ln(X)}]}$. If any of the random variables is replaced by a deterministic variable or by a constant value (${\displaystyle k}$), the previous properties remain valid considering that ${\displaystyle P[X=k]=1}$ and, therefore, ${\displaystyle E[X]=k}$.. If ${\displaystyle Z}$ is defined as a general non-linear algebraic function ${\displaystyle f}$ of a random variable ${\displaystyle X}$, then:. ${\displaystyle E[Z]=E[f(X)]\neq f(E[X])}$. Some examples of this property include:. • ${\displaystyle E[X^{2}]\neq E[X]^{2}}$. • ${\displaystyle E[1/X]\neq 1/E[X]}$. • ${\displaystyle E[e^{X}]\neq e^{E[X]}}$. • ${\displaystyle E[\ln(X)]\neq \ln(E[X])}$. The exact value of the expectation of the non-linear function will depend on the particular probability distribution of the random variable ${\displaystyle X}$.. ## Variance algebra for random variables. The variance ${\displaystyle \mathrm {Var} }$ of the random variable ${\displaystyle Z}$ resulting from an algebraic operation between random variables can be calculated using the following set of rules:. • Addition: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [X+Y]=\mathrm {Var} [X]+2\mathrm {Cov} [X,Y]+\mathrm {Var} [Y]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle \mathrm {Var} [X+Y]=\mathrm {Var} [X]+\mathrm {Var} [Y]}$.. • Subtraction: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [X-Y]=\mathrm {Var} [X]-2\mathrm {Cov} [X,Y]+\mathrm {Var} [Y]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle \mathrm {Var} [X-Y]=\mathrm {Var} [X]+\mathrm {Var} [Y]}$. That is, for independent random variables the variance is the same for additions and subtractions: ${\displaystyle \mathrm {Var} [X+Y]=\mathrm {Var} [X-Y]=\mathrm {Var} [Y-X]=\mathrm {Var} [-X-Y]}$. • Multiplication: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [XY]=\mathrm {Var} [YX]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle \mathrm {Var} [XY]=E[X^{2}]\cdot E[Y^{2}]-(E[X]\cdot E[Y])^{2}=\mathrm {Var} [X]\cdot \mathrm {Var} [Y]+\mathrm {Var} [X]\cdot (E[Y])^{2}+\mathrm {Var} [Y]\cdot (E[X])^{2}}$.. • Division: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [X/Y]=\mathrm {Var} [X\cdot (1/Y)]=\mathrm {Var} [(1/Y)\cdot X]}$. Particularly, if ${\displaystyle X}$and ${\displaystyle Y}$are independent from each other, then: ${\displaystyle \mathrm {Var} [X/Y]=E[X^{2}]\cdot E[1/Y^{2}]-(E[X]\cdot E[1/Y])^{2}=\mathrm {Var} [X]\cdot \mathrm {Var} [1/Y]+\mathrm {Var} [X]\cdot (E[1/Y])^{2}+\mathrm {Var} [1/Y]\cdot (E[X])^{2}}$.. • Exponentiation: ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [X^{Y}]=\mathrm {Var} [e^{Y\ln(X)}]}$. where ${\displaystyle \mathrm {Cov} [X,Y]=\mathrm {Cov} [Y,X]}$ represents the covariance operator between random variables ${\displaystyle X}$ and ${\displaystyle Y}$.. The variance of a random variable can also be expressed directly in terms of the covariance or in terms of the expected value:. ${\displaystyle \mathrm {Var} [X]=\mathrm {Cov} (X,X)=E[X^{2}]-E[X]^{2}}$. If any of the random variables is replaced by a deterministic variable or by a constant value (${\displaystyle k}$), the previous properties remain valid considering that ${\displaystyle P[X=k]=1}$ and ${\displaystyle E[X]=k}$, ${\displaystyle \mathrm {Var} [X]=0}$ and ${\displaystyle \mathrm {Cov} [Y,k]=0}$. Special cases are the addition and multiplication of a random variable with a deterministic variable or a constant, where:. • ${\displaystyle \mathrm {Var} [k+Y]=\mathrm {Var} [Y]}$. • ${\displaystyle \mathrm {Var} [kY]=k^{2}\mathrm {Var} [Y]}$. If ${\displaystyle Z}$ is defined as a general non-linear algebraic function ${\displaystyle f}$ of a random variable ${\displaystyle X}$, then:. ${\displaystyle \mathrm {Var} [Z]=\mathrm {Var} [f(X)]\neq f(\mathrm {Var} [X])}$. The exact value of the variance of the non-linear function will depend on the particular probability distribution of the random variable ${\displaystyle X}$.. ## Covariance algebra for random variables. The covariance (${\displaystyle \mathrm {Cov} }$) between the random variable ${\displaystyle Z}$ resulting from an algebraic operation and the random variable ${\displaystyle X}$ can be calculated using the following set of rules:. • Addition: ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [X+Y,X]=\mathrm {Var} [X]+\mathrm {Cov} [X,Y]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [X+Y,X]=\mathrm {Var} [X]}$.. • Subtraction: ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [X-Y,X]=\mathrm {Var} [X]-\mathrm {Cov} [X,Y]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [X-Y,X]=\mathrm {Var} [X]}$.. • Multiplication: ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [XY,X]=E[X^{2}Y]-E[XY]E[X]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [XY,X]=\mathrm {Var} [X]\cdot E[Y]}$.. • Division (covariance with respect to the numerator): ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [X/Y,X]=E[X^{2}/Y]-E[X/Y]E[X]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [X/Y,X]=\mathrm {Var} [X]\cdot E[1/Y]}$.. • Division (covariance with respect to the denominator): ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [Y/X,X]=E[Y]-E[Y/X]E[X]}$. If ${\displaystyle X}$ and ${\displaystyle Y}$ are independent from each other, then: ${\displaystyle \mathrm {Cov} [Y/X,X]=E[Y]\cdot (1-E[X]\cdot E[1/X])}$.
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• Exponentiation (covariance with respect to the base): ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [X^{Y},X]=E[X^{Y+1}]-E[X^{Y}]E[X]}$.. • Exponentiation (covariance with respect to the power): ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [Y^{X},X]=E[XY^{X}]-E[Y^{X}]E[X]}$.. The covariance of a random variable can also be expressed directly in terms of the expected value:. ${\displaystyle \mathrm {Cov} (X,Y)=E[XY]-E[X]E[Y]}$. If any of the random variables is replaced by a deterministic variable or by a constant value ( ${\displaystyle k}$), the previous properties remain valid considering that ${\displaystyle E[k]=k}$, ${\displaystyle \mathrm {Var} [k]=0}$ and ${\displaystyle \mathrm {Cov} [X,k]=0}$.. If ${\displaystyle Z}$ is defined as a general non-linear algebraic function ${\displaystyle f}$of a random variable ${\displaystyle X}$, then:. ${\displaystyle \mathrm {Cov} [Z,X]=\mathrm {Cov} [f(X),X]=E[Xf(X)]-E[f(X)]E[X]}$. The exact value of the variance of the non-linear function will depend on the particular probability distribution of the random variable ${\displaystyle X}$.. ## Approximations by Taylor series expansions of moments. If the moments of a certain random variable ${\displaystyle X}$are known (or can be determined by integration if the probability density function is known), then it is possible to approximate the expected value of any general non-linear function ${\displaystyle f(X)}$as a Taylor series expansion of the moments, as follows:. ${\displaystyle f(X)=\displaystyle \sum _{n=0}^{\infty }\displaystyle {\frac {1}{n!}}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }(X-\mu )^{n}}$, where ${\displaystyle \mu =E[X]}$is the mean value of ${\displaystyle X}$.. ${\displaystyle E[f(X)]=E{\biggl (}\textstyle \sum _{n=0}^{\infty }\displaystyle {1 \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }(X-\mu )^{n}{\biggr )}=\displaystyle \sum _{n=0}^{\infty }\displaystyle {1 \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }E[(X-\mu )^{n}]=\textstyle \sum _{n=0}^{\infty }\displaystyle {\frac {1}{n!}}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }\mu _{n}(X)}$, where ${\displaystyle \mu _{n}(X)=E[(X-\mu )^{n}]}$is the n-th moment of ${\displaystyle X}$ about its mean. Note that by their definition, ${\displaystyle \mu _{0}(X)=1}$ and ${\displaystyle \mu _{1}(X)=0}$. The first order term always vanishes but was kept to obtain a closed form expression.. Then,. ${\displaystyle E[f(X)]\approx \textstyle \sum _{n=0}^{n_{max}}\displaystyle {1 \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }\mu _{n}(X)}$, where the Taylor expansion is truncated after the ${\displaystyle n_{max}}$-th moment.. Particularly for functions of normal random variables, it is possible to obtain a Taylor expansion in terms of the standard normal distribution:[1]. ${\displaystyle f(X)=\textstyle \sum _{n=0}^{\infty }\displaystyle {\sigma ^{n} \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }\mu _{n}(Z)}$, where ${\displaystyle X\sim N(\mu ,\sigma ^{2})}$is a normal random variable, and ${\displaystyle Z\sim N(0,1)}$is the standard normal distribution. Thus,. ${\displaystyle E[f(X)]\approx \textstyle \sum _{n=0}^{n_{max}}\displaystyle {\sigma ^{n} \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }\mu _{n}(Z)}$, where the moments of the standard normal distribution are given by:. ${\displaystyle \mu _{n}(Z)={\begin{cases}\prod _{i=1}^{n/2}(2i-1),&{\text{if }}n{\text{ is even}}\\0,&{\text{if }}n{\text{ is odd}}\end{cases}}}$. Similarly for normal random variables, it is also possible to approximate the variance of the non-linear function as a Taylor series expansion as:. ${\displaystyle Var[f(X)]\approx \textstyle \sum _{n=1}^{n_{max}}\displaystyle {\biggl (}{\sigma ^{n} \over n!}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }{\biggr )}^{2}Var[Z^{n}]+\textstyle \sum _{n=1}^{n_{max}}\displaystyle \textstyle \sum _{m\neq n}\displaystyle {\sigma ^{n+m} \over {n!m!}}{\biggl (}{d^{n}f \over dX^{n}}{\biggr )}_{X=\mu }{\biggl (}{d^{m}f \over dX^{m}}{\biggr )}_{X=\mu }Cov[Z^{n},Z^{m}]}$, where. ${\displaystyle Var[Z^{n}]={\begin{cases}\prod _{i=1}^{n}(2i-1)-\prod _{i=1}^{n/2}(2i-1)^{2},&{\text{if }}n{\text{ is even}}\\\prod _{i=1}^{n}(2i-1),&{\text{if }}n{\text{ is odd}}\end{cases}}}$, and. ${\displaystyle Cov[Z^{n},Z^{m}]={\begin{cases}\prod _{i=1}^{(n+m)/2}(2i-1)-\prod _{i=1}^{n/2}(2i-1)\prod _{j=1}^{m/2}(2j-1),&{\text{if }}n{\text{ and }}m{\text{ are even}}\\\prod _{i=1}^{(n+m)/2}(2i-1),&{\text{if }}n{\text{ and }}m{\text{ are odd}}\\0,&{\text{otherwise}}\end{cases}}}$. ## Algebra of complex random variables. In the algebraic axiomatization of probability theory, the primary concept is not that of probability of an event, but rather that of a random variable. Probability distributions are determined by assigning an expectation to each random variable. The measurable space and the probability measure arise from the random variables and expectations by means of well-known representation theorems of analysis. One of the important features of the algebraic approach is that apparently infinite-dimensional probability distributions are not harder to formalize than finite-dimensional ones.. Random variables are assumed to have the following properties:. 1. complex constants are possible realizations of a random variable;. 2. the sum of two random variables is a random variable;. 3. the product of two random variables is a random variable;. 4. addition and multiplication of random variables are both commutative; and. 5. there is a notion of conjugation of random variables, satisfying (XY)* = Y*X* and X** = X for all random variables X,Y and coinciding with complex conjugation if X is a constant.. This means that random variables form complex commutative *-algebras. If X = X* then the random variable X is called "real".. An expectation E on an algebra A of random variables is a normalized, positive linear functional. What this means is that. 1. E[k] = k where k is a constant;. 2. E[X*X] ≥ 0 for all random variables X;. 3. E[X + Y] = E[X] + E[Y] for all random variables X and Y; and. 4. E[kX] = kE[X] if k is a constant.. One may generalize this setup, allowing the algebra to be noncommutative. This leads to other areas of noncommutative probability such as quantum probability, random matrix theory, and free probability.
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Class 6 Maths Playing With Numbers Introduction
Introduction
John‘s uncle gave him 24 muffins to distribute among his friends. This means john has 24 ÷ 1 = 24 muffins.
He wants to equally distribute it among 6 children. How will he do that?
24 ÷ 6 = 4 i.e 24 = 6 x 4
Now what if 2 more children come to his place?
How will he distribute the same number of muffins among 8 children?
24 ÷ 8 = 3 i.e 24 = 8 x 3
Suppose 4 more children visit his place at the same time. Can he distribute 24 muffins equally among all children?
Yes, he can!!! 24 ÷ 12 = 2 i.e 24 = 12 x 2
From this calculation we can see that 24 can be written as a product of two numbers in different ways as
24 = 6 × 4; 24 = 8 × 3; 24 = 12 × 2;
This means 2, 3, 4, 6, 8 and 12 are exact divisor of 24. They are known as factors of 24.
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Class 6 Maths Playing With Numbers Introduction. Introduction. John‘s uncle gave him 24 muffins to distribute among his friends. This means john has 24 ÷ 1 = 24 muffins.. He wants to equally distribute it among 6 children. How will he do that?. 24 ÷ 6 = 4 i.e 24 = 6 x 4. Now what if 2 more children come to his place?. How will he distribute the same number of muffins among 8 children?. 24 ÷ 8 = 3 i.e 24 = 8 x 3.
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Suppose 4 more children visit his place at the same time. Can he distribute 24 muffins equally among all children?. Yes, he can!!! 24 ÷ 12 = 2 i.e 24 = 12 x 2. From this calculation we can see that 24 can be written as a product of two numbers in different ways as. 24 = 6 × 4; 24 = 8 × 3; 24 = 12 × 2;. This means 2, 3, 4, 6, 8 and 12 are exact divisor of 24. They are known as factors of 24.. .
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# physics
two unknown resistors a and b are connected together. when they are connected in series their combined resistance is 15 ohm. When they are connected in parallel, their combined resistance is 3.3 ohm. What are the resistances of A and B?
1. 👍 0
2. 👎 0
3. 👁 86
1. R1+R2 = 15 Ohms
R1*R2/(R1+R2) = 3.3 Ohms
In Eq2, replace R1+R2 with 15 Ohms:
R1*R2/(15) = 3.3
R1*R2 = 49.5
R1 = 49.5/R2
In Eq1, Replace R1 with 49.5/R2:
(49.5/R2)+R2 = 15
R2^2 + 49.5 = 15R2
R2^2 - 15R2 + 49.5 = 0
R2 = 10.1 Ohms
R1 + 10.1 = 15
R1 = 4.9 Ohms
1. 👍 0
2. 👎 0
posted by Henry
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# physics. two unknown resistors a and b are connected together. when they are connected in series their combined resistance is 15 ohm. When they are connected in parallel, their combined resistance is 3.3 ohm. What are the resistances of A and B?. 1. 👍 0. 2. 👎 0. 3. 👁 86. 1. R1+R2 = 15 Ohms. R1*R2/(R1+R2) = 3.3 Ohms. In Eq2, replace R1+R2 with 15 Ohms:. R1*R2/(15) = 3.3. R1*R2 = 49.5. R1 = 49.5/R2. In Eq1, Replace R1 with 49.5/R2:. (49.5/R2)+R2 = 15. R2^2 + 49.5 = 15R2. R2^2 - 15R2 + 49.5 = 0. R2 = 10.1 Ohms. R1 + 10.1 = 15. R1 = 4.9 Ohms. 1. 👍 0. 2. 👎 0. posted by Henry. ## Similar Questions. 1. ### Physics. A certain number of identical resistors are connected in series and their total resistance is measured. When the resistors are disconnected and re-connected in parallel, their total resistance drops to one-percent of the value. asked by Charles on November 7, 2016. 4.) Three 20-Ù resistors are connected in series to a 9-V battery. What is the voltage difference across each resistor? 5.) Three resistors of 5 Ù, 10 Ù, and 2 Ù are connected in parallel with one another. What is the. asked by Amanda on April 30, 2014. 3. ### Physics (URGENT). Three wire-wound resistors have the following values: 30 ohms, 80 ohms, and 100 ohms.
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Each resistor has a voltage rating of 100V. If these three resistors are connected in series, can they be connected to a 240V circuit without. asked by Anonymous on September 24, 2013. 4. ### physics. Four 20 ohm resistors are connected in series and the combination is connected to a 20V emf device. The potential difference across any one of the resistors is: A) 1 V B) 4 V C) 5 V D) 20V E) 80 V So there are four resistors that. asked by susane (recheck thxs) on March 15, 2007. 5. ### Physics. Thanks for any help :) Four 20 ohm resistors are connected in parallel and the combination is connected to a 20V emf device. The current is: A) 0.25A B) 1.0 A C) 4.0 A D) 5.0 A E) 100 E Parallel circuits: I know that the voltage. asked by susane on March 15, 2007. 6. ### Physics. 1)Four resistors of 10.0 each are connected in parallel. A combination of four resistors of 10.0 each is connected in series along with the parallel arrangement. What is the equivalent resistance of the circuit? A)80.0 B)40.4. asked by Jon on April 9, 2008. 7. ### physics. the effective resistacce of two resistors connected in series is 100 ohm.when it connected in parallel it is 24 ohm.determine the values of two resistors. asked by yzr on February 9, 2013. 8. ### physicsc. Two 7.4 resistors are connected in parallel, as are two 4.9 resistors. These two combinations are then connected in series in a circuit with a 11 battery. What is the current in each resistor? What is the voltage across each. asked by Anonymous on February 22, 2010. 9. ### Physics. Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.4-V battery, the current from the battery is 2.13 A. When the resistors are connected in parallel to the battery, the total current from. asked by Abigail on May 11, 2015. 10. ### Physics repost. Four resistors of 10.0 each are connected in parallel. A combination of four resistors of 10.0 each is connected in series along with the parallel arrangement. What is the equivalent resistance of the circuit? A)80.0 B)40.4 C)40.0. asked by Jon on April 9, 2008. More Similar Questions.
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# 2 Easy Ways to Calculate Weighted Average in Google Sheets
Calculating the average is a common task for people working with data in Google Sheets.
And when quantities in a dataset don’t carry the same amount of importance, finding simple arithmetic mean doesn’t really suffice. In such cases, weighted arithmetic mean can be a better representation of the data.
In this weighted average Google Sheets tutorial, we will show you two easy ways in which you can calculate the weighted average in Google Sheets:
• Using the SUMPRODUCT function
• Using the AVERAGE.WEIGHTED function
So let’s get started!
## What is the Weighted Average?
The weighted average is an arithmetic mean calculated while taking into account the importance of elements in the data.
In this way, it gives a more accurate picture of the data than regular arithmetic mean.
The weighted average is often used in analyzing class performance, accounting, and statistical analytic operations.
### When to Use Weighted Average
Let us take a look at an example.
A student may take three different tests (say, a class test, a mid-term, and a final).
Here, each test carries a different amount of importance or contribution to the final grade. The class test is less important, the mid-term carries a little more importance, while the final exam carries the most importance.
In such cases, simply summing up the scores and dividing by 3 would not take into consideration the importance (or weightage) of each exam.
Therefore, it will not give an accurate representation of the student’s performance.
to calculate the weighted average if the scores are weighted as above, on the other hand, it would take into account the weightage of the individual tests. You calculate it by:
1. Multiplying each test score with its corresponding weightage.
2. Adding up each of these products
3. Dividing this sum by the sum of all the weights
So in this student’s case, the weighted average is calculated as follows:
```Weighted average = [(75 x 20) + (80 x 30) + (60 x 50)] / (20 + 30 +50)
= 6900 / 100
= 69```
This value gives a much more accurate picture of how well the student performed since it didn’t just consider the scores in individual tests, but also the importance of each score.
## How to Calculate Weighted Average In Google Sheets
• THE SUMPRODUCT function
• The AVERAGE.WEIGHTED function
We will take a look at how to use these two functions individually. For both methods, we will use the following data set:
### How to Weight Grades in Google Sheets Using the SUMPRODUCT Function
The first method involves the use of the Google Sheets SUMPRODUCT function.
The SUMPRODUCT function lets you find the sum of products of a set of variable values.
#### Syntax of the SUMPRODUCT Function
The syntax for the SUMPRODUCT function is as follows:
`SUMPRODUCT(array1, array2, ....)`
Here, array1, array2, array3, etc. are separate variables. This could be a range of cells, a list of values, or an individual column name.
For example, if we have two ranges A1:A5 and B1:B5, the function: SUMPRODUCT(A1:A5, B1:B5) will take each value of the first range (A1:A5) and multiply it with the corresponding value in the second range (B1:B5).
It will then add up all these individual products to give the final result.
So if you have the below formula
`=SUMPRODUCT(A1:A5, B1:B5)`
this is what it does in the backend in Google Sheets
`= (A1 * B1) + (A2 * B2) + (A3 * B3) + (A4 * B4) + (A5 * B5)`
The SUMPRODUCT function is great for finding the weighted average since a large part of the calculation involves finding the sum of products.
To apply it to finding the weighted average, you can specify the range containing the individual data values as array1 and the range containing the weights as array2.
You can then divide the result by the sum of the weights!
#### How to Do Weighted Average in Google Sheets Using SUMPRODUCT to Find the Weighted Average of Grades
In our example, we want to find the products of the individual test scores with their corresponding weightage, sum up all the products and then divide this sum by the sum of the weights.
Here are the steps you need to follow to apply the SUMPRODUCT function to the above example:
1. Select the cell where you want to display the weighted average (C8 in our example).
2. Type in the formula:
`=SUMPRODUCT(B2:B7, C2:C7) / SUM(C2:C7).`
3. Press the Return key.
You should see the resultant weighted average in your selected cell.
### Finding the Weighted Average using AVERAGE.WEIGHTED Google Sheets function
AVERAGE.WEIGHTED is the weighted average formula in Google Sheets.
You will not find a function like this in Excel that is dedicated solely to the purpose of finding the weighted average.
This function makes it much simpler to calculate the weighted average, compared to the SUMPRODUCT method.
#### Syntax of the AVERAGE.WEIGHTED Function
The syntax for the AVERAGE.WEIGHTED function is as follows:
`AVERAGE.WEIGHTED(values, weights, [additional values], [additional weights])`
Here,
• values are the data values you want to find the weighted average for
• weights is the range of cells containing the corresponding weights.
As you can see from the syntax, it is also possible to use multiple sets of data values and their corresponding sets of weights.
The AVERAGE.WEIGHTED function makes the calculation of weighted average much easier than SUMPRODUCT, since you only need to specify the ranges for the values and weights, without having to perform any subsequent operations.
For example, if we have a set of values in the range A1:A5 and corresponding weights in B1:B5, the function: AVERAGE.WEIGHTED(A1:A5, B1:B5) will take each value of the first range (A1:A5) and multiply it with the corresponding value in the second range (B1:B5).
It will then add up all these individual products. Finally, it will divide this final sum by the sum of the weights to give the final result.
So, if you have a formula as shown below:
`=AVERAGE.WEIGHTED(A1:A5, B1:B5)`
this is what it does in the backend in Google Sheets
`= [(A1 * B1) + (A2 * B2) + (A3 * B3) + (A4 * B4) + (A5 * B5)] / (B1+ B2 + B3 + B4 + B5)`
#### Using AVERAGE.WEIGHTED Function to Find Weighted Average
In our example, we can directly use the AVERAGE.WEIGHTED function without having to use it in combination with any other function.
Here are the steps you need to follow to apply the AVERAGE.WEIGHTEDfunction to the above example:
1. Select the cell where you want to display the weighted average (C8 in our example).
2. Type in the formula: =AVERAGE.WEIGHTED(B2:B7,C2:C7).
3. Press the Return key.
You should see the resultant weighted average in your selected cell.
Notice the simplicity of the formula. All we are using is a range of values and a range of weights.
Note: It is important to have all cells filled with a numeric value when using the AVERAGE.WEIGHTED function. If you have a cell left blank, then the formula returns an error. So, make sure you fill all blank cells with at least a ‘0’. There is no such usage barrier when using the SUMPRODUCT function, though.
### When Should You Not Use AVERAGE.WEIGHTED in Google Sheets?
• You cannot use this function when data from the corresponding column is missing
• Values in each column also need to be the same type of data (numbers, dates, etc)
## Weighted Average Google Sheets FAQ
### How Do You Calculate Weighted Average in Google Sheets?
You can use the AVERAGE.WEIGHTED function that uses the following syntax:
`=AVERAGE.WEIGHTED(values, weights, [additional values], [additional weights])`
You use cell ranges for the values and weights. The weight column must correspond to the values and is usually expressed as a percentage.
### Is Google Sheets Weighted Average the Same as SUMPRODUCT?
Both functions can calculate weighted average, but they operate under different syntax. AVERAGE.WEIGHTEDÂ is exclusive to Google Sheets and won’t work in Excel.
### What Is the Difference Between Average and Weighted Average?
Weighted average pre-defines the relative importance of each data point. On the other hand, average just takes the overall mean from the entire dataset.
### When Should I Use a Weighted Average?
You should use weighted averages when assigning more significance to certain data points. For example, a teacher may consider one assignment more significant than others when deciding overall grades. Alternatively, you could also use a moving average.
## Wrapping Up the Weighted Average Guide
In this weighted average Google Sheets tutorial, I showed you two ways to calculate the weighted average.
One using the generic SUMPRODUCT function, and another using a special Google Sheets function, WEIGHTED.AVERAGE.
The latter function is specific to Google Sheets and is dedicated specifically to calculating the weighted average so it won’t translate to Excel if you need it to.
I hope this tutorial was helpful to you.
Other Google Sheets tutorials you may like:
### A 2024 Guide to Google Sheets Date Picker
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# 2 Easy Ways to Calculate Weighted Average in Google Sheets. Calculating the average is a common task for people working with data in Google Sheets.. And when quantities in a dataset don’t carry the same amount of importance, finding simple arithmetic mean doesn’t really suffice. In such cases, weighted arithmetic mean can be a better representation of the data.. In this weighted average Google Sheets tutorial, we will show you two easy ways in which you can calculate the weighted average in Google Sheets:. • Using the SUMPRODUCT function. • Using the AVERAGE.WEIGHTED function. So let’s get started!. ## What is the Weighted Average?. The weighted average is an arithmetic mean calculated while taking into account the importance of elements in the data.. In this way, it gives a more accurate picture of the data than regular arithmetic mean.. The weighted average is often used in analyzing class performance, accounting, and statistical analytic operations.. ### When to Use Weighted Average. Let us take a look at an example.. A student may take three different tests (say, a class test, a mid-term, and a final).. Here, each test carries a different amount of importance or contribution to the final grade. The class test is less important, the mid-term carries a little more importance, while the final exam carries the most importance.. In such cases, simply summing up the scores and dividing by 3 would not take into consideration the importance (or weightage) of each exam.. Therefore, it will not give an accurate representation of the student’s performance.. to calculate the weighted average if the scores are weighted as above, on the other hand, it would take into account the weightage of the individual tests. You calculate it by:. 1. Multiplying each test score with its corresponding weightage.. 2. Adding up each of these products. 3. Dividing this sum by the sum of all the weights. So in this student’s case, the weighted average is calculated as follows:. ```Weighted average = [(75 x 20) + (80 x 30) + (60 x 50)] / (20 + 30 +50). = 6900 / 100. = 69```. This value gives a much more accurate picture of how well the student performed since it didn’t just consider the scores in individual tests, but also the importance of each score.. ## How to Calculate Weighted Average In Google Sheets. • THE SUMPRODUCT function. • The AVERAGE.WEIGHTED function. We will take a look at how to use these two functions individually. For both methods, we will use the following data set:. ### How to Weight Grades in Google Sheets Using the SUMPRODUCT Function. The first method involves the use of the Google Sheets SUMPRODUCT function.. The SUMPRODUCT function lets you find the sum of products of a set of variable values.. #### Syntax of the SUMPRODUCT Function. The syntax for the SUMPRODUCT function is as follows:. `SUMPRODUCT(array1, array2, ....)`. Here, array1, array2, array3, etc. are separate variables. This could be a range of cells, a list of values, or an individual column name.. For example, if we have two ranges A1:A5 and B1:B5, the function: SUMPRODUCT(A1:A5, B1:B5) will take each value of the first range (A1:A5) and multiply it with the corresponding value in the second range (B1:B5).. It will then add up all these individual products to give the final result.. So if you have the below formula. `=SUMPRODUCT(A1:A5, B1:B5)`. this is what it does in the backend in Google Sheets. `= (A1 * B1) + (A2 * B2) + (A3 * B3) + (A4 * B4) + (A5 * B5)`. The SUMPRODUCT function is great for finding the weighted average since a large part of the calculation involves finding the sum of products.. To apply it to finding the weighted average, you can specify the range containing the individual data values as array1 and the range containing the weights as array2.. You can then divide the result by the sum of the weights!. #### How to Do Weighted Average in Google Sheets Using SUMPRODUCT to Find the Weighted Average of Grades. In our example, we want to find the products of the individual test scores with their corresponding weightage, sum up all the products and then divide this sum by the sum of the weights.. Here are the steps you need to follow to apply the SUMPRODUCT function to the above example:. 1. Select the cell where you want to display the weighted average (C8 in our example).. 2. Type in the formula:. `=SUMPRODUCT(B2:B7, C2:C7) / SUM(C2:C7).`. 3. Press the Return key.. You should see the resultant weighted average in your selected cell.
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### Finding the Weighted Average using AVERAGE.WEIGHTED Google Sheets function. AVERAGE.WEIGHTED is the weighted average formula in Google Sheets.. You will not find a function like this in Excel that is dedicated solely to the purpose of finding the weighted average.. This function makes it much simpler to calculate the weighted average, compared to the SUMPRODUCT method.. #### Syntax of the AVERAGE.WEIGHTED Function. The syntax for the AVERAGE.WEIGHTED function is as follows:. `AVERAGE.WEIGHTED(values, weights, [additional values], [additional weights])`. Here,. • values are the data values you want to find the weighted average for. • weights is the range of cells containing the corresponding weights.. As you can see from the syntax, it is also possible to use multiple sets of data values and their corresponding sets of weights.. The AVERAGE.WEIGHTED function makes the calculation of weighted average much easier than SUMPRODUCT, since you only need to specify the ranges for the values and weights, without having to perform any subsequent operations.. For example, if we have a set of values in the range A1:A5 and corresponding weights in B1:B5, the function: AVERAGE.WEIGHTED(A1:A5, B1:B5) will take each value of the first range (A1:A5) and multiply it with the corresponding value in the second range (B1:B5).. It will then add up all these individual products. Finally, it will divide this final sum by the sum of the weights to give the final result.. So, if you have a formula as shown below:. `=AVERAGE.WEIGHTED(A1:A5, B1:B5)`. this is what it does in the backend in Google Sheets. `= [(A1 * B1) + (A2 * B2) + (A3 * B3) + (A4 * B4) + (A5 * B5)] / (B1+ B2 + B3 + B4 + B5)`. #### Using AVERAGE.WEIGHTED Function to Find Weighted Average. In our example, we can directly use the AVERAGE.WEIGHTED function without having to use it in combination with any other function.. Here are the steps you need to follow to apply the AVERAGE.WEIGHTEDfunction to the above example:. 1. Select the cell where you want to display the weighted average (C8 in our example).. 2. Type in the formula: =AVERAGE.WEIGHTED(B2:B7,C2:C7).. 3. Press the Return key.. You should see the resultant weighted average in your selected cell.. Notice the simplicity of the formula. All we are using is a range of values and a range of weights.. Note: It is important to have all cells filled with a numeric value when using the AVERAGE.WEIGHTED function. If you have a cell left blank, then the formula returns an error. So, make sure you fill all blank cells with at least a ‘0’. There is no such usage barrier when using the SUMPRODUCT function, though.. ### When Should You Not Use AVERAGE.WEIGHTED in Google Sheets?. • You cannot use this function when data from the corresponding column is missing. • Values in each column also need to be the same type of data (numbers, dates, etc). ## Weighted Average Google Sheets FAQ. ### How Do You Calculate Weighted Average in Google Sheets?. You can use the AVERAGE.WEIGHTED function that uses the following syntax:. `=AVERAGE.WEIGHTED(values, weights, [additional values], [additional weights])`. You use cell ranges for the values and weights. The weight column must correspond to the values and is usually expressed as a percentage.. ### Is Google Sheets Weighted Average the Same as SUMPRODUCT?. Both functions can calculate weighted average, but they operate under different syntax. AVERAGE.WEIGHTED is exclusive to Google Sheets and won’t work in Excel.. ### What Is the Difference Between Average and Weighted Average?. Weighted average pre-defines the relative importance of each data point. On the other hand, average just takes the overall mean from the entire dataset.. ### When Should I Use a Weighted Average?. You should use weighted averages when assigning more significance to certain data points. For example, a teacher may consider one assignment more significant than others when deciding overall grades. Alternatively, you could also use a moving average.. ## Wrapping Up the Weighted Average Guide. In this weighted average Google Sheets tutorial, I showed you two ways to calculate the weighted average.. One using the generic SUMPRODUCT function, and another using a special Google Sheets function, WEIGHTED.AVERAGE.. The latter function is specific to Google Sheets and is dedicated specifically to calculating the weighted average so it won’t translate to Excel if you need it to.. I hope this tutorial was helpful to you.. Other Google Sheets tutorials you may like:. ### A 2024 Guide to Google Sheets Date Picker. This site uses Akismet to reduce spam. Learn how your comment data is processed.. ##### Related Posts. Note that we’re supported by our audience. When you purchase through links on our site, we may earn commission at no extra cost to you.
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# Pricing financial options using Monte Carlo simulations
Fredrik Olsson
Data Scientist
Feb 11
2020
In this blogpost we will dive into the world of finance, and specifically option pricing. We are going to illustrate how we can use Monte Carlo simulations to price financial options, by looking at two examples. In both of these examples no explicit solution formula exists, making the Monte Carlo approach useful in pricing these options.
If you are not familiar with Monte Carlo simulations or financial options, or could use a reminder, I suggest that you take a look at the post Introduction to Monte Carlo simulations and option pricing where we introduced all these concepts. In that post we:
• Introduced derivative assets and financial options, especially the European call option
• Looked at our first model for stocks, namely the standard Black & Scholes model
• Introduced the concept of Monte Carlo simulations
• Looked at our first pricing example
The examples in this post are definitely more challenging mathematically, so make sure that you have a grasp on the basics from the introductory post.
Let's get started.
#### Arithmetic basket call in the standard Black & Scholes model
For our first example, we will use the standard Black & Scholes model, i.e. the same one used in the introduction post about Monte Carlo simulations and option pricing. Here the stock dynamics follow a Geometric Brownian Motion:
$dS(t) = rS(t)dt + \sigma S(t)dW(t)$
with the solution:
$S(T) = S(0)e^{(r-\frac{\sigma^2}{2})T + \sigma W(T)} = e^{\log(S(0)) + (r-\frac{\sigma^2}{2})T + \sigma W(T)}$
where $W(T)\sim N(0,T)$, so $S(T)$ has a log-normal distribution. The contract we will look at now, is very similar to a European call option, but instead of having just one underlying stock, we take the arithmetic average of several stocks ($n$ of them). This contract is called an arithmetic basket call option, and has the following payoff at time $T$:
$\Phi(S_1(T),...,S_n(T)) = \max\left((\frac{1}{n}\sum_{i=1}^n S_i(T)) - K, 0\right)$
Even if we know the distribution of every stock $S_i$, we do not know the distribution of the arithmetic average of $n$ of them (the log-normality is not preserved under summation). We can therefore not find an explicit formula like the Black & Scholes formula for European call options in this case for the option price:
$\Pi(0) = e^{-rT}\mathbb{E}\left[\max\left((\frac{1}{n}\sum_{i=1}^n S_i(T)) - K, 0\right)\right]$
Our solution will be to use Monte Carlo simulations. For simulation purposes, we will use that the log stock prices $\log(S_1(T)),...,\log(S_n(T))$ follow a multivariate normal distribution. For simplicity, we assume that all stocks have the same initial value $S(0)=s$, same volatility $\sigma$ and that the correlation $\rho$ between all the stocks are the same. The log stock prices will then follow a multivariate normal distribution with the following expected value vector $\mathbf{\mu}$ and covariance matrix $\Sigma$:
$\boldsymbol{\mu} = \begin{bmatrix} \log(s) + (r-\frac{\sigma^2}{2})T \\ \vdots \\ \log(s) + (r-\frac{\sigma^2}{2})T \end{bmatrix} \quad \text{and} \quad \boldsymbol{\Sigma} = \begin{bmatrix} \sigma^2 & \sigma^2\rho & \dots & \sigma^2\rho \\ \sigma^2\rho & \sigma^2 & \dots & \sigma^2\rho \\ \vdots & \vdots & \ddots & \vdots \\ \sigma^2\rho & \sigma^2\rho & \dots & \sigma^2 \end{bmatrix}T$
Note that $\boldsymbol{\mu}$ is an $n\times 1$-vector and $\boldsymbol{\Sigma}$ is an $n\times n$-matrix. This is useful because we can now simulate from this multivariate normal distribution and then transform the values using the exponential function to get simulations of the stocks. We thus have the following Monte Carlo sampler for estimating the price of the arithmetic basket call option:
\begin{aligned} &\textbf{for } i=1 \textbf{ to } N \\ &\quad\quad \text{draw } [\log(S_1(T)),...,\log(S_n(T))]_i \sim MVN(\boldsymbol{\mu}, \boldsymbol{\Sigma}) \\ &\quad\quad \text{transform to } [S_1(T),...,S_n(T)]_i \text{ with the exponential function} \\ &\quad\quad \text{compute arithmeric average } (\frac{1}{n}\sum_{j=1}^n S_j(T))_i \\ &\textbf{end} \\ &\Pi(0)_N^{MC} = \frac{1}{N}\sum_{i=1}^N e^{-rT}\cdot\max\left((\frac{1}{n}\sum_{j=1}^n S_j(T))_i - K, 0\right) \end{aligned}
So, assuming that we have $n=10$ underlying stocks, that $s=100$, the strike price $K=90$, the risk free interest rate $r=1\%$, the stock volatility $\sigma=0.2$, the correlation $rho=0.4$ and that the time to maturity $T=1$ year, we can implement the following Monte Carlo sampler in Python:
import numpy as np from scipy.stats import norm n = 10 # number of underlying stocks in the arithmetic basket call s = 100 # initial stock prices K = 90 # strike price sigma = 0.2 # volatility of the stocks rho = 0.4 # correlations between the stocks r = 0.01 # risk free interest rate (continuously compounded) T = 1 # time to maturity (years) mu = (np.log(s) + (r - 0.5*sigma**2)*T)*np.ones(n) # expected value vector of the log stock prices # covariance matrix for the log stock prices cov_mat = (rho*sigma**2)*np.ones((n,n)) cov_mat = cov_mat - np.diag(np.diag(cov_mat)) + np.diag((sigma**2)*np.ones(n)) cov_mat = cov_mat*T N = 100000 # size of the Monte Carlo sample # generate Monte Carlo sample mc = np.random.multivariate_normal(mu, cov_mat, N) # log stock prices mc = np.exp(mc) # stock prices mc = np.mean(mc, 1) # arithmeric mean of the stock prices mc = np.exp(-r*T)*np.maximum(mc - K, 0) # option prices price = np.mean(mc) # estimated price # confidence interval price_std = np.std(mc) / np.sqrt(N) # standard deviation of the price estimator lo = price - norm.ppf(0.975)*price_std # lower bound of confidence interval for the price hi = price + norm.ppf(0.975)*price_std # upper bound of confidence interval for the price
Running a simulation (with $N=100000$) gives us the following price estimate and 95% confidence interval for the Monte Carlo estimated price:
\begin{aligned} \Pi&(0)_N^{MC} = 12.308 \\ I_{\Pi(0)} &= (12.236, 12.381) \end{aligned}
As mentioned in the first post, we will get different values if we were to repeat the simulations. So if you try and implement this by yourself you would not get exactly the same values, but they should however be very similar.
### European call option in the Heston model
When comparing actual options prices on various financial markets with the prices given by the standard Black & Scholes model, it becomes clear that they generally do not match very well. The standard Black & Scholes model is just not complex enough to explain the stock's price process. There are several adjustments one could make to get a better model. One major drawback of the standard Black & Scholes model is that the stock volatility is constant and deterministic, while the actual stock volatility is a stochastic process just like the stock's price process. We will therefore now look at a model with stochastic stock volatility, namely the Heston model.
In the Heston model, we have dynamics given as stochastic differential equations, for both the stock and the volatility. These dynamics looks as follows:
\begin{aligned} &dS(t) = rS(t)dt + \sqrt{V(t)}S(t)(\sqrt{1-\rho^2}dW^{(1)}(t) + \rho dW^{(2)}(t)) \\ &dV(t) = \kappa(\theta - V(t))dt + \sigma\sqrt{V(t)}dW^{(2)}(t) \end{aligned}
where $S(t)$ is the stock's price process as before, and $V(t)$ is the stock's volatility process. $W^{(1)}(t)$ and $W^{(2)}(t)$ are here two independent standard Brownian Motions. Also, $r$ is the risk free interest rate, and $\rho$, $\kappa$, $\theta$ and $\sigma$ are model parameters.
We return to the European call option that we saw in the introductory post. As you might remember, this means that we want to find the option price:
$\Pi(0) = e^{-rT}\mathbb{E}[\max(S(T)-K,0)]$
In the standard Black & Scholes model we could find an explicit formula for this price. In the Heston model however, this is not the case. The reason is that we cannot find a closed form solution to the system of stochastic differential equations for $S(t)$ and $V(t)$ above. We will therefore need numerical methods and Monte Carlo simulations to generate a sample of stock prices $S(T)$ that we can use to estimate the option price.
So we cannot solve these stochastic differential equations analytically, and will instead rely on numerical methods. As always when dealing with continuous functions in numerical analysis, we are going to use some kind of discretization scheme to solve these equations. This means that instead of dealing with this as a continuous problem over the interval $[0,T]$, we are reducing this continuous interval to discrete time points $\{t_0,t_1,...,t_{n-1},t_n\}$ where we have that $t_0=0$ and $t_n=T$. We will then, using some discretization scheme, calculating the values of $S(t)$ and $V(t)$ at these time points only.
We start with the differential equation for $S(t)$. We will here use something called the Euler Scheme. For a stochastic differential equation on the form:
$dS(t) = g(S(t))dt + h(S(t))dW(t)$
where $g$ and $h$ are functions, the Euler scheme is:
$S_i = S_{i-1} + g(S_{i-1})\Delta + h(S_{i-1})\Delta W$
for $i=1,...,n$, where $\Delta=t_i-t_{i-1}$ is the size of each time step and $\Delta W=W_{t_i}-W_{t_{i-1}} \sim N(0,\Delta)$ is the increment of the standard Brownian Motion. Comparing with the equation for $S(t)$, the Euler scheme in this case becomes:
$S_i = S_{i-1} + rS_{i-1}\Delta + \sqrt{V_{i-1}}S_{i-1}(\sqrt{1-\rho^2}\Delta W^{(1)} + \rho\Delta W^{(2)})$
for $i=1,...,n$. Note that the stochastic part here are the $\Delta W$s, so the Monte Carlo aspect will be to simulate these normal random variables. Doing this scheme $N$ times, we will have a Monte Carlo sample of size $N$ of the stock prices $S(T)=S_n$. We must of course simulate the process for $V(t)$ as well. To ensure that the $V_i$s take on only positive values, we add another term to the discretization scheme. We will not go into details here but this is called the Milstein scheme (this extra term can be seen in the code in the end). If you are interested you can check out the Wikipedia page.
So, we have the following Monte Carlo sampler for estimating the price of the European call option:
\begin{aligned} &\textbf{for } i=1 \textbf{ to } N \\ &\quad\quad \textbf{for } j=1 \textbf{ to } n \\ &\quad\quad\quad\quad \text{draw } \Delta W_i^{(1)}\sim N(0,\Delta) \text{ and } \Delta W_i^{(2)}\sim N(0,\Delta) \\ &\quad\quad\quad\quad \text{compute } V_j \text{ and } S_j \text{ with respective discretization scheme} \\ &\quad\quad \textbf{end} \\ &\quad\quad \text{let } S(T)_i = S_n \\ &\textbf{end} \\ &\Pi(0)_N^{MC} = \frac{1}{N}\sum_{i=1}^N e^{-rT}\cdot\max(S(T)_i - K, 0) \end{aligned}
We assume that the initial stock price is $S(0)=100$, the initial volatility is $V(0)=0.16$, the strike price $K=90$, the risk free interest rate $r=1\%$ and the time to maturity is $T=1$ year. Also, we assume that our model parameters have the values $\kappa=10$, $\theta=0.16$, $\sigma=0.1$ and $\rho=-0.8$. Then, using $n=252$ time steps in the discretization (typically 252 trading days on a year, so daily time steps as $T=1$ year), we can implement the following Monte Carlo sampler in Python:
import numpy as np from scipy.stats import norm n = 10 # number of underlying stocks in the arithmetic basket call s = 100 # initial stock prices K = 90 # strike price sigma = 0.2 # volatility of the stocks rho = 0.4 # correlations between the stocks r = 0.01 # risk free interest rate (continuously compounded) T = 1 # time to maturity (years) mu = (np.log(s) + (r - 0.5*sigma**2)*T)*np.ones(n) # expected value vector of the log stock prices # covariance matrix for the log stock prices cov_mat = (rho*sigma**2)*np.ones((n,n)) cov_mat = cov_mat - np.diag(np.diag(cov_mat)) + np.diag((sigma**2)*np.ones(n)) cov_mat = cov_mat*T N = 100000 # size of the Monte Carlo sample # generate Monte Carlo sample mc = np.random.multivariate_normal(mu, cov_mat, N) # log stock prices mc = np.exp(mc) # stock prices mc = np.mean(mc, 1) # arithmeric mean of the stock prices mc = np.exp(-r*T)*np.maximum(mc - K, 0) # option prices price = np.mean(mc) # estimated price # confidence interval price_std = np.std(mc) / np.sqrt(N) # standard deviation of the price estimator lo = price - norm.ppf(0.975)*price_std # lower bound of confidence interval for the price hi = price + norm.ppf(0.975)*price_std # upper bound of confidence interval for the price
Running a simulation (with $N=100000$) gives us the following price estimate and 95% confidence interval for the Monte Carlo estimated price:
\begin{aligned} \Pi&(0)_N^{MC} = 21.071 \\ I_{\Pi(0)} &= (20.866, 21.276) \end{aligned}
### Conclusion
In this blogpost, we made use of the concepts introduced in Introduction to Monte Carlo simulations and option pricing to price two quite tricky financial options using Monte Carlo simulations: the arithmeric basket call option in the standard Black & Scholes model, and the European call option in the Heston model. To summarize the highlights, we have:
• Introduced the arithmetic basket call option
• Looked at an example of how we can simulate using multivariate distributions
• Introduced the Heston model for stocks
• Introduced discretization and the Euler scheme for solving stochastic differential equations
• Seen two examples of Monte Carlo samplers for estimating expected values
Published on February 11th 2020
Last updated on March 1st 2023, 10:53
Fredrik Olsson
Data Scientist
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# Pricing financial options using Monte Carlo simulations. Fredrik Olsson. Data Scientist. Feb 11. 2020. In this blogpost we will dive into the world of finance, and specifically option pricing. We are going to illustrate how we can use Monte Carlo simulations to price financial options, by looking at two examples. In both of these examples no explicit solution formula exists, making the Monte Carlo approach useful in pricing these options.. If you are not familiar with Monte Carlo simulations or financial options, or could use a reminder, I suggest that you take a look at the post Introduction to Monte Carlo simulations and option pricing where we introduced all these concepts. In that post we:. • Introduced derivative assets and financial options, especially the European call option. • Looked at our first model for stocks, namely the standard Black & Scholes model. • Introduced the concept of Monte Carlo simulations. • Looked at our first pricing example. The examples in this post are definitely more challenging mathematically, so make sure that you have a grasp on the basics from the introductory post.. Let's get started.. #### Arithmetic basket call in the standard Black & Scholes model. For our first example, we will use the standard Black & Scholes model, i.e. the same one used in the introduction post about Monte Carlo simulations and option pricing. Here the stock dynamics follow a Geometric Brownian Motion:. $dS(t) = rS(t)dt + \sigma S(t)dW(t)$. with the solution:. $S(T) = S(0)e^{(r-\frac{\sigma^2}{2})T + \sigma W(T)} = e^{\log(S(0)) + (r-\frac{\sigma^2}{2})T + \sigma W(T)}$. where $W(T)\sim N(0,T)$, so $S(T)$ has a log-normal distribution. The contract we will look at now, is very similar to a European call option, but instead of having just one underlying stock, we take the arithmetic average of several stocks ($n$ of them). This contract is called an arithmetic basket call option, and has the following payoff at time $T$:. $\Phi(S_1(T),...,S_n(T)) = \max\left((\frac{1}{n}\sum_{i=1}^n S_i(T)) - K, 0\right)$. Even if we know the distribution of every stock $S_i$, we do not know the distribution of the arithmetic average of $n$ of them (the log-normality is not preserved under summation). We can therefore not find an explicit formula like the Black & Scholes formula for European call options in this case for the option price:. $\Pi(0) = e^{-rT}\mathbb{E}\left[\max\left((\frac{1}{n}\sum_{i=1}^n S_i(T)) - K, 0\right)\right]$. Our solution will be to use Monte Carlo simulations. For simulation purposes, we will use that the log stock prices $\log(S_1(T)),...,\log(S_n(T))$ follow a multivariate normal distribution. For simplicity, we assume that all stocks have the same initial value $S(0)=s$, same volatility $\sigma$ and that the correlation $\rho$ between all the stocks are the same. The log stock prices will then follow a multivariate normal distribution with the following expected value vector $\mathbf{\mu}$ and covariance matrix $\Sigma$:. $\boldsymbol{\mu} = \begin{bmatrix} \log(s) + (r-\frac{\sigma^2}{2})T \\ \vdots \\ \log(s) + (r-\frac{\sigma^2}{2})T \end{bmatrix} \quad \text{and} \quad \boldsymbol{\Sigma} = \begin{bmatrix} \sigma^2 & \sigma^2\rho & \dots & \sigma^2\rho \\ \sigma^2\rho & \sigma^2 & \dots & \sigma^2\rho \\ \vdots & \vdots & \ddots & \vdots \\ \sigma^2\rho & \sigma^2\rho & \dots & \sigma^2 \end{bmatrix}T$. Note that $\boldsymbol{\mu}$ is an $n\times 1$-vector and $\boldsymbol{\Sigma}$ is an $n\times n$-matrix. This is useful because we can now simulate from this multivariate normal distribution and then transform the values using the exponential function to get simulations of the stocks. We thus have the following Monte Carlo sampler for estimating the price of the arithmetic basket call option:. \begin{aligned} &\textbf{for } i=1 \textbf{ to } N \\ &\quad\quad \text{draw } [\log(S_1(T)),...,\log(S_n(T))]_i \sim MVN(\boldsymbol{\mu}, \boldsymbol{\Sigma}) \\ &\quad\quad \text{transform to } [S_1(T),...,S_n(T)]_i \text{ with the exponential function} \\ &\quad\quad \text{compute arithmeric average } (\frac{1}{n}\sum_{j=1}^n S_j(T))_i \\ &\textbf{end} \\ &\Pi(0)_N^{MC} = \frac{1}{N}\sum_{i=1}^N e^{-rT}\cdot\max\left((\frac{1}{n}\sum_{j=1}^n S_j(T))_i - K, 0\right) \end{aligned}. So, assuming that we have $n=10$ underlying stocks, that $s=100$, the strike price $K=90$, the risk free interest rate $r=1\%$, the stock volatility $\sigma=0.2$, the correlation $rho=0.4$ and that the time to maturity $T=1$ year, we can implement the following Monte Carlo sampler in Python:. import numpy as np from scipy.stats import norm n = 10 # number of underlying stocks in the arithmetic basket call s = 100 # initial stock prices K = 90 # strike price sigma = 0.2 # volatility of the stocks rho = 0.4 # correlations between the stocks r = 0.01 # risk free interest rate (continuously compounded) T = 1 # time to maturity (years) mu = (np.log(s) + (r - 0.5*sigma**2)*T)*np.ones(n) # expected value vector of the log stock prices # covariance matrix for the log stock prices cov_mat = (rho*sigma**2)*np.ones((n,n)) cov_mat = cov_mat - np.diag(np.diag(cov_mat)) + np.diag((sigma**2)*np.ones(n)) cov_mat = cov_mat*T N = 100000 # size of the Monte Carlo sample # generate Monte Carlo sample mc = np.random.multivariate_normal(mu, cov_mat, N) # log stock prices mc = np.exp(mc) # stock prices mc = np.mean(mc, 1) # arithmeric mean of the stock prices mc = np.exp(-r*T)*np.maximum(mc - K, 0) # option prices price = np.mean(mc) # estimated price # confidence interval price_std = np.std(mc) / np.sqrt(N) # standard deviation of the price estimator lo = price - norm.ppf(0.975)*price_std # lower bound of confidence interval for the price hi = price + norm.ppf(0.975)*price_std # upper bound of confidence interval for the price. Running a simulation (with $N=100000$) gives us the following price estimate and 95% confidence interval for the Monte Carlo estimated price:. \begin{aligned} \Pi&(0)_N^{MC} = 12.308 \\ I_{\Pi(0)} &= (12.236, 12.381) \end{aligned}. As mentioned in the first post, we will get different values if we were to repeat the simulations. So if you try and implement this by yourself you would not get exactly the same values, but they should however be very similar.. ### European call option in the Heston model. When comparing actual options prices on various financial markets with the prices given by the standard Black & Scholes model, it becomes clear that they generally do not match very well. The standard Black & Scholes model is just not complex enough to explain the stock's price process. There are several adjustments one could make to get a better model. One major drawback of the standard Black & Scholes model is that the stock volatility is constant and deterministic, while the actual stock volatility is a stochastic process just like the stock's price process. We will therefore now look at a model with stochastic stock volatility, namely the Heston model.
|
In the Heston model, we have dynamics given as stochastic differential equations, for both the stock and the volatility. These dynamics looks as follows:. \begin{aligned} &dS(t) = rS(t)dt + \sqrt{V(t)}S(t)(\sqrt{1-\rho^2}dW^{(1)}(t) + \rho dW^{(2)}(t)) \\ &dV(t) = \kappa(\theta - V(t))dt + \sigma\sqrt{V(t)}dW^{(2)}(t) \end{aligned}. where $S(t)$ is the stock's price process as before, and $V(t)$ is the stock's volatility process. $W^{(1)}(t)$ and $W^{(2)}(t)$ are here two independent standard Brownian Motions. Also, $r$ is the risk free interest rate, and $\rho$, $\kappa$, $\theta$ and $\sigma$ are model parameters.. We return to the European call option that we saw in the introductory post. As you might remember, this means that we want to find the option price:. $\Pi(0) = e^{-rT}\mathbb{E}[\max(S(T)-K,0)]$. In the standard Black & Scholes model we could find an explicit formula for this price. In the Heston model however, this is not the case. The reason is that we cannot find a closed form solution to the system of stochastic differential equations for $S(t)$ and $V(t)$ above. We will therefore need numerical methods and Monte Carlo simulations to generate a sample of stock prices $S(T)$ that we can use to estimate the option price.. So we cannot solve these stochastic differential equations analytically, and will instead rely on numerical methods. As always when dealing with continuous functions in numerical analysis, we are going to use some kind of discretization scheme to solve these equations. This means that instead of dealing with this as a continuous problem over the interval $[0,T]$, we are reducing this continuous interval to discrete time points $\{t_0,t_1,...,t_{n-1},t_n\}$ where we have that $t_0=0$ and $t_n=T$. We will then, using some discretization scheme, calculating the values of $S(t)$ and $V(t)$ at these time points only.. We start with the differential equation for $S(t)$. We will here use something called the Euler Scheme. For a stochastic differential equation on the form:. $dS(t) = g(S(t))dt + h(S(t))dW(t)$. where $g$ and $h$ are functions, the Euler scheme is:. $S_i = S_{i-1} + g(S_{i-1})\Delta + h(S_{i-1})\Delta W$. for $i=1,...,n$, where $\Delta=t_i-t_{i-1}$ is the size of each time step and $\Delta W=W_{t_i}-W_{t_{i-1}} \sim N(0,\Delta)$ is the increment of the standard Brownian Motion. Comparing with the equation for $S(t)$, the Euler scheme in this case becomes:. $S_i = S_{i-1} + rS_{i-1}\Delta + \sqrt{V_{i-1}}S_{i-1}(\sqrt{1-\rho^2}\Delta W^{(1)} + \rho\Delta W^{(2)})$. for $i=1,...,n$. Note that the stochastic part here are the $\Delta W$s, so the Monte Carlo aspect will be to simulate these normal random variables. Doing this scheme $N$ times, we will have a Monte Carlo sample of size $N$ of the stock prices $S(T)=S_n$. We must of course simulate the process for $V(t)$ as well. To ensure that the $V_i$s take on only positive values, we add another term to the discretization scheme. We will not go into details here but this is called the Milstein scheme (this extra term can be seen in the code in the end). If you are interested you can check out the Wikipedia page.. So, we have the following Monte Carlo sampler for estimating the price of the European call option:. \begin{aligned} &\textbf{for } i=1 \textbf{ to } N \\ &\quad\quad \textbf{for } j=1 \textbf{ to } n \\ &\quad\quad\quad\quad \text{draw } \Delta W_i^{(1)}\sim N(0,\Delta) \text{ and } \Delta W_i^{(2)}\sim N(0,\Delta) \\ &\quad\quad\quad\quad \text{compute } V_j \text{ and } S_j \text{ with respective discretization scheme} \\ &\quad\quad \textbf{end} \\ &\quad\quad \text{let } S(T)_i = S_n \\ &\textbf{end} \\ &\Pi(0)_N^{MC} = \frac{1}{N}\sum_{i=1}^N e^{-rT}\cdot\max(S(T)_i - K, 0) \end{aligned}. We assume that the initial stock price is $S(0)=100$, the initial volatility is $V(0)=0.16$, the strike price $K=90$, the risk free interest rate $r=1\%$ and the time to maturity is $T=1$ year. Also, we assume that our model parameters have the values $\kappa=10$, $\theta=0.16$, $\sigma=0.1$ and $\rho=-0.8$. Then, using $n=252$ time steps in the discretization (typically 252 trading days on a year, so daily time steps as $T=1$ year), we can implement the following Monte Carlo sampler in Python:. import numpy as np from scipy.stats import norm n = 10 # number of underlying stocks in the arithmetic basket call s = 100 # initial stock prices K = 90 # strike price sigma = 0.2 # volatility of the stocks rho = 0.4 # correlations between the stocks r = 0.01 # risk free interest rate (continuously compounded) T = 1 # time to maturity (years) mu = (np.log(s) + (r - 0.5*sigma**2)*T)*np.ones(n) # expected value vector of the log stock prices # covariance matrix for the log stock prices cov_mat = (rho*sigma**2)*np.ones((n,n)) cov_mat = cov_mat - np.diag(np.diag(cov_mat)) + np.diag((sigma**2)*np.ones(n)) cov_mat = cov_mat*T N = 100000 # size of the Monte Carlo sample # generate Monte Carlo sample mc = np.random.multivariate_normal(mu, cov_mat, N) # log stock prices mc = np.exp(mc) # stock prices mc = np.mean(mc, 1) # arithmeric mean of the stock prices mc = np.exp(-r*T)*np.maximum(mc - K, 0) # option prices price = np.mean(mc) # estimated price # confidence interval price_std = np.std(mc) / np.sqrt(N) # standard deviation of the price estimator lo = price - norm.ppf(0.975)*price_std # lower bound of confidence interval for the price hi = price + norm.ppf(0.975)*price_std # upper bound of confidence interval for the price. Running a simulation (with $N=100000$) gives us the following price estimate and 95% confidence interval for the Monte Carlo estimated price:. \begin{aligned} \Pi&(0)_N^{MC} = 21.071 \\ I_{\Pi(0)} &= (20.866, 21.276) \end{aligned}. ### Conclusion. In this blogpost, we made use of the concepts introduced in Introduction to Monte Carlo simulations and option pricing to price two quite tricky financial options using Monte Carlo simulations: the arithmeric basket call option in the standard Black & Scholes model, and the European call option in the Heston model. To summarize the highlights, we have:. • Introduced the arithmetic basket call option. • Looked at an example of how we can simulate using multivariate distributions. • Introduced the Heston model for stocks. • Introduced discretization and the Euler scheme for solving stochastic differential equations. • Seen two examples of Monte Carlo samplers for estimating expected values. Published on February 11th 2020. Last updated on March 1st 2023, 10:53. Fredrik Olsson. Data Scientist.
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CMU 15-112: Fundamentals of Programming and Computer Science Writing-Session5 Practice
• These exercises will help you prepare for writing-session5, which is on Fri 27-Sep, and which will contain a randomly-chosen subset of exercises from among these.
• Unlike the hw, you may work collaboratively on these practice exercises.
• That said, during the actual writing session on Friday you must work alone, and without any notes or access to the web or other resources.
• To start:
1. Create a folder named 'writing_session5_practice'
3. Edit writing_session5_practice.py
• Do not use recursion this week.
• Do not hardcode the test cases in your solutions.
Important note: As with last week, and from now on, writing sessions will also contain very short (fill-in-the-blank or super-short code tracing) questions covering this week's course notes. Be sure to study the course notes and know them well before attending Friday's writing session! Also, be sure to look at the practice problems at the end of this page!
1. removeRowAndCol (non-destructive and destructive)
Here we will write removeRowAndCol twice -- once non-destructively, and then again destructively. Note that neither of these versions may call nor essentially duplicate the other version. So in particular, your nondestructive version may not do this:
``` L = copy.deepcopy(L)
doDestructiveVersion(L)
return L
```
Instead, do not use copy.deepcopy and directly construct the modified 2d list.
Both functions take a rectangular list L and two ints, row and col. In both cases, the goal is to obtain a version of the list that has the given row and given column removed. You may assume that row and col are both legal values (that is, they are non-negative integers that are smaller than the largest row and column, respectively). For example, the list shown to the left would lead to the result shown on the right when called with the row 1 and the column 2.
list result ```[ [ 2, 3, 4, 5], [ 8, 7, 6, 5], [ 0, 1, 2, 3] ] ``` ```[ [ 2, 3, 5], [ 0, 1, 3] ] ```
nondestructiveRemoveRowAndCol(L, row, col): the non-destructive version should return a new list, and should not modify the provided list at all.
destructiveRemoveRowAndCol(L, row, col): the destructive version should modify the original list, and should not return anything (that is, None).
Hint: writing test functions for non-destructive and destructive functions is a little different from writing ordinary test functions. Here is an example of a test case for a non-destructive function:
# This is a test case for a non-destructive function. # The input list and output list L = [ [ 2, 3, 4, 5], [ 8, 7, 6, 5], [ 0, 1, 2, 3] ] result = [ [ 2, 3, 5], [ 0, 1, 3] ] # Copy the input list so we can check it later import copy lCopy = copy.deepcopy(L) # The first assert is an ordinary test; the second is a non-destructive test assert(nondestructiveRemoveRowAndCol(L, 1, 2) == result) assert(L == lCopy) # input list should not be changed
And here is an example of a test case for a destructive function:
# This is a test case for a destructive function. # The input list and output list L = [ [ 2, 3, 4, 5], [ 8, 7, 6, 5], [ 0, 1, 2, 3] ] result = [ [ 2, 3, 5], [ 0, 1, 3] ] # The first test is an ordinary test; the second is a destructive test assert(destructiveRemoveRowAndCol(L, 1, 2) == None) assert(L == result) # input list should be changed
2. bestQuiz(a)
Write the function bestQuiz(a), which takes a rectangular 2d list of numbers that represents a gradebook, where each column represents a quiz, and each row represents a student, and each value represents that student's score on that quiz (except -1 indicates the student did not take the quiz). For example:
``` a = [ [ 88, 80, 91 ],
[ 68, 100, -1 ]
]
```
This list indicates that student0 scored 88 on quiz0, 80 on quiz1, and 91 on quiz2. Also, student1 scored 68 on quiz0, 100 on quiz1, and did not take quiz2. The function returns the quiz with the highest average. In this case, quiz0 average is 78, quiz1 average is 90, and quiz2 average is 91 (since we ignore the -1). Thus, quiz2 is the best, and so the function returns 2 in this case. You are not responsible for malformed input, except you should return None if there are no quizzes. Also, resolve ties in favor of the lower quiz number. Here is a test function for you:
def testBestQuiz(): print('Testing bestQuiz()...', end='') a = [ [ 88, 80, 91 ], [ 68, 100, -1 ]] assert(bestQuiz(a) == 2) a = [ [ 88, 80, 80 ], [ 68, 100, 100 ]] assert(bestQuiz(a) == 1) a = [ [88, -1, -1 ], [68, -1, -1 ]] assert(bestQuiz(a) == 0) a = [ [-1, -1, -1 ], [-1, -1, -1 ]] assert(bestQuiz(a) == None) print('Passed!')
Background: we can think of a 2d list in Python as a matrix in math. To add two matrices, L and M, they must have the same dimensions. Then, we loop over each row and col, and the result[row][col] is just the sum of L[row][col] and M[row][col]. For example:
L = [ [1, 2, 3], [4, 5, 6] ] M = [ [21, 22, 23], [24, 25, 26]] N = [ [1+21, 2+22, 3+23], [4+24, 5+25, 6+26]] assert(matrixAdd(L, M) == N)
With this in mind, write the function matrixAdd(L, M) that takes two rectangular 2d lists (that we will consider to be matrices) that you may assume only contain numbers, and returns a new 2d list that is the result of adding the two matrices. Return None if the two matrices cannot be added because they are of different dimensions.
4. isMostlyMagicSquare(a)
Write the function isMostlyMagicSquare(a) that takes an 2d list of numbers, which you may assume is an NxN square with N>0, and returns True if it is "mostly magic" and False otherwise, where a square is "mostly magic" if:
• Each row, each column, and each of the 2 diagonals each sum to the same total.
A completely magic square has additional restrictions (such as not allowing duplicates, and only allowing numbers from 1 to N2), which we are not enforcing here, but which you can read about here. Note: any magic square is also a "mostly magic" square, including this sample magic square:
Here is another mostly-magic square:
```[ [ 42 ]]
```
That square is 1x1 and each row, column, and diagonal sums to 42! And finally, here is a not-mostly-magic square:
```[ [ 1, 2],
[ 2, 1]]
```
Each row and each column add to 3, but one diagonal adds to 2 and the other to 4.
5. DataTable and DataColumn classes
This exercise converts a csv string (a multi-line string of comma-separated values) into a table, and then allows us to extract individual columns to do some data analysis (here, just taking the average for now).
Note: you may assume:
• the table is nonempty (so it has at least one row and at least one column)
• the table is rectangular (so each row has the same number of columns)
• the first row contains the column labels, which are strings
• the first column contains the row labels, which are strings
• all other columns contain data, which are ints, and legally formatted
Also:
• ignore empty rows, and ignore leading and trailing whitespace on any row
Here is an example of such data:
csvData = ''' Name,Hw1,Hw2,Quiz1,Quiz2 Fred,94,88,82,92 Wilma,98,80,80,100 '''
With this in mind, write the class DataTable and also the class DataColumn so the following test function passes (without hardcoding any test cases):
def testDataTableAndDataColumnClasses(): print('Testing DataTable and DataColumn classes...', end='') csvData = ''' Name,Hw1,Hw2,Quiz1,Quiz2 Fred,94,88,82,92 Wilma,98,80,80,100 ''' dataTable = DataTable(csvData) rows, cols = dataTable.getDims() assert((rows == 3) and (cols == 5)) column3 = dataTable.getColumn(3) assert(isinstance(column3, DataColumn)) assert(column3.label == 'Quiz1') assert(column3.data == [82, 80]) assert(almostEqual(column3.average(), 81)) column4 = dataTable.getColumn(4) assert(isinstance(column4, DataColumn)) assert(column4.label == 'Quiz2') assert(column4.data == [92, 100]) assert(almostEqual(column4.average(), 96)) print('Passed!')
This week's writing session will start with a physical sheet of paper containing some subset of these questions, or questions nearly identical to these questions.
```1. Which of the following correctly creates a 3x2 (3 rows, 2 columns) 2d list
without any aliases?
A) [ [0] * 2 ] * 3
B) [ [0] * 3 ] * 2
C) [ ([0] * 2) for i in range(3) ]
D) [ ([0] * 3) for i in range(2) ]
E) None of the above
2. If we had a non-ragged 3d list L (so "cubic" instead of "rectangular"),
which of the following would return all 3 dimensions:
A) len(L), len(L[0]), len(L[0,0])
B) len(L), len(L[0]), len(L[0][0])
C) len(L), len(L[0]), len(L[-1])
D) len(L[0]), len(L[0][0]), len(L[0][0][0])
E) None of the above
3. Why should we use use copy.deepcopy instead of copy.copy on 2d lists?
A) copy.copy has a bug and crashes on 2d lists
B) copy.copy creates a shallow copy with aliases to the inner lists
C) copy.copy is much slower than copy.deepcopy
D) copy.copy works the same as copy.deepcopy on 2d lists
E) None of the above
4. Why did we provide our own myDeepCopy in the notes?
A) copy.deepcopy is very slow, ours is much faster
B) copy.deepcopy is not available on some versions of Python
C) copy.deepcopy will preserve existing aliases in a 2d list, ours will not
D) copy.deepcopy is not a preferred way to copy a 2d list
E) None of the above
5. Why did we provide our own print2dList in the notes?
A) The builtin print2dList function works poorly in some cases
B) There is no builtin print2dList function and just printing a 2d list
makes it appear on a single line and also does not align the columns,
and thus makes it hard to read.
6. What happens if we reverse the order of the 'for' loops in the standard
way we loop over a 2d list L. That is, what if instead of this:
for row in range(rows):
for col in range(cols):
we did this:
for col in range(cols):
for row in range(rows):
A) We would skip some of the values in L
B) We would visit all the values in row 0 first, then row 1, and so on
C) We would visit all the values in col 0 first, then col 1, and so on
D) This code will not run at all
E) None of the above
7. Which of the following does NOT set M to a list of the values in column c
of the 10x10 2d list L?
A) M = [ L[i][c] for i in range(10) ]
B) N, M = copy.deepcopy(L), [ ]
while (N != [ ]):
M.append(N.pop(0)[c])
C) M = [ ]
for i in range(10):
M += [ L[c][i] ]
D) None of the above
```
7. Code Tracing Practice Problems
Same note as above: This week's writing session will start with a physical sheet of paper containing some subset of these questions, or questions nearly identical to these questions.
```
import copy
def ct():
# fill in each space to the right of each print statement
# with the value it prints
a = [[1]]
b = copy.copy(a)
c = copy.deepcopy(a)
b.append(2)
c.append([3])
print(a, b, c) # __________________________________
a = [[4,5]] * 2
b = copy.deepcopy(a)
a[1][1] = 6
print(a,b) # __________________________________
a = [[1,2],[3,4]]
b = a[::-1]
c = [r[::-1] for r in a]
print(b, c) # __________________________________
a = [[1,2,3],[4,5,6]]
for c in range(len(a[0])):
for r in range(c):
a[r][c] *= 10
print(a) # __________________________________
a = [[1,2],[3],[4,5,6]]
b = [ ]
for r in range(len(a)):
b += [len(a[r])]
print(b) # __________________________________
ct()
```
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CMU 15-112: Fundamentals of Programming and Computer Science Writing-Session5 Practice. • These exercises will help you prepare for writing-session5, which is on Fri 27-Sep, and which will contain a randomly-chosen subset of exercises from among these.. • Unlike the hw, you may work collaboratively on these practice exercises.. • That said, during the actual writing session on Friday you must work alone, and without any notes or access to the web or other resources.. • To start:. 1. Create a folder named 'writing_session5_practice'. 3. Edit writing_session5_practice.py. • Do not use recursion this week.. • Do not hardcode the test cases in your solutions.. Important note: As with last week, and from now on, writing sessions will also contain very short (fill-in-the-blank or super-short code tracing) questions covering this week's course notes. Be sure to study the course notes and know them well before attending Friday's writing session! Also, be sure to look at the practice problems at the end of this page!. 1. removeRowAndCol (non-destructive and destructive). Here we will write removeRowAndCol twice -- once non-destructively, and then again destructively. Note that neither of these versions may call nor essentially duplicate the other version. So in particular, your nondestructive version may not do this:. ``` L = copy.deepcopy(L). doDestructiveVersion(L). return L. ```. Instead, do not use copy.deepcopy and directly construct the modified 2d list.. Both functions take a rectangular list L and two ints, row and col. In both cases, the goal is to obtain a version of the list that has the given row and given column removed. You may assume that row and col are both legal values (that is, they are non-negative integers that are smaller than the largest row and column, respectively). For example, the list shown to the left would lead to the result shown on the right when called with the row 1 and the column 2.. list result ```[ [ 2, 3, 4, 5], [ 8, 7, 6, 5], [ 0, 1, 2, 3] ] ``` ```[ [ 2, 3, 5], [ 0, 1, 3] ] ```. nondestructiveRemoveRowAndCol(L, row, col): the non-destructive version should return a new list, and should not modify the provided list at all.. destructiveRemoveRowAndCol(L, row, col): the destructive version should modify the original list, and should not return anything (that is, None).. Hint: writing test functions for non-destructive and destructive functions is a little different from writing ordinary test functions. Here is an example of a test case for a non-destructive function:. # This is a test case for a non-destructive function. # The input list and output list L = [ [ 2, 3, 4, 5], [ 8, 7, 6, 5], [ 0, 1, 2, 3] ] result = [ [ 2, 3, 5], [ 0, 1, 3] ] # Copy the input list so we can check it later import copy lCopy = copy.deepcopy(L) # The first assert is an ordinary test; the second is a non-destructive test assert(nondestructiveRemoveRowAndCol(L, 1, 2) == result) assert(L == lCopy) # input list should not be changed. And here is an example of a test case for a destructive function:. # This is a test case for a destructive function. # The input list and output list L = [ [ 2, 3, 4, 5], [ 8, 7, 6, 5], [ 0, 1, 2, 3] ] result = [ [ 2, 3, 5], [ 0, 1, 3] ] # The first test is an ordinary test; the second is a destructive test assert(destructiveRemoveRowAndCol(L, 1, 2) == None) assert(L == result) # input list should be changed. 2. bestQuiz(a). Write the function bestQuiz(a), which takes a rectangular 2d list of numbers that represents a gradebook, where each column represents a quiz, and each row represents a student, and each value represents that student's score on that quiz (except -1 indicates the student did not take the quiz). For example:. ``` a = [ [ 88, 80, 91 ],. [ 68, 100, -1 ]. ]. ```. This list indicates that student0 scored 88 on quiz0, 80 on quiz1, and 91 on quiz2. Also, student1 scored 68 on quiz0, 100 on quiz1, and did not take quiz2. The function returns the quiz with the highest average. In this case, quiz0 average is 78, quiz1 average is 90, and quiz2 average is 91 (since we ignore the -1). Thus, quiz2 is the best, and so the function returns 2 in this case. You are not responsible for malformed input, except you should return None if there are no quizzes. Also, resolve ties in favor of the lower quiz number. Here is a test function for you:. def testBestQuiz(): print('Testing bestQuiz()...', end='') a = [ [ 88, 80, 91 ], [ 68, 100, -1 ]] assert(bestQuiz(a) == 2) a = [ [ 88, 80, 80 ], [ 68, 100, 100 ]] assert(bestQuiz(a) == 1) a = [ [88, -1, -1 ], [68, -1, -1 ]] assert(bestQuiz(a) == 0) a = [ [-1, -1, -1 ], [-1, -1, -1 ]] assert(bestQuiz(a) == None) print('Passed!'). Background: we can think of a 2d list in Python as a matrix in math. To add two matrices, L and M, they must have the same dimensions. Then, we loop over each row and col, and the result[row][col] is just the sum of L[row][col] and M[row][col]. For example:. L = [ [1, 2, 3], [4, 5, 6] ] M = [ [21, 22, 23], [24, 25, 26]] N = [ [1+21, 2+22, 3+23], [4+24, 5+25, 6+26]] assert(matrixAdd(L, M) == N). With this in mind, write the function matrixAdd(L, M) that takes two rectangular 2d lists (that we will consider to be matrices) that you may assume only contain numbers, and returns a new 2d list that is the result of adding the two matrices. Return None if the two matrices cannot be added because they are of different dimensions.. 4. isMostlyMagicSquare(a). Write the function isMostlyMagicSquare(a) that takes an 2d list of numbers, which you may assume is an NxN square with N>0, and returns True if it is "mostly magic" and False otherwise, where a square is "mostly magic" if:. • Each row, each column, and each of the 2 diagonals each sum to the same total.. A completely magic square has additional restrictions (such as not allowing duplicates, and only allowing numbers from 1 to N2), which we are not enforcing here, but which you can read about here. Note: any magic square is also a "mostly magic" square, including this sample magic square:. Here is another mostly-magic square:. ```[ [ 42 ]]. ```. That square is 1x1 and each row, column, and diagonal sums to 42! And finally, here is a not-mostly-magic square:. ```[ [ 1, 2],. [ 2, 1]]. ```. Each row and each column add to 3, but one diagonal adds to 2 and the other to 4.. 5. DataTable and DataColumn classes. This exercise converts a csv string (a multi-line string of comma-separated values) into a table, and then allows us to extract individual columns to do some data analysis (here, just taking the average for now).. Note: you may assume:. • the table is nonempty (so it has at least one row and at least one column). • the table is rectangular (so each row has the same number of columns). • the first row contains the column labels, which are strings. • the first column contains the row labels, which are strings. • all other columns contain data, which are ints, and legally formatted. Also:. • ignore empty rows, and ignore leading and trailing whitespace on any row. Here is an example of such data:. csvData = ''' Name,Hw1,Hw2,Quiz1,Quiz2 Fred,94,88,82,92 Wilma,98,80,80,100 '''. With this in mind, write the class DataTable and also the class DataColumn so the following test function passes (without hardcoding any test cases):. def testDataTableAndDataColumnClasses(): print('Testing DataTable and DataColumn classes...', end='') csvData = ''' Name,Hw1,Hw2,Quiz1,Quiz2 Fred,94,88,82,92 Wilma,98,80,80,100 ''' dataTable = DataTable(csvData) rows, cols = dataTable.getDims() assert((rows == 3) and (cols == 5)) column3 = dataTable.getColumn(3) assert(isinstance(column3, DataColumn)) assert(column3.label == 'Quiz1') assert(column3.data == [82, 80]) assert(almostEqual(column3.average(), 81)) column4 = dataTable.getColumn(4) assert(isinstance(column4, DataColumn)) assert(column4.label == 'Quiz2') assert(column4.data == [92, 100]) assert(almostEqual(column4.average(), 96)) print('Passed!'). This week's writing session will start with a physical sheet of paper containing some subset of these questions, or questions nearly identical to these questions.. ```1. Which of the following correctly creates a 3x2 (3 rows, 2 columns) 2d list. without any aliases?. A) [ [0] * 2 ] * 3. B) [ [0] * 3 ] * 2.
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C) [ ([0] * 2) for i in range(3) ]. D) [ ([0] * 3) for i in range(2) ]. E) None of the above. 2. If we had a non-ragged 3d list L (so "cubic" instead of "rectangular"),. which of the following would return all 3 dimensions:. A) len(L), len(L[0]), len(L[0,0]). B) len(L), len(L[0]), len(L[0][0]). C) len(L), len(L[0]), len(L[-1]). D) len(L[0]), len(L[0][0]), len(L[0][0][0]). E) None of the above. 3. Why should we use use copy.deepcopy instead of copy.copy on 2d lists?. A) copy.copy has a bug and crashes on 2d lists. B) copy.copy creates a shallow copy with aliases to the inner lists. C) copy.copy is much slower than copy.deepcopy. D) copy.copy works the same as copy.deepcopy on 2d lists. E) None of the above. 4. Why did we provide our own myDeepCopy in the notes?. A) copy.deepcopy is very slow, ours is much faster. B) copy.deepcopy is not available on some versions of Python. C) copy.deepcopy will preserve existing aliases in a 2d list, ours will not. D) copy.deepcopy is not a preferred way to copy a 2d list. E) None of the above. 5. Why did we provide our own print2dList in the notes?. A) The builtin print2dList function works poorly in some cases. B) There is no builtin print2dList function and just printing a 2d list. makes it appear on a single line and also does not align the columns,. and thus makes it hard to read.. 6. What happens if we reverse the order of the 'for' loops in the standard. way we loop over a 2d list L. That is, what if instead of this:. for row in range(rows):. for col in range(cols):. we did this:. for col in range(cols):. for row in range(rows):. A) We would skip some of the values in L. B) We would visit all the values in row 0 first, then row 1, and so on. C) We would visit all the values in col 0 first, then col 1, and so on. D) This code will not run at all. E) None of the above. 7. Which of the following does NOT set M to a list of the values in column c. of the 10x10 2d list L?. A) M = [ L[i][c] for i in range(10) ]. B) N, M = copy.deepcopy(L), [ ]. while (N != [ ]):. M.append(N.pop(0)[c]). C) M = [ ]. for i in range(10):. M += [ L[c][i] ]. D) None of the above. ```. 7. Code Tracing Practice Problems. Same note as above: This week's writing session will start with a physical sheet of paper containing some subset of these questions, or questions nearly identical to these questions.. ```. import copy. def ct():. # fill in each space to the right of each print statement. # with the value it prints. a = [[1]]. b = copy.copy(a). c = copy.deepcopy(a). b.append(2). c.append([3]). print(a, b, c) # __________________________________. a = [[4,5]] * 2. b = copy.deepcopy(a). a[1][1] = 6. print(a,b) # __________________________________. a = [[1,2],[3,4]]. b = a[::-1]. c = [r[::-1] for r in a]. print(b, c) # __________________________________. a = [[1,2,3],[4,5,6]]. for c in range(len(a[0])):. for r in range(c):. a[r][c] *= 10. print(a) # __________________________________. a = [[1,2],[3],[4,5,6]]. b = [ ]. for r in range(len(a)):. b += [len(a[r])]. print(b) # __________________________________. ct(). ```.
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+0
# Help.
0
317
1
+4116
Help.
Jan 29, 2018
#1
+111331
+2
√ [ 3 x^12 y^10 ]
_____________
√ [5 x^6 y ^3 ]
We are using the property that a^m / a^n = a^(m - n)
√ [ 3 x^(12 - 6) y ^ (10 - 3) ]
_______________________
√ 5
√ [ 3 x^6 y^7]
___________
√5
Now...to determine how many powers to take out and how many to leave in :
Take 6 / 2 = 3 powers of x come out
And 7/2 = 3 + 1/2 = 3 powers of y come out and 1 stays in
So we have
x^3 y^3 √ [ 3y ]
____________ rationalize the denominator by multiplying op/bottom by √5
√5
x^3 y^3 √ [ 3y ] √5
_______________
√ 5 √ 5
x^3 y^3 √ [ 15 y ]
_______________ the second answer is correct
5
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+0. # Help. . 0. 317. 1. +4116. Help.. Jan 29, 2018. #1. +111331. +2. √ [ 3 x^12 y^10 ]. _____________. √ [5 x^6 y ^3 ]. We are using the property that a^m / a^n = a^(m - n). √ [ 3 x^(12 - 6) y ^ (10 - 3) ]. _______________________. √ 5.
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√ [ 3 x^6 y^7]. ___________. √5. Now...to determine how many powers to take out and how many to leave in :. Take 6 / 2 = 3 powers of x come out. And 7/2 = 3 + 1/2 = 3 powers of y come out and 1 stays in. So we have. x^3 y^3 √ [ 3y ]. ____________ rationalize the denominator by multiplying op/bottom by √5. √5. x^3 y^3 √ [ 3y ] √5. _______________. √ 5 √ 5. x^3 y^3 √ [ 15 y ]. _______________ the second answer is correct. 5. Jan 29, 2018.
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https://proofwiki.org/wiki/Form_of_Geometric_Sequence_of_Integers/Corollary
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# Form of Geometric Sequence of Integers/Corollary
## Corollary to Form of Geometric Sequence of Integers
Let $p$ and $q$ be integers.
Then the finite sequence $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ defined as:
$a_j = p^j q^{n - j}$
is a geometric sequence whose common ratio is $\dfrac p q$.
## Proof
Let the greatest common divisor of $p$ and $q$ be $d$.
$p = d r$
$q = d s$
where $r$ and $s$ are coprime integers.
Thus:
$a_j = p^j q^{n - j}$
$\ds a_j$ $=$ $\ds p^j q^{n - j}$ $\ds$ $=$ $\ds \paren {d r}^j \paren {d s}^{n - j}$ $\ds$ $=$ $\ds d^n r^j s^{n - j}$
and so by Form of Geometric Sequence of Integers it follows that $P$ is a geometric sequence whose common ratio is $\dfrac r s$.
Then:
$\ds \dfrac r s$ $=$ $\ds \paren {\dfrac p d} / \paren {\dfrac q d}$ $\ds$ $=$ $\ds \dfrac p d \dfrac d q$ $\ds$ $=$ $\ds \dfrac p q$
Hence the result.
$\blacksquare$
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# Form of Geometric Sequence of Integers/Corollary. ## Corollary to Form of Geometric Sequence of Integers. Let $p$ and $q$ be integers.. Then the finite sequence $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ defined as:. $a_j = p^j q^{n - j}$. is a geometric sequence whose common ratio is $\dfrac p q$.. ## Proof. Let the greatest common divisor of $p$ and $q$ be $d$.. $p = d r$. $q = d s$.
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where $r$ and $s$ are coprime integers.. Thus:. $a_j = p^j q^{n - j}$. $\ds a_j$ $=$ $\ds p^j q^{n - j}$ $\ds$ $=$ $\ds \paren {d r}^j \paren {d s}^{n - j}$ $\ds$ $=$ $\ds d^n r^j s^{n - j}$. and so by Form of Geometric Sequence of Integers it follows that $P$ is a geometric sequence whose common ratio is $\dfrac r s$.. Then:. $\ds \dfrac r s$ $=$ $\ds \paren {\dfrac p d} / \paren {\dfrac q d}$ $\ds$ $=$ $\ds \dfrac p d \dfrac d q$ $\ds$ $=$ $\ds \dfrac p q$. Hence the result.. $\blacksquare$.
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# math
posted by on .
solve for X ?
1) 8-3x < 14
-3x/-3 < 6/-3
x=-2
2) 7x -3 > 4x +3
3x -3> 3
3x/3 >6/3
x>2
3) x-5<3 and 2x > -8
x<8 x>-4
4) x+ 7 <3 or 5x > 30
x<-4 x> 6
are they correct..
• math - ,
1. The solution is an inequality. Also when you multiply or divide by a negative number, the inequality is reversed.
from you step:
-3x/-3 < 6/-3
Since you are dividing both sides by a negative number the inequality is flipped:
x > -2
2. OK
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# math. posted by on .. solve for X ?. 1) 8-3x < 14. -3x/-3 < 6/-3. x=-2. 2) 7x -3 > 4x +3. 3x -3> 3. 3x/3 >6/3. x>2. 3) x-5<3 and 2x > -8. x<8 x>-4. 4) x+ 7 <3 or 5x > 30.
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x<-4 x> 6. are they correct... • math - ,. 1. The solution is an inequality. Also when you multiply or divide by a negative number, the inequality is reversed.. from you step:. -3x/-3 < 6/-3. Since you are dividing both sides by a negative number the inequality is flipped:. x > -2. 2. OK.
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Divide 36 by 4 to find the quotient.
Sure! Here's a brief introduction for your blog post:
Welcome to Warren Institute! Today, we'll explore the fundamental concept of division as we answer the question: How many times does 4 go into 36? Understanding division is crucial in building a strong foundation in mathematics, and we'll dive deep into this problem to uncover the principles behind it. Through clear explanations and illustrative examples, we aim to demystify the process of division and equip our readers with the confidence to tackle similar challenges. Join us on this mathematical journey as we unravel the mysteries of division!
Understanding Division
Division is a fundamental concept in mathematics and is essential for solving real-world problems. It involves separating a quantity into equal parts or groups. When discussing "how many times does 4 go into 36," it's important to understand the basics of division and its application in various scenarios.
Applying the Division Algorithm
The division algorithm states that for any two integers a and b, with b ≠ 0, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < |b|. This algorithm is crucial in determining how many times a number can be divided into another and the remainder left over.
Strategies for Division
When addressing "how many times does 4 go into 36," students can utilize different strategies such as long division, repeated subtraction, or using manipulatives to understand the concept visually. Each strategy offers a unique approach to solving the division problem and helps reinforce the understanding of division as repeated subtraction.
Real-life Applications of Division
Understanding how many times 4 goes into 36 extends beyond the classroom and into real-world applications. Whether it's dividing resources, sharing items equally, or calculating rates and proportions, division plays a vital role in everyday life. By exploring practical examples, students can see the relevance and significance of mastering division skills.
How can I teach students to divide 36 by 4 using different strategies in a mathematics education setting?
You can teach students to divide 36 by 4 using different strategies such as long division, repeated subtraction, and grouping.
What are some effective ways to help students understand the concept of division when they are learning how many times 4 goes into 36?
One effective way to help students understand the concept of division when learning how many times 4 goes into 36 is to use manipulatives and visual aids to demonstrate the division process. This can include using blocks, counters, or pictures to show how 36 can be divided into 4 equal groups.
How can I incorporate real-life examples or manipulatives to help students grasp the concept of dividing 36 by 4 in a mathematics education lesson?
You can use real-life examples such as sharing 36 cookies among 4 friends or dividing a 36-piece puzzle into 4 equal parts. Additionally, you can use manipulatives like counters, base-10 blocks, or fraction strips to visually demonstrate the division process.
What misconceptions or challenges might students face when learning how many times 4 goes into 36, and how can educators address these in a mathematics education context?
One common misconception is that students might struggle with the concept of division and understanding how many times 4 goes into 36. They may also face challenges in memorizing multiplication facts. Educators can address these by providing visual aids, real-life examples, and hands-on activities to help students understand the concept of division and reinforce multiplication skills. Additionally, breaking down the problem into smaller steps and using manipulatives can support student comprehension.
Are there any engaging activities or games that can be used to reinforce the understanding of dividing 36 by 4 in a mathematics education classroom?
Yes, math bingo or division board games can be engaging activities to reinforce the understanding of dividing 36 by 4 in a mathematics education classroom.
In conclusion, understanding the concept of division and how many times 4 goes into 36 is crucial for building a strong foundation in mathematics education. By mastering this fundamental skill, students can develop their problem-solving abilities and pave the way for more advanced mathematical concepts. It is essential for educators to provide ample opportunities for students to practice and grasp the concept of division to ensure their success in mathematics education.
See also Understanding Molecular Orbital Theory: Bonding & Antibonding MOs and Bond Order
If you want to know other articles similar to Divide 36 by 4 to find the quotient. you can visit the category General Education.
Michaell Miller
Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all.
Go up
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Divide 36 by 4 to find the quotient.. Sure! Here's a brief introduction for your blog post:. Welcome to Warren Institute! Today, we'll explore the fundamental concept of division as we answer the question: How many times does 4 go into 36? Understanding division is crucial in building a strong foundation in mathematics, and we'll dive deep into this problem to uncover the principles behind it. Through clear explanations and illustrative examples, we aim to demystify the process of division and equip our readers with the confidence to tackle similar challenges. Join us on this mathematical journey as we unravel the mysteries of division!. Understanding Division. Division is a fundamental concept in mathematics and is essential for solving real-world problems. It involves separating a quantity into equal parts or groups. When discussing "how many times does 4 go into 36," it's important to understand the basics of division and its application in various scenarios.. Applying the Division Algorithm. The division algorithm states that for any two integers a and b, with b ≠ 0, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < |b|. This algorithm is crucial in determining how many times a number can be divided into another and the remainder left over.. Strategies for Division. When addressing "how many times does 4 go into 36," students can utilize different strategies such as long division, repeated subtraction, or using manipulatives to understand the concept visually. Each strategy offers a unique approach to solving the division problem and helps reinforce the understanding of division as repeated subtraction.. Real-life Applications of Division. Understanding how many times 4 goes into 36 extends beyond the classroom and into real-world applications. Whether it's dividing resources, sharing items equally, or calculating rates and proportions, division plays a vital role in everyday life. By exploring practical examples, students can see the relevance and significance of mastering division skills.. How can I teach students to divide 36 by 4 using different strategies in a mathematics education setting?. You can teach students to divide 36 by 4 using different strategies such as long division, repeated subtraction, and grouping.. What are some effective ways to help students understand the concept of division when they are learning how many times 4 goes into 36?. One effective way to help students understand the concept of division when learning how many times 4 goes into 36 is to use manipulatives and visual aids to demonstrate the division process. This can include using blocks, counters, or pictures to show how 36 can be divided into 4 equal groups.
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How can I incorporate real-life examples or manipulatives to help students grasp the concept of dividing 36 by 4 in a mathematics education lesson?. You can use real-life examples such as sharing 36 cookies among 4 friends or dividing a 36-piece puzzle into 4 equal parts. Additionally, you can use manipulatives like counters, base-10 blocks, or fraction strips to visually demonstrate the division process.. What misconceptions or challenges might students face when learning how many times 4 goes into 36, and how can educators address these in a mathematics education context?. One common misconception is that students might struggle with the concept of division and understanding how many times 4 goes into 36. They may also face challenges in memorizing multiplication facts. Educators can address these by providing visual aids, real-life examples, and hands-on activities to help students understand the concept of division and reinforce multiplication skills. Additionally, breaking down the problem into smaller steps and using manipulatives can support student comprehension.. Are there any engaging activities or games that can be used to reinforce the understanding of dividing 36 by 4 in a mathematics education classroom?. Yes, math bingo or division board games can be engaging activities to reinforce the understanding of dividing 36 by 4 in a mathematics education classroom.. In conclusion, understanding the concept of division and how many times 4 goes into 36 is crucial for building a strong foundation in mathematics education. By mastering this fundamental skill, students can develop their problem-solving abilities and pave the way for more advanced mathematical concepts. It is essential for educators to provide ample opportunities for students to practice and grasp the concept of division to ensure their success in mathematics education.. See also Understanding Molecular Orbital Theory: Bonding & Antibonding MOs and Bond Order. If you want to know other articles similar to Divide 36 by 4 to find the quotient. you can visit the category General Education.. Michaell Miller. Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all.. Go up.
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https://stats.stackexchange.com/questions/544053/why-i-cannot-generate-random-numbers-having-a-truncated-lognormal-distribution
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# Why I cannot generate random numbers having a truncated lognormal distribution?
My deduction is:
When the distribution is truncated, a normalization factor should be introduced: $$$$g(x) = \frac{C}{x\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{\ln{x}-\mu}{\sigma}\right)^2},$$$$ where $$C$$ is the factor. Integrating $$g(x)$$ over the interval $$[x_1, x_2]$$ ($$x_1 \geq 0$$), we have $$$$\frac{C}{\sigma\sqrt{2\pi}}\int_{x_1}^{x_2}\frac{1}{x}e^{-\frac{1}{2}\left(\frac{\ln{x}-\mu}{\sigma}\right)^2}\mathrm{d}x = 1. \label{CDF_truncated}$$$$ To determine $$C$$, let $$t = \frac{\ln{x}-\mu}{\sigma}$$, so $$x = e^{\sigma t + \mu}$$ and $$\mathrm{d}x = \sigma e^{\sigma t + \mu}\mathrm{d}t$$. The interval becomes $$\left[t_1, t_2\right]$$ which is given by $$$$t_1 = \frac{\ln{x_1}-\mu}{\sigma} \text{ and } t_2 = \frac{\ln{x_2}-\mu}{\sigma}. \notag$$$$ Hence, the inregration can be rewritten as $$$$\frac{C}{\sigma\sqrt{2\pi}}\int_{t_1}^{t_2} \sigma\cdot e^\mu \cdot e^{-\sigma t - \mu}\cdot e^{-\frac{t^2}{2} + \sigma t}\mathrm{d}t = 1, \notag$$$$ $$$$\Rightarrow \frac{C}{\sqrt{2\pi}}\int_{t_1}^{t_2} e^{-\frac{t^2}{2}}\mathrm{d}t = 1. \notag$$$$
Next, let $$m = \frac{t}{\sqrt{2}}$$ and $$t = \sqrt{2}m$$. So, $$\mathrm{d}t=\sqrt{2}\mathrm{d}m$$. The Eintegration can be further simplified into $$$$\frac{C}{\sqrt{2\pi}}\int_{m_1}^{m_2} \sqrt{2}e^{-m^2}\mathrm{d}m = 1, \notag$$$$ $$$$\Rightarrow \frac{C}{\sqrt{\pi}}\int_{m_1}^{m_2} e^{-m^2}\mathrm{d}m = 1. \notag$$$$. So, the $$C$$ is $$$$C = \frac{-2}{[{erf}(m_1)-{erf}(m_2)]},$$$$ where $${erf}(\cdot)$$ is the error function. If $$x_1 = 0 \text{ and } x_2 = +\infty$$, then $$t_1 = -\infty \text{ and } t_2 = +\infty$$, and $$m_1 = -\infty \text{ and } m_2 = +\infty$$, and $${erf}(m_1) \text{ and } {erf}(m_1)$$ are -1 and 1, respectively, $$C = 1$$ which leads to a regular non-truncated PDF.
Now, let us generate random log-normal variables from the standard uniform distributions. $$$$p = C\int_{x_1}^{x}\frac{1}{x\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{\ln{x}-\mu}{\sigma}\right)^2}\mathrm{d}x.$$$$ where $$p$$ is a random number within [0, 1]. In a similar way, $$$$p = \frac{C}{ \sqrt{\pi}}\int_{m_1}^{m}e^{-m^2}\mathrm{d}m, \notag$$$$
$$$$\Rightarrow m = {erfinv}\left({{erf}(m_1) - \frac{2 p }{C}}\right), \notag$$$$ and $$m = \frac{\frac{\ln{x}-\mu}{\sigma}}{\sqrt{2}}$$, $$erfinv(\cdot)$$ is the inverse of the error function.
I write the following MATLAB script to generate random lognormal variables, with given expection and variance. But I got NaN.
clc;
close all;
clear all;
p = rand;
mean = 6; %given expection
variance = 0.5; % given variance
x1 = 1.5;
x2 = 15;
mu_1 = log(mean * mean / ((variance + mean * mean)^0.5)); %determine mean
sigma_1 = (log(1 + (variance) / (mean * mean)))^0.5; %determine std. deviation
t1 = (log(x1) - mu_1) / sigma_1;
t2 = (log(x2) - mu_1) / sigma_1;
m1 = t1 / (2^0.5);
m2 = t2 / (2^0.5);
C = -2 / (erf(m1) - erf(m2));
m = erfinv(erf(m1) - 2 * p / C);
m =
NaN
what's wrong with my deduction? or my code?
Update------------------------------------------
according to the comments of @jbowman, I write the following MATLAB code:
clc;
close all;
clear all;
mean = 6; %given expection
variance = 9; % given variance
x1 = 1.5;
x2 = 15;
mu_1 = log(mean * mean / ((variance + mean * mean)^0.5)); %determine mean
sigma_1 = (log(1 + (variance) / (mean * mean)))^0.5; %determine std. deviation
% then, convert normal to std. normal
% Any point (x) from a normal distribution can be converted to the standard normal distribution (z) with the formula z = (x-mean) / standard deviation
l = (log(x1) - mu_1) / sigma_1;
u = (log(x2) - mu_1) / sigma_1;
pd = makedist('Normal', 'mu', 0, 'sigma', 1);
p_l = cdf(pd, l);
p_u = cdf(pd, u);
array_rand_num = [];
nf = 1000;
for i = 1:1:nf
x = rand;
x_prime = p_l + (p_u - p_l) * x;
z = icdf(pd, x_prime);
p = exp(z * sigma_1 + mu_1);
array_rand_num(i) = p;
end
figure(1)
nbins = 30;
histfit(array_rand_num, nbins, 'lognormal');
xticks(0:1:20);
pd = fitdist(array_rand_num', 'lognormal');
fitting_Ex = exp(pd.mu + pd.sigma^2 * 0.5)
fitting_Dx = exp(2 * pd.mu + pd.sigma^2) * (exp(pd.sigma^2) - 1)
%since the pdf is truncated, the fitting Ex and Dx are not exactly equal to
%the input ones
• Why not just generate random variables from the untruncated lognormal distribution, then throw away the ones that are below / above the lower / upper bounds respectively? Commented Sep 9, 2021 at 0:27
• @jbowman This is indeed an alternative way, also, could be my last resort. I just want a more scientific method. Commented Sep 9, 2021 at 0:36
• Rather than throw anything away, you can generate a suitably truncated uniform (by simple rescaling to the corresponding truncation points), apply the inverse Gaussian to convert to truncated normal, and then exponentiate (or if you have the inverse cdf of a lognormal, you may be able to do the last two steps in one and it might save a little overhead) Commented Sep 9, 2021 at 1:41
• stats.stackexchange.com/questions/56747/…
– whuber
Commented Sep 9, 2021 at 12:39
It is a little easier notationally to work directly with the standard Normal distribution than with the lognormal distribution and error functions, and the conversion from the random number generated from a truncated standard Normal distribution to one generated from the desired truncated Lognormal distribution should be clear.
Let us label the lower and upper bounds (converted to the standard Normal distribution appropriately) as $$(l, u)$$ respectively. Then define $$p_l = \Phi(l)$$ and $$p_u = \Phi(u)$$, the values of the cumulative density function at $$l$$ and $$u$$. Generate random numbers $$z$$ as follows:
1. $$x \leftarrow U(0,1)$$
2. $$x' \leftarrow p_l + (p_u - p_l)x$$
3. $$z \leftarrow \Phi^{-1}(x')$$
The second assignment creates $$x' \sim U(p_l, p_u)$$, and running this through the inverse CDF function of a standard Normal variate will produce a random number that has a standard Normal distribution truncated at $$(l, u)$$.
Of course, if you have access to a function that will calculate the inverse CDF of a Lognormal distribution with specified parameters, you can work directly with that in step 3, saving yourself any effort of converting from the Lognormal to the standard Normal and back again.
• So, that means: I have to convert the $X \sim LN(\mu_1, \sigma_1)$ to $Y\sim N(\mu_2, \sigma_2)$ and then to $Z \sim N(0, 1)$. $l = (ln(x1) - \mu_2) / \sigma_2$. $\Phi$ is the CDF of $N(0, 1)$. Am I right? Commented Sep 9, 2021 at 1:26
• The parameters of the lognormal are (typically) the same as those of the Normal. More simply, $z = (\ln(x)-\mu) / \sigma$, where $x \sim LN(\mu, \sigma)$ and $z \sim N(0,1)$. Commented Sep 9, 2021 at 1:29
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# Why I cannot generate random numbers having a truncated lognormal distribution?. My deduction is:. When the distribution is truncated, a normalization factor should be introduced: $$$$g(x) = \frac{C}{x\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{\ln{x}-\mu}{\sigma}\right)^2},$$$$ where $$C$$ is the factor. Integrating $$g(x)$$ over the interval $$[x_1, x_2]$$ ($$x_1 \geq 0$$), we have $$$$\frac{C}{\sigma\sqrt{2\pi}}\int_{x_1}^{x_2}\frac{1}{x}e^{-\frac{1}{2}\left(\frac{\ln{x}-\mu}{\sigma}\right)^2}\mathrm{d}x = 1. \label{CDF_truncated}$$$$ To determine $$C$$, let $$t = \frac{\ln{x}-\mu}{\sigma}$$, so $$x = e^{\sigma t + \mu}$$ and $$\mathrm{d}x = \sigma e^{\sigma t + \mu}\mathrm{d}t$$. The interval becomes $$\left[t_1, t_2\right]$$ which is given by $$$$t_1 = \frac{\ln{x_1}-\mu}{\sigma} \text{ and } t_2 = \frac{\ln{x_2}-\mu}{\sigma}. \notag$$$$ Hence, the inregration can be rewritten as $$$$\frac{C}{\sigma\sqrt{2\pi}}\int_{t_1}^{t_2} \sigma\cdot e^\mu \cdot e^{-\sigma t - \mu}\cdot e^{-\frac{t^2}{2} + \sigma t}\mathrm{d}t = 1, \notag$$$$ $$$$\Rightarrow \frac{C}{\sqrt{2\pi}}\int_{t_1}^{t_2} e^{-\frac{t^2}{2}}\mathrm{d}t = 1. \notag$$$$. Next, let $$m = \frac{t}{\sqrt{2}}$$ and $$t = \sqrt{2}m$$. So, $$\mathrm{d}t=\sqrt{2}\mathrm{d}m$$. The Eintegration can be further simplified into $$$$\frac{C}{\sqrt{2\pi}}\int_{m_1}^{m_2} \sqrt{2}e^{-m^2}\mathrm{d}m = 1, \notag$$$$ $$$$\Rightarrow \frac{C}{\sqrt{\pi}}\int_{m_1}^{m_2} e^{-m^2}\mathrm{d}m = 1. \notag$$$$. So, the $$C$$ is $$$$C = \frac{-2}{[{erf}(m_1)-{erf}(m_2)]},$$$$ where $${erf}(\cdot)$$ is the error function. If $$x_1 = 0 \text{ and } x_2 = +\infty$$, then $$t_1 = -\infty \text{ and } t_2 = +\infty$$, and $$m_1 = -\infty \text{ and } m_2 = +\infty$$, and $${erf}(m_1) \text{ and } {erf}(m_1)$$ are -1 and 1, respectively, $$C = 1$$ which leads to a regular non-truncated PDF.. Now, let us generate random log-normal variables from the standard uniform distributions. $$$$p = C\int_{x_1}^{x}\frac{1}{x\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{\ln{x}-\mu}{\sigma}\right)^2}\mathrm{d}x.$$$$ where $$p$$ is a random number within [0, 1]. In a similar way, $$$$p = \frac{C}{ \sqrt{\pi}}\int_{m_1}^{m}e^{-m^2}\mathrm{d}m, \notag$$$$. $$$$\Rightarrow m = {erfinv}\left({{erf}(m_1) - \frac{2 p }{C}}\right), \notag$$$$ and $$m = \frac{\frac{\ln{x}-\mu}{\sigma}}{\sqrt{2}}$$, $$erfinv(\cdot)$$ is the inverse of the error function.. I write the following MATLAB script to generate random lognormal variables, with given expection and variance. But I got NaN.. clc;. close all;. clear all;. p = rand;. mean = 6; %given expection. variance = 0.5; % given variance. x1 = 1.5;. x2 = 15;. mu_1 = log(mean * mean / ((variance + mean * mean)^0.5)); %determine mean. sigma_1 = (log(1 + (variance) / (mean * mean)))^0.5; %determine std. deviation. t1 = (log(x1) - mu_1) / sigma_1;. t2 = (log(x2) - mu_1) / sigma_1;. m1 = t1 / (2^0.5);. m2 = t2 / (2^0.5);. C = -2 / (erf(m1) - erf(m2));. m = erfinv(erf(m1) - 2 * p / C);. m =. NaN. what's wrong with my deduction? or my code?. Update------------------------------------------. according to the comments of @jbowman, I write the following MATLAB code:. clc;. close all;. clear all;. mean = 6; %given expection. variance = 9; % given variance. x1 = 1.5;. x2 = 15;. mu_1 = log(mean * mean / ((variance + mean * mean)^0.5)); %determine mean. sigma_1 = (log(1 + (variance) / (mean * mean)))^0.5; %determine std. deviation. % then, convert normal to std. normal.
|
% Any point (x) from a normal distribution can be converted to the standard normal distribution (z) with the formula z = (x-mean) / standard deviation. l = (log(x1) - mu_1) / sigma_1;. u = (log(x2) - mu_1) / sigma_1;. pd = makedist('Normal', 'mu', 0, 'sigma', 1);. p_l = cdf(pd, l);. p_u = cdf(pd, u);. array_rand_num = [];. nf = 1000;. for i = 1:1:nf. x = rand;. x_prime = p_l + (p_u - p_l) * x;. z = icdf(pd, x_prime);. p = exp(z * sigma_1 + mu_1);. array_rand_num(i) = p;. end. figure(1). nbins = 30;. histfit(array_rand_num, nbins, 'lognormal');. xticks(0:1:20);. pd = fitdist(array_rand_num', 'lognormal');. fitting_Ex = exp(pd.mu + pd.sigma^2 * 0.5). fitting_Dx = exp(2 * pd.mu + pd.sigma^2) * (exp(pd.sigma^2) - 1). %since the pdf is truncated, the fitting Ex and Dx are not exactly equal to. %the input ones. • Why not just generate random variables from the untruncated lognormal distribution, then throw away the ones that are below / above the lower / upper bounds respectively? Commented Sep 9, 2021 at 0:27. • @jbowman This is indeed an alternative way, also, could be my last resort. I just want a more scientific method. Commented Sep 9, 2021 at 0:36. • Rather than throw anything away, you can generate a suitably truncated uniform (by simple rescaling to the corresponding truncation points), apply the inverse Gaussian to convert to truncated normal, and then exponentiate (or if you have the inverse cdf of a lognormal, you may be able to do the last two steps in one and it might save a little overhead) Commented Sep 9, 2021 at 1:41. • stats.stackexchange.com/questions/56747/…. – whuber. Commented Sep 9, 2021 at 12:39. It is a little easier notationally to work directly with the standard Normal distribution than with the lognormal distribution and error functions, and the conversion from the random number generated from a truncated standard Normal distribution to one generated from the desired truncated Lognormal distribution should be clear.. Let us label the lower and upper bounds (converted to the standard Normal distribution appropriately) as $$(l, u)$$ respectively. Then define $$p_l = \Phi(l)$$ and $$p_u = \Phi(u)$$, the values of the cumulative density function at $$l$$ and $$u$$. Generate random numbers $$z$$ as follows:. 1. $$x \leftarrow U(0,1)$$. 2. $$x' \leftarrow p_l + (p_u - p_l)x$$. 3. $$z \leftarrow \Phi^{-1}(x')$$. The second assignment creates $$x' \sim U(p_l, p_u)$$, and running this through the inverse CDF function of a standard Normal variate will produce a random number that has a standard Normal distribution truncated at $$(l, u)$$.. Of course, if you have access to a function that will calculate the inverse CDF of a Lognormal distribution with specified parameters, you can work directly with that in step 3, saving yourself any effort of converting from the Lognormal to the standard Normal and back again.. • So, that means: I have to convert the $X \sim LN(\mu_1, \sigma_1)$ to $Y\sim N(\mu_2, \sigma_2)$ and then to $Z \sim N(0, 1)$. $l = (ln(x1) - \mu_2) / \sigma_2$. $\Phi$ is the CDF of $N(0, 1)$. Am I right? Commented Sep 9, 2021 at 1:26. • The parameters of the lognormal are (typically) the same as those of the Normal. More simply, $z = (\ln(x)-\mu) / \sigma$, where $x \sim LN(\mu, \sigma)$ and $z \sim N(0,1)$. Commented Sep 9, 2021 at 1:29.
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https://gmatclub.com/forum/the-distance-from-point-a-to-point-b-is-15-meters-and-the-169750.html
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# The distance from point A to point B is 15 meters, and the
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The distance from point A to point B is 15 meters, and the [#permalink]
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03 Apr 2014, 13:22
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The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?
A. 0 < d < 24
B. 1 < d < 25
C. 2 < d < 26
D. 3 < d < 27
E. 4 < d < 28
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Posts: 51185
Re: The distance from point A to point B is 15 meters, and the [#permalink]
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20 Jun 2015, 23:54
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Bunuel wrote:
The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?
A. 0 < d < 24
B. 1 < d < 25
C. 2 < d < 26
D. 3 < d < 27
E. 4 < d < 28
If point C in on line AB, between points A and B, then the distance between B and C will be 15 - 11 = 4 meters (least possible distance).
If point C in on line AB, to the left of point A, then the distance between B and C will be 15 + 11 = 26 meters (greatest possible distance).
Therefore $$4 \le AC \le 26$$. Correct option must cover this entire range. Only option D offers the range which has all possible values of AC.
Attachment:
M31-51.png [ 1.66 KiB | Viewed 7736 times ]
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
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03 Apr 2014, 16:12
I don't get this. Distance between C and B will be shortest, if both these points are on the same side of point A. Distance will be the farthest, if B and C are on opposite side of point A.
so d should be between, 4 and 26. Not sure which option to choose now!
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
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03 Apr 2014, 19:02
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Yes. I also arrived at d between 4 and 26. Going by the options given I would go for D.
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
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03 Apr 2014, 21:19
Bunuel wrote:
The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?
A. 0 < d < 24
B. 1 < d < 25
C. 2 < d < 26
D. 3 < d < 27
E. 4 < d < 28
point C can be either in line between A and B, opposite direction of B or out of this line, in this case, we get triangle ABC. If it is in line AB then BC=4 meters, if opposite direction, then BC=26 meters, if out of line BC should be less than AB+AC<26 meters. So, maximum BC is less than 26. I think it is C
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
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03 Apr 2014, 21:22
1
Bunuel wrote:
The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?
A. 0 < d < 24
B. 1 < d < 25
C. 2 < d < 26
D. 3 < d < 27
E. 4 < d < 28
The distance has to vary between 4 and 26 depending upon whether B and C fall on the same side of A or they are on the opposite sides of A. The sequence could be A----B----C or B----A----C. From the former case we get least distance between B and C = 4 and from the later we get max distance as 38. So, I would go for D, as 3< d < 27 means 4<= d <=26
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
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04 Apr 2014, 02:38
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Answer = C. 2 < d < 26
Addition of 2 sides of a triangle has to be MORE than the 3rd side
d has to be less than 26
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
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07 May 2014, 02:43
Bunuel wrote:
The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?
A. 0 < d < 24
B. 1 < d < 25
C. 2 < d < 26
D. 3 < d < 27
E. 4 < d < 28
Bunuel my thought process...
There are three possibilities
a) Point C is between A and B .. In that case d = 15 - 11 = 4m
b) Point C is aware from A and B but on same line . In that case d = 15 + 11 = 26m
c) Point C is any arbitrary point thus forming a triangle ABC . Now sum of 2 sides should be greater than third side. So d < 26m. So possible values look like 4 < d<26...
None of the option and now I am confused
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
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07 May 2014, 03:08
himanshujovi wrote:
Bunuel wrote:
The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?
A. 0 < d < 24
B. 1 < d < 25
C. 2 < d < 26
D. 3 < d < 27
E. 4 < d < 28
Bunuel my thought process...
There are three possibilities
a) Point C is between A and B .. In that case d = 15 - 11 = 4m
b) Point C is aware from A and B but on same line . In that case d = 15 + 11 = 26m
c) Point C is any arbitrary point thus forming a triangle ABC . Now sum of 2 sides should be greater than third side. So d < 26m. So possible values look like 4 < d<26...
None of the option and now I am confused
Attachment:
Untitled.png [ 14.15 KiB | Viewed 6790 times ]
Here's is my take
We have 3 cases here as you pointed out. Now which range covers all the 3 options should be the answer.
Going by that logic, Ans should be D cause we don't which case are we looking at so Correct answer will cover all the possible options.
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
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07 May 2014, 03:08
himanshujovi wrote:
Bunuel wrote:
The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?
A. 0 < d < 24
B. 1 < d < 25
C. 2 < d < 26
D. 3 < d < 27
E. 4 < d < 28
Bunuel my thought process...
There are three possibilities
a) Point C is between A and B .. In that case d = 15 - 11 = 4m
b) Point C is aware from A and B but on same line . In that case d = 15 + 11 = 26m
c) Point C is any arbitrary point thus forming a triangle ABC . Now sum of 2 sides should be greater than third side. So d < 26m. So possible values look like 4 < d<26...
None of the option and now I am confused
Let me ask you a question: doesn't option D covers all the possible values of d?
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
### Show Tags
22 Jun 2015, 05:11
Hello Bunuel,
In the case where the lines form a triangle, option D will include 26 too - isnt this wrong?
The third side of the triangle should be between difference of two ( 15-11) and summation of two (11+15).
Am I doing something wrong?
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aimtoteach
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
### Show Tags
22 Jun 2015, 05:12
aimtoteach wrote:
Hello Bunuel,
In the case where the lines form a triangle, option D will include 26 too - isnt this wrong?
The third side of the triangle should be between difference of two ( 15-11) and summation of two (11+15).
Am I doing something wrong?
Who said that A, B and C must form a triangle?
_________________
Manager
Joined: 17 Aug 2015
Posts: 100
Re: The distance from point A to point B is 15 meters, and the [#permalink]
### Show Tags
29 May 2016, 21:51
Bunuel wrote:
aimtoteach wrote:
Hello Bunuel,
In the case where the lines form a triangle, option D will include 26 too - isnt this wrong?
The third side of the triangle should be between difference of two ( 15-11) and summation of two (11+15).
Am I doing something wrong?
Who said that A, B and C must form a triangle?
Bunnuel,
Are you saying essentially that they can form a triangle and can also be on the same straight line?
So If they are on the same line then either d=4 or d=26.
But if they are on a triangle then d> 4 and d<26.
And so combining the two possibilities, we get 4<=d<=26
Math Expert
Joined: 02 Sep 2009
Posts: 51185
Re: The distance from point A to point B is 15 meters, and the [#permalink]
### Show Tags
30 May 2016, 11:47
ajdse22 wrote:
Bunuel wrote:
aimtoteach wrote:
Hello Bunuel,
In the case where the lines form a triangle, option D will include 26 too - isnt this wrong?
The third side of the triangle should be between difference of two ( 15-11) and summation of two (11+15).
Am I doing something wrong?
Who said that A, B and C must form a triangle?
Bunnuel,
Are you saying essentially that they can form a triangle and can also be on the same straight line?
So If they are on the same line then either d=4 or d=26.
But if they are on a triangle then d> 4 and d<26.
And so combining the two possibilities, we get 4<=d<=26
Yes, the points can be on the same line. Check here: the-distance-from-point-a-to-point-b-is-15-meters-and-the-169750.html#p1540270
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Re: The distance from point A to point B is 15 meters, and the [#permalink]
### Show Tags
16 Nov 2017, 23:15
Bunuel wrote:
The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?
A. 0 < d < 24
B. 1 < d < 25
C. 2 < d < 26
D. 3 < d < 27
E. 4 < d < 28
It's between 4 and 26 inclusive. But I think it must be given in the question that d is an integer. Only then D stands
_________________
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Re: The distance from point A to point B is 15 meters, and the &nbs [#permalink] 16 Nov 2017, 23:15
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GMAT Question of the Day - Daily to your Mailbox; hard ones only. It is currently 13 Dec 2018, 12:14. ### GMAT Club Daily Prep. #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.. Customized. for You. we will pick new questions that match your level based on your Timer History. Track. every week, we’ll send you an estimated GMAT score based on your performance. Practice. Pays. we will pick new questions that match your level based on your Timer History. ## Events & Promotions. ###### Events & Promotions in December. PrevNext. SuMoTuWeThFrSa. 2526272829301. 2345678. 9101112131415. 16171819202122. 23242526272829. 303112345. Open Detailed Calendar. • ### GMATbuster's Weekly GMAT Quant Quiz, Tomorrow, Saturday at 9 AM PST. December 14, 2018. December 14, 2018. 09:00 AM PST. 10:00 AM PST. 10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.. • ### The winning strategy for 700+ on the GMAT. December 13, 2018. December 13, 2018. 08:00 AM PST. 09:00 AM PST. What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.. # The distance from point A to point B is 15 meters, and the. Author Message. TAGS:. ### Hide Tags. Math Expert. Joined: 02 Sep 2009. Posts: 51185. The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 03 Apr 2014, 13:22. 1. 7. 00:00. Difficulty:. 75% (hard). Question Stats:. 51% (01:47) correct 49% (01:55) wrong based on 240 sessions. ### HideShow timer Statistics. The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?. A. 0 < d < 24. B. 1 < d < 25. C. 2 < d < 26. D. 3 < d < 27. E. 4 < d < 28. _________________. Math Expert. Joined: 02 Sep 2009. Posts: 51185. Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 20 Jun 2015, 23:54. 3. 2. Bunuel wrote:. The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?. A. 0 < d < 24. B. 1 < d < 25. C. 2 < d < 26. D. 3 < d < 27. E. 4 < d < 28. If point C in on line AB, between points A and B, then the distance between B and C will be 15 - 11 = 4 meters (least possible distance).. If point C in on line AB, to the left of point A, then the distance between B and C will be 15 + 11 = 26 meters (greatest possible distance).. Therefore $$4 \le AC \le 26$$. Correct option must cover this entire range. Only option D offers the range which has all possible values of AC.. Attachment:. M31-51.png [ 1.66 KiB | Viewed 7736 times ]. _________________. ##### General Discussion. Intern. Joined: 27 Jan 2014. Posts: 26. Location: United States. Schools: Stanford '16 (D). Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 03 Apr 2014, 16:12. I don't get this. Distance between C and B will be shortest, if both these points are on the same side of point A. Distance will be the farthest, if B and C are on opposite side of point A.. so d should be between, 4 and 26. Not sure which option to choose now!. Manager. Joined: 08 Sep 2010. Posts: 61. Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 03 Apr 2014, 19:02. 2. Yes. I also arrived at d between 4 and 26. Going by the options given I would go for D.. Director. Joined: 23 Jan 2013. Posts: 568. Schools: Cambridge'16. Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 03 Apr 2014, 21:19. Bunuel wrote:. The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?. A. 0 < d < 24. B. 1 < d < 25. C. 2 < d < 26. D. 3 < d < 27. E. 4 < d < 28. point C can be either in line between A and B, opposite direction of B or out of this line, in this case, we get triangle ABC. If it is in line AB then BC=4 meters, if opposite direction, then BC=26 meters, if out of line BC should be less than AB+AC<26 meters. So, maximum BC is less than 26. I think it is C. Manager. Joined: 25 Mar 2014. Posts: 137. Location: India. Concentration: Operations, Finance. GMAT Date: 05-10-2015. GPA: 3.51. WE: Programming (Computer Software). Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 03 Apr 2014, 21:22. 1. Bunuel wrote:. The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?. A. 0 < d < 24. B. 1 < d < 25. C. 2 < d < 26. D. 3 < d < 27. E. 4 < d < 28. The distance has to vary between 4 and 26 depending upon whether B and C fall on the same side of A or they are on the opposite sides of A. The sequence could be A----B----C or B----A----C. From the former case we get least distance between B and C = 4 and from the later we get max distance as 38. So, I would go for D, as 3< d < 27 means 4<= d <=26. _________________. Please give Kudos to the post if you liked.. SVP. Status: The Best Or Nothing. Joined: 27 Dec 2012. Posts: 1825. Location: India. Concentration: General Management, Technology. WE: Information Technology (Computer Software). Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 04 Apr 2014, 02:38. 2. 2. Answer = C. 2 < d < 26. Addition of 2 sides of a triangle has to be MORE than the 3rd side. d has to be less than 26. Attachments. tr.jpg [ 6.62 KiB | Viewed 6997 times ]. _________________. Kindly press "+1 Kudos" to appreciate. Manager. Joined: 28 Apr 2014. Posts: 217.
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Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 07 May 2014, 02:43. Bunuel wrote:. The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?. A. 0 < d < 24. B. 1 < d < 25. C. 2 < d < 26. D. 3 < d < 27. E. 4 < d < 28. Bunuel my thought process.... There are three possibilities. a) Point C is between A and B .. In that case d = 15 - 11 = 4m. b) Point C is aware from A and B but on same line . In that case d = 15 + 11 = 26m. c) Point C is any arbitrary point thus forming a triangle ABC . Now sum of 2 sides should be greater than third side. So d < 26m. So possible values look like 4 < d<26.... None of the option and now I am confused. Director. Joined: 25 Apr 2012. Posts: 684. Location: India. GPA: 3.21. Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 07 May 2014, 03:08. himanshujovi wrote:. Bunuel wrote:. The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?. A. 0 < d < 24. B. 1 < d < 25. C. 2 < d < 26. D. 3 < d < 27. E. 4 < d < 28. Bunuel my thought process.... There are three possibilities. a) Point C is between A and B .. In that case d = 15 - 11 = 4m. b) Point C is aware from A and B but on same line . In that case d = 15 + 11 = 26m. c) Point C is any arbitrary point thus forming a triangle ABC . Now sum of 2 sides should be greater than third side. So d < 26m. So possible values look like 4 < d<26.... None of the option and now I am confused. Attachment:. Untitled.png [ 14.15 KiB | Viewed 6790 times ]. Here's is my take. We have 3 cases here as you pointed out. Now which range covers all the 3 options should be the answer.. Going by that logic, Ans should be D cause we don't which case are we looking at so Correct answer will cover all the possible options.. _________________. “If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”. Math Expert. Joined: 02 Sep 2009. Posts: 51185. Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 07 May 2014, 03:08. himanshujovi wrote:. Bunuel wrote:. The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?. A. 0 < d < 24. B. 1 < d < 25. C. 2 < d < 26. D. 3 < d < 27. E. 4 < d < 28. Bunuel my thought process.... There are three possibilities. a) Point C is between A and B .. In that case d = 15 - 11 = 4m. b) Point C is aware from A and B but on same line . In that case d = 15 + 11 = 26m. c) Point C is any arbitrary point thus forming a triangle ABC . Now sum of 2 sides should be greater than third side. So d < 26m. So possible values look like 4 < d<26.... None of the option and now I am confused. Let me ask you a question: doesn't option D covers all the possible values of d?. _________________. Current Student. Status: GMAT Date: 10/08/15. Joined: 17 Jul 2014. Posts: 87. Location: United States (MA). Concentration: Human Resources, Strategy. GMAT 1: 640 Q48 V35. GPA: 3.5. WE: Human Resources (Consumer Products). Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 22 Jun 2015, 05:11. Hello Bunuel,. In the case where the lines form a triangle, option D will include 26 too - isnt this wrong?. The third side of the triangle should be between difference of two ( 15-11) and summation of two (11+15).. Am I doing something wrong?. _________________. Thanks,. aimtoteach. ~~~~~~~~~~~~~~~~~. Please give Kudos if you find this post useful.. Math Expert. Joined: 02 Sep 2009. Posts: 51185. Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 22 Jun 2015, 05:12. aimtoteach wrote:. Hello Bunuel,. In the case where the lines form a triangle, option D will include 26 too - isnt this wrong?. The third side of the triangle should be between difference of two ( 15-11) and summation of two (11+15).. Am I doing something wrong?. Who said that A, B and C must form a triangle?. _________________. Manager. Joined: 17 Aug 2015. Posts: 100. Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 29 May 2016, 21:51. Bunuel wrote:. aimtoteach wrote:. Hello Bunuel,. In the case where the lines form a triangle, option D will include 26 too - isnt this wrong?. The third side of the triangle should be between difference of two ( 15-11) and summation of two (11+15).. Am I doing something wrong?. Who said that A, B and C must form a triangle?. Bunnuel,. Are you saying essentially that they can form a triangle and can also be on the same straight line?. So If they are on the same line then either d=4 or d=26.. But if they are on a triangle then d> 4 and d<26.. And so combining the two possibilities, we get 4<=d<=26. Math Expert. Joined: 02 Sep 2009. Posts: 51185. Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 30 May 2016, 11:47. ajdse22 wrote:. Bunuel wrote:. aimtoteach wrote:. Hello Bunuel,. In the case where the lines form a triangle, option D will include 26 too - isnt this wrong?. The third side of the triangle should be between difference of two ( 15-11) and summation of two (11+15).. Am I doing something wrong?. Who said that A, B and C must form a triangle?. Bunnuel,. Are you saying essentially that they can form a triangle and can also be on the same straight line?. So If they are on the same line then either d=4 or d=26.. But if they are on a triangle then d> 4 and d<26.. And so combining the two possibilities, we get 4<=d<=26. Yes, the points can be on the same line. Check here: the-distance-from-point-a-to-point-b-is-15-meters-and-the-169750.html#p1540270. _________________. Senior Manager. Joined: 03 Apr 2013. Posts: 274. Location: India. Concentration: Marketing, Finance. GMAT 1: 740 Q50 V41. GPA: 3. Re: The distance from point A to point B is 15 meters, and the [#permalink]. ### Show Tags. 16 Nov 2017, 23:15. Bunuel wrote:. The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If d is the distance, in meters, between points B and C, then which of the following must be correct?. A. 0 < d < 24. B. 1 < d < 25. C. 2 < d < 26. D. 3 < d < 27. E. 4 < d < 28. It's between 4 and 26 inclusive. But I think it must be given in the question that d is an integer. Only then D stands. _________________. Spread some love..Like = +1 Kudos. Re: The distance from point A to point B is 15 meters, and the &nbs [#permalink] 16 Nov 2017, 23:15. Display posts from previous: Sort by.
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https://www.reddit.com/r/CasualMath/comments/17dzzj/find_the_necessary_conditions_for_these_two/
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This is an archived post. You won't be able to vote or comment.
[–] 2 points3 points (3 children)
I bruteforced the solutions out of curiosity and got the following sets of numbers:
a = |0| b = |0| x = |0| y = |0| z = |0
a = |0| b = |1| x = |0| y = |0| z = |1
a = |0| b = |2| x = |0| y = |0| z = |4
a = |0| b = |3| x = |0| y = |0| z = |9
a = |1| b = |0| x = |1| y = |0| z = |0
a = |1| b = |1| x = |1| y = |2| z = |1
a = |1| b = |2| x = |1| y = |4| z = |4
a = |1| b = |3| x = |1| y = |6| z = |9
a = |2| b = |0| x = |4| y = |0| z = |0
a = |2| b = |1| x = |4| y = |4| z = |1
a = |2| b = |2| x = |4| y = |8| z = |4
a = |3| b = |0| x = |9| y = |0| z = |0
a = |3| b = |1| x = |9| y = |6| z = |1
Does that seem right?
[–][S] 2 points3 points (2 children)
The solutions seem right, but it reminded me that I probably should have expanded the criteria to allow for a thousands place in case there were numbers that squared to something at least 1000.
From what you have, there are some things I can extrapolate.
• a <=3, b <=3,
Because if either were at least 4, then the one of the squares would be 4 digits or more.
• a+b<=4
I don't have a good reason for this conclusion, other than the previous conclusion and the fact that 2,3 or 3,2 don't work.
[–] 1 point2 points (0 children)
The first square that exceeds three digits is 32, since x2 = 103 at ten times the square root of ten, or about 31.6. This reduces the space of solutions, since the right side only goes from 0 to 999 where the left side goes from 0 to 9801 (992 ). I bet there are more numbers that satisfy both equations with a thousands place. These are all very amenable to brute force calculation as /u/kcoPkcoP pointed out, since all of the operations are quite inexpensive and the search space is quite small.
[–] 1 point2 points (0 children)
it seems you want the first digit of your answer to be the square of the first digit of the inputs (true any time the digit squared is one digit, so a and b must be less than or equal to 3), the third digit must be the square of the second digit of your input (same constraints, since it is symmetric), and the second digit must be twice the first times second digits (so 2ab must also be one digit, which is what causes 2,3 to fail).
this is just saying (10a + b)2 = 100a + 20ab + b2, and if each of those terms stays politely in it's own digit without carrying over to the next, the whole thing will be "reversible"
[–] 0 points1 point (0 children)
{1,2,3,10,11,12,13,20,21,22,30,31,100,101,102,103,110,111,112,113,120,121,122,130,200,201,202,210,211,212,220,221,300,301,310,311,1000,1001,1002,1003,1010,1011,1012,1013,1020,1021,1022,1030,1031,1100,1101,1102,1103,1110,1111,1112,1113,1120,1121,1122,1130,1200,1201,1202,1210,1211,1212,1220,1300,1301,2000,2001,2002,2010,2011,2012,2020,2021,2022,2100,2101,2102,2110,2111,2120,2121,2200,2201,2202,2210,2211,3000,3001,3010,3011,3100,3101,3110,3111,10000}
That's all the ingegers up to 10000 satisfying x2 = rev(rev(x)2)
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This is an archived post. You won't be able to vote or comment.. [–] 2 points3 points (3 children). I bruteforced the solutions out of curiosity and got the following sets of numbers:. a = |0| b = |0| x = |0| y = |0| z = |0. a = |0| b = |1| x = |0| y = |0| z = |1. a = |0| b = |2| x = |0| y = |0| z = |4. a = |0| b = |3| x = |0| y = |0| z = |9. a = |1| b = |0| x = |1| y = |0| z = |0. a = |1| b = |1| x = |1| y = |2| z = |1. a = |1| b = |2| x = |1| y = |4| z = |4. a = |1| b = |3| x = |1| y = |6| z = |9. a = |2| b = |0| x = |4| y = |0| z = |0. a = |2| b = |1| x = |4| y = |4| z = |1. a = |2| b = |2| x = |4| y = |8| z = |4. a = |3| b = |0| x = |9| y = |0| z = |0. a = |3| b = |1| x = |9| y = |6| z = |1. Does that seem right?.
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[–][S] 2 points3 points (2 children). The solutions seem right, but it reminded me that I probably should have expanded the criteria to allow for a thousands place in case there were numbers that squared to something at least 1000.. From what you have, there are some things I can extrapolate.. • a <=3, b <=3,. Because if either were at least 4, then the one of the squares would be 4 digits or more.. • a+b<=4. I don't have a good reason for this conclusion, other than the previous conclusion and the fact that 2,3 or 3,2 don't work.. [–] 1 point2 points (0 children). The first square that exceeds three digits is 32, since x2 = 103 at ten times the square root of ten, or about 31.6. This reduces the space of solutions, since the right side only goes from 0 to 999 where the left side goes from 0 to 9801 (992 ). I bet there are more numbers that satisfy both equations with a thousands place. These are all very amenable to brute force calculation as /u/kcoPkcoP pointed out, since all of the operations are quite inexpensive and the search space is quite small.. [–] 1 point2 points (0 children). it seems you want the first digit of your answer to be the square of the first digit of the inputs (true any time the digit squared is one digit, so a and b must be less than or equal to 3), the third digit must be the square of the second digit of your input (same constraints, since it is symmetric), and the second digit must be twice the first times second digits (so 2ab must also be one digit, which is what causes 2,3 to fail).. this is just saying (10a + b)2 = 100a + 20ab + b2, and if each of those terms stays politely in it's own digit without carrying over to the next, the whole thing will be "reversible". [–] 0 points1 point (0 children). {1,2,3,10,11,12,13,20,21,22,30,31,100,101,102,103,110,111,112,113,120,121,122,130,200,201,202,210,211,212,220,221,300,301,310,311,1000,1001,1002,1003,1010,1011,1012,1013,1020,1021,1022,1030,1031,1100,1101,1102,1103,1110,1111,1112,1113,1120,1121,1122,1130,1200,1201,1202,1210,1211,1212,1220,1300,1301,2000,2001,2002,2010,2011,2012,2020,2021,2022,2100,2101,2102,2110,2111,2120,2121,2200,2201,2202,2210,2211,3000,3001,3010,3011,3100,3101,3110,3111,10000}. That's all the ingegers up to 10000 satisfying x2 = rev(rev(x)2).
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https://gottwurfelt.com/2013/07/17/the-probability-of-catching-four-foul-balls/
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# The probability of catching four foul balls
Greg Van Niel caught four foul balls at Sunday’s Cleveland Indians game.
ESPN reported that this is a one-in-a-trillion event – a number due to Ideal Seat, which I’ll take to mean that this guy had a one-in-a-trillion chance of catching four fouls. This is immediately suspicious to me. Total MLB attendance last year was about 75 million, so a one in a trillion event should happen once every thirteen thousand years. The fact that it happened, given that we’ve had way less than thirteen thousand years of baseball, is evidence that this computation was done incorrectly.
Somewhat surprisingly, given how small the number is, it actually seems to be an overestimate. I’ll assume that their numbers are correct: 30 balls enter the stands in an average game, and there are 30,000 fans at that game. Say I’m one of those fans. Let’s assume that all foul balls are hit independently, and that they’re equally likely to be caught by any person in the stands. The probability that exactly four balls will be hit to me are ${30 \choose 4} p^4 (1-p)^(30-4)$, where $p = 1/30000$. This is about $3.38 \times 10^{-14}$, or one in thirty trillion. (The probably that five or more balls will be hit to me is orders of magnitude lower than that.)
IdealSeat also claims that two fans caught two foul balls in the same game last year. I suspect that there’s some massive underreporting going on here, because the same analysis gives that the probability that I’ll get two balls is ${30 \choose 2} p^2 (1-p)^(30-4)$, which is about one in two million. So this should have happened 35 to 40 times last year – it’s just that most of the people who it happened to didn’t bother telling anybody! (Other than their friends, who probably didn’t believe them.)
What’s wrong with the one in a trillion, or one in thirty trillion, numbers?
• They assume that all foul balls are uniformly distributed over all the seats. This is patently untrue. Some seats by definition can’t receive a foul ball, because they’re in fair territory. Some seats, although they can theoretically receive a foul ball, just won’t. Ideal Seat has a heatmap of foul ball locations at Safeco Field in Seattle — basically the closer you are to home plate, the better your chances. Your chances of getting a foul ball drop off much faster with height than with horizontal distance. In addition, aisle seats are more likely to be the closest seat to where a ball lands than adjacent non-aisle seats.
• They assume that all foul ball locations are independent. I don’t know if there’s data on this, but batters have tendencies on where they hit balls in play; they should have tendencies on where they hit foul balls as well.
• They assume that a person can only get foul balls hit to their seat. This might be true in, say, San Francisco (where most games sell out), but it’s not true in Oakland (where there are plenty of empty seats). Van Niel’s section looks pretty full in the pictures, though. But Van Niel himself admits at least one of the balls wasn’t hit right to him.
All I can say for sure is that these drive the chances up – so the probability of catching four foul balls in a single game is probably a good deal higher than one in a trillion.
## 2 thoughts on “The probability of catching four foul balls”
1. Sergio says:
I think that this probability, should be calculated as the probability that at least two people in a room of n, have the same birthday. The probability that the same person captures more than two balls in a given game (yes, the probability that this happens in one game) should be calculated using 30000 instead of 365 and n = 30. However, I don’t know how to proceed with more than four balls.
What do you think?
Hope I am not terrible wrong
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# The probability of catching four foul balls. Greg Van Niel caught four foul balls at Sunday’s Cleveland Indians game.. ESPN reported that this is a one-in-a-trillion event – a number due to Ideal Seat, which I’ll take to mean that this guy had a one-in-a-trillion chance of catching four fouls. This is immediately suspicious to me. Total MLB attendance last year was about 75 million, so a one in a trillion event should happen once every thirteen thousand years. The fact that it happened, given that we’ve had way less than thirteen thousand years of baseball, is evidence that this computation was done incorrectly.. Somewhat surprisingly, given how small the number is, it actually seems to be an overestimate. I’ll assume that their numbers are correct: 30 balls enter the stands in an average game, and there are 30,000 fans at that game. Say I’m one of those fans. Let’s assume that all foul balls are hit independently, and that they’re equally likely to be caught by any person in the stands. The probability that exactly four balls will be hit to me are ${30 \choose 4} p^4 (1-p)^(30-4)$, where $p = 1/30000$. This is about $3.38 \times 10^{-14}$, or one in thirty trillion. (The probably that five or more balls will be hit to me is orders of magnitude lower than that.). IdealSeat also claims that two fans caught two foul balls in the same game last year. I suspect that there’s some massive underreporting going on here, because the same analysis gives that the probability that I’ll get two balls is ${30 \choose 2} p^2 (1-p)^(30-4)$, which is about one in two million. So this should have happened 35 to 40 times last year – it’s just that most of the people who it happened to didn’t bother telling anybody! (Other than their friends, who probably didn’t believe them.). What’s wrong with the one in a trillion, or one in thirty trillion, numbers?. • They assume that all foul balls are uniformly distributed over all the seats. This is patently untrue. Some seats by definition can’t receive a foul ball, because they’re in fair territory.
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Some seats, although they can theoretically receive a foul ball, just won’t. Ideal Seat has a heatmap of foul ball locations at Safeco Field in Seattle — basically the closer you are to home plate, the better your chances. Your chances of getting a foul ball drop off much faster with height than with horizontal distance. In addition, aisle seats are more likely to be the closest seat to where a ball lands than adjacent non-aisle seats.. • They assume that all foul ball locations are independent. I don’t know if there’s data on this, but batters have tendencies on where they hit balls in play; they should have tendencies on where they hit foul balls as well.. • They assume that a person can only get foul balls hit to their seat. This might be true in, say, San Francisco (where most games sell out), but it’s not true in Oakland (where there are plenty of empty seats). Van Niel’s section looks pretty full in the pictures, though. But Van Niel himself admits at least one of the balls wasn’t hit right to him.. All I can say for sure is that these drive the chances up – so the probability of catching four foul balls in a single game is probably a good deal higher than one in a trillion.. ## 2 thoughts on “The probability of catching four foul balls”. 1. Sergio says:. I think that this probability, should be calculated as the probability that at least two people in a room of n, have the same birthday. The probability that the same person captures more than two balls in a given game (yes, the probability that this happens in one game) should be calculated using 30000 instead of 365 and n = 30. However, I don’t know how to proceed with more than four balls.. What do you think?. Hope I am not terrible wrong.
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You are Here: Home >< Maths
1. A garage door is 4m longer that it is wide and 1 m higher than it is wide.
The longest pole which can fit in side the garage is 7.5m long.
How wide is the garage?
Okay, I get the height to be x+1 , length to be x+4 and then the width to be x
This involves quadratics but surely the length equals x+4 as it fits in the garage? Meaning that x, the width is 3.5m?
But this is wrong according to the answers..
any ideas?
2. the longest distance will be a diagonal. try some 3d pythagoras.
3. (Original post by Phredd)
the longest distance will be a diagonal. try some 3d pythagoras.
so it would be (x+4)^2 + (x+1)^2 = 7.5?
so it would be (x+4)^2 + (x+1)^2 = 7.5?
That's a diagonal across one side. Don't forget you are in 3 dimensions so you can have an even longer diagonal that involves the lengths of all 3 sides!
5. (Original post by davros)
That's a diagonal across one side. Don't forget you are in 3 dimensions so you can have an even longer diagonal that involves the lengths of all 3 sides!
ehhh I am not exactly sure what you mean :/
ehhh I am not exactly sure what you mean :/
Have you done 3-D co-ordinates
Or any 3-D Pythagoras
For example , can you find the distance between (1, 2, 4) and (3, 6, 5)
7. (Original post by TenOfThem)
Have you done 3-D co-ordinates
Or any 3-D Pythagoras
For example , can you find the distance between (1, 2, 4) and (3, 6, 5)
nope I have never done that... wasnt on my GCSE course and I have not done it at alevel
nope I have never done that... wasnt on my GCSE course and I have not done it at alevel
Pythagoras in 2D:
Pythogaras in 3D:
It certainly was part of your GCSE course.
nope I have never done that... wasnt on my GCSE course and I have not done it at alevel
Ok
Well the diagonal in 3_D space is given by
nope I have never done that... wasnt on my GCSE course and I have not done it at alevel
Are you self-teaching or do you have a teacher who can go through this with you?
There's basically a very simple extension of Pythagoras's theorem from 2 dimensions to 3 dimensions that will help you solve this, but you probably need to do some practice questions with it to see what's going on.
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x Turn on thread page Beta. You are Here: Home >< Maths. 1. A garage door is 4m longer that it is wide and 1 m higher than it is wide.. The longest pole which can fit in side the garage is 7.5m long.. How wide is the garage?. Okay, I get the height to be x+1 , length to be x+4 and then the width to be x. This involves quadratics but surely the length equals x+4 as it fits in the garage? Meaning that x, the width is 3.5m?. But this is wrong according to the answers... any ideas?. 2. the longest distance will be a diagonal. try some 3d pythagoras.. 3. (Original post by Phredd). the longest distance will be a diagonal. try some 3d pythagoras.. so it would be (x+4)^2 + (x+1)^2 = 7.5?. so it would be (x+4)^2 + (x+1)^2 = 7.5?. That's a diagonal across one side. Don't forget you are in 3 dimensions so you can have an even longer diagonal that involves the lengths of all 3 sides!. 5. (Original post by davros). That's a diagonal across one side. Don't forget you are in 3 dimensions so you can have an even longer diagonal that involves the lengths of all 3 sides!. ehhh I am not exactly sure what you mean :/. ehhh I am not exactly sure what you mean :/. Have you done 3-D co-ordinates. Or any 3-D Pythagoras. For example , can you find the distance between (1, 2, 4) and (3, 6, 5). 7. (Original post by TenOfThem). Have you done 3-D co-ordinates. Or any 3-D Pythagoras. For example , can you find the distance between (1, 2, 4) and (3, 6, 5). nope I have never done that... wasnt on my GCSE course and I have not done it at alevel.
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nope I have never done that... wasnt on my GCSE course and I have not done it at alevel. Pythagoras in 2D:. Pythogaras in 3D:. It certainly was part of your GCSE course.. nope I have never done that... wasnt on my GCSE course and I have not done it at alevel. Ok. Well the diagonal in 3_D space is given by. nope I have never done that... wasnt on my GCSE course and I have not done it at alevel. Are you self-teaching or do you have a teacher who can go through this with you?. There's basically a very simple extension of Pythagoras's theorem from 2 dimensions to 3 dimensions that will help you solve this, but you probably need to do some practice questions with it to see what's going on.. TSR Support Team. We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.. This forum is supported by:. Updated: March 22, 2013. Today on TSR. ### Any tips for freshers.... who are introverts?. ### More snow?!. Discussions on TSR. • Latest. Poll. Useful resources. ### Maths Forum posting guidelines. Not sure where to post? Read the updated guidelines here. ### How to use LaTex. Writing equations the easy way. ### Study habits of A* students. Top tips from students who have already aced their exams. ## Groups associated with this forum:. View associated groups. Discussions on TSR. • Latest. The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.. Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE.
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# What is the focus and Directrix?
## What is the focus and Directrix?
A parabola is set of all points in a plane which are an equal distance away from a given point and given line. The point is called the focus of the parabola, and the line is called the directrix . If the axis of symmetry of a parabola is vertical, the directrix is a horizontal line .
## How do you find the focus and Directrix in standard form?
The standard form is (x – h)2 = 4p (y – k), where the focus is (h, k + p) and the directrix is y = k – p. If the parabola is rotated so that its vertex is (h,k) and its axis of symmetry is parallel to the x-axis, it has an equation of (y – k)2 = 4p (x – h), where the focus is (h + p, k) and the directrix is x = h – p.
## What is ellipse equation?
The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form. (x−h)2b2+(y−k)2a2=1 ( x − h ) 2 b 2 + ( y − k ) 2 a 2 = 1.
## Where is the focus of a parabola?
A parabola is set of all points in a plane which are an equal distance away from a given point and given line. The point is called the focus of the parabola and the line is called the directrix. The focus lies on the axis of symmetry of the parabola.
## How do you find the focus and directrix of a parabola on a calculator?
Parabola focus and directrix All you have to do is to use the following equations: Focus x-coordinate: x₀ = – b/(2a) ; Focus y-coordinate: y₀ = c – (b² – 1)/(4a) ; and. Directrix equation: y = c – (b² + 1)/(4a) .
## How do you find the vertex of a focus and Directrix?
In the equations y=a(x-h)^2+k and x=a(y-k)^2+h where a=1/4p , (h,k ) represents vertex and p is the distance between focus and vertex . In the above equations (a,b) represents FOCUS not VERTEX . In the first one directrix is y=k and in second one directrix is x=k .
## What is p value in parabola?
The absolute value of p is the distance between the vertex and the focus and the distance between the vertex and the directrix. (The sign on p tells me which way the parabola faces.) Since the focus and directrix are two units apart, then this distance has to be one unit, so | p | = 1.
## How do you find the vertex in standard form?
y=a(x−h)(x−h)+ky=ax2−2ahx+ah2+k . This means that in the standard form, y=ax2+bx+c , the expression −b2a gives the x -coordinate of the vertex.
## How do you find standard form?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.
## How do you convert standard form to vertex form on a calculator?
How to convert from the standard form to the vertex form?
1. Write the parabola equation in the standard form: y = a*x² + b*x + c ;
2. Extract a from the first two terms: y = a * (x² + b/a * x) + c ;
3. Complete the square for the expressions with x .
## How do you write quadratic equations in standard form?
A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable.
## How are quadratic equations used in real life?
Quadratic equations are actually used in everyday life, as when calculating areas, determining a product’s profit or formulating the speed of an object. Quadratic equations refer to equations with at least one squared variable, with the most standard form being ax² + bx + c = 0.
## What does a quadratic equation look like on a graph?
The graph of a quadratic function is a U-shaped curve called a parabola. The sign on the coefficient a of the quadratic function affects whether the graph opens up or down. The x-intercepts are the points at which the parabola crosses the x-axis.
## How do you determine if a quadratic equation opens up or down?
There is an easy way to tell whether the graph of a quadratic function opens upward or downward: if the leading coefficient is greater than zero, the parabola opens upward, and if the leading coefficient is less than zero, the parabola opens downward.
## How do you plot a quadratic graph?
Have a go
1. Click to see a step-by-step slideshow.
2. Step 1 – The x axis goes from –2 to 2.
3. Step 2 – Create a table for the x and y values that you will calculate to plot the graph.
4. Step 3 – Find the values for y.
5. Step 4 – Repeat this process for the remaining values, where x = -1, x = 0, x = 1 and x = 2.
## How do you graph a standard form equation?
First, find the intercepts by setting y and then x equal to zero. This is pretty straightforward since the line is already in standard form. Plot the x and y-intercepts, which in this case is (9, 0) and (0, 6) and draw the line on the graph paper!
## How do you convert Y intercept to standard form?
To convert from slope intercept form y = mx + b to standard form Ax + By + C = 0, let m = A/B, collect all terms on the left side of the equation and multiply by the denominator B to get rid of the fraction.
## How do you rewrite an equation in standard form?
The standard form of a linear equation is Ax+By=C A x + B y = C . Rewrite the equation with the sides flipped. Move all terms containing variables to the left side of the equation. Subtract y y from both sides of the equation.
## What are the rules of standard form?
The Standard Form for a linear equation in two variables, x and y, is usually given as Ax + By = C where, if at all possible, A, B, and C are integers, and A is non-negative, and, A, B, and C have no common factors other than 1.
## What are standard form numbers?
Any number that we can write as a decimal number, between 1.0 and 10.0, multiplied by a power of 10, is said to be in standard form. 1.98 ✕ 10¹³; 0.76 ✕ 10¹³ are examples of numbers in standard form.
## Can a in standard form be negative?
Standard Form of a Linear Equation A shouldn’t be negative, A and B shouldn’t both be zero, and A, B and C should be integers.
#### Andrew
Andrey is a coach, sports writer and editor. He is mainly involved in weightlifting. He also edits and writes articles for the IronSet blog where he shares his experiences. Andrey knows everything from warm-up to hard workout.
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# What is the focus and Directrix?. ## What is the focus and Directrix?. A parabola is set of all points in a plane which are an equal distance away from a given point and given line. The point is called the focus of the parabola, and the line is called the directrix . If the axis of symmetry of a parabola is vertical, the directrix is a horizontal line .. ## How do you find the focus and Directrix in standard form?. The standard form is (x – h)2 = 4p (y – k), where the focus is (h, k + p) and the directrix is y = k – p. If the parabola is rotated so that its vertex is (h,k) and its axis of symmetry is parallel to the x-axis, it has an equation of (y – k)2 = 4p (x – h), where the focus is (h + p, k) and the directrix is x = h – p.. ## What is ellipse equation?. The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form. (x−h)2b2+(y−k)2a2=1 ( x − h ) 2 b 2 + ( y − k ) 2 a 2 = 1.. ## Where is the focus of a parabola?. A parabola is set of all points in a plane which are an equal distance away from a given point and given line. The point is called the focus of the parabola and the line is called the directrix. The focus lies on the axis of symmetry of the parabola.. ## How do you find the focus and directrix of a parabola on a calculator?. Parabola focus and directrix All you have to do is to use the following equations: Focus x-coordinate: x₀ = – b/(2a) ; Focus y-coordinate: y₀ = c – (b² – 1)/(4a) ; and. Directrix equation: y = c – (b² + 1)/(4a) .. ## How do you find the vertex of a focus and Directrix?. In the equations y=a(x-h)^2+k and x=a(y-k)^2+h where a=1/4p , (h,k ) represents vertex and p is the distance between focus and vertex . In the above equations (a,b) represents FOCUS not VERTEX . In the first one directrix is y=k and in second one directrix is x=k .. ## What is p value in parabola?. The absolute value of p is the distance between the vertex and the focus and the distance between the vertex and the directrix. (The sign on p tells me which way the parabola faces.) Since the focus and directrix are two units apart, then this distance has to be one unit, so | p | = 1.. ## How do you find the vertex in standard form?. y=a(x−h)(x−h)+ky=ax2−2ahx+ah2+k . This means that in the standard form, y=ax2+bx+c , the expression −b2a gives the x -coordinate of the vertex.. ## How do you find standard form?. The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.. ## How do you convert standard form to vertex form on a calculator?. How to convert from the standard form to the vertex form?. 1. Write the parabola equation in the standard form: y = a*x² + b*x + c ;. 2. Extract a from the first two terms: y = a * (x² + b/a * x) + c ;. 3. Complete the square for the expressions with x .. ## How do you write quadratic equations in standard form?. A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable.. ## How are quadratic equations used in real life?. Quadratic equations are actually used in everyday life, as when calculating areas, determining a product’s profit or formulating the speed of an object.
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Quadratic equations refer to equations with at least one squared variable, with the most standard form being ax² + bx + c = 0.. ## What does a quadratic equation look like on a graph?. The graph of a quadratic function is a U-shaped curve called a parabola. The sign on the coefficient a of the quadratic function affects whether the graph opens up or down. The x-intercepts are the points at which the parabola crosses the x-axis.. ## How do you determine if a quadratic equation opens up or down?. There is an easy way to tell whether the graph of a quadratic function opens upward or downward: if the leading coefficient is greater than zero, the parabola opens upward, and if the leading coefficient is less than zero, the parabola opens downward.. ## How do you plot a quadratic graph?. Have a go. 1. Click to see a step-by-step slideshow.. 2. Step 1 – The x axis goes from –2 to 2.. 3. Step 2 – Create a table for the x and y values that you will calculate to plot the graph.. 4. Step 3 – Find the values for y.. 5. Step 4 – Repeat this process for the remaining values, where x = -1, x = 0, x = 1 and x = 2.. ## How do you graph a standard form equation?. First, find the intercepts by setting y and then x equal to zero. This is pretty straightforward since the line is already in standard form. Plot the x and y-intercepts, which in this case is (9, 0) and (0, 6) and draw the line on the graph paper!. ## How do you convert Y intercept to standard form?. To convert from slope intercept form y = mx + b to standard form Ax + By + C = 0, let m = A/B, collect all terms on the left side of the equation and multiply by the denominator B to get rid of the fraction.. ## How do you rewrite an equation in standard form?. The standard form of a linear equation is Ax+By=C A x + B y = C . Rewrite the equation with the sides flipped. Move all terms containing variables to the left side of the equation. Subtract y y from both sides of the equation.. ## What are the rules of standard form?. The Standard Form for a linear equation in two variables, x and y, is usually given as Ax + By = C where, if at all possible, A, B, and C are integers, and A is non-negative, and, A, B, and C have no common factors other than 1.. ## What are standard form numbers?. Any number that we can write as a decimal number, between 1.0 and 10.0, multiplied by a power of 10, is said to be in standard form. 1.98 ✕ 10¹³; 0.76 ✕ 10¹³ are examples of numbers in standard form.. ## Can a in standard form be negative?. Standard Form of a Linear Equation A shouldn’t be negative, A and B shouldn’t both be zero, and A, B and C should be integers.. #### Andrew. Andrey is a coach, sports writer and editor. He is mainly involved in weightlifting. He also edits and writes articles for the IronSet blog where he shares his experiences. Andrey knows everything from warm-up to hard workout.
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# Quiz Discussion
Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar's age at present?
Course Name: Quantitative Aptitude
• 1]
16 years
• 2]
18 years
• 3]
20 years
• 4]
None of these
##### Solution
No Solution Present Yet
#### Top 5 Similar Quiz - Based On AI&ML
Quiz Recommendation System API Link - https://fresherbell-quiz-api.herokuapp.com/fresherbell_quiz_api
# Quiz
1
Discuss
The ages of Shakti and Kanti are in the ratio of 8 : 7 respectively. After 10 years, the ratio of their ages will be 13 : 12. What is the difference between their ages ?
• 1] 2 Years
• 2] 4 Years
• 3] 8 Years
• 4] 6 Years
• 5] None of these
##### Solution
2
Discuss
The age of a father 10 years ago was thrice the age of his son. 10 years hence , the father's age will be twice that of his son. The ratio of their present age is = ?
• 1] 8 : 5
• 2] 7 : 3
• 3] 9 : 5
• 4] 5 : 2
##### Solution
3
Discuss
A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B?
• 1]
7
• 2]
8
• 3]
9
• 4]
10
##### Solution
4
Discuss
Ayesha's father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents ?
• 1]
2 years
• 2]
4 years
• 3]
6 years
• 4]
8 years
##### Solution
5
Discuss
The respective ratio between the present age Aarti and Savita is 5 : x. Aartl is 9 years younger than Jahnavi. Jahnavi's age after 9 years will be 33 years. The difference between Savita's and Aarti's age is same as the present age of Jahnavi. What will come in place of x?
• 1]
21
• 2]
37
• 3]
17
• 4]
13
##### Solution
Jahnavi's present age= 33 - 9 = 24 years
Aarti's present age= 24 - 9 = 15 years
Aarti : Savita= 5 : x = 15 : 3x
Therefore, Savita present age= 3x years
Therefore, 3x - 15 = 24
=> 3x = 24 + 15 = 39
=> x = 393393 = 13
6
Discuss
Ratio of ages of three persons is 4:7:9, Eight years ago, the sum of their ages was 56. Find their present ages
• 1]
16,35,36
• 2]
12,28,36
• 3]
16,28,27
• 4]
16,28,36
##### Solution
7
Discuss
Farah got married 8 years ago, Today her age is $$1\frac{2}{7}$$ times her age at the time of her marriage. At present her daughter's age is one-sixth of her age. What was her daughter's age 3 years ago ?
• 1] 4 Years
• 2] 3 Years
• 3] 6 Years
• 4] Cannot be determind
• 5] None of these
##### Solution
8
Discuss
The ratio between the parents ages of A and B is 5 : 3 respectively. The ratio between A's age 4 years ago and B's 4 years hence is 1 : 1. What is the ratio between A's age 4 years hence and B's age 4 years ago ?
• 1] 1 : 3
• 2] 3 : 1
• 3] 2 : 1
• 4] 4 : 1
• 5] None of these
##### Solution
9
Discuss
The ages of Nitish and Vinnee are in the ratio 6 : 5 respectively. After 9 years the ratio of their ages will be 9 : 8. What is the difference in their ages now ?
• 1] 3 Years
• 2] 5 Years
• 3] 7 Years
• 4] 9 Years
• 5] None of these
##### Solution
10
Discuss
The age of a man 10 years ago was thrice the age of his son. 10 years hence, the man's age will be twice the age of his son. The ratio of their present ages is = ?
• 1] 5 : 2
• 2] 7 : 3
• 3] 9 : 2
• 4] 13 : 4
# Quiz
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# Quiz Discussion. Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar's age at present?. Course Name: Quantitative Aptitude. • 1]. 16 years. • 2]. 18 years. • 3]. 20 years. • 4]. None of these. ##### Solution. No Solution Present Yet. #### Top 5 Similar Quiz - Based On AI&ML. Quiz Recommendation System API Link - https://fresherbell-quiz-api.herokuapp.com/fresherbell_quiz_api. # Quiz. 1. Discuss. The ages of Shakti and Kanti are in the ratio of 8 : 7 respectively. After 10 years, the ratio of their ages will be 13 : 12. What is the difference between their ages ?. • 1] 2 Years. • 2] 4 Years. • 3] 8 Years. • 4] 6 Years. • 5] None of these. ##### Solution. 2. Discuss. The age of a father 10 years ago was thrice the age of his son. 10 years hence , the father's age will be twice that of his son. The ratio of their present age is = ?. • 1] 8 : 5. • 2] 7 : 3. • 3] 9 : 5. • 4] 5 : 2. ##### Solution. 3. Discuss. A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B?. • 1]. 7. • 2]. 8. • 3]. 9. • 4]. 10. ##### Solution. 4. Discuss. Ayesha's father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents ?. • 1]. 2 years. • 2]. 4 years. • 3]. 6 years. • 4]. 8 years. ##### Solution. 5. Discuss. The respective ratio between the present age Aarti and Savita is 5 : x. Aartl is 9 years younger than Jahnavi. Jahnavi's age after 9 years will be 33 years. The difference between Savita's and Aarti's age is same as the present age of Jahnavi. What will come in place of x?. • 1].
|
21. • 2]. 37. • 3]. 17. • 4]. 13. ##### Solution. Jahnavi's present age= 33 - 9 = 24 years. Aarti's present age= 24 - 9 = 15 years. Aarti : Savita= 5 : x = 15 : 3x. Therefore, Savita present age= 3x years. Therefore, 3x - 15 = 24. => 3x = 24 + 15 = 39. => x = 393393 = 13. 6. Discuss. Ratio of ages of three persons is 4:7:9, Eight years ago, the sum of their ages was 56. Find their present ages. • 1]. 16,35,36. • 2]. 12,28,36. • 3]. 16,28,27. • 4]. 16,28,36. ##### Solution. 7. Discuss. Farah got married 8 years ago, Today her age is $$1\frac{2}{7}$$ times her age at the time of her marriage. At present her daughter's age is one-sixth of her age. What was her daughter's age 3 years ago ?. • 1] 4 Years. • 2] 3 Years. • 3] 6 Years. • 4] Cannot be determind. • 5] None of these. ##### Solution. 8. Discuss. The ratio between the parents ages of A and B is 5 : 3 respectively. The ratio between A's age 4 years ago and B's 4 years hence is 1 : 1. What is the ratio between A's age 4 years hence and B's age 4 years ago ?. • 1] 1 : 3. • 2] 3 : 1. • 3] 2 : 1. • 4] 4 : 1. • 5] None of these. ##### Solution. 9. Discuss. The ages of Nitish and Vinnee are in the ratio 6 : 5 respectively. After 9 years the ratio of their ages will be 9 : 8. What is the difference in their ages now ?. • 1] 3 Years. • 2] 5 Years. • 3] 7 Years. • 4] 9 Years. • 5] None of these. ##### Solution. 10. Discuss. The age of a man 10 years ago was thrice the age of his son. 10 years hence, the man's age will be twice the age of his son. The ratio of their present ages is = ?. • 1] 5 : 2. • 2] 7 : 3. • 3] 9 : 2. • 4] 13 : 4. # Quiz.
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How many is
Conversion between units of measurement
You can easily convert 4 miles into yards using each unit definition:
Miles
5280 ft = 1609.344 m
Yards
yard = 3 ft = 0.9144 m
With this information, you can calculate the quantity of yards 4 miles is equal to.
## ¿How many yd are there in 4 mi?
In 4 mi there are 7040 yd.
Which is the same to say that 4 miles is 7040 yards.
Four miles equals to seven thousand forty yards. *Approximation
### ¿What is the inverse calculation between 1 yard and 4 miles?
Performing the inverse calculation of the relationship between units, we obtain that 1 yard is 0.00014204545 times 4 miles.
A yard is zero times four miles. *Approximation
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How many is. Conversion between units of measurement. You can easily convert 4 miles into yards using each unit definition:. Miles. 5280 ft = 1609.344 m. Yards. yard = 3 ft = 0.9144 m. With this information, you can calculate the quantity of yards 4 miles is equal to.. ## ¿How many yd are there in 4 mi?.
|
In 4 mi there are 7040 yd.. Which is the same to say that 4 miles is 7040 yards.. Four miles equals to seven thousand forty yards. *Approximation. ### ¿What is the inverse calculation between 1 yard and 4 miles?. Performing the inverse calculation of the relationship between units, we obtain that 1 yard is 0.00014204545 times 4 miles.. A yard is zero times four miles. *Approximation.
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http://math.stackexchange.com/questions/54603/group-multiplication-table
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# Group Multiplication Table
I'm currently trying to learn abstract algebra myself, and the following is a quote from the book I am using, "A set of equations, involving only the generators and their inverses, is called a set of defining equations for $G$ if these equations completely determine the multiplication table of $G$."
Then the book proceeds to give an example: "Let $G$ be the group $\{e, a, b, b^{2}, ab, ab^{2} \}$ whose generators $a$ and $b$ satisfy the equations $a^{2} = e$, $b^{3} = e$, and $ba = ab^{2}$." And claims that the three equations determine the multiplication table of $G$.
So I worked out the multiplication table and displayed it below.
• When they say, "completely determine the multiplication table of $G$," does that mean the product of two elements can be simplified to another element? For example, $(ab^{2})(ab^{2}) = ab(ba)bb = ab(ab^{2})b^{2} = abab(b^{3}) = a(ba)b = a(ab^{2})b = aab^{3} = e.$
• I also don't see how inverses are used in determining the multiplication table in this case. I've only used substitution in this case. Can someone explain why inverses might be important?
• How did the author know that only 3 equations were enough to determine the multiplication table? And why did he choose those equations?
• Also what is the significance of determining a multiplication table for elements of a group?
-
What text is this? – Pete L. Clark Jul 30 '11 at 14:35
@Pete: This is from the 2nd edition of Pinter's Abstract Algebra on pg 47-48. – Student Jul 30 '11 at 14:42
I am missing something but why is $b(ab^2) = a$? Should it not be $ab$? – user38268 Aug 1 '11 at 12:24
@D Lim : Indeed, the table is wrong: look at the row for $b$ and the column for $ab^2$. A typo I suppose. – yatima2975 Aug 1 '11 at 14:01
1. "Completely determine" means two things simultaneously. Firstly, "determine" here means that each entry in the multiplication table can be filled in by one of those six elements, which is basically what you said. Secondly, "completely" means that there is no ambiguity: some entries in the multiplication table can be worked out in different ways, and no matter how many different ways you try to work some given entry out, you will get the same thing.
2. It's entirely conceivable that you might be faced with an equation like $ab^{-1} = cd^{-1}$. No matter how hard you try to get rid of the inverses, you can't.
3. The author knew because this is a very standard group called $D_6$ (sometimes called $D_3$). You'll know it too once you meet the dihedral groups. But that's not a good answer. Another way to motivate this: notice that every element that he claims lives in that six-element set is of the form $a^ib^j$, where $i$ is 0 or 1 and $j$ is 0, 1 or 2. But what about $ba$, or other things that don't fit into this form? Our group axioms tell us that, since we have an element $b$ and an element $a$, we should be able to multiply them. So we need to know what to do when we have something of the 'wrong' form. And these wrong forms can be broken up into three types:
• $a^i$ where $i > 1$ or $i < 0$,
• $b^j$ where $j > 2$ or $j < 0$,
• $ba$ (multiplied the 'wrong' way round).
These three relations tell us exactly what to do in each of those situations.
4. Exactly the same reason children write out their times tables. It shows you what's going on! It gives you much more of a clue of the structure of the group than you might expect at this point; on the other hand, you're right to be suspicious of them, because nobody uses them after a certain level. But pedagogically they're very important.
-
Fun fact to know: in 1992 Ales Drápal proved that if two finite groups agree on 89% on their multiplication tables, the groups must be isomorphic! He conjectured that the same holds true if the tables agree on 75% of their entries. The conjecture has not been proved yet. See also Groups St. Andrews 2001 at Oxford, featuring the paper of Drápal On the distance of 2-groups and 3-groups.
-
Interesting! I take it 75% is sharp because of $C_4$ versus $C_2 \times C_2$? Do you have an electronic version that's not behind a paywall, by any chance? – yatima2975 Aug 1 '11 at 14:08
I have the book (two volumes). Maybe a math library near you has it too. – Nicky Hekster Aug 1 '11 at 21:10
Maybe I'll venture out to the campus one day, but then I'll have to remember to look for it... It may be easier to prove it myself - 89% is 8/9, I suppose? That should be a hint! – yatima2975 Aug 1 '11 at 21:23
@yatima2975, I looked it up, the "seeding" paper is A. Drápal, How far apart can the group multiplication tables be?, European J. Combin. 13 (1992), no. 5, 335–343. If you Google on the title you will find a slew of other interesting papers. And yes, the 89% is 8/9 ... – Nicky Hekster Aug 4 '11 at 11:35
It has been proved for 78%, no? arxiv.org/pdf/1107.0133 – user641 Jan 6 '12 at 9:44
The use of inverses (or more precisely, the use of negative exponents) is superfluous in the case of finite groups. In a finite group, every element has finite order. In other words, for every element $a$, there is a positive integer $k$ such that $a^k=e$. Hence instead of writing $a^{-1}$, one can write $a^{k-1}$, which is the same thing. In your example $b^{-1}=b^2$.
On the other hand, as soon as your group has elements of infinite order, it is no longer possible to write inverses as positive powers of the elements, and so the "$-1$" notation may become necessary. This can only happen in groups of infinite size.
The three relations in your example tell you
1. what the order of $a$ is,
2. what the order of $b$ is,
3. how $a$ and $b$ move past each other.
With two generators, this is enough to determine the results of arbitrarily complicated multiplications, so you won't need additional relations.
In groups of large size, it may be impractical to write out the multiplication table. What's important is being able to produce the multiplication table in principle. The group is only defined once you've specified how to multiply arbitrary elements of the group. The multiplication table is one way to do that, but if you can prove that your relations allow you to compute arbitrary products, that's fine too.
-
1. Yes, the author means that the product of any two elements can always be determined by referencing and manipulating the facts from defining set of equations.
2. Inverses aren't relevant in the given example because they aren't present. If the author replaced the first equation with the equivalent form $a = a^{-1}$, then you could say they're relevant. It's all just a matter of acceptable presentation.
3. I might not be familiar enough with abstract algebra to know this, but my guess is that the author actually started with the defining set of equations and then swiftly established a multiplication table. It's also possible they just tinkered with the algebra a bit until they found an acceptable example for the text - this sort of thing is absolutely trivial for people who know this stuff inside and out.
4. It presents the group's entire multiplicative structure for easy reference.
-
"this sort of thing is absolutely trivial for people who know this stuff inside and out" does not seem particularly useful :-) – Srivatsan Jul 30 '11 at 15:02
A group multiplication table is always a latin square. A latin square is a group multiplication table if it satisfies associativity. So technically you can find all groups of order $n$ by finding all associative $n\times n$ latin squares.
-
I am sure nicer answers will be posted, but I will post my attempt at explanation, though I am a group theory amateur myself. (Sorry for the extremely long answer!) Also check the wikipedia page for more details and for precise definitions/theorems.
This is an example of a "presentation" of a group. A few definitions will be useful for the discussion. The elements $a$ and $b$ are called the "generators" of the group, and the "equations" $a^2 = e$, $b^3 = e$ and $ba = ab^2$ are called "relations". A "word" is a string involving the generators and their inverses (e.g.: $a^{-2} b^{-4} bab^2$). Note that it is sometimes straightforward to rewrite a word in terms of just the generators. For instance, in this example, we can rewrite $a^{-1}$ as $a$ and $b^{-1}$ as $b^2$; when we do this $bab^2 a^{-2} b^{-6}$ becomes $b a b^2a^{2} b^{12}$.
Completely determining the multiplication table
Yes, it means that the product of two elements can be expressed as one of these elements themselves. This can be done by simplifying the product by repeatedly using the relations $a^2 = e, b^3 = e, ba = ab^2$. (In simple examples such as this, you can usually do routing group operations just by inspection.)
Where are the inverses? Inverses are automatically and implicitly defined by the relations that you have prescribed. For instance, $a^{-1} = a$ and $b^{-1} = b^2$. Why? Because the relation $b^3 = e$ implies that multiplying $b$ by $b^2$ (to the right or to the left) gives you $e$. So, $b^2$ must be the inverse of $b$. You can similarly check that $a^{-1}$ is $a$. (@WillO's comment that you can always write $b^{-1}$ as a power of $b$ in a finite group is relevant here.)
To compute the inverse of words (like $bab^2 a^{-2} b^{-6}$), you can use the identity $(xy)^{-1} = y^{-1} x^{-1}$. For example, we have $(bab^2 a^{-2} b^{-6})^{-1} = b^{6} a^2 b^{-2} a^{-1} b^{-1}$. This can, of course, be simplified further using the usual $a^{-1}=a$ and $b^{-1} = b^2$ trick.
Why 3 equations? In one sense, $3$ equations is not sacred. You can take any set of generators and any set of relations between the generators, and you will end up with some group. For instance, taking the generators to be $a$ and $b$, and the set of relations to be the empty set, you will get what is called the free group on $\{a, b\}$.
But not all examples that we obtain this way are nice; for instance, the group may turn out to be infinite or even uncountable. If you want to express your finite group in terms of a presentation, that is also quite easy. We will come to this point later.
Why is a multiplication table significant?
Pedagogical value. A multiplication table is a proof that the group that the author defines is in fact a group. I remember that when I first read about representing groups using relations, I used to hand-compute the whole table, and check the group axioms. Though tedious even for medium-sized groups, the exercise helped me intuitively understand the idea of group presentations and also taught me some simple tricks in simplifying the products of elements of a finitely generated group.
As a finite presentation. The multiplication table of a finite group $G$ also automatically gives you a finite presentation of the group. (I found this point in the Wikipedia page while researching for writing this answer.) Simply define the elements of $G$ as the generators, and for each entry $g_i g_j = g_k$ in the multiplication table, add a relation $g_i g_j g_k^{-1} = e$. (In this sense, finite presentation is a generalization of finite number of elements.) But of course, out of the $|G| \times |G|$ relations that we can write this way, only a few are useful; the remaining can be deduced from other relations. For example, you show how to deduce the relation $(ab^2)(ab^2)e^{-1} = e$ from the usual relations $a^2 = e$, $b^3=e$ and $ba=ab^2$.
-
To amplify what others have said about the multiplication table, by inspection we can see this group doesn't satisfy commutation, it isn't an idempotent group, it has neutral (or identity element) of e, and the inverse of each element can also get determined readily.
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# Group Multiplication Table. I'm currently trying to learn abstract algebra myself, and the following is a quote from the book I am using, "A set of equations, involving only the generators and their inverses, is called a set of defining equations for $G$ if these equations completely determine the multiplication table of $G$.". Then the book proceeds to give an example: "Let $G$ be the group $\{e, a, b, b^{2}, ab, ab^{2} \}$ whose generators $a$ and $b$ satisfy the equations $a^{2} = e$, $b^{3} = e$, and $ba = ab^{2}$." And claims that the three equations determine the multiplication table of $G$.. So I worked out the multiplication table and displayed it below.. • When they say, "completely determine the multiplication table of $G$," does that mean the product of two elements can be simplified to another element? For example, $(ab^{2})(ab^{2}) = ab(ba)bb = ab(ab^{2})b^{2} = abab(b^{3}) = a(ba)b = a(ab^{2})b = aab^{3} = e.$. • I also don't see how inverses are used in determining the multiplication table in this case. I've only used substitution in this case. Can someone explain why inverses might be important?. • How did the author know that only 3 equations were enough to determine the multiplication table? And why did he choose those equations?. • Also what is the significance of determining a multiplication table for elements of a group?. -. What text is this? – Pete L. Clark Jul 30 '11 at 14:35. @Pete: This is from the 2nd edition of Pinter's Abstract Algebra on pg 47-48. – Student Jul 30 '11 at 14:42. I am missing something but why is $b(ab^2) = a$? Should it not be $ab$? – user38268 Aug 1 '11 at 12:24. @D Lim : Indeed, the table is wrong: look at the row for $b$ and the column for $ab^2$. A typo I suppose. – yatima2975 Aug 1 '11 at 14:01. 1. "Completely determine" means two things simultaneously. Firstly, "determine" here means that each entry in the multiplication table can be filled in by one of those six elements, which is basically what you said. Secondly, "completely" means that there is no ambiguity: some entries in the multiplication table can be worked out in different ways, and no matter how many different ways you try to work some given entry out, you will get the same thing.. 2. It's entirely conceivable that you might be faced with an equation like $ab^{-1} = cd^{-1}$. No matter how hard you try to get rid of the inverses, you can't.. 3. The author knew because this is a very standard group called $D_6$ (sometimes called $D_3$). You'll know it too once you meet the dihedral groups. But that's not a good answer. Another way to motivate this: notice that every element that he claims lives in that six-element set is of the form $a^ib^j$, where $i$ is 0 or 1 and $j$ is 0, 1 or 2. But what about $ba$, or other things that don't fit into this form? Our group axioms tell us that, since we have an element $b$ and an element $a$, we should be able to multiply them. So we need to know what to do when we have something of the 'wrong' form. And these wrong forms can be broken up into three types:. • $a^i$ where $i > 1$ or $i < 0$,. • $b^j$ where $j > 2$ or $j < 0$,. • $ba$ (multiplied the 'wrong' way round).. These three relations tell us exactly what to do in each of those situations.. 4. Exactly the same reason children write out their times tables. It shows you what's going on! It gives you much more of a clue of the structure of the group than you might expect at this point; on the other hand, you're right to be suspicious of them, because nobody uses them after a certain level. But pedagogically they're very important.. -. Fun fact to know: in 1992 Ales Drápal proved that if two finite groups agree on 89% on their multiplication tables, the groups must be isomorphic! He conjectured that the same holds true if the tables agree on 75% of their entries. The conjecture has not been proved yet. See also Groups St. Andrews 2001 at Oxford, featuring the paper of Drápal On the distance of 2-groups and 3-groups.. -. Interesting! I take it 75% is sharp because of $C_4$ versus $C_2 \times C_2$? Do you have an electronic version that's not behind a paywall, by any chance? – yatima2975 Aug 1 '11 at 14:08. I have the book (two volumes). Maybe a math library near you has it too. – Nicky Hekster Aug 1 '11 at 21:10. Maybe I'll venture out to the campus one day, but then I'll have to remember to look for it... It may be easier to prove it myself - 89% is 8/9, I suppose? That should be a hint! – yatima2975 Aug 1 '11 at 21:23. @yatima2975, I looked it up, the "seeding" paper is A. Drápal, How far apart can the group multiplication tables be?, European J. Combin. 13 (1992), no. 5, 335–343. If you Google on the title you will find a slew of other interesting papers. And yes, the 89% is 8/9 ... – Nicky Hekster Aug 4 '11 at 11:35. It has been proved for 78%, no? arxiv.org/pdf/1107.0133 – user641 Jan 6 '12 at 9:44. The use of inverses (or more precisely, the use of negative exponents) is superfluous in the case of finite groups. In a finite group, every element has finite order. In other words, for every element $a$, there is a positive integer $k$ such that $a^k=e$. Hence instead of writing $a^{-1}$, one can write $a^{k-1}$, which is the same thing. In your example $b^{-1}=b^2$.. On the other hand, as soon as your group has elements of infinite order, it is no longer possible to write inverses as positive powers of the elements, and so the "$-1$" notation may become necessary.
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This can only happen in groups of infinite size.. The three relations in your example tell you. 1. what the order of $a$ is,. 2. what the order of $b$ is,. 3. how $a$ and $b$ move past each other.. With two generators, this is enough to determine the results of arbitrarily complicated multiplications, so you won't need additional relations.. In groups of large size, it may be impractical to write out the multiplication table. What's important is being able to produce the multiplication table in principle. The group is only defined once you've specified how to multiply arbitrary elements of the group. The multiplication table is one way to do that, but if you can prove that your relations allow you to compute arbitrary products, that's fine too.. -. 1. Yes, the author means that the product of any two elements can always be determined by referencing and manipulating the facts from defining set of equations.. 2. Inverses aren't relevant in the given example because they aren't present. If the author replaced the first equation with the equivalent form $a = a^{-1}$, then you could say they're relevant. It's all just a matter of acceptable presentation.. 3. I might not be familiar enough with abstract algebra to know this, but my guess is that the author actually started with the defining set of equations and then swiftly established a multiplication table. It's also possible they just tinkered with the algebra a bit until they found an acceptable example for the text - this sort of thing is absolutely trivial for people who know this stuff inside and out.. 4. It presents the group's entire multiplicative structure for easy reference.. -. "this sort of thing is absolutely trivial for people who know this stuff inside and out" does not seem particularly useful :-) – Srivatsan Jul 30 '11 at 15:02. A group multiplication table is always a latin square. A latin square is a group multiplication table if it satisfies associativity. So technically you can find all groups of order $n$ by finding all associative $n\times n$ latin squares.. -. I am sure nicer answers will be posted, but I will post my attempt at explanation, though I am a group theory amateur myself. (Sorry for the extremely long answer!) Also check the wikipedia page for more details and for precise definitions/theorems.. This is an example of a "presentation" of a group. A few definitions will be useful for the discussion. The elements $a$ and $b$ are called the "generators" of the group, and the "equations" $a^2 = e$, $b^3 = e$ and $ba = ab^2$ are called "relations". A "word" is a string involving the generators and their inverses (e.g.: $a^{-2} b^{-4} bab^2$). Note that it is sometimes straightforward to rewrite a word in terms of just the generators. For instance, in this example, we can rewrite $a^{-1}$ as $a$ and $b^{-1}$ as $b^2$; when we do this $bab^2 a^{-2} b^{-6}$ becomes $b a b^2a^{2} b^{12}$.. Completely determining the multiplication table. Yes, it means that the product of two elements can be expressed as one of these elements themselves. This can be done by simplifying the product by repeatedly using the relations $a^2 = e, b^3 = e, ba = ab^2$. (In simple examples such as this, you can usually do routing group operations just by inspection.). Where are the inverses? Inverses are automatically and implicitly defined by the relations that you have prescribed. For instance, $a^{-1} = a$ and $b^{-1} = b^2$. Why? Because the relation $b^3 = e$ implies that multiplying $b$ by $b^2$ (to the right or to the left) gives you $e$. So, $b^2$ must be the inverse of $b$. You can similarly check that $a^{-1}$ is $a$. (@WillO's comment that you can always write $b^{-1}$ as a power of $b$ in a finite group is relevant here.). To compute the inverse of words (like $bab^2 a^{-2} b^{-6}$), you can use the identity $(xy)^{-1} = y^{-1} x^{-1}$. For example, we have $(bab^2 a^{-2} b^{-6})^{-1} = b^{6} a^2 b^{-2} a^{-1} b^{-1}$. This can, of course, be simplified further using the usual $a^{-1}=a$ and $b^{-1} = b^2$ trick.. Why 3 equations? In one sense, $3$ equations is not sacred. You can take any set of generators and any set of relations between the generators, and you will end up with some group. For instance, taking the generators to be $a$ and $b$, and the set of relations to be the empty set, you will get what is called the free group on $\{a, b\}$.. But not all examples that we obtain this way are nice; for instance, the group may turn out to be infinite or even uncountable. If you want to express your finite group in terms of a presentation, that is also quite easy. We will come to this point later.. Why is a multiplication table significant?. Pedagogical value. A multiplication table is a proof that the group that the author defines is in fact a group. I remember that when I first read about representing groups using relations, I used to hand-compute the whole table, and check the group axioms. Though tedious even for medium-sized groups, the exercise helped me intuitively understand the idea of group presentations and also taught me some simple tricks in simplifying the products of elements of a finitely generated group.. As a finite presentation. The multiplication table of a finite group $G$ also automatically gives you a finite presentation of the group. (I found this point in the Wikipedia page while researching for writing this answer.) Simply define the elements of $G$ as the generators, and for each entry $g_i g_j = g_k$ in the multiplication table, add a relation $g_i g_j g_k^{-1} = e$. (In this sense, finite presentation is a generalization of finite number of elements.) But of course, out of the $|G| \times |G|$ relations that we can write this way, only a few are useful; the remaining can be deduced from other relations. For example, you show how to deduce the relation $(ab^2)(ab^2)e^{-1} = e$ from the usual relations $a^2 = e$, $b^3=e$ and $ba=ab^2$.. -. To amplify what others have said about the multiplication table, by inspection we can see this group doesn't satisfy commutation, it isn't an idempotent group, it has neutral (or identity element) of e, and the inverse of each element can also get determined readily.. -.
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### Sample Problem
Use the Law of Sines to solve this triangle. Round the lengths of the sides to three decimal places.
C= °
Side b= ft
Side c= ft
#### Solution
C=180°-35°-44°=101°
Side b=15/sin 44 * (sin 35)=12.385
Side c=15/sin 44 * (sin 101)=21.197
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### Sample Problem. Use the Law of Sines to solve this triangle. Round the lengths of the sides to three decimal places.. C= °. Side b= ft. Side c= ft.
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#### Solution. C=180°-35°-44°=101°. Side b=15/sin 44 * (sin 35)=12.385. Side c=15/sin 44 * (sin 101)=21.197.
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# MA.912.FL.1.2
Extend previous knowledge of ratios and proportional relationships to solve real-world problems involving money and business.
### Examples
Example: A local grocery stores sells trail mix for \$1.75 per pound. If the grocery store spends \$0.82 on each pound of mix, how much will the store gain in gross profit if they sell 6.4 pounds in one day?
Example: If Juan makes \$25.00 per hour and works 40 hours per week, what is his annual salary?
General Information
Subject Area: Mathematics (B.E.S.T.)
Strand: Financial Literacy
Status: State Board Approved
## Benchmark Instructional Guide
### Terms from the K-12 Glossary
• Proportional relationships
### Vertical Alignment
Previous Benchmarks
Next Benchmarks
### Purpose and Instructional Strategies
In grade 7, students have worked with ratios and proportional relationships to solve real-world problems. In Math for Data and Financial Literacy, students utilize their understanding of proportional relationships and problem solving to solve real-world problems involving money and business, including currency exchange, taxes and simple interest. Students also use concepts of proportionality when working with relative frequencies in data.
• Throughout instruction, it will be important to help students connect the mathematical concepts to everyday experiences (MTR.7.1) as they validate conclusions by comparing them to a given situation.
• Instruction for this benchmark includes opportunities to compare two different proportional relationships to each other. Allow various methods for solving, encouraging discussion and analysis of efficient and effective solutions (MTR.4.1).
• Students should understand the difference between flat rate taxes and other tax rates. Flat rate taxes are proportional tax rates such as sales tax. Comparatively, a progressive tax is one that can vary such as an income-based tax rate increasing when income increases.
### Common Misconceptions or Errors
• Students may not understand the difference between an additive comparison and a multiplicative comparison. To help address this misconception, instruction includes the understanding that proportions are multiplicative comparisons.
• Students may incorrectly set up proportions with one of the ratios having incorrect numbers in the numerator and denominator.
• Students may need information to clarify flat tax rates versus other tax rates. When working with students related to business and taxes in this benchmark include discussions on flat tax rates.
• An online clothing store, B’s Boutique sells 24 pieces of clothing every 30 minutes, and an online athletic store sells 5 shirts every 12 minutes.
• Part A. Estimate how long it will take each store to sell 100 pieces of clothing.
• Part B. Once an online store sells 200 items, the parent company for the online site gives the selling company a bonus of 20% of their sales. Will either store make a 20% bonus in a 24-hour period?
• At a local farm in Ruskin, Florida, a box of tomatoes sells for \$8.50.
• Part A. How many boxes would they need to sell to reach the sales goal of \$8,000?
• Part B. If the farm currently employs 10 workers who can fill 285 boxes in 3 days, how many days will it take them to fill enough boxes for the farm to make more than \$8,000 on the sales of the boxes of tomatoes?
### Instructional Items
Instructional Item 1
• At We Play Sports store, they try to keep their inventory at a ratio of 7 to 4 for used equipment to new equipment. If the total inventory amount for the new equipment is \$50,250, what is the amount of inventory for the used equipment?
Instructional Item 2
• The Easy Connect company makes straps to keep luggage closed. They make some with a snap lock and some with locks that are either a combination lock or work with a special luggage app. Based on sales, they typically make 20% with a snap lock, 55% with a combination lock and 25% with an app lock. If they typically sell 1,250 straps a month, how many would they expect to sell that have combination locking straps?
*The strategies, tasks and items included in the B1G-M are examples and should not be considered comprehensive.
## Related Courses
This benchmark is part of these courses.
1200400: Foundational Skills in Mathematics 9-12 (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current))
1200388: Mathematics for Data and Financial Literacy Honors (Specifically in versions: 2022 and beyond (current))
1700600: GEAR Up 1 (Specifically in versions: 2020 - 2022, 2022 and beyond (current))
1700610: GEAR Up 2 (Specifically in versions: 2020 - 2022, 2022 and beyond (current))
1700620: GEAR Up 3 (Specifically in versions: 2020 - 2022, 2022 and beyond (current))
1700630: GEAR Up 4 (Specifically in versions: 2020 - 2022, 2022 and beyond (current))
1200384: Mathematics for Data and Financial Literacy (Specifically in versions: 2022 and beyond (current))
7912120: Access Mathematics for Data and Financial Literacy (Specifically in versions: 2022 - 2023, 2023 and beyond (current))
1209315: Mathematics for ACT and SAT (Specifically in versions: 2022 and beyond (current))
2102305: Economics and Personal Finance Honors (Specifically in versions: 2023 and beyond (current))
## Related Access Points
Alternate version of this benchmark for students with significant cognitive disabilities.
MA.912.FL.1.AP.2: Solve simple real-world problems involving money using ratios or proportions.
## Related Resources
Vetted resources educators can use to teach the concepts and skills in this benchmark.
## Lesson Plan
Efficient Storage:
The topic of this MEA is work and power. Students will be assigned the task of hiring employees to complete a given task. In order to make a decision as to which candidates to hire, the students initially must calculate the required work. The power each potential employee is capable of, the days they are available to work, the percentage of work-shifts they have missed over the past 12 months, and the hourly pay rate each worker commands will be provided to assist in the decision process. Full- and/or part-time positions are available. Through data analysis, the students will need to evaluate which factors are most significant in the hiring process. For instance, some groups may prioritize speed of work, while others prioritize cost or availability/dependability.
Type: Lesson Plan
## STEM Lessons - Model Eliciting Activity
Efficient Storage:
The topic of this MEA is work and power. Students will be assigned the task of hiring employees to complete a given task. In order to make a decision as to which candidates to hire, the students initially must calculate the required work. The power each potential employee is capable of, the days they are available to work, the percentage of work-shifts they have missed over the past 12 months, and the hourly pay rate each worker commands will be provided to assist in the decision process. Full- and/or part-time positions are available. Through data analysis, the students will need to evaluate which factors are most significant in the hiring process. For instance, some groups may prioritize speed of work, while others prioritize cost or availability/dependability.
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# MA.912.FL.1.2. Extend previous knowledge of ratios and proportional relationships to solve real-world problems involving money and business.. ### Examples. Example: A local grocery stores sells trail mix for \$1.75 per pound. If the grocery store spends \$0.82 on each pound of mix, how much will the store gain in gross profit if they sell 6.4 pounds in one day?. Example: If Juan makes \$25.00 per hour and works 40 hours per week, what is his annual salary?. General Information. Subject Area: Mathematics (B.E.S.T.). Strand: Financial Literacy. Status: State Board Approved. ## Benchmark Instructional Guide. ### Terms from the K-12 Glossary. • Proportional relationships. ### Vertical Alignment. Previous Benchmarks. Next Benchmarks. ### Purpose and Instructional Strategies. In grade 7, students have worked with ratios and proportional relationships to solve real-world problems. In Math for Data and Financial Literacy, students utilize their understanding of proportional relationships and problem solving to solve real-world problems involving money and business, including currency exchange, taxes and simple interest. Students also use concepts of proportionality when working with relative frequencies in data.. • Throughout instruction, it will be important to help students connect the mathematical concepts to everyday experiences (MTR.7.1) as they validate conclusions by comparing them to a given situation.. • Instruction for this benchmark includes opportunities to compare two different proportional relationships to each other. Allow various methods for solving, encouraging discussion and analysis of efficient and effective solutions (MTR.4.1).. • Students should understand the difference between flat rate taxes and other tax rates. Flat rate taxes are proportional tax rates such as sales tax. Comparatively, a progressive tax is one that can vary such as an income-based tax rate increasing when income increases.. ### Common Misconceptions or Errors. • Students may not understand the difference between an additive comparison and a multiplicative comparison. To help address this misconception, instruction includes the understanding that proportions are multiplicative comparisons.. • Students may incorrectly set up proportions with one of the ratios having incorrect numbers in the numerator and denominator.. • Students may need information to clarify flat tax rates versus other tax rates. When working with students related to business and taxes in this benchmark include discussions on flat tax rates.. • An online clothing store, B’s Boutique sells 24 pieces of clothing every 30 minutes, and an online athletic store sells 5 shirts every 12 minutes.. • Part A. Estimate how long it will take each store to sell 100 pieces of clothing.. • Part B. Once an online store sells 200 items, the parent company for the online site gives the selling company a bonus of 20% of their sales. Will either store make a 20% bonus in a 24-hour period?. • At a local farm in Ruskin, Florida, a box of tomatoes sells for \$8.50.. • Part A. How many boxes would they need to sell to reach the sales goal of \$8,000?. • Part B. If the farm currently employs 10 workers who can fill 285 boxes in 3 days, how many days will it take them to fill enough boxes for the farm to make more than \$8,000 on the sales of the boxes of tomatoes?. ### Instructional Items. Instructional Item 1.
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• At We Play Sports store, they try to keep their inventory at a ratio of 7 to 4 for used equipment to new equipment. If the total inventory amount for the new equipment is \$50,250, what is the amount of inventory for the used equipment?. Instructional Item 2. • The Easy Connect company makes straps to keep luggage closed. They make some with a snap lock and some with locks that are either a combination lock or work with a special luggage app. Based on sales, they typically make 20% with a snap lock, 55% with a combination lock and 25% with an app lock. If they typically sell 1,250 straps a month, how many would they expect to sell that have combination locking straps?. *The strategies, tasks and items included in the B1G-M are examples and should not be considered comprehensive.. ## Related Courses. This benchmark is part of these courses.. 1200400: Foundational Skills in Mathematics 9-12 (Specifically in versions: 2014 - 2015, 2015 - 2022, 2022 and beyond (current)). 1200388: Mathematics for Data and Financial Literacy Honors (Specifically in versions: 2022 and beyond (current)). 1700600: GEAR Up 1 (Specifically in versions: 2020 - 2022, 2022 and beyond (current)). 1700610: GEAR Up 2 (Specifically in versions: 2020 - 2022, 2022 and beyond (current)). 1700620: GEAR Up 3 (Specifically in versions: 2020 - 2022, 2022 and beyond (current)). 1700630: GEAR Up 4 (Specifically in versions: 2020 - 2022, 2022 and beyond (current)). 1200384: Mathematics for Data and Financial Literacy (Specifically in versions: 2022 and beyond (current)). 7912120: Access Mathematics for Data and Financial Literacy (Specifically in versions: 2022 - 2023, 2023 and beyond (current)). 1209315: Mathematics for ACT and SAT (Specifically in versions: 2022 and beyond (current)). 2102305: Economics and Personal Finance Honors (Specifically in versions: 2023 and beyond (current)). ## Related Access Points. Alternate version of this benchmark for students with significant cognitive disabilities.. MA.912.FL.1.AP.2: Solve simple real-world problems involving money using ratios or proportions.. ## Related Resources. Vetted resources educators can use to teach the concepts and skills in this benchmark.. ## Lesson Plan. Efficient Storage:. The topic of this MEA is work and power. Students will be assigned the task of hiring employees to complete a given task. In order to make a decision as to which candidates to hire, the students initially must calculate the required work. The power each potential employee is capable of, the days they are available to work, the percentage of work-shifts they have missed over the past 12 months, and the hourly pay rate each worker commands will be provided to assist in the decision process. Full- and/or part-time positions are available. Through data analysis, the students will need to evaluate which factors are most significant in the hiring process. For instance, some groups may prioritize speed of work, while others prioritize cost or availability/dependability.. Type: Lesson Plan. ## STEM Lessons - Model Eliciting Activity. Efficient Storage:. The topic of this MEA is work and power. Students will be assigned the task of hiring employees to complete a given task. In order to make a decision as to which candidates to hire, the students initially must calculate the required work. The power each potential employee is capable of, the days they are available to work, the percentage of work-shifts they have missed over the past 12 months, and the hourly pay rate each worker commands will be provided to assist in the decision process. Full- and/or part-time positions are available. Through data analysis, the students will need to evaluate which factors are most significant in the hiring process. For instance, some groups may prioritize speed of work, while others prioritize cost or availability/dependability.
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www.delorie.com/gnu/docs/calc/calc_83.html search
GNU Emacs Calc 2.02 Manual
[ < ] [ > ] [ << ] [ Up ] [ >> ] [Top] [Contents] [Index] [ ? ]
### 3.7.49 Algebra Tutorial Exercise 2
Suppose our roots are [a, b, c]. We want a polynomial which is zero when x is any of these values. The trivial polynomial x-a is zero when x=a, so the product (x-a)(x-b)(x-c) will do the job. We can use a c x to write this in a more familiar form.
```1: 34 x - 24 x^3 1: [1.19023, -1.19023, 0] . . r 2 a P x RET ```
```1: [x - 1.19023, x + 1.19023, x] 1: (x - 1.19023) (x + 1.19023) x . . V M ' x-\$ RET V R * ```
```1: x^3 - 1.41666 x 1: 34 x - 24 x^3 . . a c x RET 24 n * a x ```
Sure enough, our answer (multiplied by a suitable constant) is the same as the original polynomial.
webmaster donations bookstore delorie software privacy Copyright © 2003 by The Free Software Foundation Updated Jun 2003
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www.delorie.com/gnu/docs/calc/calc_83.html search. GNU Emacs Calc 2.02 Manual. [ < ] [ > ] [ << ] [ Up ] [ >> ] [Top] [Contents] [Index] [ ? ]. ### 3.7.49 Algebra Tutorial Exercise 2. Suppose our roots are [a, b, c]. We want a polynomial which is zero when x is any of these values. The trivial polynomial x-a is zero when x=a, so the product (x-a)(x-b)(x-c) will do the job. We can use a c x to write this in a more familiar form.
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```1: 34 x - 24 x^3 1: [1.19023, -1.19023, 0] . . r 2 a P x RET ```. ```1: [x - 1.19023, x + 1.19023, x] 1: (x - 1.19023) (x + 1.19023) x . . V M ' x-\$ RET V R * ```. ```1: x^3 - 1.41666 x 1: 34 x - 24 x^3 . . a c x RET 24 n * a x ```. Sure enough, our answer (multiplied by a suitable constant) is the same as the original polynomial.. webmaster donations bookstore delorie software privacy Copyright © 2003 by The Free Software Foundation Updated Jun 2003.
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# 1 kilometer equals how many meters?
Updated: 12/24/2022
Wiki User
14y ago
1kilometer=1000 meters
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14y ago
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Q: 1 kilometer equals how many meters?
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Related questions
### -1 Kilometer equals how many meters?
1 kilometer is 1000 meters
### Kilometer equals how many meters?
1000 meters equals to 1kilometer
### How many meters are equals to 7 kilometers?
A: 7,000 meters since 1 kilometer equals 1,000 meters
### 1 kl equals how many meters?
1 kilometer = 1,000 meters
### A Kilometer equals how many meters?
0.001 kilometers = 1 meter. 1/1000 of a kilometer
### 7.3 kilometers equals how many meters?
7300 meters 1 kilometer = 1000 meters 1 meter = 0.001 kilometer
### Kilometer equals how meters?
1 kilometer is 1,000 meters.
### 250 meters equals how many killometers?
250 meters is 0.25 or 1/4 of a kilometer. There are 1000 meters in a kilometer.
### 7000 meters is equals to how many kilometers?
7 km 1 kilometer = 1000 meters 1 meter = 0.001 kilometer
### 7 kilometers and 8 meters equals 6 kilometers and how many meters?
1008 meters 1 kilometer = 1000 meters 1 meter = 0.001 kilometer
### 52 meters equals how many kilometers?
1 kilometer = 1000 meters 52 meters = 52/1000 kilometrs = 0.052 kilometer
### How many meters are there in halve a kilometer?
1/2km equals 500m
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0. # 1 kilometer equals how many meters?. Updated: 12/24/2022. Wiki User. 14y ago. 1kilometer=1000 meters. Wiki User. 14y ago. Earn +20 pts. Q: 1 kilometer equals how many meters?. Submit. Still have questions?. Related questions. ### -1 Kilometer equals how many meters?. 1 kilometer is 1000 meters. ### Kilometer equals how many meters?. 1000 meters equals to 1kilometer. ### How many meters are equals to 7 kilometers?. A: 7,000 meters since 1 kilometer equals 1,000 meters. ### 1 kl equals how many meters?.
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1 kilometer = 1,000 meters. ### A Kilometer equals how many meters?. 0.001 kilometers = 1 meter. 1/1000 of a kilometer. ### 7.3 kilometers equals how many meters?. 7300 meters 1 kilometer = 1000 meters 1 meter = 0.001 kilometer. ### Kilometer equals how meters?. 1 kilometer is 1,000 meters.. ### 250 meters equals how many killometers?. 250 meters is 0.25 or 1/4 of a kilometer. There are 1000 meters in a kilometer.. ### 7000 meters is equals to how many kilometers?. 7 km 1 kilometer = 1000 meters 1 meter = 0.001 kilometer. ### 7 kilometers and 8 meters equals 6 kilometers and how many meters?. 1008 meters 1 kilometer = 1000 meters 1 meter = 0.001 kilometer. ### 52 meters equals how many kilometers?. 1 kilometer = 1000 meters 52 meters = 52/1000 kilometrs = 0.052 kilometer. ### How many meters are there in halve a kilometer?. 1/2km equals 500m.
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<< problem 610 - Roman Numerals II Friend numbers - problem 612 >>
# Problem 611: Hallway of square steps
Peter moves in a hallway with N+1 doors consecutively numbered from 0 through N.
All doors are initially closed. Peter starts in front of door 0, and repeatedly performs the following steps:
• First, he walks a positive square number of doors away from his position.
• Then he walks another, larger square number of doors away from his new position.
• He toggles the door he faces (opens it if closed, closes it if open).
• And finally returns to door 0.
We call an action any sequence of those steps. Peter never performs the exact same action twice,
and makes sure to perform all possible actions that don't bring him past the last door.
Let F(N) be the number of doors that are open after Peter has performed all possible actions.
You are given that F(5) = 1, F(100) = 27, F(1000) = 233 and F(10^6) = 112168.
Find F(10^12).
# Very inefficient solution
My code needs more than 60 seconds to find the correct result. (scroll down to the benchmark section)
Apparantly a much smarter algorithm exists - or my implementation is just inefficient.
# My Algorithm
Yes, I got to admit: I just used brute force ... it solved the problem in under an hour and thus was my first time
where I solved the most recent problem and became one of the 100 fastest solvers.
Peter first walks to door i^2, takes a breath and then continues to door i^2 + j^2 where i > 0 and j > i.
He toggles the door's state (closed → open, open → closed).
All I need is to figure out a fast way to know how many doors have an odd number of ways to be represented as a sum of two squares.
I read a few websites about "sum of two squares" (e.g. en.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares) but could deduce a simple counting formula.
I process doors in chunks of 10^7 doors each, which requires 1.2 MByte per segment (std::vector<bool>).
Then I iterate over all "relevant" pairs i^2 + j^2 and negate the bit at that position.
The term "relevant" means that i^2 + j^2 has to be a value in the current segment: from <= i^2 + j^2 < to
Even though I have to iterate over all i, I can restrict j to "match" the current segment:
j_{min} = \lceil sqrt{from - i^2} \rceil
## Alternative Approaches
As mentioned above, Fermat's theorem is the way to go. You need a prime-counting function for primes 4k+1.
I have prime-counting functions in my toolbox but they need to be modified to eliminate 4k+3 primes.
## Note
From the perspective of a software engineer (and that's what I do for a living) I did the right thing:
instead of spending more than an hour on further reading, typing, debugging, etc., I kept the computer running for 49 minutes
(with #define PARALLEL and numCores = 6) and got the correct result.
That's not the proper scientific way to solve a problem ... but a sound real-world approach !
# Interactive test
You can submit your own input to my program and it will be instantly processed at my server:
Input data (separated by spaces or newlines):
This is equivalent to
echo 1000000 | ./611
Output:
Note: the original problem's input 1000000000000 cannot be entered
because just copying results is a soft skill reserved for idiots.
(this interactive test is still under development, computations will be aborted after one second)
# My code
… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.
#include <iostream>
#include <vector>
#include <cmath>
// count open doors in a segment "from"-"to" (including "from", but excluding "to"
unsigned long long bruteForce(unsigned long long from, unsigned long long to)
{
// allocate enough RAM
auto size = to - from;
std::vector<bool> doors(size, 0);
// note: char instead of bool is about 15% faster but needs 700% more memory
// look at all relevant squares i^2 + j^2
for (unsigned long long i = 1; 2*i*i < to; i++)
{
// find smallest j such that i^2 + j^2 >= from
auto minJ = i + 1;
if (from > i*i + minJ*minJ)
minJ = ceil(sqrt(from - i*i));
// process all j until i^2 + j^2 >= to
for (auto j = minJ; ; j++)
{
auto index = i*i + j*j;
// range check
if (index >= to)
break;
index -= from;
// and "flip" that door
doors[index] = !doors[index];
}
}
// open doors => lowest bit is set
unsigned long long result = 0;
for (auto x : doors)
result += x & 1; // x is odd => door is open
return result;
}
int main()
{
// 10^7 doors per segment
const auto SliceSize = 100000000ULL;
// 10^12 doors in total
auto limit = 1000000000000ULL;
std::cin >> limit;
// total number of segments
auto numSlices = limit / SliceSize; // => 10^12 / 10^7 = 10^5 by default
if (numSlices == 0)
numSlices++;
// will contain the final result
unsigned long long sum = 0;
// 0 => all cores, 1 => one CPU, 2 => 2 CPUs, ... I used 6 on my 4+4-core system
#define PARALLEL
#ifdef PARALLEL
auto numCores = 6;
#pragma omp parallel for num_threads(numCores) reduction(+:sum) schedule(dynamic)
#endif
for (unsigned int i = 0; i < numSlices; i++)
{
// compute lower/upper limit of current segment
auto from = i * SliceSize;
auto to = from + SliceSize; // "to" is excluded
if (to >= limit)
to = limit + 1; // extend last segment to include "to" as well
// process segment
auto current = bruteForce(from, to);
sum += current;
// time echo "1000000000000" | ./611 <= numCores was set to 6
// real 48m34.024s
// user 290m21.924s
// sys 0m49.327s
// and for numCores = 1 (roughly half as fast)
// real 94m51.945s
// user 94m42.514s
// sys 0m8.803s
}
std::cout << sum << std::endl;
return 0;
}
This solution contains 14 empty lines, 25 comments and 7 preprocessor commands.
# Benchmark
The correct solution to the original Project Euler problem was found in 5692 seconds (exceeding the limit of 60 seconds).
The code can be accelerated with OpenMP but the timings refer to the single-threaded version on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 15 MByte.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)
See here for a comparison of all solutions.
Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.
# Changelog
October 10, 2017 submitted solution
# Heatmap
Please click on a problem's number to open my solution to that problem:
green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently
I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
Look at my progress and performance pages to get more details.
<< problem 610 - Roman Numerals II Friend numbers - problem 612 >>
more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
thanks to the KaTeX team for their great typesetting library !
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<< problem 610 - Roman Numerals II Friend numbers - problem 612 >>. # Problem 611: Hallway of square steps. Peter moves in a hallway with N+1 doors consecutively numbered from 0 through N.. All doors are initially closed. Peter starts in front of door 0, and repeatedly performs the following steps:. • First, he walks a positive square number of doors away from his position.. • Then he walks another, larger square number of doors away from his new position.. • He toggles the door he faces (opens it if closed, closes it if open).. • And finally returns to door 0.. We call an action any sequence of those steps. Peter never performs the exact same action twice,. and makes sure to perform all possible actions that don't bring him past the last door.. Let F(N) be the number of doors that are open after Peter has performed all possible actions.. You are given that F(5) = 1, F(100) = 27, F(1000) = 233 and F(10^6) = 112168.. Find F(10^12).. # Very inefficient solution. My code needs more than 60 seconds to find the correct result. (scroll down to the benchmark section). Apparantly a much smarter algorithm exists - or my implementation is just inefficient.. # My Algorithm. Yes, I got to admit: I just used brute force ... it solved the problem in under an hour and thus was my first time. where I solved the most recent problem and became one of the 100 fastest solvers.. Peter first walks to door i^2, takes a breath and then continues to door i^2 + j^2 where i > 0 and j > i.. He toggles the door's state (closed → open, open → closed).. All I need is to figure out a fast way to know how many doors have an odd number of ways to be represented as a sum of two squares.. I read a few websites about "sum of two squares" (e.g. en.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares) but could deduce a simple counting formula.. I process doors in chunks of 10^7 doors each, which requires 1.2 MByte per segment (std::vector<bool>).. Then I iterate over all "relevant" pairs i^2 + j^2 and negate the bit at that position.. The term "relevant" means that i^2 + j^2 has to be a value in the current segment: from <= i^2 + j^2 < to. Even though I have to iterate over all i, I can restrict j to "match" the current segment:. j_{min} = \lceil sqrt{from - i^2} \rceil. ## Alternative Approaches. As mentioned above, Fermat's theorem is the way to go. You need a prime-counting function for primes 4k+1.. I have prime-counting functions in my toolbox but they need to be modified to eliminate 4k+3 primes.. ## Note. From the perspective of a software engineer (and that's what I do for a living) I did the right thing:. instead of spending more than an hour on further reading, typing, debugging, etc., I kept the computer running for 49 minutes. (with #define PARALLEL and numCores = 6) and got the correct result.. That's not the proper scientific way to solve a problem ... but a sound real-world approach !. # Interactive test. You can submit your own input to my program and it will be instantly processed at my server:. Input data (separated by spaces or newlines):. This is equivalent to. echo 1000000 | ./611. Output:. Note: the original problem's input 1000000000000 cannot be entered. because just copying results is a soft skill reserved for idiots.. (this interactive test is still under development, computations will be aborted after one second). # My code. … was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.. #include <iostream>. #include <vector>. #include <cmath>. // count open doors in a segment "from"-"to" (including "from", but excluding "to". unsigned long long bruteForce(unsigned long long from, unsigned long long to). {. // allocate enough RAM. auto size = to - from;. std::vector<bool> doors(size, 0);. // note: char instead of bool is about 15% faster but needs 700% more memory. // look at all relevant squares i^2 + j^2. for (unsigned long long i = 1; 2*i*i < to; i++). {. // find smallest j such that i^2 + j^2 >= from. auto minJ = i + 1;. if (from > i*i + minJ*minJ). minJ = ceil(sqrt(from - i*i));. // process all j until i^2 + j^2 >= to. for (auto j = minJ; ; j++). {. auto index = i*i + j*j;. // range check. if (index >= to). break;. index -= from;. // and "flip" that door. doors[index] = !doors[index];. }.
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}. // open doors => lowest bit is set. unsigned long long result = 0;. for (auto x : doors). result += x & 1; // x is odd => door is open. return result;. }. int main(). {. // 10^7 doors per segment. const auto SliceSize = 100000000ULL;. // 10^12 doors in total. auto limit = 1000000000000ULL;. std::cin >> limit;. // total number of segments. auto numSlices = limit / SliceSize; // => 10^12 / 10^7 = 10^5 by default. if (numSlices == 0). numSlices++;. // will contain the final result. unsigned long long sum = 0;. // 0 => all cores, 1 => one CPU, 2 => 2 CPUs, ... I used 6 on my 4+4-core system. #define PARALLEL. #ifdef PARALLEL. auto numCores = 6;. #pragma omp parallel for num_threads(numCores) reduction(+:sum) schedule(dynamic). #endif. for (unsigned int i = 0; i < numSlices; i++). {. // compute lower/upper limit of current segment. auto from = i * SliceSize;. auto to = from + SliceSize; // "to" is excluded. if (to >= limit). to = limit + 1; // extend last segment to include "to" as well. // process segment. auto current = bruteForce(from, to);. sum += current;. // time echo "1000000000000" | ./611 <= numCores was set to 6. // real 48m34.024s. // user 290m21.924s. // sys 0m49.327s. // and for numCores = 1 (roughly half as fast). // real 94m51.945s. // user 94m42.514s. // sys 0m8.803s. }. std::cout << sum << std::endl;. return 0;. }. This solution contains 14 empty lines, 25 comments and 7 preprocessor commands.. # Benchmark. The correct solution to the original Project Euler problem was found in 5692 seconds (exceeding the limit of 60 seconds).. The code can be accelerated with OpenMP but the timings refer to the single-threaded version on an Intel® Core™ i7-2600K CPU @ 3.40GHz.. Peak memory usage was about 15 MByte.. (compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL). See here for a comparison of all solutions.. Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.. # Changelog. October 10, 2017 submitted solution. # Heatmap. Please click on a problem's number to open my solution to that problem:. green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently. I stopped working on Project Euler problems around the time they released 617.. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200. 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300. 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400. 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500. 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600. 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700. 701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800. 801 802 803 804 805 806 807 808. The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and. I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.. My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.. Look at my progress and performance pages to get more details.. << problem 610 - Roman Numerals II Friend numbers - problem 612 >>. more about me can be found on my homepage, especially in my coding blog.. some names mentioned on this site may be trademarks of their respective owners.. thanks to the KaTeX team for their great typesetting library !.
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## Household Water Use, Algebra
Assignment Help:
During a recent rainy day, 72 mm of rain was recorded. Calculate how many litres of water will be collected from the roof and stored in the water tank. Remember
The volume V of water in the tank, initially full, remaining after (d) days is given by the equation:
V=10 000-235 d
(b) How much is left from a full tank after 30 days of no rain?
(c) From the equation, determine how many litres of water the tank holds when full.
(d) From the equation, determine how many litres of water the householders use per day
(e) When will the volume of the tank be reduced to 2500 litres?
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a+2+3(2a-1)
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#### bill
1/12/2013 4:08:34 AM
hi
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## Household Water Use, Algebra. Assignment Help:. During a recent rainy day, 72 mm of rain was recorded. Calculate how many litres of water will be collected from the roof and stored in the water tank. Remember. The volume V of water in the tank, initially full, remaining after (d) days is given by the equation:. V=10 000-235 d. (b) How much is left from a full tank after 30 days of no rain?. (c) From the equation, determine how many litres of water the tank holds when full.. (d) From the equation, determine how many litres of water the householders use per day. (e) When will the volume of the tank be reduced to 2500 litres?. #### Square roots, The formula v=(the square root of)64h gives the velocity v in.... The formula v=(the square root of)64h gives the velocity v in feet per second of an object that has fallen h feet. Find the velocity of an object that has fallen 76 feet. Round to. #### Expressions, (2a 3b 0 ) – 3 C 2. (2a 3b 0 ) – 3 C 2. #### Solve for v, -1 1/2+ v = -3 3/10. -1 1/2+ v = -3 3/10. #### Shapes, I know im not in the exact grade yet but i would like to know how i.... I know im not in the exact grade yet but i would like to know how it works to be ahead of time.
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#### Distance - rate problems, Distance/Rate Problems These are some standar.... Distance/Rate Problems These are some standard problems which most people think about while they think about Algebra word problems. The standard formula which we will be using. #### Exponential and logarithm functions, In this section we will look at expone.... In this section we will look at exponential & logarithm functions. Both of these functions are extremely important and have to be understood through anyone who is going on to late. #### Rational root theorem, If the rational number x= b/ c is a zero of the n th.... If the rational number x= b/ c is a zero of the n th degree polynomial, P ( x ) = sx n + ...........+ t Where every the coefficients are. a+2+3(2a-1). #### Distribution Property, I really need help in this question (-5d + 1)(-2) I .... I really need help in this question (-5d + 1)(-2) I am really confused. #### Coordinates for the point - graphing, Coordinates for the point The li.... Coordinates for the point The listed first number is the x-coordinate of the point and the second number listed is the y-coordinate of the point. The ordered pair for any spec. #### bill. 1/12/2013 4:08:34 AM. hi. ### Write Your Message!. #### Assured A++ Grade. Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!.
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# How do you graph y=-4x+3?
Nov 15, 2016
graph{y=-4x+3 [-5.17, 4.83, 0.59, 5.59]}
See below for explanation.
#### Explanation:
The equation is in the form $y = m x + b$, where $m$ is slope and $b$ is the y-intercept.
Slope is in the form of $\frac{r i s e}{r u n}$, or y-distance over x-distance. A negative slope means you go DOWN the y-axis, but the x-axis still goes rightwards.
Your y-intercept, $b = 3$, means you have a point at $\left(0 , 3\right)$.
Now, your slope is $m = - 4$, or $- \frac{4}{1}$, meaning you go down 4 units and right 1 unit. Use this to graph the next point down, which should be $\left(1 , - 1\right)$.
Connect these two points, and you have your graph.
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# How do you graph y=-4x+3?. Nov 15, 2016. graph{y=-4x+3 [-5.17, 4.83, 0.59, 5.59]}. See below for explanation.. #### Explanation:. The equation is in the form $y = m x + b$, where $m$ is slope and $b$ is the y-intercept.. Slope is in the form of $\frac{r i s e}{r u n}$, or y-distance over x-distance.
|
A negative slope means you go DOWN the y-axis, but the x-axis still goes rightwards.. Your y-intercept, $b = 3$, means you have a point at $\left(0 , 3\right)$.. Now, your slope is $m = - 4$, or $- \frac{4}{1}$, meaning you go down 4 units and right 1 unit. Use this to graph the next point down, which should be $\left(1 , - 1\right)$.. Connect these two points, and you have your graph.
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# Prime Numbers
Printable View
• Feb 9th 2007, 02:26 AM
Ideasman
Prime Numbers
Prove that if p, a prime number, is in the arithmetic progression of 3n + 1, n = 1, 2, 3, ..., then its also in the arithmetic progression of 6n + 1, n = 1, 2, 3, ....
• Feb 9th 2007, 04:30 AM
topsquark
Quote:
Originally Posted by Ideasman
Prove that if p, a prime number, is in the arithmetic progression of 3n + 1, n = 1, 2, 3, ..., then its also in the arithmetic progression of 6n + 1, n = 1, 2, 3, ....
It doesn't matter whether the number is prime or not. The entire series 6n + 1 is contained within 3n + 1. To see this suppose we have a number k = 6n + 1. What is n' for k = 3n' + 1?
k = 6n + 1 = 3n' + 1
6n + 1 = 3n' + 1
6n = 3n'
n' = 2n.
Since there is an n' for every n, thus every number 6n + 1 can be written in the form 3n' + 1.
-Dan
• Feb 9th 2007, 01:53 PM
AfterShock
Quote:
Originally Posted by topsquark
It doesn't matter whether the number is prime or not. The entire series 6n + 1 is contained within 3n + 1. To see this suppose we have a number k = 6n + 1. What is n' for k = 3n' + 1?
k = 6n + 1 = 3n' + 1
6n + 1 = 3n' + 1
6n = 3n'
n' = 2n.
Since there is an n' for every n, thus every number 6n + 1 can be written in the form 3n' + 1.
-Dan
You have to be careful, Dan.
You have:
{6n + 1 : n is a pos integer} "is a subset of (although not equal) " {3n + 1 : n is a pos int}
Yes I will agree with that;
Looking at the 3n + 1 set;
Given a prime in the largest set, we want to know if it a prime in the smaller set, too.
We know p > 2
3n + 1 is odd; therefore, 3n must be even
If the above is prime, then n has to be even = 2k
3n + 1 = 3(2k) + 1 = 3k + 1
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# Prime Numbers. Printable View. • Feb 9th 2007, 02:26 AM. Ideasman. Prime Numbers. Prove that if p, a prime number, is in the arithmetic progression of 3n + 1, n = 1, 2, 3, ..., then its also in the arithmetic progression of 6n + 1, n = 1, 2, 3, ..... • Feb 9th 2007, 04:30 AM. topsquark. Quote:. Originally Posted by Ideasman. Prove that if p, a prime number, is in the arithmetic progression of 3n + 1, n = 1, 2, 3, ..., then its also in the arithmetic progression of 6n + 1, n = 1, 2, 3, ..... It doesn't matter whether the number is prime or not. The entire series 6n + 1 is contained within 3n + 1. To see this suppose we have a number k = 6n + 1. What is n' for k = 3n' + 1?. k = 6n + 1 = 3n' + 1. 6n + 1 = 3n' + 1. 6n = 3n'. n' = 2n.. Since there is an n' for every n, thus every number 6n + 1 can be written in the form 3n' + 1.. -Dan. • Feb 9th 2007, 01:53 PM. AfterShock.
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Quote:. Originally Posted by topsquark. It doesn't matter whether the number is prime or not. The entire series 6n + 1 is contained within 3n + 1. To see this suppose we have a number k = 6n + 1. What is n' for k = 3n' + 1?. k = 6n + 1 = 3n' + 1. 6n + 1 = 3n' + 1. 6n = 3n'. n' = 2n.. Since there is an n' for every n, thus every number 6n + 1 can be written in the form 3n' + 1.. -Dan. You have to be careful, Dan.. You have:. {6n + 1 : n is a pos integer} "is a subset of (although not equal) " {3n + 1 : n is a pos int}. Yes I will agree with that;. Looking at the 3n + 1 set;. Given a prime in the largest set, we want to know if it a prime in the smaller set, too.. We know p > 2. 3n + 1 is odd; therefore, 3n must be even. If the above is prime, then n has to be even = 2k. 3n + 1 = 3(2k) + 1 = 3k + 1.
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+0
# need help
0
66
1
If a cuboid of dimensions 10 cm×15 cm×5 cm is cut to form cubes of sides 5 cm, then what is the difference between the sum of the surface areas of these cubes and the surface area of the original cuboid?
Jan 12, 2020
#1
+21753
+2
Each smaller cube has a side area of 5x5 = 25 cm^2
in the original solid you can see 22 of these sides 22x25 = 550 cm2
Each smaller cube has 6 sides 6 x 25 =150 cm2 and there are 6 of them = 900 cm2
I think you can calc the diff !
Jan 12, 2020
edited by ElectricPavlov Jan 12, 2020
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+0. # need help. 0. 66. 1. If a cuboid of dimensions 10 cm×15 cm×5 cm is cut to form cubes of sides 5 cm, then what is the difference between the sum of the surface areas of these cubes and the surface area of the original cuboid?. Jan 12, 2020. #1. +21753.
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+2. Each smaller cube has a side area of 5x5 = 25 cm^2. in the original solid you can see 22 of these sides 22x25 = 550 cm2. Each smaller cube has 6 sides 6 x 25 =150 cm2 and there are 6 of them = 900 cm2. I think you can calc the diff !. Jan 12, 2020. edited by ElectricPavlov Jan 12, 2020.
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https://www.coursehero.com/tutors-problems/Statistics-and-Probability/482531-562-A-certain-airplane-has-two-independent-alternators-to-provide-ele/
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View the step-by-step solution to:
# 62 A certain airplane has two independent alternators to provide electrical power. The probability that a given alternator will fail on a 1-hour...
5.62 A certain airplane has two independent alternators to provide electrical power. The probability
that a given alternator will fail on a 1-hour flight is .02. What is the probability that (a) both will
fail? (b) Neither will fail? (c) One or the other will fail? Show all steps carefully.
5.70 The probability is 1 in 4,000,000 that a single auto trip in the United States will result in a
fatality. Over a lifetime, an average U.S. driver takes 50,000 trips. (a) What is the probability of
a fatal accident over a lifetime? Explain your reasoning carefully. Hint: Assume independent
events. Why might the assumption of independence be violated? (b) Why might a driver be
tempted not to use a seat belt “just on this trip”?
8.46 A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on
a very accurate scale. The results in grams were
3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477
(a) Construct a 90 percent confidence interval for the true mean weight. (b) What sample size
would be necessary to estimate the true weight with an error of } 0.03 grams with 90 percent confidence?
(c) Discuss the factors which might cause variation in the weight of Tootsie Rolls during
manufacture. (Data are from a project by MBA student Henry Scussel.)
8.62 In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be
engaged in safety and security-related jobs, and found that 1,143 were positive. (a) Construct a
95 percent confidence interval for the population proportion of positive drug tests. (b) Why is the
normality assumption not a problem, despite the very small value of p? (Data are from Flying 120,
no. 11 [November 1993], p. 31.)
### Why Join Course Hero?
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View the step-by-step solution to:. # 62 A certain airplane has two independent alternators to provide electrical power. The probability that a given alternator will fail on a 1-hour.... 5.62 A certain airplane has two independent alternators to provide electrical power. The probability. that a given alternator will fail on a 1-hour flight is .02. What is the probability that (a) both will. fail? (b) Neither will fail? (c) One or the other will fail? Show all steps carefully.. 5.70 The probability is 1 in 4,000,000 that a single auto trip in the United States will result in a. fatality. Over a lifetime, an average U.S. driver takes 50,000 trips. (a) What is the probability of. a fatal accident over a lifetime? Explain your reasoning carefully. Hint: Assume independent. events. Why might the assumption of independence be violated? (b) Why might a driver be. tempted not to use a seat belt “just on this trip”?. 8.46 A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on. a very accurate scale. The results in grams were. 3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477. (a) Construct a 90 percent confidence interval for the true mean weight.
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(b) What sample size. would be necessary to estimate the true weight with an error of } 0.03 grams with 90 percent confidence?. (c) Discuss the factors which might cause variation in the weight of Tootsie Rolls during. manufacture. (Data are from a project by MBA student Henry Scussel.). 8.62 In 1992, the FAA conducted 86,991 pre-employment drug tests on job applicants who were to be. engaged in safety and security-related jobs, and found that 1,143 were positive. (a) Construct a. 95 percent confidence interval for the population proportion of positive drug tests. (b) Why is the. normality assumption not a problem, despite the very small value of p? (Data are from Flying 120,. no. 11 [November 1993], p. 31.). ### Why Join Course Hero?. Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.. ### -. Educational Resources. • ### -. Study Documents. Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.. Browse Documents.
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NIACL AO Mains– Quantitative Aptitude Questions Day- 24
Dear Readers, Bank Exam Race for the Year 2019 is already started, To enrich your preparation here we have providing new series of Practice Questions on Quantitative Aptitude – Section. Candidates those who are preparing for NIACL AO Mains 2019 Exams can practice these questions daily and make your preparation effective.
[WpProQuiz 5332]
Direction (1- 5): Find the wrong term in the following number series?
1) 324, 476, 712, 988, 1289
a) 712
b) 988
c) 1289
d) 476
e) 324
2) 2400, 1200, 800, 600, 460, 400
a) 460
b) 800
c) 1200
d) 600
e) 400
3) 420, 2516, 504, 2016, 672, 1344
a) 504
b) 1344
c) 672
d) 2516
e) 2016
4) 33, 16, 17, 24, 50, 123, 370.5
a) 24
b) 123
c) 370.5
d) 50
e) 33
5) 360, 1036, 460, 944, 503, 868
a) 1036
b) 944
c) 503
d) 868
e) 360
Directions (Q. 6 – 10): Study the following information carefully and answer the given questions:
The following bar graph shows the total number of newspaper readers (In lakhs) of 6 different states in 3 different years.
6) Find the difference between the total number of male newspaper readers of state C in the year 2017 to that of total number of female newspaper readers of state D in the year 2018, if the ratio between the male to that of female newspaper readers of state C in the year 2017 and that of state D in the year 2018 is 8 : 7 and 10 : 7 respectively?
a) 2 lakhs
b) 8 lakhs
c) 5 lakhs
d) 3 lakhs
e) None of these
7) Find the ratio between the total number of newspaper readers in the year 2016 to that of 2018 in all the given states together?
a) 7: 11
b) 8: 13
c) 21: 32
d) 123: 157
e) None of these
8) Total number of newspaper readers of State B in all the given years together is approximately what percentage of total number of newspaper readers of State E in all the given years together?
a) 87 %
b) 115 %
c) 72 %
d) 103 %
e) 66 %
9) Find the average number of newspaper readers from state G, H and D together in all the given years, if the total number of newspaper readers from state G and H in all the given years is 20 % more than the total number of newspaper readers from state E and 20 % less than the total number of newspaper readers from state E respectively?
a) 196.66 lakhs
b) 178.33 lakhs
c) 180.5 lakhs
d) 175 lakhs
e) None of these
10) In which state has second lowest number of newspaper readers in all the given years together?
a) State A
b) State C
c) State E
d) State D
e) State B
Directions (Q. 1 – 5):
The correct series is,
324Â Â Â Â Â Â Â Â Â Â Â Â 476Â Â Â Â Â Â Â Â Â Â Â Â 712Â Â Â Â Â Â Â Â Â Â Â Â 990Â Â Â Â Â Â Â Â Â Â Â Â 1289
         152            236            278            299
                  84              42              21
The difference of difference is, ÷ 2
The wrong term is, 988
The correct series is,
2400, 1200, 800, 600, 480, 400
The pattern is, *1/2, *2/3, *3/4, *4/5, *5/6
The wrong term is, 460
The correct series is,
420, 2520, 504, 2016, 672, 1344
The pattern is, *6, ÷ 5, *4, ÷ 3, *2,…
The wrong term is, 2516
The correct series is,
33, 16, 17, 24, 50, 122.5, 370.5
The pattern is, *0.5 – 0.5, *1 + 1, *1.5 – 1.5, *2 + 2, *2.5 – 2.5, *3 + 3,…
The wrong term is, 123
The correct series is,
360, 1036, 460, 944, 544, 868
The pattern is, +262, -242, +222, -202, +182,…
The wrong term is, 503
Directions (Q. 6 – 10):
The total number of male newspaper readers of state C in the year 2017
= > (75/15)*8 = 40 lakhs
The total number of female newspaper readers of state D in the year 2018
= > (85/17)*7 = 35 lakhs
Required difference = 40 – 35 = 5 lakhs
The total number of newspaper readers in the year 2016 in all the given states together
= > 55 + 45 + 60 + 40 + 50 + 65 = 315 lakhs
The total number of newspaper readers in the year 2018 in all the given states together
= > 80 + 75 + 90 + 85 + 70 + 95 = 495 lakhs
Required ratio = 315: 495 = 7: 11
Total number of newspaper readers of State B in all the given years together
= > 45 + 65 + 75 = 185 lakhs
Total number of newspaper readers of State E in all the given years together
= > 50 + 60 + 70 = 180 lakhs
Required % = (185/180)*100 = 102.77 = 103 %
The total number of newspaper readers from state G, H and D together in all the given years
= > 180*(120/100) + 180*(80/100) + 175
= > 216 + 144 + 175 = 535 lakhs
Required average = 535/3 = 178.33 lakhs
State A = > 55 + 70 + 80 = 205 lakhs
State B = > 45 + 65 + 75 = 185 lakhs
State C = > 60 + 75 + 90 = 225 lakhs
State D = > 40 + 50 + 85 = 175 lakhs
State E = > 50 + 60 + 70 = 180 lakhs
State F = > 65 + 80 + 95 = 240 lakhs
State E has second lowest number of newspaper readers.
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NIACL AO Mains– Quantitative Aptitude Questions Day- 24. Dear Readers, Bank Exam Race for the Year 2019 is already started, To enrich your preparation here we have providing new series of Practice Questions on Quantitative Aptitude – Section. Candidates those who are preparing for NIACL AO Mains 2019 Exams can practice these questions daily and make your preparation effective.. [WpProQuiz 5332]. Direction (1- 5): Find the wrong term in the following number series?. 1) 324, 476, 712, 988, 1289. a) 712. b) 988. c) 1289. d) 476. e) 324. 2) 2400, 1200, 800, 600, 460, 400. a) 460. b) 800. c) 1200. d) 600. e) 400. 3) 420, 2516, 504, 2016, 672, 1344. a) 504. b) 1344. c) 672. d) 2516. e) 2016. 4) 33, 16, 17, 24, 50, 123, 370.5. a) 24. b) 123. c) 370.5. d) 50. e) 33. 5) 360, 1036, 460, 944, 503, 868. a) 1036. b) 944. c) 503. d) 868. e) 360. Directions (Q. 6 – 10): Study the following information carefully and answer the given questions:. The following bar graph shows the total number of newspaper readers (In lakhs) of 6 different states in 3 different years.. 6) Find the difference between the total number of male newspaper readers of state C in the year 2017 to that of total number of female newspaper readers of state D in the year 2018, if the ratio between the male to that of female newspaper readers of state C in the year 2017 and that of state D in the year 2018 is 8 : 7 and 10 : 7 respectively?. a) 2 lakhs. b) 8 lakhs. c) 5 lakhs. d) 3 lakhs. e) None of these. 7) Find the ratio between the total number of newspaper readers in the year 2016 to that of 2018 in all the given states together?. a) 7: 11. b) 8: 13. c) 21: 32. d) 123: 157. e) None of these. 8) Total number of newspaper readers of State B in all the given years together is approximately what percentage of total number of newspaper readers of State E in all the given years together?. a) 87 %. b) 115 %. c) 72 %. d) 103 %. e) 66 %. 9) Find the average number of newspaper readers from state G, H and D together in all the given years, if the total number of newspaper readers from state G and H in all the given years is 20 % more than the total number of newspaper readers from state E and 20 % less than the total number of newspaper readers from state E respectively?. a) 196.66 lakhs. b) 178.33 lakhs. c) 180.5 lakhs. d) 175 lakhs.
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e) None of these. 10) In which state has second lowest number of newspaper readers in all the given years together?. a) State A. b) State C. c) State E. d) State D. e) State B. Directions (Q. 1 – 5):. The correct series is,. 324            476            712            990            1289.          152            236            278            299.                   84              42              21. The difference of difference is, ÷ 2. The wrong term is, 988. The correct series is,. 2400, 1200, 800, 600, 480, 400. The pattern is, *1/2, *2/3, *3/4, *4/5, *5/6. The wrong term is, 460. The correct series is,. 420, 2520, 504, 2016, 672, 1344. The pattern is, *6, ÷ 5, *4, ÷ 3, *2,…. The wrong term is, 2516. The correct series is,. 33, 16, 17, 24, 50, 122.5, 370.5. The pattern is, *0.5 – 0.5, *1 + 1, *1.5 – 1.5, *2 + 2, *2.5 – 2.5, *3 + 3,…. The wrong term is, 123. The correct series is,. 360, 1036, 460, 944, 544, 868. The pattern is, +262, -242, +222, -202, +182,…. The wrong term is, 503. Directions (Q. 6 – 10):. The total number of male newspaper readers of state C in the year 2017. = > (75/15)*8 = 40 lakhs. The total number of female newspaper readers of state D in the year 2018. = > (85/17)*7 = 35 lakhs. Required difference = 40 – 35 = 5 lakhs. The total number of newspaper readers in the year 2016 in all the given states together. = > 55 + 45 + 60 + 40 + 50 + 65 = 315 lakhs. The total number of newspaper readers in the year 2018 in all the given states together. = > 80 + 75 + 90 + 85 + 70 + 95 = 495 lakhs. Required ratio = 315: 495 = 7: 11. Total number of newspaper readers of State B in all the given years together. = > 45 + 65 + 75 = 185 lakhs. Total number of newspaper readers of State E in all the given years together. = > 50 + 60 + 70 = 180 lakhs. Required % = (185/180)*100 = 102.77 = 103 %. The total number of newspaper readers from state G, H and D together in all the given years. = > 180*(120/100) + 180*(80/100) + 175. = > 216 + 144 + 175 = 535 lakhs. Required average = 535/3 = 178.33 lakhs. State A = > 55 + 70 + 80 = 205 lakhs. State B = > 45 + 65 + 75 = 185 lakhs. State C = > 60 + 75 + 90 = 225 lakhs. State D = > 40 + 50 + 85 = 175 lakhs. State E = > 50 + 60 + 70 = 180 lakhs. State F = > 65 + 80 + 95 = 240 lakhs. State E has second lowest number of newspaper readers.
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https://www.geeksforgeeks.org/sum-of-maximum-and-minimum-prime-factor-of-every-number-in-the-array/
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# Sum of Maximum and Minimum prime factor of every number in the Array
Given an array arr[], the task is to find the sum of the maximum and the minimum prime factor of every number in the given array.
Examples:
Input: arr[] = {15}
Output:
The maximum and the minimum prime factors
of 15 are 5 and 3 respectively.
Input: arr[] = {5, 10, 15, 20, 25, 30}
Output: 10 7 8 7 10 7
Approach: The idea is to use Sieve of Eratosthenes to precompute all the minimum and maximum prime factors of every number and store it in two arrays. After this precomputation, the sum of the minimum and the maximum prime factor can be found in constant time.
Below is the implementation of the above approach:
## CPP
`// C++ implementation of the approach` `#include ``using` `namespace` `std;` `const` `int` `MAX = 100000;` `// max_prime[i] represent maximum prime``// number that divides the number i``int` `max_prime[MAX];` `// min_prime[i] represent minimum prime``// number that divides the number i``int` `min_prime[MAX];` `// Function to store the minimum prime factor``// and the maximum prime factor in two arrays``void` `sieve(``int` `n)``{`` ``for` `(``int` `i = 2; i <= n; ++i) {` ` ``// Check for prime number`` ``// if min_prime[i]>0,`` ``// then it is not a prime number`` ``if` `(min_prime[i] > 0) {`` ``continue``;`` ``}` ` ``// if i is a prime number`` ``// min_prime number that divide prime number`` ``// and max_prime number that divide prime number`` ``// is the number itself.`` ``min_prime[i] = i;`` ``max_prime[i] = i;` ` ``int` `j = i + i;` ` ``while` `(j <= n) {`` ``if` `(min_prime[j] == 0) {` ` ``// If this number is being visited`` ``// for first time then this divisor`` ``// must be the smallest prime number`` ``// that divides this number`` ``min_prime[j] = i;`` ``}` ` ``// Update prime number till`` ``// last prime number that divides this number` ` ``// The last prime number that`` ``// divides this number will be maximum.`` ``max_prime[j] = i;`` ``j += i;`` ``}`` ``}``}` `// Function to find the sum of the minimum``// and the maximum prime factors of every``// number from the given array``void` `findSum(``int` `arr[], ``int` `n)``{` ` ``// Pre-calculation`` ``sieve(MAX);` ` ``// For every element of the given array`` ``for` `(``int` `i = 0; i < n; i++) {` ` ``// The sum of its smallest`` ``// and largest prime factor`` ``int` `sum = min_prime[arr[i]]`` ``+ max_prime[arr[i]];` ` ``cout << sum << ``" "``;`` ``}``}` `// Driver code``int` `main()``{`` ``int` `arr[] = { 5, 10, 15, 20, 25, 30 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);` ` ``findSum(arr, n);` ` ``return` `0;``}`
## Java
`// Java implementation of the approach ``class` `GFG``{`` ` ` ``static` `int` `MAX = ``100000``; `` ` ` ``// max_prime[i] represent maximum prime `` ``// number that divides the number i `` ``static` `int` `max_prime[] = ``new` `int``[MAX + ``1``]; `` ` ` ``// min_prime[i] represent minimum prime `` ``// number that divides the number i `` ``static` `int` `min_prime[] = ``new` `int``[MAX + ``1``]; `` ` ` ``// Function to store the minimum prime factor `` ``// and the maximum prime factor in two arrays `` ``static` `void` `sieve(``int` `n) `` ``{ `` ``for` `(``int` `i = ``2``; i <= n; ++i)`` ``{ `` ` ` ``// Check for prime number `` ``// if min_prime[i] > 0, `` ``// then it is not a prime number `` ``if` `(min_prime[i] > ``0``)`` ``{ `` ``continue``; `` ``} `` ` ` ``// if i is a prime number `` ``// min_prime number that divide prime number `` ``// and max_prime number that divide prime number `` ``// is the number itself. `` ``min_prime[i] = i; `` ``max_prime[i] = i; `` ` ` ``int` `j = i + i; `` ` ` ``while` `(j <= n)`` ``{ `` ``if` `(min_prime[j] == ``0``)`` ``{ `` ` ` ``// If this number is being visited `` ``// for first time then this divisor `` ``// must be the smallest prime number `` ``// that divides this number `` ``min_prime[j] = i; `` ``} `` ` ` ``// Update prime number till `` ``// last prime number that divides this number `` ` ` ``// The last prime number that `` ``// divides this number will be maximum. `` ``max_prime[j] = i; `` ``j += i; `` ``} `` ``} `` ``} `` ` ` ``// Function to find the sum of the minimum `` ``// and the maximum prime factors of every `` ``// number from the given array `` ``static` `void` `findSum(``int` `arr[], ``int` `n) `` ``{ `` ` ` ``// Pre-calculation `` ``sieve(MAX); `` ` ` ``// For every element of the given array `` ``for` `(``int` `i = ``0``; i < n; i++)`` ``{ `` ` ` ``// The sum of its smallest `` ``// and largest prime factor `` ``int` `sum = min_prime[arr[i]] `` ``+ max_prime[arr[i]]; `` ` ` ``System.out.print(sum + ``" "``); `` ``} `` ``} `` ` ` ``// Driver code `` ``public` `static` `void` `main (String[] args)`` ``{ `` ``int` `arr[] = { ``5``, ``10``, ``15``, ``20``, ``25``, ``30` `}; `` ``int` `n = arr.length ;`` ` ` ``findSum(arr, n); `` ` ` ``} ``}` `// This code is contributed by AnkitRai01`
## Python
`# Python3 implementation of the approach``MAX` `=` `100000` `# max_prime[i] represent maximum prime``# number that divides the number i``max_prime ``=` `[``0``]``*``(``MAX` `+` `1``)` `# min_prime[i] represent minimum prime``# number that divides the number i``min_prime ``=` `[``0``]``*``(``MAX` `+` `1``)` `# Function to store the minimum prime factor``# and the maximum prime factor in two arrays``def` `sieve(n):`` ``for` `i ``in` `range``(``2``, n ``+` `1``):` ` ``# Check for prime number`` ``# if min_prime[i]>0,`` ``# then it is not a prime number`` ``if` `(min_prime[i] > ``0``):`` ``continue` ` ``# if i is a prime number`` ``# min_prime number that divide prime number`` ``# and max_prime number that divide prime number`` ``# is the number itself.`` ``min_prime[i] ``=` `i`` ``max_prime[i] ``=` `i` ` ``j ``=` `i ``+` `i` ` ``while` `(j <``=` `n):`` ``if` `(min_prime[j] ``=``=` `0``):` ` ``# If this number is being visited`` ``# for first time then this divisor`` ``# must be the smallest prime number`` ``# that divides this number`` ``min_prime[j] ``=` `i` ` ``# Update prime number till`` ``# last prime number that divides this number` ` ``# The last prime number that`` ``# divides this number will be maximum.`` ``max_prime[j] ``=` `i`` ``j ``+``=` `i` `# Function to find the sum of the minimum``# and the maximum prime factors of every``# number from the given array``def` `findSum(arr, n):` ` ``# Pre-calculation`` ``sieve(``MAX``)` ` ``# For every element of the given array`` ``for` `i ``in` `range``(n):` ` ``# The sum of its smallest`` ``# and largest prime factor`` ``sum` `=` `min_prime[arr[i]] ``+` `max_prime[arr[i]]` ` ``print``(``sum``, end ``=` `" "``)` `# Driver code``arr ``=` `[``5``, ``10``, ``15``, ``20``, ``25``, ``30``]``n ``=` `len``(arr)` `findSum(arr, n)` `# This code is contributed by mohit kumar 29`
## C#
`// C# implementation of the approach ``using` `System;` `class` `GFG``{`` ` ` ``static` `int` `MAX = 100000; `` ` ` ``// max_prime[i] represent maximum prime `` ``// number that divides the number i `` ``static` `int` `[]max_prime = ``new` `int``[MAX + 1]; `` ` ` ``// min_prime[i] represent minimum prime `` ``// number that divides the number i `` ``static` `int` `[]min_prime = ``new` `int``[MAX + 1]; `` ` ` ``// Function to store the minimum prime factor `` ``// and the maximum prime factor in two arrays `` ``static` `void` `sieve(``int` `n) `` ``{ `` ``for` `(``int` `i = 2; i <= n; ++i)`` ``{ `` ` ` ``// Check for prime number `` ``// if min_prime[i] > 0, `` ``// then it is not a prime number `` ``if` `(min_prime[i] > 0)`` ``{ `` ``continue``; `` ``} `` ` ` ``// if i is a prime number `` ``// min_prime number that divide prime number `` ``// and max_prime number that divide prime number `` ``// is the number itself. `` ``min_prime[i] = i; `` ``max_prime[i] = i; `` ` ` ``int` `j = i + i; `` ` ` ``while` `(j <= n)`` ``{ `` ``if` `(min_prime[j] == 0)`` ``{ `` ` ` ``// If this number is being visited `` ``// for first time then this divisor `` ``// must be the smallest prime number `` ``// that divides this number `` ``min_prime[j] = i; `` ``} `` ` ` ``// Update prime number till `` ``// last prime number that divides this number `` ` ` ``// The last prime number that `` ``// divides this number will be maximum. `` ``max_prime[j] = i; `` ``j += i; `` ``} `` ``} `` ``} `` ` ` ``// Function to find the sum of the minimum `` ``// and the maximum prime factors of every `` ``// number from the given array `` ``static` `void` `findSum(``int` `[]arr, ``int` `n) `` ``{ `` ` ` ``// Pre-calculation `` ``sieve(MAX); `` ` ` ``// For every element of the given array `` ``for` `(``int` `i = 0; i < n; i++)`` ``{ `` ` ` ``// The sum of its smallest `` ``// and largest prime factor `` ``int` `sum = min_prime[arr[i]] `` ``+ max_prime[arr[i]]; `` ` ` ``Console.Write(sum + ``" "``); `` ``} `` ``} `` ` ` ``// Driver code `` ``public` `static` `void` `Main(String[] args)`` ``{ `` ``int` `[]arr = { 5, 10, 15, 20, 25, 30 }; `` ``int` `n = arr.Length ;`` ` ` ``findSum(arr, n); `` ` ` ``} ``}` `// This code is contributed by 29AjayKumar`
## Javascript
``
Output:
`10 7 8 7 10 7`
Time Complexity: O(n2)
Auxiliary Space: O(100000)
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# Sum of Maximum and Minimum prime factor of every number in the Array. Given an array arr[], the task is to find the sum of the maximum and the minimum prime factor of every number in the given array.. Examples:. Input: arr[] = {15}. Output:. The maximum and the minimum prime factors. of 15 are 5 and 3 respectively.. Input: arr[] = {5, 10, 15, 20, 25, 30}. Output: 10 7 8 7 10 7. Approach: The idea is to use Sieve of Eratosthenes to precompute all the minimum and maximum prime factors of every number and store it in two arrays. After this precomputation, the sum of the minimum and the maximum prime factor can be found in constant time.. Below is the implementation of the above approach:. ## CPP. `// C++ implementation of the approach` `#include ``using` `namespace` `std;` `const` `int` `MAX = 100000;` `// max_prime[i] represent maximum prime``// number that divides the number i``int` `max_prime[MAX];` `// min_prime[i] represent minimum prime``// number that divides the number i``int` `min_prime[MAX];` `// Function to store the minimum prime factor``// and the maximum prime factor in two arrays``void` `sieve(``int` `n)``{`` ``for` `(``int` `i = 2; i <= n; ++i) {` ` ``// Check for prime number`` ``// if min_prime[i]>0,`` ``// then it is not a prime number`` ``if` `(min_prime[i] > 0) {`` ``continue``;`` ``}` ` ``// if i is a prime number`` ``// min_prime number that divide prime number`` ``// and max_prime number that divide prime number`` ``// is the number itself.`` ``min_prime[i] = i;`` ``max_prime[i] = i;` ` ``int` `j = i + i;` ` ``while` `(j <= n) {`` ``if` `(min_prime[j] == 0) {` ` ``// If this number is being visited`` ``// for first time then this divisor`` ``// must be the smallest prime number`` ``// that divides this number`` ``min_prime[j] = i;`` ``}` ` ``// Update prime number till`` ``// last prime number that divides this number` ` ``// The last prime number that`` ``// divides this number will be maximum.`` ``max_prime[j] = i;`` ``j += i;`` ``}`` ``}``}` `// Function to find the sum of the minimum``// and the maximum prime factors of every``// number from the given array``void` `findSum(``int` `arr[], ``int` `n)``{` ` ``// Pre-calculation`` ``sieve(MAX);` ` ``// For every element of the given array`` ``for` `(``int` `i = 0; i < n; i++) {` ` ``// The sum of its smallest`` ``// and largest prime factor`` ``int` `sum = min_prime[arr[i]]`` ``+ max_prime[arr[i]];` ` ``cout << sum << ``" "``;`` ``}``}` `// Driver code``int` `main()``{`` ``int` `arr[] = { 5, 10, 15, 20, 25, 30 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);` ` ``findSum(arr, n);` ` ``return` `0;``}`. ## Java. `// Java implementation of the approach ``class` `GFG``{`` ` ` ``static` `int` `MAX = ``100000``; `` ` ` ``// max_prime[i] represent maximum prime `` ``// number that divides the number i `` ``static` `int` `max_prime[] = ``new` `int``[MAX + ``1``]; `` ` ` ``// min_prime[i] represent minimum prime `` ``// number that divides the number i `` ``static` `int` `min_prime[] = ``new` `int``[MAX + ``1``]; `` ` ` ``// Function to store the minimum prime factor `` ``// and the maximum prime factor in two arrays `` ``static` `void` `sieve(``int` `n) `` ``{ `` ``for` `(``int` `i = ``2``; i <= n; ++i)`` ``{ `` ` ` ``// Check for prime number `` ``// if min_prime[i] > 0, `` ``// then it is not a prime number `` ``if` `(min_prime[i] > ``0``)`` ``{ `` ``continue``; `` ``} `` ` ` ``// if i is a prime number `` ``// min_prime number that divide prime number `` ``// and max_prime number that divide prime number `` ``// is the number itself. `` ``min_prime[i] = i; `` ``max_prime[i] = i; `` ` ` ``int` `j = i + i; `` ` ` ``while` `(j <= n)`` ``{ `` ``if` `(min_prime[j] == ``0``)`` ``{ `` ` ` ``// If this number is being visited `` ``// for first time then this divisor `` ``// must be the smallest prime number `` ``// that divides this number `` ``min_prime[j] = i; `` ``} `` ` ` ``// Update prime number till `` ``// last prime number that divides this number `` ` ` ``// The last prime number that `` ``// divides this number will be maximum.
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`` ``max_prime[j] = i; `` ``j += i; `` ``} `` ``} `` ``} `` ` ` ``// Function to find the sum of the minimum `` ``// and the maximum prime factors of every `` ``// number from the given array `` ``static` `void` `findSum(``int` `arr[], ``int` `n) `` ``{ `` ` ` ``// Pre-calculation `` ``sieve(MAX); `` ` ` ``// For every element of the given array `` ``for` `(``int` `i = ``0``; i < n; i++)`` ``{ `` ` ` ``// The sum of its smallest `` ``// and largest prime factor `` ``int` `sum = min_prime[arr[i]] `` ``+ max_prime[arr[i]]; `` ` ` ``System.out.print(sum + ``" "``); `` ``} `` ``} `` ` ` ``// Driver code `` ``public` `static` `void` `main (String[] args)`` ``{ `` ``int` `arr[] = { ``5``, ``10``, ``15``, ``20``, ``25``, ``30` `}; `` ``int` `n = arr.length ;`` ` ` ``findSum(arr, n); `` ` ` ``} ``}` `// This code is contributed by AnkitRai01`. ## Python. `# Python3 implementation of the approach``MAX` `=` `100000` `# max_prime[i] represent maximum prime``# number that divides the number i``max_prime ``=` `[``0``]``*``(``MAX` `+` `1``)` `# min_prime[i] represent minimum prime``# number that divides the number i``min_prime ``=` `[``0``]``*``(``MAX` `+` `1``)` `# Function to store the minimum prime factor``# and the maximum prime factor in two arrays``def` `sieve(n):`` ``for` `i ``in` `range``(``2``, n ``+` `1``):` ` ``# Check for prime number`` ``# if min_prime[i]>0,`` ``# then it is not a prime number`` ``if` `(min_prime[i] > ``0``):`` ``continue` ` ``# if i is a prime number`` ``# min_prime number that divide prime number`` ``# and max_prime number that divide prime number`` ``# is the number itself.`` ``min_prime[i] ``=` `i`` ``max_prime[i] ``=` `i` ` ``j ``=` `i ``+` `i` ` ``while` `(j <``=` `n):`` ``if` `(min_prime[j] ``=``=` `0``):` ` ``# If this number is being visited`` ``# for first time then this divisor`` ``# must be the smallest prime number`` ``# that divides this number`` ``min_prime[j] ``=` `i` ` ``# Update prime number till`` ``# last prime number that divides this number` ` ``# The last prime number that`` ``# divides this number will be maximum.`` ``max_prime[j] ``=` `i`` ``j ``+``=` `i` `# Function to find the sum of the minimum``# and the maximum prime factors of every``# number from the given array``def` `findSum(arr, n):` ` ``# Pre-calculation`` ``sieve(``MAX``)` ` ``# For every element of the given array`` ``for` `i ``in` `range``(n):` ` ``# The sum of its smallest`` ``# and largest prime factor`` ``sum` `=` `min_prime[arr[i]] ``+` `max_prime[arr[i]]` ` ``print``(``sum``, end ``=` `" "``)` `# Driver code``arr ``=` `[``5``, ``10``, ``15``, ``20``, ``25``, ``30``]``n ``=` `len``(arr)` `findSum(arr, n)` `# This code is contributed by mohit kumar 29`. ## C#. `// C# implementation of the approach ``using` `System;` `class` `GFG``{`` ` ` ``static` `int` `MAX = 100000; `` ` ` ``// max_prime[i] represent maximum prime `` ``// number that divides the number i `` ``static` `int` `[]max_prime = ``new` `int``[MAX + 1]; `` ` ` ``// min_prime[i] represent minimum prime `` ``// number that divides the number i `` ``static` `int` `[]min_prime = ``new` `int``[MAX + 1]; `` ` ` ``// Function to store the minimum prime factor `` ``// and the maximum prime factor in two arrays `` ``static` `void` `sieve(``int` `n) `` ``{ `` ``for` `(``int` `i = 2; i <= n; ++i)`` ``{ `` ` ` ``// Check for prime number `` ``// if min_prime[i] > 0, `` ``// then it is not a prime number `` ``if` `(min_prime[i] > 0)`` ``{ `` ``continue``; `` ``} `` ` ` ``// if i is a prime number `` ``// min_prime number that divide prime number `` ``// and max_prime number that divide prime number `` ``// is the number itself. `` ``min_prime[i] = i; `` ``max_prime[i] = i; `` ` ` ``int` `j = i + i; `` ` ` ``while` `(j <= n)`` ``{ `` ``if` `(min_prime[j] == 0)`` ``{ `` ` ` ``// If this number is being visited `` ``// for first time then this divisor `` ``// must be the smallest prime number `` ``// that divides this number `` ``min_prime[j] = i; `` ``} `` ` ` ``// Update prime number till `` ``// last prime number that divides this number `` ` ` ``// The last prime number that `` ``// divides this number will be maximum. `` ``max_prime[j] = i; `` ``j += i; `` ``} `` ``} `` ``} `` ` ` ``// Function to find the sum of the minimum `` ``// and the maximum prime factors of every `` ``// number from the given array `` ``static` `void` `findSum(``int` `[]arr, ``int` `n) `` ``{ `` ` ` ``// Pre-calculation `` ``sieve(MAX); `` ` ` ``// For every element of the given array `` ``for` `(``int` `i = 0; i < n; i++)`` ``{ `` ` ` ``// The sum of its smallest `` ``// and largest prime factor `` ``int` `sum = min_prime[arr[i]] `` ``+ max_prime[arr[i]]; `` ` ` ``Console.Write(sum + ``" "``); `` ``} `` ``} `` ` ` ``// Driver code `` ``public` `static` `void` `Main(String[] args)`` ``{ `` ``int` `[]arr = { 5, 10, 15, 20, 25, 30 }; `` ``int` `n = arr.Length ;`` ` ` ``findSum(arr, n); `` ` ` ``} ``}` `// This code is contributed by 29AjayKumar`. ## Javascript. ``. Output:. `10 7 8 7 10 7`. Time Complexity: O(n2). Auxiliary Space: O(100000). Previous. Next.
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# algebra
posted by deli
Consider all 3-term geometric sequences with first term 1 and with common ratio the square of an integer between 1 and 1000. How many of these 1000 geometric sequences have the property that the sum of the 3 terms is prime?
1. Reiny
I did this question here
http://www.jiskha.com/display.cgi?id=1367223190
Somebody called "tsong" disagreed, but clearly did not read my solution, since I "proved" that there is only one such case,
when the GS is 1 , 1, 1
that is, when the common ratio is 1
then 1 + 1^2 + 1^4 = 3 , which is a prime number
conclusion: there is only ONE such sequence.
## Similar Questions
1. ### math...
how can I tell if a sequences is airthmetic, geometric or neither?
2. ### check!
If the third and ninth temr of a geometric series with a positive common ratio are -3 and -192 repectively, determine the value of the first term a Let a_1 be denoted by 'a' (first terms) then, a_3=r^2 a And, a_9=r^8 a Thus, r^8 a=-192 …
3. ### Math *URGENT
Please give the answers and solutions for each. 1.If the second term is 2 and the seventh term of a geometric sequence is 64, find the 12th term. 2. Which term if the geometric sequence 18,54,162,486,... is 3,188,646?
4. ### Maths
Eric thinks of 2 sequences.One is geometric and the other arithmetic.Both sequences start with the number 3.The common ratio of the geometric sequence is the same as the common difference of the arithmetic sequence.If the 6-th term …
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1..The first 2 terms of a geometric progression are the same as the first two terms of an arithmetic progression.The first term is 12 and is greater than the second term.The sum of the first 3 terms od the arithmetic progression is …
6. ### Maths
Consider all 3-term geometric sequences with first term 1 and with common ratio the square of an integer between 1 and 1000. How many of these 1000 geometric sequences have the property that the sum of the 3 terms is prime?
7. ### Math
Consider all 3-term geometric sequences with first term 1 and with common ratio the square of an integer between 1 and 1000 (inclusive). How many of these 1000 geometric sequences have the property that the sum of the 3 terms is prime?
8. ### arithmetic
1. The first and last term of an A.P are, a and l respectively, show that the sum of nth term from the beginning and nth term from the end is a + l. 2. If mth term of an A.P be 1/n and nth term be 1/m, then show that its mnth term …
9. ### arithmetic
1. The first and last term of an A.P are, a and l respectively, show that the sum of nth term from the beginning and nth term from the end is a + l. 2. If mth term of an A.P be 1/n and nth term be 1/m, then show that its mnth term …
10. ### math C2 sequences and series
The eight,fourth and second terms of an arithmetic progression form the first three terms of a geometric series. The arithmetic progression has first term A and common difference d, and the geometric progression has first term G and …
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# algebra. posted by deli. Consider all 3-term geometric sequences with first term 1 and with common ratio the square of an integer between 1 and 1000. How many of these 1000 geometric sequences have the property that the sum of the 3 terms is prime?. 1. Reiny. I did this question here. http://www.jiskha.com/display.cgi?id=1367223190. Somebody called "tsong" disagreed, but clearly did not read my solution, since I "proved" that there is only one such case,. when the GS is 1 , 1, 1. that is, when the common ratio is 1. then 1 + 1^2 + 1^4 = 3 , which is a prime number. conclusion: there is only ONE such sequence.. ## Similar Questions. 1. ### math.... how can I tell if a sequences is airthmetic, geometric or neither?. 2. ### check!. If the third and ninth temr of a geometric series with a positive common ratio are -3 and -192 repectively, determine the value of the first term a Let a_1 be denoted by 'a' (first terms) then, a_3=r^2 a And, a_9=r^8 a Thus, r^8 a=-192 …. 3. ### Math *URGENT. Please give the answers and solutions for each. 1.If the second term is 2 and the seventh term of a geometric sequence is 64, find the 12th term. 2. Which term if the geometric sequence 18,54,162,486,... is 3,188,646?. 4. ### Maths. Eric thinks of 2 sequences.One is geometric and the other arithmetic.Both sequences start with the number 3.The common ratio of the geometric sequence is the same as the common difference of the arithmetic sequence.If the 6-th term ….
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5. ### Maths. 1..The first 2 terms of a geometric progression are the same as the first two terms of an arithmetic progression.The first term is 12 and is greater than the second term.The sum of the first 3 terms od the arithmetic progression is …. 6. ### Maths. Consider all 3-term geometric sequences with first term 1 and with common ratio the square of an integer between 1 and 1000. How many of these 1000 geometric sequences have the property that the sum of the 3 terms is prime?. 7. ### Math. Consider all 3-term geometric sequences with first term 1 and with common ratio the square of an integer between 1 and 1000 (inclusive). How many of these 1000 geometric sequences have the property that the sum of the 3 terms is prime?. 8. ### arithmetic. 1. The first and last term of an A.P are, a and l respectively, show that the sum of nth term from the beginning and nth term from the end is a + l. 2. If mth term of an A.P be 1/n and nth term be 1/m, then show that its mnth term …. 9. ### arithmetic. 1. The first and last term of an A.P are, a and l respectively, show that the sum of nth term from the beginning and nth term from the end is a + l. 2. If mth term of an A.P be 1/n and nth term be 1/m, then show that its mnth term …. 10. ### math C2 sequences and series. The eight,fourth and second terms of an arithmetic progression form the first three terms of a geometric series. The arithmetic progression has first term A and common difference d, and the geometric progression has first term G and …. More Similar Questions.
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# math
posted by .
the square root of 1000 divided by 2 then multiplied by 100 to the fith power then minus 6 then timed by two equals?
• math -
Use brackets to show exactly what you want
The way you wrote it, I see it as:
(([(√1000)/2]*100^5)-6)*2
Is this really what you want?
## Similar Questions
1. ### MATH
how do you simplify square root of 27 minus square root of 49 all divided by square root of 3. I have gotten two different answers when I tried to check it. It kind of looks like (27-49) (3) except all the numbers have their own square …
2. ### Algebra
I am a bit confused about conjugates in algebra. I am supposed to multiply 8/(the square root of two + 4) by the square root of 2 -4. My math book then goes on to simplify this to (8 times the square root of two minus 32) divided by …
3. ### Algebra (urgent!)
I am a bit confused about conjugates in algebra. I am supposed to multiply 8/(the square root of two + 4) by the square root of 2 -4. My math book then goes on to simplify this to (8 times the square root of two minus 32) divided by …
4. ### Math
Rationalize the denominator. Assume that all expressions under radicals represent positive numbers. The square root of 3y^4 to the third root divided by the square root of 15x^4 to the third root. Here is what I have so far: I multiplied. …
5. ### math
1.0 Four square root of seven open bracket two square root of three plus three square root of six minus square root of seven close bracket. 2.0 open bracket three square root of seven minus square root of two close bracket squared. …
6. ### math
Simply: (a) Four multiply by square root of seven open bracket two multiply by the square root of three plus three multiply by the square root of six minus the square root of seven close bracket. (b) open bracket three multiply by …
7. ### math
square root of 3 to the second power minus 5 divided by 8
8. ### math
what is the work of the square root of 3 to the second power minus 5 divided by 8
9. ### math
Here's my question: Determine if each system has one, two, or no solutions. (Show your work - 3 marks each) a)y=2x^2-2x+1 and y=3x-5 b)y=x^2+3x-16 and y=-x^2-8x-18 Here's my work on the question: a) x=(-b plus/minus sign (square root …
10. ### linear equations
Given: square root of 5 minus x is equals to x plus 1. Show that the solution to the above equation lies in the interval of negative 1less than or equals to x less than or equals to 5. Solve the equation. Solve the equation negative …
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# math. posted by .. the square root of 1000 divided by 2 then multiplied by 100 to the fith power then minus 6 then timed by two equals?. • math -. Use brackets to show exactly what you want. The way you wrote it, I see it as:. (([(√1000)/2]*100^5)-6)*2. Is this really what you want?. ## Similar Questions. 1. ### MATH. how do you simplify square root of 27 minus square root of 49 all divided by square root of 3. I have gotten two different answers when I tried to check it. It kind of looks like (27-49) (3) except all the numbers have their own square …. 2. ### Algebra. I am a bit confused about conjugates in algebra. I am supposed to multiply 8/(the square root of two + 4) by the square root of 2 -4. My math book then goes on to simplify this to (8 times the square root of two minus 32) divided by …. 3. ### Algebra (urgent!). I am a bit confused about conjugates in algebra. I am supposed to multiply 8/(the square root of two + 4) by the square root of 2 -4. My math book then goes on to simplify this to (8 times the square root of two minus 32) divided by …. 4. ### Math. Rationalize the denominator. Assume that all expressions under radicals represent positive numbers. The square root of 3y^4 to the third root divided by the square root of 15x^4 to the third root.
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Here is what I have so far: I multiplied. …. 5. ### math. 1.0 Four square root of seven open bracket two square root of three plus three square root of six minus square root of seven close bracket. 2.0 open bracket three square root of seven minus square root of two close bracket squared. …. 6. ### math. Simply: (a) Four multiply by square root of seven open bracket two multiply by the square root of three plus three multiply by the square root of six minus the square root of seven close bracket. (b) open bracket three multiply by …. 7. ### math. square root of 3 to the second power minus 5 divided by 8. 8. ### math. what is the work of the square root of 3 to the second power minus 5 divided by 8. 9. ### math. Here's my question: Determine if each system has one, two, or no solutions. (Show your work - 3 marks each) a)y=2x^2-2x+1 and y=3x-5 b)y=x^2+3x-16 and y=-x^2-8x-18 Here's my work on the question: a) x=(-b plus/minus sign (square root …. 10. ### linear equations. Given: square root of 5 minus x is equals to x plus 1. Show that the solution to the above equation lies in the interval of negative 1less than or equals to x less than or equals to 5. Solve the equation. Solve the equation negative …. More Similar Questions.
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# Uncountable connected graph has an infinite degree vertex
I am reading Dixon's Permutation Groups book and am currently looking at the section about graphs. I am a bit stuck on the exercise 2.3.1 which basically asks to prove that if $$G$$ is a connected graph whose vertex set is uncountable, then there's a vertex with infinite degree. My attempt at the solution is the following:
Take an arbitrary vertex $$v$$. Since $$G$$ is connected, we can reach any other vertex in finitely many steps. Hence there's an integer $$n$$ such that any path from $$v$$ to any other vertex has length smaller than $$n$$. Also, since every vertex has a finite degree, if we run BFS on this graph from $$v$$, we will reach every vertex in finite amount of time, since each recursive call will be finite, and there will be at most $$n$$ such calls in "depth", which implies that there are finitely many vertices in this graph.
Now clearly I did something wrong, since I "proved" a stronger statement, namely that if $$G$$ is any connected infinite graph, it has a vertex of infinite degree, which is not true (I think), that is, I never used the fact that $$V(G)$$ is uncountable.
• Think of the graph of the integers with each one connected one above and one below. This is an infinite connected graph with no infinite vertex. Where does your proof fail? Commented Nov 30, 2021 at 14:59
• Right, so my assumption that all the paths are finite is wrong? So essentially, the only condition for a graph to be connected is that for any pair of vertices, there's a finite path between them? ie. I got my quantifiers in the wrong order :D Commented Nov 30, 2021 at 15:22
• Your assumption that the paths are finite is correct, but there is no upper bound to the length of the path. It is like all natural numbers are finite, but there is a countable infinity of them. On any given path there is an upper bound. Commented Nov 30, 2021 at 15:59
Your proof only needs a minor tweak to reflect the fact that there is no $$n$$ maximum path length available.
A graph that has no vertex of infinite degree has a maximum vertex degree, say $$d_{\small M}$$. This also means that, from a given vertex, the set of vertices at distance $$i$$, $$U_i$$ is finite; specifically $$|U_i| \le {d_{{}_M}}^i$$. We can thus number the vertices in blocks based on distance from an arbitrary start point, showing that the vertices are countable - a contradiction as required.
• 'A graph that has no vertex of infinite degree has a maximum vertex degree'. This is not true. We can for example let $G$ be the one-dimensional lattice and give vertex $n$ (for $n$ positive) an amount of $n$ extra children, which are not connected to any other vertex. Commented Jun 6, 2023 at 11:45
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# Uncountable connected graph has an infinite degree vertex. I am reading Dixon's Permutation Groups book and am currently looking at the section about graphs. I am a bit stuck on the exercise 2.3.1 which basically asks to prove that if $$G$$ is a connected graph whose vertex set is uncountable, then there's a vertex with infinite degree. My attempt at the solution is the following:. Take an arbitrary vertex $$v$$. Since $$G$$ is connected, we can reach any other vertex in finitely many steps. Hence there's an integer $$n$$ such that any path from $$v$$ to any other vertex has length smaller than $$n$$. Also, since every vertex has a finite degree, if we run BFS on this graph from $$v$$, we will reach every vertex in finite amount of time, since each recursive call will be finite, and there will be at most $$n$$ such calls in "depth", which implies that there are finitely many vertices in this graph.. Now clearly I did something wrong, since I "proved" a stronger statement, namely that if $$G$$ is any connected infinite graph, it has a vertex of infinite degree, which is not true (I think), that is, I never used the fact that $$V(G)$$ is uncountable.. • Think of the graph of the integers with each one connected one above and one below. This is an infinite connected graph with no infinite vertex. Where does your proof fail? Commented Nov 30, 2021 at 14:59. • Right, so my assumption that all the paths are finite is wrong? So essentially, the only condition for a graph to be connected is that for any pair of vertices, there's a finite path between them? ie.
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I got my quantifiers in the wrong order :D Commented Nov 30, 2021 at 15:22. • Your assumption that the paths are finite is correct, but there is no upper bound to the length of the path. It is like all natural numbers are finite, but there is a countable infinity of them. On any given path there is an upper bound. Commented Nov 30, 2021 at 15:59. Your proof only needs a minor tweak to reflect the fact that there is no $$n$$ maximum path length available.. A graph that has no vertex of infinite degree has a maximum vertex degree, say $$d_{\small M}$$. This also means that, from a given vertex, the set of vertices at distance $$i$$, $$U_i$$ is finite; specifically $$|U_i| \le {d_{{}_M}}^i$$. We can thus number the vertices in blocks based on distance from an arbitrary start point, showing that the vertices are countable - a contradiction as required.. • 'A graph that has no vertex of infinite degree has a maximum vertex degree'. This is not true. We can for example let $G$ be the one-dimensional lattice and give vertex $n$ (for $n$ positive) an amount of $n$ extra children, which are not connected to any other vertex. Commented Jun 6, 2023 at 11:45.
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# Evaluate $(1+2i)(3+4i)$
The product of one plus two times imaginary unit and three plus four times imaginary unit is the given mathematical expression in this math problem.
$(1+2i)(3+4i)$
The product of $1$ plus $2i$ and $3$ plus $4i$ is a mathematical representation for the multiplication of them. Therefore, the above mathematical expression can be written as follows.
$=\,\,\,$ $(1+2i) \times (3+4i)$
### Trick to avoid confusion in multiplication
Multiplying the terms in the complex number $3$ plus $4i$ by another complex number $1$ plus $2i$ confuses some learners. So, let’s denote the complex number $1$ plus $2i$ by a variable. In this problem, the complex number $1+2i$ is denoted by a variable $z$.
$=\,\,\,$ $z \times (3+4i)$
### Multiply the terms by their coefficient
The complex $3+4i$ is a binomial and its terms are multiplied by a variable $z$. So, the terms $3$ and $4i$ can be multiplied by the variable $z$ as per the distributive property of multiplication over the addition.
$=\,\,\,$ $z \times 3$ $+$ $z \times 4i$
The use of variable $z$ is over. So, replace the variable $z$ by its actual value in the above mathematical expression.
$=\,\,\,$ $(1+2i) \times 3$ $+$ $(1+2i) \times 4i$
According to the commutative property of multiplication, the positions of the factors can be changed in each term.
$=\,\,\,$ $3 \times (1+2i)$ $+$ $4i \times (1+2i)$
Now, use the distributive property one more time in each term to distribute the coefficient over the addition of the terms.
$=\,\,\,$ $3 \times 1$ $+$ $3 \times 2i$ $+$ $4i \times 1$ $+$ $4i \times 2i$
### Simplify the Mathematical expression
The complex number $1$ plus $2i$ is multiplied by another complex number $3$ plus $4i$. The multiplication of them formed a mathematical expression. Now, it is time to simplify the mathematical expression to find the product of the given complex numbers $1$ plus $2i$ and $3$ plus $4i$.
$=\,\,\,$ $3$ $+$ $6i$ $+$ $4i$ $+$ $8i^2$
Second and third terms are like terms in the above mathematical expression. Therefore, add the like terms $6i$ and $4i$ to find the sum of them.
$=\,\,\,$ $3$ $+$ $10i$ $+$ $8i^2$
$=\,\,\,$ $3$ $+$ $10i$ $+$ $8 \times i^2$
According to the complex numbers, the square of imaginary unit is negative one.
$=\,\,\,$ $3$ $+$ $10i$ $+$ $8 \times (-1)$
$=\,\,\,$ $3$ $+$ $10i$ $-$ $8$
Now, use the commutative property to write the terms in an order for our convenience.
$=\,\,\,$ $3$ $-$ $8$ $+$ $10i$
Look at the first and second terms. They are numbers. So, subtract the number $8$ from number $3$ to find difference of them.
$=\,\,\,$ $-5+10i$
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# Evaluate $(1+2i)(3+4i)$. The product of one plus two times imaginary unit and three plus four times imaginary unit is the given mathematical expression in this math problem.. $(1+2i)(3+4i)$. The product of $1$ plus $2i$ and $3$ plus $4i$ is a mathematical representation for the multiplication of them. Therefore, the above mathematical expression can be written as follows.. $=\,\,\,$ $(1+2i) \times (3+4i)$. ### Trick to avoid confusion in multiplication. Multiplying the terms in the complex number $3$ plus $4i$ by another complex number $1$ plus $2i$ confuses some learners. So, let’s denote the complex number $1$ plus $2i$ by a variable. In this problem, the complex number $1+2i$ is denoted by a variable $z$.. $=\,\,\,$ $z \times (3+4i)$. ### Multiply the terms by their coefficient. The complex $3+4i$ is a binomial and its terms are multiplied by a variable $z$. So, the terms $3$ and $4i$ can be multiplied by the variable $z$ as per the distributive property of multiplication over the addition.. $=\,\,\,$ $z \times 3$ $+$ $z \times 4i$. The use of variable $z$ is over. So, replace the variable $z$ by its actual value in the above mathematical expression.. $=\,\,\,$ $(1+2i) \times 3$ $+$ $(1+2i) \times 4i$. According to the commutative property of multiplication, the positions of the factors can be changed in each term.. $=\,\,\,$ $3 \times (1+2i)$ $+$ $4i \times (1+2i)$. Now, use the distributive property one more time in each term to distribute the coefficient over the addition of the terms.. $=\,\,\,$ $3 \times 1$ $+$ $3 \times 2i$ $+$ $4i \times 1$ $+$ $4i \times 2i$. ### Simplify the Mathematical expression. The complex number $1$ plus $2i$ is multiplied by another complex number $3$ plus $4i$. The multiplication of them formed a mathematical expression. Now, it is time to simplify the mathematical expression to find the product of the given complex numbers $1$ plus $2i$ and $3$ plus $4i$.
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$=\,\,\,$ $3$ $+$ $6i$ $+$ $4i$ $+$ $8i^2$. Second and third terms are like terms in the above mathematical expression. Therefore, add the like terms $6i$ and $4i$ to find the sum of them.. $=\,\,\,$ $3$ $+$ $10i$ $+$ $8i^2$. $=\,\,\,$ $3$ $+$ $10i$ $+$ $8 \times i^2$. According to the complex numbers, the square of imaginary unit is negative one.. $=\,\,\,$ $3$ $+$ $10i$ $+$ $8 \times (-1)$. $=\,\,\,$ $3$ $+$ $10i$ $-$ $8$. Now, use the commutative property to write the terms in an order for our convenience.. $=\,\,\,$ $3$ $-$ $8$ $+$ $10i$. Look at the first and second terms. They are numbers. So, subtract the number $8$ from number $3$ to find difference of them.. $=\,\,\,$ $-5+10i$. ###### Math Questions. The math problems with solutions to learn how to solve a problem.. Learn solutions. Practice now. ###### Math Videos. The math videos tutorials with visual graphics to learn every concept.. Watch now. ###### Subscribe us. Get the latest math updates from the Math Doubts by subscribing us.
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# The Speed of an Ordinary Train is X Km per Hr and that of an Express Train is (X + 25) Km per Hr. (1) Find the Time Taken by Each Train to Cover 300 Km. (2) If the Ordinary Train Takes 2 Hr - ICSE Class 10 - Mathematics
#### Question
The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr.
(1) Find the time taken by each train to cover 300 km.
(2) If the ordinary train takes 2 hrs more than the express train;calculate speed of the express train.
#### Solution
1) Speed of ordinary train = x km/hr
Speed of express train = (x + 25) km/hr
Distance = 300 km
We know
"Time" = "Distance"/"Speed"
∴ Time taken by ordinary train to cover 300 km = 300/x hrs
Time taken by express train to cover 300 km = 300/(x + 25) hrs
2) Given that the ordinary train takes 2 hours more than the express train to cover the distance.
Therefore,
300/x - 300/(x + 25) = 2
(300x + 7500 - 300x)/(x(x + 25)) = 2
7500 = 2x^2 + 50x
2x^2 + 50x - 7500 = 0
x^2 + 25x - 3750 = 0
x^2 + 75x - 50x - 3750 = 0
x(x + 75) - 50(x + 75) = 0
(x + 75)(x - 50) = 0
x = -75, 50
But, speed cannot be negative. So, x = 50.
∴ Speed of the express train = (x + 25) km/hr = 75 km/hr
Is there an error in this question or solution?
#### APPEARS IN
Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current)
Chapter 6: Solving (simple) Problems (Based on Quadratic Equations)
Ex.6C | Q: 1
#### Video TutorialsVIEW ALL [5]
Solution The Speed of an Ordinary Train is X Km per Hr and that of an Express Train is (X + 25) Km per Hr. (1) Find the Time Taken by Each Train to Cover 300 Km. (2) If the Ordinary Train Takes 2 Hr Concept: Quadratic Equations.
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Share. Books Shortlist. # The Speed of an Ordinary Train is X Km per Hr and that of an Express Train is (X + 25) Km per Hr. (1) Find the Time Taken by Each Train to Cover 300 Km. (2) If the Ordinary Train Takes 2 Hr - ICSE Class 10 - Mathematics. #### Question. The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr.. (1) Find the time taken by each train to cover 300 km.. (2) If the ordinary train takes 2 hrs more than the express train;calculate speed of the express train.. #### Solution. 1) Speed of ordinary train = x km/hr. Speed of express train = (x + 25) km/hr. Distance = 300 km. We know. "Time" = "Distance"/"Speed". ∴ Time taken by ordinary train to cover 300 km = 300/x hrs. Time taken by express train to cover 300 km = 300/(x + 25) hrs. 2) Given that the ordinary train takes 2 hours more than the express train to cover the distance.. Therefore,. 300/x - 300/(x + 25) = 2. (300x + 7500 - 300x)/(x(x + 25)) = 2.
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7500 = 2x^2 + 50x. 2x^2 + 50x - 7500 = 0. x^2 + 25x - 3750 = 0. x^2 + 75x - 50x - 3750 = 0. x(x + 75) - 50(x + 75) = 0. (x + 75)(x - 50) = 0. x = -75, 50. But, speed cannot be negative. So, x = 50.. ∴ Speed of the express train = (x + 25) km/hr = 75 km/hr. Is there an error in this question or solution?. #### APPEARS IN. Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current). Chapter 6: Solving (simple) Problems (Based on Quadratic Equations). Ex.6C | Q: 1. #### Video TutorialsVIEW ALL [5]. Solution The Speed of an Ordinary Train is X Km per Hr and that of an Express Train is (X + 25) Km per Hr. (1) Find the Time Taken by Each Train to Cover 300 Km. (2) If the Ordinary Train Takes 2 Hr Concept: Quadratic Equations.. S.
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https://www.educents.com/minion-math-centers.html?mc_cid=67b5293cc0&mc_eid=8f0902ae95
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Save 10% off sitewide with code HOLIDAY. Shop Now >
# Minion Math Centers
\$4.99
• ## Product Type: Printable
### Product Description
Add some fun to your math practice with these cute little Minions! Use these hands-on math centers to practice and reinforce key math skills.
These cute and fun minions, which your students already love, are the perfect character for these engaging math centers. This packet includes seven different math centers or resources, some of which can be used in multiple ways. The activities include:
Use this worksheet to practice addition skills. Students cut out the minions at the bottom of the page to use as counters. Students can use the counters to help them solve the addition problems.
2. Minion Subtraction (to 12)
Use this worksheet to practice subtraction skills. Students cut out the minions at the bottom of the page to use as counters. Students can use the counters to help them solve the subtraction problems.
3. Minions Making Ten
Use this worksheet to practice decomposing the number 10. The minion at the top has the number ten on him to represent the whole. The students determine the two parts to use to make ten. For students that are still developing the concept of decomposing numbers, I would suggest using some type of manipulative to help them see how to make ten in two different parts.
4. Minion Measurement
The minions are ready to print and use. Simply cut out each minion and use in a math station with your choice of standard or nonstandard measurement tools. Students could also use these minions to put in order by size or even to make conclusions on which would weigh more.
5. Minion Money Match
This memory type of game is ready to print and use. Simply print and cut the matching cards. Students will place the cards face down and try to find the three matching cards. One card represents the front of the coin. One card represents the back of the coin. The final card represents the value of the coin. Students could also use these cards to play a card game like Go Fish. “Do you have the back of a dime?”
6. Missing Minion Numbers
Use this worksheet to help your students master number order. On this two page worksheet students will determine which number the minion is covering up. The student will write that number in the box at the end of the line. The missing numbers start with 00zero to twenty counting by ones. I have also included counting by tens and counting backwards to help them truly master number order.
7. Minion Number Cards
Use these numbers cards for a variety of activities. There are three sets of cards included: counting by tens to one hundred; counting by fives to one hundred; counting by ones. Students can practice ordering numbers with any of the sets. Use the number cards to play an addition or subtraction game by drawing two cards and adding them together or subtracting. Students can also use the number cards to practice comparing numbers by determine which is greater than or less than. Bonus: the counting by ones cards go to thirty-one so they can also be used as calendar numbers. :-)
### Leave a Review!
OUTSTANDING
Aug 05, 2015
My grandson is loving working with the minions and my wife is having so much fun helping him!
Terrell - Teacher
Fun
Aug 04, 2015
My son doesn't know he's doing school work .
Jeanine
MATH STATIONS!
Jul 31, 2015
The kindergarten kids will LOVE using your Minion Math Centers when they are in math stations. Educational and fun! GREAT PRODUCT!
Firstname
So cute!!!
Jul 31, 2015
My kids love, love, love these characters and it really kept their attention throughout the lessons. Prepping the lesson was easy for me but I do recommend cutting the pieces yourself so your kids can focus more on the math part rather than the semi arts and crafts. Or make it a game on its own.
Tammy - Teacher
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Save 10% off sitewide with code HOLIDAY. Shop Now >. # Minion Math Centers. \$4.99. • ## Product Type: Printable. ### Product Description. Add some fun to your math practice with these cute little Minions! Use these hands-on math centers to practice and reinforce key math skills.. These cute and fun minions, which your students already love, are the perfect character for these engaging math centers. This packet includes seven different math centers or resources, some of which can be used in multiple ways. The activities include:. Use this worksheet to practice addition skills. Students cut out the minions at the bottom of the page to use as counters. Students can use the counters to help them solve the addition problems.. 2. Minion Subtraction (to 12). Use this worksheet to practice subtraction skills. Students cut out the minions at the bottom of the page to use as counters. Students can use the counters to help them solve the subtraction problems.. 3. Minions Making Ten. Use this worksheet to practice decomposing the number 10. The minion at the top has the number ten on him to represent the whole. The students determine the two parts to use to make ten. For students that are still developing the concept of decomposing numbers, I would suggest using some type of manipulative to help them see how to make ten in two different parts.. 4. Minion Measurement. The minions are ready to print and use. Simply cut out each minion and use in a math station with your choice of standard or nonstandard measurement tools. Students could also use these minions to put in order by size or even to make conclusions on which would weigh more.. 5. Minion Money Match. This memory type of game is ready to print and use. Simply print and cut the matching cards. Students will place the cards face down and try to find the three matching cards. One card represents the front of the coin. One card represents the back of the coin. The final card represents the value of the coin. Students could also use these cards to play a card game like Go Fish.
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“Do you have the back of a dime?”. 6. Missing Minion Numbers. Use this worksheet to help your students master number order. On this two page worksheet students will determine which number the minion is covering up. The student will write that number in the box at the end of the line. The missing numbers start with 00zero to twenty counting by ones. I have also included counting by tens and counting backwards to help them truly master number order.. 7. Minion Number Cards. Use these numbers cards for a variety of activities. There are three sets of cards included: counting by tens to one hundred; counting by fives to one hundred; counting by ones. Students can practice ordering numbers with any of the sets. Use the number cards to play an addition or subtraction game by drawing two cards and adding them together or subtracting. Students can also use the number cards to practice comparing numbers by determine which is greater than or less than. Bonus: the counting by ones cards go to thirty-one so they can also be used as calendar numbers. :-). ### Leave a Review!. OUTSTANDING. Aug 05, 2015. My grandson is loving working with the minions and my wife is having so much fun helping him!. Terrell - Teacher. Fun. Aug 04, 2015. My son doesn't know he's doing school work .. Jeanine. MATH STATIONS!. Jul 31, 2015. The kindergarten kids will LOVE using your Minion Math Centers when they are in math stations. Educational and fun! GREAT PRODUCT!. Firstname. So cute!!!. Jul 31, 2015. My kids love, love, love these characters and it really kept their attention throughout the lessons. Prepping the lesson was easy for me but I do recommend cutting the pieces yourself so your kids can focus more on the math part rather than the semi arts and crafts. Or make it a game on its own.. Tammy - Teacher.
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https://mathandmultimedia.com/2011/07/11/angles-of-parallelograms/
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# Discovering properties of angles of parallelograms
Consider the parallelogram below. We mark the bottom-left angle with a and the bottom-right angle with b to denote their measures as shown in Figure 1.
Figure 1
We create another parallelogram with same size and the shame shape, but of different color. We place the two parallelograms side by side as shown in Figure 2. The angles a and b is a linear pair; therefore, (1) a + b = 180˚
Figure 2
We create two more parallelograms and place them as shown in Figure 3 to form a bigger parallelogram. Angle a in the red parallelogram and the adjacent angle in the blue parallelogram is also a linear pair; therefore, the upper-left angle in the blue parallelogram also equals b (see Figure 4). In effect, the the upper-right angle in the yellow parallelogram also equals the measure of angle a.
Figure 3
In Figure 4, the parallelograms meet at a point with no gaps and overlaps which means that all the angle measures add up to 360 degrees. This is supported algebraically by (1):
(a + b) + (a + b) = 2(a+b) = 360˚.
It is clear that given the measures of the angles in Figure 3, it is possible to give the measures the other angles as shown in Figure 4. We name the angles at the opposite side of the intersecting lines or segments vertical angles.
Figure 4
From Figure 4, we can conclude that
1. The consecutive angles of a parallelogram add up to 180 degrees;
2. Vertical angles are congruent;
3. The opposite angles of a parallelogram are congruent; and
4. The angle sum of a parallelogram is 360 degrees.
Notice that the bigger parallelograms – the one containing two parallelograms and the one containing four smaller parallelograms – also measure (a + b) + (a + b) = 360˚ .
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# Discovering properties of angles of parallelograms. Consider the parallelogram below. We mark the bottom-left angle with a and the bottom-right angle with b to denote their measures as shown in Figure 1.. Figure 1. We create another parallelogram with same size and the shame shape, but of different color. We place the two parallelograms side by side as shown in Figure 2. The angles a and b is a linear pair; therefore, (1) a + b = 180˚. Figure 2. We create two more parallelograms and place them as shown in Figure 3 to form a bigger parallelogram. Angle a in the red parallelogram and the adjacent angle in the blue parallelogram is also a linear pair; therefore, the upper-left angle in the blue parallelogram also equals b (see Figure 4). In effect, the the upper-right angle in the yellow parallelogram also equals the measure of angle a.. Figure 3. In Figure 4, the parallelograms meet at a point with no gaps and overlaps which means that all the angle measures add up to 360 degrees. This is supported algebraically by (1):. (a + b) + (a + b) = 2(a+b) = 360˚.. It is clear that given the measures of the angles in Figure 3, it is possible to give the measures the other angles as shown in Figure 4.
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We name the angles at the opposite side of the intersecting lines or segments vertical angles.. Figure 4. From Figure 4, we can conclude that. 1. The consecutive angles of a parallelogram add up to 180 degrees;. 2. Vertical angles are congruent;. 3. The opposite angles of a parallelogram are congruent; and. 4. The angle sum of a parallelogram is 360 degrees.. Notice that the bigger parallelograms – the one containing two parallelograms and the one containing four smaller parallelograms – also measure (a + b) + (a + b) = 360˚ .
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http://www.xpmath.com/forums/showthread.php?s=fdc380e6137922716b714a57ce1cd3bc&p=19416
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could someone help me with this assignment in Algebra? - XP Math - Forums
XP Math - Forums could someone help me with this assignment in Algebra?
User Name Remember Me? Password
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07-25-2007 #1 MarkAngelo Guest Posts: n/a could someone help me with this assignment in Algebra? 1) (x + 2y + 2z)sqare2) (2x + 3y - 4z + 3w)square3) (xsquare - 2x + 3)square4) (2x + 3yz + w)sqare5) (x + y - 3z)square6) (x + y)to the 4thplease do include your solutions, I would really appreciate the helps. T_T
07-25-2007 #2 de4th Guest Posts: n/a It's the same process for all of them... just expand and simplify.(x+2y+2z)^2= x^2 + 2xy + 2xz + 2xy + 4y^2 + 4yz + 2xz + 4yz + 4z^2= 4y^2 + 4z^2 + x^2 + 4xy + 4xz + 8yz...You can do the rest =), good luck.
07-25-2007 #3 gp4rts Guest Posts: n/a Multiply each term in the expression by the whole expression and add like terms. For examplex + 2y + 2z)^2 = (x + 2y + 2z)*(x + 2y + 2z)Multiplying the first term (x) by the full expression (x + 2y + 2z) getx^2 + 2xy + 2xz (a)Now do the same with the second term (2y)2xy + 4*y^2 + 4*y*z (b)and finally the last term (2z)2xz + 4zy + 4z^2 (c)Then add them all [(a) + (b) + (c)] together and collect like terms.
07-25-2007 #4 DBSII Guest Posts: n/a 1) (x+2y+2z)(x+2y+2z) (x^2+2xy+2xz)+(2xy+4y^2+4zy)+(2xz+4zy+4z^2) x^2+4xy+4xz+4y^2+8zy+4z^2
07-25-2007 #5 handsomeboy1702 Guest Posts: n/a (x + 2y + 2z)^2 = x^2 + 4y^2 + 4z^2 + 4xy + 4xz + 8yz(2x + 3y - 4z + 3w)^2 = 4x^2 + 9y^2 + 16z^2 + 9w^2 + 12xy - 16xz + 12xw - 24yz +18yx - 24zw(x^2 - 2x + 3)^2 = x^4 + 4x^2 + 9 - 4x^3 + 6x^2 - 12x = x^4 - 4x^3 + 10x^2 - 12x + 9(2x + 3yz + w)^2 = 4x^2 + 9y^2z^2 + w^2 + 12xyz + 4xw + 6yzw(x + y - 3z)^2 = x^2 + y^2 + 9z^2 + 2xy -6xz - 6yz(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4
07-25-2007 #6 sean_mccully Guest Posts: n/a Hhm perhaps you need the definition of a square, a square is an equation to the power of 2. 2 squared is 4,the square root of 4 is 2 (2*2=4). Given above the answer to number 1 is (x + 2y + 2z)*(x + 2y + 2z) =yum alegbra!!!!!!
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could someone help me with this assignment in Algebra? - XP Math - Forums. XP Math - Forums could someone help me with this assignment in Algebra?. User Name Remember Me? Password. Register Arcade Members List Mark Forums Read. Thread Tools Display Modes. 07-25-2007 #1 MarkAngelo Guest Posts: n/a could someone help me with this assignment in Algebra? 1) (x + 2y + 2z)sqare2) (2x + 3y - 4z + 3w)square3) (xsquare - 2x + 3)square4) (2x + 3yz + w)sqare5) (x + y - 3z)square6) (x + y)to the 4thplease do include your solutions, I would really appreciate the helps. T_T. 07-25-2007 #2 de4th Guest Posts: n/a It's the same process for all of them... just expand and simplify.(x+2y+2z)^2= x^2 + 2xy + 2xz + 2xy + 4y^2 + 4yz + 2xz + 4yz + 4z^2= 4y^2 + 4z^2 + x^2 + 4xy + 4xz + 8yz...You can do the rest =), good luck.. 07-25-2007 #3 gp4rts Guest Posts: n/a Multiply each term in the expression by the whole expression and add like terms. For examplex + 2y + 2z)^2 = (x + 2y + 2z)*(x + 2y + 2z)Multiplying the first term (x) by the full expression (x + 2y + 2z) getx^2 + 2xy + 2xz (a)Now do the same with the second term (2y)2xy + 4*y^2 + 4*y*z (b)and finally the last term (2z)2xz + 4zy + 4z^2 (c)Then add them all [(a) + (b) + (c)] together and collect like terms.. 07-25-2007 #4 DBSII Guest Posts: n/a 1) (x+2y+2z)(x+2y+2z) (x^2+2xy+2xz)+(2xy+4y^2+4zy)+(2xz+4zy+4z^2) x^2+4xy+4xz+4y^2+8zy+4z^2.
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07-25-2007 #5 handsomeboy1702 Guest Posts: n/a (x + 2y + 2z)^2 = x^2 + 4y^2 + 4z^2 + 4xy + 4xz + 8yz(2x + 3y - 4z + 3w)^2 = 4x^2 + 9y^2 + 16z^2 + 9w^2 + 12xy - 16xz + 12xw - 24yz +18yx - 24zw(x^2 - 2x + 3)^2 = x^4 + 4x^2 + 9 - 4x^3 + 6x^2 - 12x = x^4 - 4x^3 + 10x^2 - 12x + 9(2x + 3yz + w)^2 = 4x^2 + 9y^2z^2 + w^2 + 12xyz + 4xw + 6yzw(x + y - 3z)^2 = x^2 + y^2 + 9z^2 + 2xy -6xz - 6yz(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4. 07-25-2007 #6 sean_mccully Guest Posts: n/a Hhm perhaps you need the definition of a square, a square is an equation to the power of 2. 2 squared is 4,the square root of 4 is 2 (2*2=4). Given above the answer to number 1 is (x + 2y + 2z)*(x + 2y + 2z) =yum alegbra!!!!!!. Thread Tools Display Modes Linear Mode. Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules. Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Welcome XP Math News Off-Topic Discussion Mathematics XP Math Games Worksheets Homework Help Problems Library Math Challenges. All times are GMT -4. The time now is 07:42 PM.. Contact Us - XP Math - Forums - Archive - Privacy Statement - Top.
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https://www.hackmath.net/en/math-problem/5000
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# Cutting circles
From the square 1 m side we have to cut the circles with a radius of 10 cm. How many discs we cut and how many percent will be waste?
Correct result:
n = 25
p = 21.4602 %
#### Solution:
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#### You need to know the following knowledge to solve this word math problem:
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# Cutting circles. From the square 1 m side we have to cut the circles with a radius of 10 cm. How many discs we cut and how many percent will be waste?. Correct result:. n = 25. p = 21.4602 %. #### Solution:. We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!. Tips to related online calculators. Our percentage calculator will help you quickly calculate various typical tasks with percentages.. #### You need to know the following knowledge to solve this word math problem:. We encourage you to watch this tutorial video on this math problem:. ## Next similar math problems:. • Percentage of waste. In a square plate with side 75 cm we cut 4 same circles. Calculate the percentage of waste.. • Cutting square. From a square with a side of 30 cm, we cut the circle with the highest possible diameter. How many percents of the square content is this circle?. • Waste. How many percents are waste from a circular plate with a radius of 1 m from which we cut a square with the highest area?. • Glass mosaic. How many dm2 glass is nessesary to produc 97 slides of a regular 6-gon, whose side has length 21 cm? Assume that cutting glass waste is 10%.. • Square metal sheet. Four squares of 300 mm side were cut out from a square sheet metal plate with a side of 0,7 m. Express the fraction and the percentage of waste from the square metal sheet.. The cone-shaped lampshade has a diameter of 30 cm and a height of 10 cm. How many cm2 of material will we need when we 10% is waste?. • Annulus. Two concentric circles form an annulus of width 10 cm. The radius of the smaller circle is 20 cm. Calculate the content area of annulus.. • Tiles. The pupils of the building school have to calculate how many roof tiles they will need to cover the roof of the house in the shape of two rectangles measuring 10 x 7.2 m.
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One roof tile covers a rectangular area of 22 cmx32 cm. How many tiles will they nee. • Concentric circles. There is given a circle K with a radius r = 8 cm. How large must a radius have a smaller concentric circle that divides the circle K into two parts with the same area?. • Two annuluses. The area of the annular circle formed by two circles with a common center is 100 cm2. The radius of the outer circle is equal to twice the radius of the inner circle. Determine the outside circle radius in centimeters.. • Square and circles. The square in the picture has a side length of a = 20 cm. Circular arcs have centers at the vertices of the square. Calculate the areas of the colored unit. Express area using side a.. • Circles. The areas of the two circles are in the ratio 2:20. The larger circle has a diameter 20. Calculate the radius of the smaller circle.. • Tent. Pyramid-shaped tent has a base square with a side length of 2 m and a height 1.7 m. How many meters of canvas is nneded to make it if for a waste should be added 10%?. • Concentric circles and chord. In a circle with a diameter d = 10 cm, a chord with a length of 6 cm is constructed. What radius have the concentric circle while touch this chord?. • Two circles. Two circles with the same radius r = 1 are given. The center of the second circle lies on the circumference of the first. What is the area of a square inscribed in the intersection of given circles?. • Square side. If we enlarge the square side a = 5m, its area will increase by 10,25%. How many percent will the side of the square increase? How many percent will it increase the circumference of the square?. • Four circles. 1) Calculate the circle radius if its area is 400 cm square 2) Calculate the radius of the circle whose circumference is 400 cm. 3) Calculate circle circumference if its area is 400 cm square 4) Calculate the circle's area if perimeter 400 cm.
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https://www.reference.com/web?q=speed+vs+velocity+physics&qo=contentPageRelatedSearch&o=600605&l=dir
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Web Results
www.physicsclassroom.com/.../1DKin/Lesson-1/Speed-and-Velocity
The physics teacher walked a distance of 12 meters in 24 seconds; thus, her average speed was 0.50 m/s. However, since her displacement is 0 meters, her average velocity is 0 m/s. Remember that the displacement refers to the change in position and the velocity is based upon this position change. In this case of the teacher's motion, there is a position change of 0 meters and thus an average .....
www.miniphysics.com/speed-vs-velocity.html
If either magnitude or direction is changing, the velocity will not remain constant. Hence, for an object to have a constant velocity, it is necessary for it to be moving in a straight line at a constant speed. On the other hand, speed can remain constant if direction is changed and magnitude is kept constant, as speed is a scalar quantity.
physics.info/velocity
Speed and velocity are related in much the same way that distance and displacement are related. Speed is a scalar and velocity is a vector. Speed gets the symbol v (italic) and velocity gets the symbol v (boldface). Average values get a bar over the symbol.
This is a simple physics tutorial aimed at high school students in grades 9 and 10 (GCSE, HSC) exploring what the difference is between speed and velocity. Speed tells us how far an object or ...
www.diffen.com/difference/Acceleration_vs_Velocity
If an object is moving at constant speed in a circular motion -- such as a satellite orbiting the earth -- it is said to be accelerating because change in direction of motion means its velocity is changing even if speed may be constant. (See Speed vs Velocity) This is called centripetal (directed towards the center) acceleration. On the other ...
I'm getting confused with speed and velocity. I know that speed does not tell direction and velocity does but here's the question: A person walks first at a constant speed of 5.00m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00m/s.
www.ducksters.com/science/physics/speed_and_velocity.php
Although speed and velocity are often used interchangeably in everyday life, they represent different quantities in physics. What is speed? Speed is a measurement of how fast an object moves relative to a reference point.
www.diffen.com/difference/Speed_vs_Velocity
Speed is the rate of change of motion, i.e. distance moved by an object in a specified time irrespective of direction. Velocity is speed with respect to direction. Speed is a scalar quantity while velocity is a vector. Speed is the rate at which an object covers a specified distance. It is a scalar ...
www.difference.wiki/speed-vs-velocity
Speed and velocity are the two interrelated terms that are often used interchangeably, but the one knowing basics of physics can easily differentiate it between both these terms. While differentiating it between both these terms, informally, we can say that velocity is the speed with a direction.
www.physicsclassroom.com/mmedia/kinema/trip.cfm
Average vs. Instantaneous Speed. During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car reveals information about the instantaneous speed of your car.
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Web Results. www.physicsclassroom.com/.../1DKin/Lesson-1/Speed-and-Velocity. The physics teacher walked a distance of 12 meters in 24 seconds; thus, her average speed was 0.50 m/s. However, since her displacement is 0 meters, her average velocity is 0 m/s. Remember that the displacement refers to the change in position and the velocity is based upon this position change. In this case of the teacher's motion, there is a position change of 0 meters and thus an average ...... www.miniphysics.com/speed-vs-velocity.html. If either magnitude or direction is changing, the velocity will not remain constant. Hence, for an object to have a constant velocity, it is necessary for it to be moving in a straight line at a constant speed. On the other hand, speed can remain constant if direction is changed and magnitude is kept constant, as speed is a scalar quantity.. physics.info/velocity. Speed and velocity are related in much the same way that distance and displacement are related. Speed is a scalar and velocity is a vector. Speed gets the symbol v (italic) and velocity gets the symbol v (boldface). Average values get a bar over the symbol.. This is a simple physics tutorial aimed at high school students in grades 9 and 10 (GCSE, HSC) exploring what the difference is between speed and velocity. Speed tells us how far an object or .... www.diffen.com/difference/Acceleration_vs_Velocity. If an object is moving at constant speed in a circular motion -- such as a satellite orbiting the earth -- it is said to be accelerating because change in direction of motion means its velocity is changing even if speed may be constant. (See Speed vs Velocity) This is called centripetal (directed towards the center) acceleration. On the other .... I'm getting confused with speed and velocity. I know that speed does not tell direction and velocity does but here's the question: A person walks first at a constant speed of 5.00m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00m/s.
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www.ducksters.com/science/physics/speed_and_velocity.php. Although speed and velocity are often used interchangeably in everyday life, they represent different quantities in physics. What is speed? Speed is a measurement of how fast an object moves relative to a reference point.. www.diffen.com/difference/Speed_vs_Velocity. Speed is the rate of change of motion, i.e. distance moved by an object in a specified time irrespective of direction. Velocity is speed with respect to direction. Speed is a scalar quantity while velocity is a vector. Speed is the rate at which an object covers a specified distance. It is a scalar .... www.difference.wiki/speed-vs-velocity. Speed and velocity are the two interrelated terms that are often used interchangeably, but the one knowing basics of physics can easily differentiate it between both these terms. While differentiating it between both these terms, informally, we can say that velocity is the speed with a direction.. www.physicsclassroom.com/mmedia/kinema/trip.cfm. Average vs. Instantaneous Speed. During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car reveals information about the instantaneous speed of your car.. Related Search. Related Search.
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# Current research topics
My research activities are centered around Decision Theory. More precisely, I am interested in the mathematical models underlying our decision making processes. To make it short, consider I ask you whether you want to play to a certain game (decision $$d_1$$) or not (decision $$d_2$$). If I do not tell you which game it is, it will be difficult for you to tell me which decision you would prefer. Now I explain to you that playing the game simply consists in giving you $100. You will certainly tell me that you prefer decision $$d_1$$) to $$d_2$$, which could be denoted mathematically by $$d_1 \succsim d_2$$, where $$\succsim$$ is a binary relation over decisions. One question that naturally arises is: why did you prefer $$d_1$$ to $$d_2$$? Simply because the consequence (or outcome) of taking this decision, i.e. winning$100, seems more appealing than that of decision $$d_2$$, i.e. winning nothing. Hence $$\succsim$$, our preference relation over decisions, reflects our preferences over the consequences of the decisions. This explains why one aspect of my researches concerns the mathematical models representing preference relations over consequence sets.
Let us now change the above game as follows: instead of giving you $100, I toss a coin, and if the head obtains, I give you$100, but if the tail obtains, you lose $200. Now which decision do you prefer? Here, if the coin is fair, you will probably prefer $$d_2$$ to $$d_1$$ because there is a substantial chance that $$d_1$$ results in losing$200. However if the coin is such that the probability that the head obtains is 99.99%, you may prefer $$d_1$$ because the chance of losing \$200 is very small. This means that preference relation $$\succsim$$ does not only take into account your preferences over outcomes, but it also takes into account the uncertainty about the realisation of the outcomes. Probabilities are a popular model of uncertainty, and since the late 80's and Pearl's seminal book, graphical models have been widely used for their encoding and their computation. This explains why my second research field concerns graphical models. It turns out that graphical models also prove useful for preference encodings.
# Representation of the Decision Maker's Preferences
Mathematically, modeling preferences over a set of decisions or over a set of consequences merely amounts to finding a binary relation over pairs decisions or pairs of consequences. For instance, assume that a Decision Maker (DM) has some preferences over a set $$X$$ = {eat some chicken, eat some fish, eat an apple pie, eat a soup}. For any pair $$(x,y)$$ of elements of $$X$$, the following may happen:
1. the Decision Maker may not be able to compare $$x$$ and $$y$$, i.e., she cannot say whether she prefers $$x$$ to $$y$$ or $$y$$ to $$x$$. This may be the case for $$x$$="eat a soup" and $$y$$="eat an apple pie" since one is a starter and the other is a dessert.
2. the Decision Maker may prefer $$x$$ to $$y$$ or $$y$$ to $$x$$, or even be indifferent between $$x$$ and $$y$$.
Mathematically, the Decision Maker's preferences may be represented by a binary relation $$\succsim$$ on $$X \times X$$ such that $$x \succsim y$$ means the DM prefers $$x$$ to $$y$$ or is indifferent between $$x$$ and $$y$$. Thus $$x$$ and $$y$$ being incomparable is equivalent to Not ($$x \succsim y$$) and Not ($$y \succsim x$$). The strict preference of $$x$$ over $$y$$ can be represented by $$x \succsim y$$ and Not ($$y \succsim x$$), and the indifference between $$x$$ and $$y$$ is captured by $$x \succsim y$$ and $$y \succsim x$$.
The qualitative aspect of binary relation $$\succsim$$ makes it not particularly well suited for fast computation, in particular if one wishes to find the best decision under a given set of constraints. Hence, in practical situations, this relation needs often be encoded numerically. One of the most popular encodings is that of utility functions, a.k.a. utilities. The idea is to use a mapping $$u : X \mapsto \mathbb{R}$$, the set of the real numbers, assigning numbers to elements of $$X$$ such that the higher the preferred. More formally, $$u$$ is defined as: $x \succsim y \Longleftrightarrow u(x) \geq u(y), \forall\ x,y \in X.$
In practical situations, the DM's decisions take into account multiple conflicting objectives, hence the outcome set is a multidimensional space and the outcomes are tuples (of attributes) $$x = (x_1,\ldots,x_n)$$. Thus utilities are functions of tuples. This is a major problem when they are to be constructed. Indeed, as each Decision Maker has her own preferences, utilities differ from one DM to another. So the only way to construct the DM's utlities is to ask her some questions. But when questions involve to many different attributes, they become too complicated for the human brain to handle and the DM is not able to give accurate answers. In order to simplify the questions, assumptions on the structure of the utilities need be made. One of the most popular is the additive decomposition: $u(x_1,\ldots,x_n) = u_1(x_1) + \cdots + u_n(x_n).$
During my PhD thesis, I have studied testable conditions on the DM preference relation ensuring the existence of additively decomposable utility functions (additive utilities for short). I am still interested in decompositions of utilities, but I now study more complicated decompositions such as generalized additive decompositions (GAI): instead of splitting the overall utility into pieces, each one depending only on one attribute, GAI splits it into utilities over some (possibly overlapping) sets of attributes. For instance~: $u(a,b,c,d,e,f) = u_1(a,b,c) + u_2(c,d,e) + u_3(d,e,f).$
# Graphical Models
Among graphical models, I am especially interested in Bayesian networks (BN) and GAI-nets. The former are powerful tools of the Artificial Intelligence community enabling fast computation of probabilities (marginal, conditional, a priori or a posteriori probabilities). To achieve this result, BN use graphical structures that encode knowledge about the decomposition of joint probability distributions of interest. GAI-nets are rather similar in spirit, except that instead of decomposing a probability distribution over random variables, they decompose attributes or criteria of preference tuples.
Mainly my current activities on graphical models can be divided into three classes:
• Until now, I was mainly concerned with the improvement of BN computation algorithms. In the literature, roughly two main approaches emerged: directed and undirected methods. It is commonly thought that undirected methods outperform directed ones. One of my goals was to unify both types of methods to show that directed methods can compete with undirected ones. Using the advantages of each kind of method, this unification enabled the improvement of all the algorithms.
• As I mentioned, Bayesian networks are graphical structures. These represent the decomposition of joint probabilities. Hence they are specific for each problem of interest. In theory, if you possess a sufficiently good database of values of the random variables you are interested in, computer programs should be able to learn the graph structure of the BN (using for instance statistical independence tests). In practice, however, such programs do not always produce high quality graphs. This explains why I am working on learning methods.
• My last main research activity concerns GAI-nets. The idea is to take advantage of some decomposition of the Decision Maker's preference relations to be able to elicit efficiently his/her preferences, and to answer quickly to questions such as "does the Decision Maker prefers this alternative to that one?" or "what is the preferred alternative of the Decision Maker?". Efficient elicitation procedures could prove useful for instance for web shopping sites.
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# Current research topics. My research activities are centered around Decision Theory. More precisely, I am interested in the mathematical models underlying our decision making processes. To make it short, consider I ask you whether you want to play to a certain game (decision $$d_1$$) or not (decision $$d_2$$). If I do not tell you which game it is, it will be difficult for you to tell me which decision you would prefer. Now I explain to you that playing the game simply consists in giving you $100. You will certainly tell me that you prefer decision $$d_1$$) to $$d_2$$, which could be denoted mathematically by $$d_1 \succsim d_2$$, where $$\succsim$$ is a binary relation over decisions. One question that naturally arises is: why did you prefer $$d_1$$ to $$d_2$$? Simply because the consequence (or outcome) of taking this decision, i.e. winning$100, seems more appealing than that of decision $$d_2$$, i.e. winning nothing. Hence $$\succsim$$, our preference relation over decisions, reflects our preferences over the consequences of the decisions. This explains why one aspect of my researches concerns the mathematical models representing preference relations over consequence sets.. Let us now change the above game as follows: instead of giving you $100, I toss a coin, and if the head obtains, I give you$100, but if the tail obtains, you lose $200. Now which decision do you prefer? Here, if the coin is fair, you will probably prefer $$d_2$$ to $$d_1$$ because there is a substantial chance that $$d_1$$ results in losing$200. However if the coin is such that the probability that the head obtains is 99.99%, you may prefer $$d_1$$ because the chance of losing \$200 is very small. This means that preference relation $$\succsim$$ does not only take into account your preferences over outcomes, but it also takes into account the uncertainty about the realisation of the outcomes. Probabilities are a popular model of uncertainty, and since the late 80's and Pearl's seminal book, graphical models have been widely used for their encoding and their computation. This explains why my second research field concerns graphical models. It turns out that graphical models also prove useful for preference encodings.. # Representation of the Decision Maker's Preferences. Mathematically, modeling preferences over a set of decisions or over a set of consequences merely amounts to finding a binary relation over pairs decisions or pairs of consequences. For instance, assume that a Decision Maker (DM) has some preferences over a set $$X$$ = {eat some chicken, eat some fish, eat an apple pie, eat a soup}. For any pair $$(x,y)$$ of elements of $$X$$, the following may happen:. 1. the Decision Maker may not be able to compare $$x$$ and $$y$$, i.e., she cannot say whether she prefers $$x$$ to $$y$$ or $$y$$ to $$x$$. This may be the case for $$x$$="eat a soup" and $$y$$="eat an apple pie" since one is a starter and the other is a dessert.. 2. the Decision Maker may prefer $$x$$ to $$y$$ or $$y$$ to $$x$$, or even be indifferent between $$x$$ and $$y$$.. Mathematically, the Decision Maker's preferences may be represented by a binary relation $$\succsim$$ on $$X \times X$$ such that $$x \succsim y$$ means the DM prefers $$x$$ to $$y$$ or is indifferent between $$x$$ and $$y$$. Thus $$x$$ and $$y$$ being incomparable is equivalent to Not ($$x \succsim y$$) and Not ($$y \succsim x$$). The strict preference of $$x$$ over $$y$$ can be represented by $$x \succsim y$$ and Not ($$y \succsim x$$), and the indifference between $$x$$ and $$y$$ is captured by $$x \succsim y$$ and $$y \succsim x$$.. The qualitative aspect of binary relation $$\succsim$$ makes it not particularly well suited for fast computation, in particular if one wishes to find the best decision under a given set of constraints. Hence, in practical situations, this relation needs often be encoded numerically. One of the most popular encodings is that of utility functions, a.k.a. utilities.
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The idea is to use a mapping $$u : X \mapsto \mathbb{R}$$, the set of the real numbers, assigning numbers to elements of $$X$$ such that the higher the preferred. More formally, $$u$$ is defined as: $x \succsim y \Longleftrightarrow u(x) \geq u(y), \forall\ x,y \in X.$. In practical situations, the DM's decisions take into account multiple conflicting objectives, hence the outcome set is a multidimensional space and the outcomes are tuples (of attributes) $$x = (x_1,\ldots,x_n)$$. Thus utilities are functions of tuples. This is a major problem when they are to be constructed. Indeed, as each Decision Maker has her own preferences, utilities differ from one DM to another. So the only way to construct the DM's utlities is to ask her some questions. But when questions involve to many different attributes, they become too complicated for the human brain to handle and the DM is not able to give accurate answers. In order to simplify the questions, assumptions on the structure of the utilities need be made. One of the most popular is the additive decomposition: $u(x_1,\ldots,x_n) = u_1(x_1) + \cdots + u_n(x_n).$. During my PhD thesis, I have studied testable conditions on the DM preference relation ensuring the existence of additively decomposable utility functions (additive utilities for short). I am still interested in decompositions of utilities, but I now study more complicated decompositions such as generalized additive decompositions (GAI): instead of splitting the overall utility into pieces, each one depending only on one attribute, GAI splits it into utilities over some (possibly overlapping) sets of attributes. For instance~: $u(a,b,c,d,e,f) = u_1(a,b,c) + u_2(c,d,e) + u_3(d,e,f).$. # Graphical Models. Among graphical models, I am especially interested in Bayesian networks (BN) and GAI-nets. The former are powerful tools of the Artificial Intelligence community enabling fast computation of probabilities (marginal, conditional, a priori or a posteriori probabilities). To achieve this result, BN use graphical structures that encode knowledge about the decomposition of joint probability distributions of interest. GAI-nets are rather similar in spirit, except that instead of decomposing a probability distribution over random variables, they decompose attributes or criteria of preference tuples.. Mainly my current activities on graphical models can be divided into three classes:. • Until now, I was mainly concerned with the improvement of BN computation algorithms. In the literature, roughly two main approaches emerged: directed and undirected methods. It is commonly thought that undirected methods outperform directed ones. One of my goals was to unify both types of methods to show that directed methods can compete with undirected ones. Using the advantages of each kind of method, this unification enabled the improvement of all the algorithms.. • As I mentioned, Bayesian networks are graphical structures. These represent the decomposition of joint probabilities. Hence they are specific for each problem of interest. In theory, if you possess a sufficiently good database of values of the random variables you are interested in, computer programs should be able to learn the graph structure of the BN (using for instance statistical independence tests). In practice, however, such programs do not always produce high quality graphs. This explains why I am working on learning methods.. • My last main research activity concerns GAI-nets. The idea is to take advantage of some decomposition of the Decision Maker's preference relations to be able to elicit efficiently his/her preferences, and to answer quickly to questions such as "does the Decision Maker prefers this alternative to that one?" or "what is the preferred alternative of the Decision Maker?". Efficient elicitation procedures could prove useful for instance for web shopping sites.
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# How do you calculate the derivative for h(t) = t^3 + 2e^t?
Mar 2, 2018
h´(t)=3t^2+2e^t
#### Explanation:
Because:
A constant is "neutral" for derivative (linearity)
the derivative of ${x}^{n}$ is n·x^(n-1)
And The derivative of ${e}^{x}$ is itself, ${e}^{x}$
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# How do you calculate the derivative for h(t) = t^3 + 2e^t?. Mar 2, 2018. h´(t)=3t^2+2e^t. #### Explanation:. Because:.
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A constant is "neutral" for derivative (linearity). the derivative of ${x}^{n}$ is n·x^(n-1). And The derivative of ${e}^{x}$ is itself, ${e}^{x}$.
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https://zdroweporadniki.pl/how-to-find/demystifying-the-mechanics-how-to-find-domain/
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# Demystifying the Mechanics: How to Find Domain
Understanding how to find the domain of a function is essential in solving various math problems. Whether you’re a student, professional, or simply someone seeking to enhance your mathematical knowledge, this fundamental concept is vital. By mastering the domain, you gain valuable insights into the behavior and restrictions of a function, enabling you to unlock a world of mathematical possibilities. In this article, we will unravel the mystery behind determining the domain of a function, providing you with the tools necessary to confidently navigate the realm of mathematics.
## Understanding the Concept of the Domain
In mathematics, the concept of the domain refers to the set of all possible input values for a function. It determines the range of values that the function can take on and is crucial for understanding the behavior of functions. The domain is typically represented by the variable x in function notation, such as f(x).
To demystify the mechanics of finding the domain, it is important to first grasp the idea that the domain restricts the values that can be plugged into a function. Some functions may have restrictions based on the nature of their mathematical operations or the properties of the functions themselves.
Identifying the domain of a function is essential for many reasons, such as determining the existence of inverses, specifying applicable ranges of comparison, or ensuring valid solutions in certain applications like optimization or computer simulations.
Now let’s delve into the various types of functions and how to identify their respective domains.
## Identifying the Type of Function
Functions can take on different forms and exhibit diverse behaviors. Before finding the domain of a function, it is crucial to identify the type of function being encountered. This can help in determining the appropriate methods and considerations for finding the domain.
Common types of functions include basic functions, functions with fractions or square roots, polynomial and rational functions, inequalities and absolute values, exponential and logarithmic functions, trigonometric functions, and piecewise functions. Each type has its own unique characteristics, and understanding these characteristics is crucial for finding the domain effectively.
## Finding the Domain of Basic Functions
Basic functions, such as linear functions, quadratic functions, and constant functions, have relatively simple domains. For linear functions of the form f(x) = mx + b, the domain is the set of all real numbers since there are no restrictions on x. Similarly, constant functions like f(x) = c also have a domain of all real numbers.
Quadratic functions, on the other hand, may have some restrictions depending on the specific form of the equation. For example, the domain of a quadratic function given by f(x) = ax^2 + bx + c is typically all real numbers unless it contains square roots or fractions, which we will discuss in the next section.
To summarize, when dealing with basic functions, the domain is often all real numbers unless indicated otherwise.
## Handling Functions with Fractions or Square Roots
Functions that involve fractions or square roots require additional considerations when determining their domains. These functions may have limitations due to the presence of denominators or the square root operation, which may lead to undefined or complex values.
For functions with fractions, we need to exclude any values of x that would result in a zero denominator. To find the domain, we set the denominator equal to zero and solve for x. The values obtained are the ones we need to exclude from the domain.
Similarly, for functions with square roots, we must ensure that the expression inside the square root is non-negative, as square roots of negative numbers are imaginary. We set the expression under the square root greater than or equal to zero and solve for x.
It is important to remember that the domain of these functions may also depend on any additional restrictions specified in the problem or context.
Key Points to Remember:
• For functions with fractions, exclude any values of x that make the denominator zero.
• For functions with square roots, ensure that the expression inside the square root is non-negative.
• ## Dealing with Polynomial and Rational Functions
Polynomial and rational functions can have more complex domains, depending on their equation forms. A polynomial function is a function made up of terms with non-negative integer exponents. For example, f(x) = 2x^3 – 5x^2 + 3x – 2 is a polynomial function.
The domain of a polynomial function is typically all real numbers, unless there are square roots, denominators, or even radical functions involved. In such cases, we need to follow the steps mentioned earlier for handling functions with fractions or square roots.
Rational functions, on the other hand, are functions defined as a ratio of two polynomial functions. The domain of a rational function is determined by both the denominator and any square roots or fractions present. Therefore, we need to follow the procedures described in the previous sections to find the domain of rational functions.
Continued in the next message…
Inspired by this? Share the article with your friends!
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# Demystifying the Mechanics: How to Find Domain. Understanding how to find the domain of a function is essential in solving various math problems. Whether you’re a student, professional, or simply someone seeking to enhance your mathematical knowledge, this fundamental concept is vital. By mastering the domain, you gain valuable insights into the behavior and restrictions of a function, enabling you to unlock a world of mathematical possibilities. In this article, we will unravel the mystery behind determining the domain of a function, providing you with the tools necessary to confidently navigate the realm of mathematics.. ## Understanding the Concept of the Domain. In mathematics, the concept of the domain refers to the set of all possible input values for a function. It determines the range of values that the function can take on and is crucial for understanding the behavior of functions. The domain is typically represented by the variable x in function notation, such as f(x).. To demystify the mechanics of finding the domain, it is important to first grasp the idea that the domain restricts the values that can be plugged into a function. Some functions may have restrictions based on the nature of their mathematical operations or the properties of the functions themselves.. Identifying the domain of a function is essential for many reasons, such as determining the existence of inverses, specifying applicable ranges of comparison, or ensuring valid solutions in certain applications like optimization or computer simulations.. Now let’s delve into the various types of functions and how to identify their respective domains.. ## Identifying the Type of Function. Functions can take on different forms and exhibit diverse behaviors. Before finding the domain of a function, it is crucial to identify the type of function being encountered. This can help in determining the appropriate methods and considerations for finding the domain.. Common types of functions include basic functions, functions with fractions or square roots, polynomial and rational functions, inequalities and absolute values, exponential and logarithmic functions, trigonometric functions, and piecewise functions. Each type has its own unique characteristics, and understanding these characteristics is crucial for finding the domain effectively.. ## Finding the Domain of Basic Functions. Basic functions, such as linear functions, quadratic functions, and constant functions, have relatively simple domains. For linear functions of the form f(x) = mx + b, the domain is the set of all real numbers since there are no restrictions on x. Similarly, constant functions like f(x) = c also have a domain of all real numbers.. Quadratic functions, on the other hand, may have some restrictions depending on the specific form of the equation. For example, the domain of a quadratic function given by f(x) = ax^2 + bx + c is typically all real numbers unless it contains square roots or fractions, which we will discuss in the next section.
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To summarize, when dealing with basic functions, the domain is often all real numbers unless indicated otherwise.. ## Handling Functions with Fractions or Square Roots. Functions that involve fractions or square roots require additional considerations when determining their domains. These functions may have limitations due to the presence of denominators or the square root operation, which may lead to undefined or complex values.. For functions with fractions, we need to exclude any values of x that would result in a zero denominator. To find the domain, we set the denominator equal to zero and solve for x. The values obtained are the ones we need to exclude from the domain.. Similarly, for functions with square roots, we must ensure that the expression inside the square root is non-negative, as square roots of negative numbers are imaginary. We set the expression under the square root greater than or equal to zero and solve for x.. It is important to remember that the domain of these functions may also depend on any additional restrictions specified in the problem or context.. Key Points to Remember:. • For functions with fractions, exclude any values of x that make the denominator zero.. • For functions with square roots, ensure that the expression inside the square root is non-negative.. • ## Dealing with Polynomial and Rational Functions. Polynomial and rational functions can have more complex domains, depending on their equation forms. A polynomial function is a function made up of terms with non-negative integer exponents. For example, f(x) = 2x^3 – 5x^2 + 3x – 2 is a polynomial function.. The domain of a polynomial function is typically all real numbers, unless there are square roots, denominators, or even radical functions involved. In such cases, we need to follow the steps mentioned earlier for handling functions with fractions or square roots.. Rational functions, on the other hand, are functions defined as a ratio of two polynomial functions. The domain of a rational function is determined by both the denominator and any square roots or fractions present. Therefore, we need to follow the procedures described in the previous sections to find the domain of rational functions.. Continued in the next message…. Inspired by this? Share the article with your friends!.
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http://best-excel-tutorial.com/57-vba-tutorial/59-tips-and-tricks/259-integral-function
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# How to use the integral function?
The integration is part of the important concepts that associate with mathematic, and is part of the main operations in calculus. It has to be given the function f of a variable x that is real, as well as the interval which is a, b, of that real line, and this is the definite integral:
This has been informally defined as signed area of region that is in x/y-plane, and its boundary by graph of f, x-axis, as well as vertical lines x that is a, and the x that is equal to b. The term known as integral could also refer to related notion of antiderivative, which is a function F, and whose derivative is the function of that is given. In such case, an indefinite integral is defined, and also written like this:
You are going to work with integral of a simple calculation following this simple steps below.
## Data Layout
You should decide the dx value, which is 0,1. This is showing in the picture above.
Click on the column that is beside the x^2, and labeled as number 1. Click on insert tab, the one labeled as number 2.
Click on the equation, which is marked in red. Do not click on the arrow, just on the equation itself.
Click on the integral (labeled number 1) showing once you have clicked on the equation in previous step, and then click on the kind of integral you would like, in this case, the one labeled number 2 is chosen.
Place the value that are relevant to the integral function. As you could see in the one marked in red, and labeled as number 1 is going to have a detail of your choice. On top of the f there is a, while on the end, there is a 0, and in the middle there is an x. You should click on the line (where number 2 is marked), and expand the whole row to fit the integral equation that has been inserted.
Tip: you should type in the x2 (marked in the number 1), using an insert symbol (which is in the insert tab).
## Integral Calculation
Add 0 (in this case in a7) with the 0,1 – the dx.
Information: If you see any error calculating, you should change from 0,1 to 0.1 to continue.
Click on the small square beside the number 1, using your mouse. Drag it down to the end, as you wish. In this case, it is dragged to a20.
Click on the column in the row of the ^2, and add the content in “a” row to multiply with ^2. Repeat the step 2 – 2 with the A row, as it is in this case. Check the picture for clarification.
Type in the columns that are relevant to the integral you are calculating. You would see the marked columns, as well as some added number.
Click on the integral you have just calculated, and now click on the small square that is in the marked area and on the left side of the one labeled number 1, and drag it down to save yourself the stress of calculating one after the other.
## Design (Optional)
Mark the rows you would like to display in a specific design.
Click on the INSERT tab, the one labeled as number 1, and then choose the type of chart you would like to display, which is labeled as number 2.
Choose the design you would prefer for your integral chart.
Insert chart of your integral function and write the title you would like to give the chart.
The chart is ready.
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# How to use the integral function?. The integration is part of the important concepts that associate with mathematic, and is part of the main operations in calculus. It has to be given the function f of a variable x that is real, as well as the interval which is a, b, of that real line, and this is the definite integral:. This has been informally defined as signed area of region that is in x/y-plane, and its boundary by graph of f, x-axis, as well as vertical lines x that is a, and the x that is equal to b. The term known as integral could also refer to related notion of antiderivative, which is a function F, and whose derivative is the function of that is given. In such case, an indefinite integral is defined, and also written like this:. You are going to work with integral of a simple calculation following this simple steps below.. ## Data Layout. You should decide the dx value, which is 0,1. This is showing in the picture above.. Click on the column that is beside the x^2, and labeled as number 1. Click on insert tab, the one labeled as number 2.. Click on the equation, which is marked in red. Do not click on the arrow, just on the equation itself.. Click on the integral (labeled number 1) showing once you have clicked on the equation in previous step, and then click on the kind of integral you would like, in this case, the one labeled number 2 is chosen.. Place the value that are relevant to the integral function. As you could see in the one marked in red, and labeled as number 1 is going to have a detail of your choice. On top of the f there is a, while on the end, there is a 0, and in the middle there is an x. You should click on the line (where number 2 is marked), and expand the whole row to fit the integral equation that has been inserted.. Tip: you should type in the x2 (marked in the number 1), using an insert symbol (which is in the insert tab).
|
## Integral Calculation. Add 0 (in this case in a7) with the 0,1 – the dx.. Information: If you see any error calculating, you should change from 0,1 to 0.1 to continue.. Click on the small square beside the number 1, using your mouse. Drag it down to the end, as you wish. In this case, it is dragged to a20.. Click on the column in the row of the ^2, and add the content in “a” row to multiply with ^2. Repeat the step 2 – 2 with the A row, as it is in this case. Check the picture for clarification.. Type in the columns that are relevant to the integral you are calculating. You would see the marked columns, as well as some added number.. Click on the integral you have just calculated, and now click on the small square that is in the marked area and on the left side of the one labeled number 1, and drag it down to save yourself the stress of calculating one after the other.. ## Design (Optional). Mark the rows you would like to display in a specific design.. Click on the INSERT tab, the one labeled as number 1, and then choose the type of chart you would like to display, which is labeled as number 2.. Choose the design you would prefer for your integral chart.. Insert chart of your integral function and write the title you would like to give the chart.. The chart is ready.
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https://curious.com/williamspaniel/a-trick-with-weakly-dominated-strategies/in/game-theory-101
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STEM
# Game Theory 101
151 CQ
36 Lessons
Previous...
• Playing
Free
4 CQ
31. A Trick with Weakly Dominated Strategies
A lesson with William Spaniel
Want to dominate game theory? Work through the "Take or Share" game and learn a trick that will have you second guessing your mom's advice on sharing.
Want to dominate game theory? Work through the "Take or Share" game and learn a trick that will have you second guessing your mom's advice on sharing.
• Playing
Free
4 CQ
32. Rock, Paper, Scissors Game
A lesson with William Spaniel
How can you win Rock, Paper, Scissors? Not with a pure strategy, that’s for sure! Learn the theory behind the game, and see why you just can't beat randomness.
How can you win Rock, Paper, Scissors? Not with a pure strategy, that’s for sure! Learn the theory behind the game, and see why you just can't beat randomness.
• Playing
Free
5 CQ
33. Symmetric Zero Sum Games
A lesson with William Spaniel
Rock, Paper... Symmetry? Learn about symmetric zero sum games: game theory scenarios in which each player's expected utility at equilibrium must equal zero.
Rock, Paper... Symmetry? Learn about symmetric zero sum games: game theory scenarios in which each player's expected utility at equilibrium must equal zero.
• Playing
Free
5 CQ
34. Modified Rock, Paper, Scissors Game
A lesson with William Spaniel
Solve a modified rock, paper, scissors game theory scenario, and along the way, learn about the zero sum theorem and mixed strategy Nash equilibria.
Solve a modified rock, paper, scissors game theory scenario, and along the way, learn about the zero sum theorem and mixed strategy Nash equilibria.
• Playing
Free
5 CQ
35. Mixing Among Three Strategies
A lesson with William Spaniel
Game theory needn't mix you up! Use this algorithm to solve for the mixed strategy Nash equilibrium in games with three possible strategies.
Game theory needn't mix you up! Use this algorithm to solve for the mixed strategy Nash equilibrium in games with three possible strategies.
More...
• Recommended Recommended
• History & In Progress History
• Browse Library
• Most Popular Library
### Get Personalized Recommendations
Let us help you figure out what to learn! By taking a short interview you’ll be able to specify your learning interests and goals, so we can recommend the perfect courses and lessons to try next.
You don't have any lessons in your history.
Just find something that looks interesting and start learning!
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STEM. # Game Theory 101. 151 CQ. 36 Lessons. Previous.... • Playing. Free. 4 CQ. 31. A Trick with Weakly Dominated Strategies. A lesson with William Spaniel. Want to dominate game theory? Work through the "Take or Share" game and learn a trick that will have you second guessing your mom's advice on sharing.. Want to dominate game theory? Work through the "Take or Share" game and learn a trick that will have you second guessing your mom's advice on sharing.. • Playing. Free. 4 CQ. 32. Rock, Paper, Scissors Game. A lesson with William Spaniel. How can you win Rock, Paper, Scissors? Not with a pure strategy, that’s for sure! Learn the theory behind the game, and see why you just can't beat randomness.. How can you win Rock, Paper, Scissors? Not with a pure strategy, that’s for sure! Learn the theory behind the game, and see why you just can't beat randomness.. • Playing. Free. 5 CQ. 33. Symmetric Zero Sum Games. A lesson with William Spaniel. Rock, Paper... Symmetry? Learn about symmetric zero sum games: game theory scenarios in which each player's expected utility at equilibrium must equal zero.
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Rock, Paper... Symmetry? Learn about symmetric zero sum games: game theory scenarios in which each player's expected utility at equilibrium must equal zero.. • Playing. Free. 5 CQ. 34. Modified Rock, Paper, Scissors Game. A lesson with William Spaniel. Solve a modified rock, paper, scissors game theory scenario, and along the way, learn about the zero sum theorem and mixed strategy Nash equilibria.. Solve a modified rock, paper, scissors game theory scenario, and along the way, learn about the zero sum theorem and mixed strategy Nash equilibria.. • Playing. Free. 5 CQ. 35. Mixing Among Three Strategies. A lesson with William Spaniel. Game theory needn't mix you up! Use this algorithm to solve for the mixed strategy Nash equilibrium in games with three possible strategies.. Game theory needn't mix you up! Use this algorithm to solve for the mixed strategy Nash equilibrium in games with three possible strategies.. More.... • Recommended Recommended. • History & In Progress History. • Browse Library. • Most Popular Library. ### Get Personalized Recommendations. Let us help you figure out what to learn! By taking a short interview you’ll be able to specify your learning interests and goals, so we can recommend the perfect courses and lessons to try next.. You don't have any lessons in your history.. Just find something that looks interesting and start learning!.
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+0
# plz help
0
31
1
The three angles of a triangle have measures angle A = x + 2y degrees, angle B = 6x - 5y degrees, and angle C = 8x + 3y degrees. Find x (in degrees).
Oct 10, 2020
#1
+27725
+1
In a triangle the 3 angles have to sum to 180 so add them up and equate them to 180:
x+2y + 6x-5y + 8x+3y = 180 when yo add them the y's will be gone....solve for x
Oct 10, 2020
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+0. # plz help. 0. 31. 1. The three angles of a triangle have measures angle A = x + 2y degrees, angle B = 6x - 5y degrees, and angle C = 8x + 3y degrees. Find x (in degrees).. Oct 10, 2020.
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#1. +27725. +1. In a triangle the 3 angles have to sum to 180 so add them up and equate them to 180:. x+2y + 6x-5y + 8x+3y = 180 when yo add them the y's will be gone....solve for x. Oct 10, 2020.
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1. Determining inconsistent system
Is there anyway to determine at face-value whether or not this linear system of equations is consistent or not? Consistent being that it has at least one solution. For example...
I would really appreciate help!!!
- Olivia
2. Re: Determining inconsistent system
If a square matrix of coefficients is non-singular there will be a unique solution.
A non-singular matrix will have a non-zero determinant.
This matrix is fairly sparse so computing the determinant via minors shouldn't be that difficult.
Of course you can just dump it into Wolfram as well. It should be
Det[{{1,0,3,0},{0,1,0,-3},{0,-2,3,2},{3,0,0,7}}]
on the command line
3. Re: Determining inconsistent system
I don't know what you mean by "determine at face value". If you mean "just look at it an immediately know", I can't help you. I am just not that good. I would have to actually do the calculations!
The system of equation is
$\displaystyle x_1+ 0x_2+ 3x_3+ 0x_4= 2$
$\displaystyle 0x_1+ x_2+ 0x_3- 3x_4= 2$
$\displaystyle 0x_1- 2x_2+ 3x_3+ 2x_4= 1$
$\displaystyle 3x_1+ 0x_2+ 0x_3+ 3x_4= -5$.
As romek said, that is fairly sparce- there are a whole lot of 0s in there.
We can get even more 0s by:
1) Subtract 3 times the first equation from the fourth and add twice the second equation to the third:
$\displaystyle x_1+ 0x_2+ 3x_3+ 0x_4= 2$
$\displaystyle 0x_1+ x_2+ 0x_3- 3x_4= 3$
$\displaystyle 0x_1+ 0x_2+ x_3- 4x_4= 7$
$\displaystyle 0x_1+ 0x_2- 9x_3+ 7x_4= -11$.
That is almost triangular.
2) Just add three times the third of those equations to the fourth to get
$\displaystyle x_1+ 0x_2+ 3x_3+ 0x_4= 2$
$\displaystyle 0x_1+ x_2+ 0x_3- 3x_4= 3$
$\displaystyle 0x_1+ 0x_2+ x_3- 4x_4= 7$
$\displaystyle 0x_1+ 0x_2+ 0x_3- 5x_4= 10$.
Which is triangular and it is easy to see that is a consistent set of equations. In fact, it is easy to solve the system by "back-substituting".
4. Re: Determining inconsistent system
ok ok thank you
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1. Determining inconsistent system. Is there anyway to determine at face-value whether or not this linear system of equations is consistent or not? Consistent being that it has at least one solution. For example.... I would really appreciate help!!!. - Olivia. 2. Re: Determining inconsistent system. If a square matrix of coefficients is non-singular there will be a unique solution.. A non-singular matrix will have a non-zero determinant.. This matrix is fairly sparse so computing the determinant via minors shouldn't be that difficult.. Of course you can just dump it into Wolfram as well. It should be. Det[{{1,0,3,0},{0,1,0,-3},{0,-2,3,2},{3,0,0,7}}]. on the command line. 3. Re: Determining inconsistent system. I don't know what you mean by "determine at face value". If you mean "just look at it an immediately know", I can't help you. I am just not that good. I would have to actually do the calculations!. The system of equation is. $\displaystyle x_1+ 0x_2+ 3x_3+ 0x_4= 2$.
|
$\displaystyle 0x_1+ x_2+ 0x_3- 3x_4= 2$. $\displaystyle 0x_1- 2x_2+ 3x_3+ 2x_4= 1$. $\displaystyle 3x_1+ 0x_2+ 0x_3+ 3x_4= -5$.. As romek said, that is fairly sparce- there are a whole lot of 0s in there.. We can get even more 0s by:. 1) Subtract 3 times the first equation from the fourth and add twice the second equation to the third:. $\displaystyle x_1+ 0x_2+ 3x_3+ 0x_4= 2$. $\displaystyle 0x_1+ x_2+ 0x_3- 3x_4= 3$. $\displaystyle 0x_1+ 0x_2+ x_3- 4x_4= 7$. $\displaystyle 0x_1+ 0x_2- 9x_3+ 7x_4= -11$.. That is almost triangular.. 2) Just add three times the third of those equations to the fourth to get. $\displaystyle x_1+ 0x_2+ 3x_3+ 0x_4= 2$. $\displaystyle 0x_1+ x_2+ 0x_3- 3x_4= 3$. $\displaystyle 0x_1+ 0x_2+ x_3- 4x_4= 7$. $\displaystyle 0x_1+ 0x_2+ 0x_3- 5x_4= 10$.. Which is triangular and it is easy to see that is a consistent set of equations. In fact, it is easy to solve the system by "back-substituting".. 4. Re: Determining inconsistent system. ok ok thank you.
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http://maths.gopract.com/Calc/Trignometry/Cylinder
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# Calculator For Cylinder
Calculation Option
Provide r (radius), h (height) Provide r (radius), V (volume) Provide r (radius), L (lateral surface area) Provide h (height), V (volume) Provide h (height), L (lateral surface area) Radius r : Height h :
r =
height
h =
Volume
V =
Top surface area
T =
Lateral surface area
L =
Base surface area
B =
Total surface area
A =
##### Cylinder
h=height
V=Volume
L=Lateral surface area
B=Base surface area
T=Top surface area
A=Total surface area
### What Is Cylinder?
A cylinder is a three-dimensional shape consisting of two parallel circular bases which joined by a curved surface. The line segment joining the two centers is the axis which denotes the height of the cylinder.
### What Are Calculation Option Available In Cylinder?
Select Which value you are providing (radius (r) and height (h) , radius (r) and Volume (V), radius (r) and Lateral surface area (L) , height (h) and Volume (V) , height (h) and Lateral surface area (L)) input value and click on Calculate.
A cylinder is a three-dimensional shape consisting of two parallel circular bases, joined by a curved surface. The line segment joining the two centers is the axis which denotes the height of the cylinder. To Calculate all the properties of Cylinder this calculator is helpfull.Using the following formula :-
Volume of a cylinder :
V = pr2h
Where
V = Volume
h = height
Lateral surface area of a cylinder:
L = 2prh
Where
L = Lateral surface area
h = height
Base surface area of a cylinder:
B = pr2
Where
B = Base surface area
Top surface area of a cylinder:
T = pr2
Where
T = Top surface area
Total surface area of a cylinder:
A = L + B + T
Where
A = Total surface area
L = Lateral surface area
B = Base surface area
T = Top surface area
### How to use Cylinder Calculator?
Step by step procedure for Cylinder Calculator is as follows.
Step 1: Select which value you will be providing? Dropdown list has ( radius (r) and height (h) , radius (r) and Volume (V), radius (r) and Lateral surface area (L) , height (h) and Volume (V) , height (h) and Lateral surface area (L) ) options.
Step 2: Input appropriate value as per selected type in Step 1.
Step 3: Now click on "Calculate" button to get result.
You will get radius (r) , height (h) , Volume (V) , Lateral surface area (L) , Base surface area (T), Base surface area (B), Total surface area (A) as per your value input.
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# Calculator For Cylinder. Calculation Option. Provide r (radius), h (height) Provide r (radius), V (volume) Provide r (radius), L (lateral surface area) Provide h (height), V (volume) Provide h (height), L (lateral surface area) Radius r : Height h :. r =. height. h =. Volume. V =. Top surface area. T =. Lateral surface area. L =. Base surface area. B =. Total surface area. A =. ##### Cylinder. h=height. V=Volume. L=Lateral surface area. B=Base surface area. T=Top surface area. A=Total surface area. ### What Is Cylinder?. A cylinder is a three-dimensional shape consisting of two parallel circular bases which joined by a curved surface. The line segment joining the two centers is the axis which denotes the height of the cylinder.. ### What Are Calculation Option Available In Cylinder?. Select Which value you are providing (radius (r) and height (h) , radius (r) and Volume (V), radius (r) and Lateral surface area (L) , height (h) and Volume (V) , height (h) and Lateral surface area (L)) input value and click on Calculate.. A cylinder is a three-dimensional shape consisting of two parallel circular bases, joined by a curved surface. The line segment joining the two centers is the axis which denotes the height of the cylinder. To Calculate all the properties of Cylinder this calculator is helpfull.Using the following formula :-. Volume of a cylinder :.
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V = pr2h. Where. V = Volume. h = height. Lateral surface area of a cylinder:. L = 2prh. Where. L = Lateral surface area. h = height. Base surface area of a cylinder:. B = pr2. Where. B = Base surface area. Top surface area of a cylinder:. T = pr2. Where. T = Top surface area. Total surface area of a cylinder:. A = L + B + T. Where. A = Total surface area. L = Lateral surface area. B = Base surface area. T = Top surface area. ### How to use Cylinder Calculator?. Step by step procedure for Cylinder Calculator is as follows.. Step 1: Select which value you will be providing? Dropdown list has ( radius (r) and height (h) , radius (r) and Volume (V), radius (r) and Lateral surface area (L) , height (h) and Volume (V) , height (h) and Lateral surface area (L) ) options.. Step 2: Input appropriate value as per selected type in Step 1.. Step 3: Now click on "Calculate" button to get result.. You will get radius (r) , height (h) , Volume (V) , Lateral surface area (L) , Base surface area (T), Base surface area (B), Total surface area (A) as per your value input.
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https://byjus.com/question-answer/a-train-first-travels-for-30-min-with-a-velocity-30-kmh-1-and-then-1/
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Question
# A train first travels for 30 min with a velocity 30 kmh−1 and then for 40 min with a velocity 40 kmh−1 in the same direction, calculate the average velocity of the train.
A
45.82 kmh1
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B
90.47 kmh1
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C
35.71 kmh1
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D
70.11 kmh1
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Open in App
Solution
## The correct option is A 35.71 kmh−1Total time of journey t=t1+t2 ⇒t=30+4060=76h∴ Average velocity =Total displacement travelled (S)Total time of journey (t)Total displacement in t1 is s1:s1=3060×30⇒s1=15 kmTotal displacement in t2 is s2:s2=4060×40⇒s2=26.66 kmTotal displacement s in time t:s=41.66 kmVavg=s1+s2tVavg=41.6676Vavg=35.71 kmh−1
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1. You visited us 1 times! Enjoying our articles? Unlock Full Access!. Question. # A train first travels for 30 min with a velocity 30 kmh−1 and then for 40 min with a velocity 40 kmh−1 in the same direction, calculate the average velocity of the train.. A. 45.82 kmh1. No worries! We‘ve got your back. Try BYJU‘S free classes today!. B. 90.47 kmh1. No worries! We‘ve got your back. Try BYJU‘S free classes today!. C. 35.71 kmh1. Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses. D.
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70.11 kmh1. No worries! We‘ve got your back. Try BYJU‘S free classes today!. Open in App. Solution. ## The correct option is A 35.71 kmh−1Total time of journey t=t1+t2 ⇒t=30+4060=76h∴ Average velocity =Total displacement travelled (S)Total time of journey (t)Total displacement in t1 is s1:s1=3060×30⇒s1=15 kmTotal displacement in t2 is s2:s2=4060×40⇒s2=26.66 kmTotal displacement s in time t:s=41.66 kmVavg=s1+s2tVavg=41.6676Vavg=35.71 kmh−1. Suggest Corrections. 0. Join BYJU'S Learning Program. Related Videos. Velocity. PHYSICS. Watch in App. Explore more. Join BYJU'S Learning Program.
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http://community.boredofstudies.org/14/mathematics-extension-2/386318/4u-polynomial-question-help.html
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Thread: 4U Polynomial question HELP !!
1. 4U Polynomial question HELP !!
If the polynomial x^3+3ax^2+3bx+c=0 has a double root, show that the double root is (c-ab)/(2(a^2-b)) , given that a^2 doesnt equal to b
2. Re: 4U Polynomial question HELP !!
I am gonna do it a bit later, a bit busy, but have you tried 3rd derivative getting it to be 0 then testing against second derivative and subbing in the original function?
3. Re: 4U Polynomial question HELP !!
But since the root has a multiplicity of 2 can it be found in the third derivative?
4. Re: 4U Polynomial question HELP !!
I read third for some reason im blind
5. Re: 4U Polynomial question HELP !!
yeah okay can you do it now?
6. Re: 4U Polynomial question HELP !!
can't do the question right now but i assume that this is how you do it:
since x^3 + 3ax^2 + 3bx + c = 0 has a double root then this means that p(x) = x^3 + 3ax^2 + 3bx + c has a double zero. let this double zero be y. then p(y) = 0 and p'(y) = 0. this means that y^3 + 3ay^2 + 3by + c = 0 and 3y^2 + 6ay + 3b = 0 -> y^2 + 2ay + b = 0. now we solve this last equation for the double root, y. u will get 2 solutions (since quadratic), then sub each presumed double root into the original equation (p(y)). whichever results in p(y) = 0 is the double root. good luck!
7. Re: 4U Polynomial question HELP !!
Notice that the required answer includes the constant term $c$. That should be a hint - if we were to differentiate the polynomial, we would lose $c$, so not only would we have to expand the cube of a binomial (which can be really messy), we would also have to rewrite the result to include $c$ in order to find the double root, so let's try a different approach.
Let $P(x) = x^3 + 3ax^2+3bx+c$ and let the roots be $\alpha, \, \alpha , \,\beta$.
By the relationships between the roots and co-efficients,
$2\alpha + \beta = -3a$
$\alpha^2 + 2\alpha\beta = 3b$
$\alpha^2 \beta = -c$
Rearranging,
$a = - \frac{2\alpha + \beta}{3}$
$b = \frac{\alpha^2 + 2\alpha\beta}{3}$
$c=-\alpha^2 \beta$
Now:
$c - ab = \frac{(2\alpha+\beta)(\alpha^2 + 2\alpha\beta)-9\alpha^2\beta}{9} = \frac{2\alpha(\alpha - \beta)^2}{9}$
$2(a^2-b) = 2 \cdot \frac{(2\alpha + \beta)^2-3(\alpha^2 + 2\alpha\beta)}{9} = \frac{2(\alpha - \beta)^2}{9}$
Therefore,
$\frac{c-ab}{2(a^2-b)} = \alpha, \quad (a^2 - b \neq 0)$
8. Re: 4U Polynomial question HELP !!
While the method above is indeed valid, one could also solve the problem using calculus:
Let P(x)=x3+3ax2+3bx+c.
Then P'(x)=3x2+6ax+3b.
For there to be a double root, there must exist some x such that P(x)=0 and P'(x)=0.
P'(x)=0 iff 3x2+6ax+3b=0, i.e. x2+2ax+b=0.
Now, P(x)=0 for the same value of x,
so x3+3ax2+3bx+c=0,
i.e. x(x2+2ax+b)+ax2+2bx+c=0.
x(x2+2ax+b)+a(x2+2ax+b)-2a2x-ab+2bx+c=0.
But x2+2ax+b=0 (proven above),
so -2a2x-ab+2bx+c=0,
i.e. x(2b-2a2)+c-ab=0.
x=(c-ab)/(2(a2-b)) as a2-b≠0.
Hence, the double root must be equal to (c-ab)/(2(a2-b)) as required.
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Thread: 4U Polynomial question HELP !!. 1. 4U Polynomial question HELP !!. If the polynomial x^3+3ax^2+3bx+c=0 has a double root, show that the double root is (c-ab)/(2(a^2-b)) , given that a^2 doesnt equal to b. 2. Re: 4U Polynomial question HELP !!. I am gonna do it a bit later, a bit busy, but have you tried 3rd derivative getting it to be 0 then testing against second derivative and subbing in the original function?. 3. Re: 4U Polynomial question HELP !!. But since the root has a multiplicity of 2 can it be found in the third derivative?. 4. Re: 4U Polynomial question HELP !!. I read third for some reason im blind. 5. Re: 4U Polynomial question HELP !!. yeah okay can you do it now?. 6. Re: 4U Polynomial question HELP !!. can't do the question right now but i assume that this is how you do it:. since x^3 + 3ax^2 + 3bx + c = 0 has a double root then this means that p(x) = x^3 + 3ax^2 + 3bx + c has a double zero. let this double zero be y. then p(y) = 0 and p'(y) = 0. this means that y^3 + 3ay^2 + 3by + c = 0 and 3y^2 + 6ay + 3b = 0 -> y^2 + 2ay + b = 0. now we solve this last equation for the double root, y. u will get 2 solutions (since quadratic), then sub each presumed double root into the original equation (p(y)). whichever results in p(y) = 0 is the double root. good luck!. 7. Re: 4U Polynomial question HELP !!. Notice that the required answer includes the constant term $c$. That should be a hint - if we were to differentiate the polynomial, we would lose $c$, so not only would we have to expand the cube of a binomial (which can be really messy), we would also have to rewrite the result to include $c$ in order to find the double root, so let's try a different approach.. Let $P(x) = x^3 + 3ax^2+3bx+c$ and let the roots be $\alpha, \, \alpha , \,\beta$.. By the relationships between the roots and co-efficients,. $2\alpha + \beta = -3a$. $\alpha^2 + 2\alpha\beta = 3b$. $\alpha^2 \beta = -c$. Rearranging,. $a = - \frac{2\alpha + \beta}{3}$.
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$b = \frac{\alpha^2 + 2\alpha\beta}{3}$. $c=-\alpha^2 \beta$. Now:. $c - ab = \frac{(2\alpha+\beta)(\alpha^2 + 2\alpha\beta)-9\alpha^2\beta}{9} = \frac{2\alpha(\alpha - \beta)^2}{9}$. $2(a^2-b) = 2 \cdot \frac{(2\alpha + \beta)^2-3(\alpha^2 + 2\alpha\beta)}{9} = \frac{2(\alpha - \beta)^2}{9}$. Therefore,. $\frac{c-ab}{2(a^2-b)} = \alpha, \quad (a^2 - b \neq 0)$. 8. Re: 4U Polynomial question HELP !!. While the method above is indeed valid, one could also solve the problem using calculus:. Let P(x)=x3+3ax2+3bx+c.. Then P'(x)=3x2+6ax+3b.. For there to be a double root, there must exist some x such that P(x)=0 and P'(x)=0.. P'(x)=0 iff 3x2+6ax+3b=0, i.e. x2+2ax+b=0.. Now, P(x)=0 for the same value of x,. so x3+3ax2+3bx+c=0,. i.e. x(x2+2ax+b)+ax2+2bx+c=0.. x(x2+2ax+b)+a(x2+2ax+b)-2a2x-ab+2bx+c=0.. But x2+2ax+b=0 (proven above),. so -2a2x-ab+2bx+c=0,. i.e. x(2b-2a2)+c-ab=0.. x=(c-ab)/(2(a2-b)) as a2-b≠0.. Hence, the double root must be equal to (c-ab)/(2(a2-b)) as required.. Users Browsing this Thread. There are currently 1 users browsing this thread. (0 members and 1 guests). Posting Permissions. • You may not post new threads. • You may not post replies. • You may not post attachments. • You may not edit your posts. •.
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http://mathhelpforum.com/calculus/182761-common-ratio.html
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1. ## common ratio
Given that $m, -4, m+15$ are the fourth, sixth and eighth terms of a geometric progression that has a first term which is positive, find the common ratio.
$\frac{-4}{m}=2r$
$r=\frac{-2}{m}$
but answer is $-\frac{1}{2}$
2. Originally Posted by Punch
Given that $m, -4, m+15$ are the fourth, sixth and eighth terms of a geometric progression that has a first term which is positive, find the common ratio.
$\frac{-4}{m}=2r$
shouldn't it be r^2 instead of 2r
$r=\frac{-2}{m}$
but answer is $-\frac{1}{2}$
...
3. Originally Posted by abhishekkgp
...
sorry,
$\frac{-4}{m}=r^2$
$r=-\frac{2}{\sqrt{m}}$
but ans still differs from given ans
4. Hello, Punch!
$\text{Given that: }\:m,\:-4,\:m+15\,\text{ are the fourth, sixth and eighth terms of a}$
$\text{geometric progression that has a positive first term, find the common ratio.}$
We have: . $\begin{Bmatrix}a_4 &=& ar^3 &=& m & [1] \\ a_6 &=& ar^5 &=& -4 & [2] \\ a_8 &=& ar^7 &=& m+15 & [3] \end{Bmatrix}$
$\begin{array}{ccccccccccc}\text{Divide [2] by [1]:} & \dfrac{ar^5}{ar^3} &=& -\dfrac{4}{m} & \Rightarrow & r^2 &=& -\dfrac{4}{m} & [5] \\ \\[-3mm] \text{Divide [3] by [2]:} & \dfrac{ar^7}{ar^5} &=& \dfrac{m+15}{-4} & \Rightarrow & r^2 &=& \dfrac{m+15}{-4} & [5] \end{array}$
Equate [4] and [5]: . $-\frac{4}{m} \:=\:\frac{m+15}{-4} \quad\Rightarrow\quad m^2 + 15m - 16 \:=\:0$
. . $(m - 1)(m + 16) \:=\:0 \quad\Rightarrow\quad m \:=\:1,\:\text{-}16$
$m = 1:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{1} \quad\text{No real solution}$
$m = \text{-}16:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{\text{-}16} \quad\Rightarrow\quad r \:=\:\pm\tfrac{1}{2}$
$\text{If }m = \text{-}16,\:r = \tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& \text{-}64 \end{Bmatrix} \;\;\text{ but }a_1\text{ is positive.}$
$\text{If }m = \text{-}16,\:r = \text{-}\tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& +64 \end{Bmatrix} \;\;\text{ There!}$
Therefore, the common ratio is: . $r \,=\,\text{-}\tfrac{1}{2}$
5. Originally Posted by Soroban
Hello, Punch!
We have: . $\begin{Bmatrix}a_4 &=& ar^3 &=& m & [1] \\ a_6 &=& ar^5 &=& -4 & [2] \\ a_8 &=& ar^7 &=& m+15 & [3] \end{Bmatrix}$
$\begin{array}{ccccccccccc}\text{Divide [2] by [1]:} & \dfrac{ar^5}{ar^3} &=& -\dfrac{4}{m} & \Rightarrow & r^2 &=& -\dfrac{4}{m} & [5] \\ \\[-3mm] \text{Divide [3] by [2]:} & \dfrac{ar^7}{ar^5} &=& \dfrac{m+15}{-4} & \Rightarrow & r^2 &=& \dfrac{m+15}{-4} & [5] \end{array}$
Equate [4] and [5]: . $-\frac{4}{m} \:=\:\frac{m+15}{-4} \quad\Rightarrow\quad m^2 + 15m - 16 \:=\:0$
. . $(m - 1)(m + 16) \:=\:0 \quad\Rightarrow\quad m \:=\:1,\:\text{-}16$
$m = 1:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{1} \quad\text{No real solution}$
$m = \text{-}16:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{\text{-}16} \quad\Rightarrow\quad r \:=\:\pm\tfrac{1}{2}$
$\text{If }m = \text{-}16,\:r = \tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& \text{-}64 \end{Bmatrix} \;\;\text{ but }a_1\text{ is positive.}$
$\text{If }m = \text{-}16,\:r = \text{-}\tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& +64 \end{Bmatrix} \;\;\text{ There!}$
Therefore, the common ratio is: . $r \,=\,\text{-}\tfrac{1}{2}$
that was a lot of work, i understood it fully! thanks.
but just to point out a minor mistake, you should have wrote: substitute in [5] because there isnt an equation [4]
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1. ## common ratio. Given that $m, -4, m+15$ are the fourth, sixth and eighth terms of a geometric progression that has a first term which is positive, find the common ratio.. $\frac{-4}{m}=2r$. $r=\frac{-2}{m}$. but answer is $-\frac{1}{2}$. 2. Originally Posted by Punch. Given that $m, -4, m+15$ are the fourth, sixth and eighth terms of a geometric progression that has a first term which is positive, find the common ratio.. $\frac{-4}{m}=2r$. shouldn't it be r^2 instead of 2r. $r=\frac{-2}{m}$. but answer is $-\frac{1}{2}$. .... 3. Originally Posted by abhishekkgp. .... sorry,. $\frac{-4}{m}=r^2$. $r=-\frac{2}{\sqrt{m}}$. but ans still differs from given ans. 4. Hello, Punch!. $\text{Given that: }\:m,\:-4,\:m+15\,\text{ are the fourth, sixth and eighth terms of a}$. $\text{geometric progression that has a positive first term, find the common ratio.}$.
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We have: . $\begin{Bmatrix}a_4 &=& ar^3 &=& m & [1] \\ a_6 &=& ar^5 &=& -4 & [2] \\ a_8 &=& ar^7 &=& m+15 & [3] \end{Bmatrix}$. $\begin{array}{ccccccccccc}\text{Divide [2] by [1]:} & \dfrac{ar^5}{ar^3} &=& -\dfrac{4}{m} & \Rightarrow & r^2 &=& -\dfrac{4}{m} & [5] \\ \\[-3mm] \text{Divide [3] by [2]:} & \dfrac{ar^7}{ar^5} &=& \dfrac{m+15}{-4} & \Rightarrow & r^2 &=& \dfrac{m+15}{-4} & [5] \end{array}$. Equate [4] and [5]: . $-\frac{4}{m} \:=\:\frac{m+15}{-4} \quad\Rightarrow\quad m^2 + 15m - 16 \:=\:0$. . . $(m - 1)(m + 16) \:=\:0 \quad\Rightarrow\quad m \:=\:1,\:\text{-}16$. $m = 1:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{1} \quad\text{No real solution}$. $m = \text{-}16:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{\text{-}16} \quad\Rightarrow\quad r \:=\:\pm\tfrac{1}{2}$. $\text{If }m = \text{-}16,\:r = \tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& \text{-}64 \end{Bmatrix} \;\;\text{ but }a_1\text{ is positive.}$. $\text{If }m = \text{-}16,\:r = \text{-}\tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& +64 \end{Bmatrix} \;\;\text{ There!}$. Therefore, the common ratio is: . $r \,=\,\text{-}\tfrac{1}{2}$. 5. Originally Posted by Soroban. Hello, Punch!. We have: . $\begin{Bmatrix}a_4 &=& ar^3 &=& m & [1] \\ a_6 &=& ar^5 &=& -4 & [2] \\ a_8 &=& ar^7 &=& m+15 & [3] \end{Bmatrix}$. $\begin{array}{ccccccccccc}\text{Divide [2] by [1]:} & \dfrac{ar^5}{ar^3} &=& -\dfrac{4}{m} & \Rightarrow & r^2 &=& -\dfrac{4}{m} & [5] \\ \\[-3mm] \text{Divide [3] by [2]:} & \dfrac{ar^7}{ar^5} &=& \dfrac{m+15}{-4} & \Rightarrow & r^2 &=& \dfrac{m+15}{-4} & [5] \end{array}$. Equate [4] and [5]: . $-\frac{4}{m} \:=\:\frac{m+15}{-4} \quad\Rightarrow\quad m^2 + 15m - 16 \:=\:0$. . . $(m - 1)(m + 16) \:=\:0 \quad\Rightarrow\quad m \:=\:1,\:\text{-}16$. $m = 1:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{1} \quad\text{No real solution}$. $m = \text{-}16:\;\text{Substitute into [4]: }\:r^2 \:=\:-\frac{4}{\text{-}16} \quad\Rightarrow\quad r \:=\:\pm\tfrac{1}{2}$. $\text{If }m = \text{-}16,\:r = \tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& \text{-}64 \end{Bmatrix} \;\;\text{ but }a_1\text{ is positive.}$. $\text{If }m = \text{-}16,\:r = \text{-}\tfrac{1}{2},\text{ we have: }\begin{Bmatrix} a_8 &=& \text{-}1 \\ a_6 &=& \text{-}4 \\ a_4 &=& \text{-}16 \\ a_2 &=& \text{-}32 \\ a_1 &=& +64 \end{Bmatrix} \;\;\text{ There!}$. Therefore, the common ratio is: . $r \,=\,\text{-}\tfrac{1}{2}$. that was a lot of work, i understood it fully! thanks.. but just to point out a minor mistake, you should have wrote: substitute in [5] because there isnt an equation [4].
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http://slideplayer.com/slide/3281308/
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# Lecture 9 : Balanced Search Trees Bong-Soo Sohn Assistant Professor School of Computer Science and Engineering Chung-Ang University.
## Presentation on theme: "Lecture 9 : Balanced Search Trees Bong-Soo Sohn Assistant Professor School of Computer Science and Engineering Chung-Ang University."— Presentation transcript:
Lecture 9 : Balanced Search Trees Bong-Soo Sohn Assistant Professor School of Computer Science and Engineering Chung-Ang University
Balanced Binary Search Tree In Binary Search Tree Average and maximum search times will be minimized when BST is maintained as a complete tree at all times. : O (lg N) If not balanced, the search time degrades to O(N) Idea : Keep BST as balanced as possible
AVL tree "height-balanced“ binary tree the height of the left and right subtrees of every node differ by at most one 3 25 6 47 3 5 29 14 3 12 14 7 8 11 10
Non-AVL tree example 6 49 357 8
Balance factor BF = (height of right subtree - height of left subtree) So, BF = -1, 0 or +1 for an AVL tree. 0 00 0 0 +1 -2 0+1 0
AVL tree rebalancing When the AVL property is lost we can rebalance the tree via one of four rotations Single right rotation Single left rotation Double left rotation Double right rotation
Single Left Rotation (SLR) when A is unbalanced to the left and B is left-heavy A BT3 T1T2 B T1A T2T3 SLR at A
Single Right Rotation (SRR) when A is unbalanced to the right and B is right-heavy A T1B T2T3 B A T1T2 SRR at A
Double Left Rotation (DLR) When C is unbalanced to left And A is right heavy C AT4 T1B T2T3 C BT4 AT3 T1T2 B AC T1T2T3T4 SLR at A SRR at C
Double Right Rotation (DRR) When C is unbalanced to right And A is left heavy A T1B T2C T3T4 B AC T1T2T3T4 A T1C BT4 T2T3 SRR at CSRR at A
Insertion in AVL tree An AVL tree may become out of balance in two basic situations After inserting a node in the right subtree of the right child After inserting a node in the left subtree of the right child
insertion Insertion of a node in the right subtree of the right child Involves SLR at node P
insertion Insertion of a node in the left subtree of the right child Involves DRR :
insertion In each case the tree is rebalanced locally insertion requires one single or one double rotation O(constant) for rebalancing Cost of search for insert is still O(lg N) Experiments have shown that 53% of insertions do not bring the tree out of balance
Deletion in AVL tree Not covered here Requires O(lg N) rotations in worst case
example Given the following AVL tree, insert the node with value 9: 6 312 4 713 10
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# Lecture 9 : Balanced Search Trees Bong-Soo Sohn Assistant Professor School of Computer Science and Engineering Chung-Ang University.. ## Presentation on theme: "Lecture 9 : Balanced Search Trees Bong-Soo Sohn Assistant Professor School of Computer Science and Engineering Chung-Ang University."— Presentation transcript:. Lecture 9 : Balanced Search Trees Bong-Soo Sohn Assistant Professor School of Computer Science and Engineering Chung-Ang University. Balanced Binary Search Tree In Binary Search Tree Average and maximum search times will be minimized when BST is maintained as a complete tree at all times. : O (lg N) If not balanced, the search time degrades to O(N) Idea : Keep BST as balanced as possible. AVL tree "height-balanced“ binary tree the height of the left and right subtrees of every node differ by at most one 3 25 6 47 3 5 29 14 3 12 14 7 8 11 10. Non-AVL tree example 6 49 357 8. Balance factor BF = (height of right subtree - height of left subtree) So, BF = -1, 0 or +1 for an AVL tree. 0 00 0 0 +1 -2 0+1 0. AVL tree rebalancing When the AVL property is lost we can rebalance the tree via one of four rotations Single right rotation Single left rotation Double left rotation Double right rotation. Single Left Rotation (SLR) when A is unbalanced to the left and B is left-heavy A BT3 T1T2 B T1A T2T3 SLR at A.
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Single Right Rotation (SRR) when A is unbalanced to the right and B is right-heavy A T1B T2T3 B A T1T2 SRR at A. Double Left Rotation (DLR) When C is unbalanced to left And A is right heavy C AT4 T1B T2T3 C BT4 AT3 T1T2 B AC T1T2T3T4 SLR at A SRR at C. Double Right Rotation (DRR) When C is unbalanced to right And A is left heavy A T1B T2C T3T4 B AC T1T2T3T4 A T1C BT4 T2T3 SRR at CSRR at A. Insertion in AVL tree An AVL tree may become out of balance in two basic situations After inserting a node in the right subtree of the right child After inserting a node in the left subtree of the right child. insertion Insertion of a node in the right subtree of the right child Involves SLR at node P. insertion Insertion of a node in the left subtree of the right child Involves DRR :. insertion In each case the tree is rebalanced locally insertion requires one single or one double rotation O(constant) for rebalancing Cost of search for insert is still O(lg N) Experiments have shown that 53% of insertions do not bring the tree out of balance. Deletion in AVL tree Not covered here Requires O(lg N) rotations in worst case. example Given the following AVL tree, insert the node with value 9: 6 312 4 713 10. Similar presentations.
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# Three electric bulbs with markings 100 V-150 W, 100V-400W and 100V-500W are connected in series along power supply of 100V. Calculate the current in the circuit through each bulb.
Video Solution
Text Solution
Generated By DoubtnutGPT
## To solve the problem of finding the current in a circuit with three electric bulbs connected in series, we will follow these steps:Step 1: Determine the Resistance of Each BulbThe resistance of each bulb can be calculated using the formula:R=V2Pwhere V is the voltage rating of the bulb and P is the power rating.1. For the first bulb (100 V, 150 W): R1=1002150=10000150=66.67Ω2. For the second bulb (100 V, 400 W): R2=1002400=10000400=25Ω3. For the third bulb (100 V, 500 W): R3=1002500=10000500=20ΩStep 2: Calculate the Total Resistance in the CircuitSince the bulbs are connected in series, the total resistance Req is the sum of the individual resistances:Req=R1+R2+R3Substituting the values we calculated:Req=66.67+25+20=111.67ΩStep 3: Calculate the Current in the CircuitUsing Ohm's Law, the current I in the circuit can be calculated using the formula:I=VReqwhere V is the total voltage supplied (100 V):I=100111.67≈0.896AFinal AnswerThe current flowing through each bulb in the circuit is approximately:I≈0.9A---
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Updated on:7/8/2024
### Knowledge Check
• Question 1 - Select One
## Two electric bulbs marked 25W-220V and 100W-220V are connected in series to a 440 V supply. Which of the bulbs will fuse ?
ABoth
B100 W
C25 W
DNeither
• Question 2 - Select One
## Two electric bulbs marked 25W-220V and 100W-220V are connected in series to a 440 V supply. Which of the bulbs will fuse ?
Aboth
B100W
C25W
Dneither
• Question 3 - Select One
## Two electic bulbs marked 25W−220V and 100W−220V are connected in series to a 440 V supply. Which of the bulbs will fuse?
ABoth
B100 W
C25 W
DNeither
Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc
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# Three electric bulbs with markings 100 V-150 W, 100V-400W and 100V-500W are connected in series along power supply of 100V. Calculate the current in the circuit through each bulb.. Video Solution. Text Solution. Generated By DoubtnutGPT. ## To solve the problem of finding the current in a circuit with three electric bulbs connected in series, we will follow these steps:Step 1: Determine the Resistance of Each BulbThe resistance of each bulb can be calculated using the formula:R=V2Pwhere V is the voltage rating of the bulb and P is the power rating.1. For the first bulb (100 V, 150 W): R1=1002150=10000150=66.67Ω2. For the second bulb (100 V, 400 W): R2=1002400=10000400=25Ω3. For the third bulb (100 V, 500 W): R3=1002500=10000500=20ΩStep 2: Calculate the Total Resistance in the CircuitSince the bulbs are connected in series, the total resistance Req is the sum of the individual resistances:Req=R1+R2+R3Substituting the values we calculated:Req=66.67+25+20=111.67ΩStep 3: Calculate the Current in the CircuitUsing Ohm's Law, the current I in the circuit can be calculated using the formula:I=VReqwhere V is the total voltage supplied (100 V):I=100111.67≈0.896AFinal AnswerThe current flowing through each bulb in the circuit is approximately:I≈0.9A---. |. Updated on:7/8/2024. ### Knowledge Check. • Question 1 - Select One. ## Two electric bulbs marked 25W-220V and 100W-220V are connected in series to a 440 V supply. Which of the bulbs will fuse ?. ABoth. B100 W. C25 W. DNeither. • Question 2 - Select One. ## Two electric bulbs marked 25W-220V and 100W-220V are connected in series to a 440 V supply.
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Which of the bulbs will fuse ?. Aboth. B100W. C25W. Dneither. • Question 3 - Select One. ## Two electic bulbs marked 25W−220V and 100W−220V are connected in series to a 440 V supply. Which of the bulbs will fuse?. ABoth. B100 W. C25 W. DNeither. Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc. NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams.. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation.
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https://stats.stackexchange.com/questions/104749/why-is-this-likelihood-function-equal-to-the-noise-pdf
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# Why is this likelihood function equal to the noise PDF?
My professor has this slide up here:
Here, $y$ is an observed signal. $H$ is a deterministic transformation, which is assumed known. $f$ is the original signal (which we dont know), and $w$ is random gaussian noise. We are trying to recover $f$.
I understand everything, except for, why $p(\mathbf{w})$ = $p(\mathbf{y}|\mathbf{f})$.
That is, I understand that the multidimensional noise PDF is given by the above expression.
But why is that expression, ALSO equal to the likelihood function, $\mathbf{y}$, given $\mathbf{f}$? I'm not seeing this...
• Where does it say p(w)=Likelihoodfunction? The function p(y|f) is a conditional probability density. Jun 25, 2014 at 17:38
• @emcor What do you mean? p(w) = p(y|f) as seen above, and p(y|f) is the likelihood function as given here Jun 25, 2014 at 17:41
• The likelihood is a function of the parameters,, so notation like "$p(w)$" clearly does not refer to a likelihood. Unless a distribution is assumed for $f$, though, "$p(y|f)$" is not a conditional probability density, either: it merely refers to the probability density of $y$ as it depends on the parameters $f$. By assuming $W=Y-Hf$ has a Gaussian distribution, all you have to do is plug $y-Hf$ into the formula for a (multivariate) Gaussian density. Fixing $y$, $H$, and $C_{ww}$, it becomes a function of $f$: in that sense it's a likelihood.
– whuber
Jun 25, 2014 at 17:55
• We have to do some careful interpreting because the notation is sloppy. Apparently the model is $Y=Hf+W$ where $W$ is a random vector-valued variable. This makes $Y$ a random variable, too. Given any value of $f$, any realization of $Y$, which is written $y$, corresponds to a realization $y-Hf$ of $W$. The probability density of that realization is given by the equation. The right hand side $\Lambda$ is a function of $(y,H,C_{ww},f)$. If you assume values for $H$ and $C_{ww}$, and are given the data $y$, $f$ remains the only variable and you can study how $\Lambda$ depends on $f$.
– whuber
Jun 25, 2014 at 18:10
• Your edits support my suppositions about how to read the slide. The $f$ you are trying to recover plays the role of unknown parameters; everything else is either known or assumed. Thus the likelihood will be considered a function of $f$ and you will later find values of $f$ that make the likelihood as large as possible. You might go even further and deduce confidence limits for your estimates of $f$ by studying how the likelihood varies as you vary $f$ around its maximizing value. You might possibly even adopt a "prior distribution" for $f$, but that would not alter the present interpretation.
– whuber
Jun 25, 2014 at 18:44
$p(w)=p(y|f)$:
It is because of what is annotated in red on the slide, you have $w$ and $y$ linked as:
$w=y-Hf$
,so $p(w)=p(y-Hf)$ as well.
If $H$ and $f$ are held constant, $y$ is the only random variable which determines the probability:
$p(w)=p(y-Hf|f,H)=p(y|f)$.
I assume he omits $H$ because it is defined as constant anyways, so the probability is no longer dependent on $f$ neither on $H$.
He then correctly substitutes $'w=y-Hf'$ into the Gaussian density of $w$.
• One point worth addressing is the potential confusion of data with random variables. If the right hand side is called a "likelihood," then "$y$" must refer to data (that is, a realization), not to a random variable. Furthermore, if we accept that calling the RHS a likelihood was intentional, then we must emphasize its dependence on $f$ rather than dismiss it.
– whuber
Jun 25, 2014 at 18:24
• Thank you Emcor. FYI, I edited my question to give more details on the background setup. That said, unfortunately I still find myself somewhat puzzled by why, exactly, p(w) = p(y|f). Specifically, I am not sure why you are saying that f is held constant, when we are in fact trying to find it... Jun 25, 2014 at 18:26
• "Likelihood" is the density evaluated at a datapoint, which we have in p(w) as already clear. As we also have p(w)=p(y-Hf), it is a notational convention to write the parameter to be maximized in the Maximum Likelihood Function as $f(x|\theta)$. This convention might be confusing here. Jun 25, 2014 at 18:32
• Creatron, I believe you may be overthinking this. In a formula like $2-x^2$, $x$ has some definite but unknown value. Your circumstance is no different conceptually. You will estimate $f$ by maximizing the likelihood, just as you would consider varying the unknown $x$ to maximize $2-x^2$, even though the quantity $x$ refers to is whatever it is and that doesn't vary at all. For more about how the likelihood works I will refer you once more to the link I gave in an earlier comment: please read the thread at stats.stackexchange.com/questions/2641.
– whuber
Jun 25, 2014 at 18:35
• @whuber Thanks whuber, let me study that link and digest. Jun 25, 2014 at 19:55
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# Why is this likelihood function equal to the noise PDF?. My professor has this slide up here:. Here, $y$ is an observed signal. $H$ is a deterministic transformation, which is assumed known. $f$ is the original signal (which we dont know), and $w$ is random gaussian noise. We are trying to recover $f$.. I understand everything, except for, why $p(\mathbf{w})$ = $p(\mathbf{y}|\mathbf{f})$.. That is, I understand that the multidimensional noise PDF is given by the above expression.. But why is that expression, ALSO equal to the likelihood function, $\mathbf{y}$, given $\mathbf{f}$? I'm not seeing this.... • Where does it say p(w)=Likelihoodfunction? The function p(y|f) is a conditional probability density. Jun 25, 2014 at 17:38. • @emcor What do you mean? p(w) = p(y|f) as seen above, and p(y|f) is the likelihood function as given here Jun 25, 2014 at 17:41. • The likelihood is a function of the parameters,, so notation like "$p(w)$" clearly does not refer to a likelihood. Unless a distribution is assumed for $f$, though, "$p(y|f)$" is not a conditional probability density, either: it merely refers to the probability density of $y$ as it depends on the parameters $f$. By assuming $W=Y-Hf$ has a Gaussian distribution, all you have to do is plug $y-Hf$ into the formula for a (multivariate) Gaussian density. Fixing $y$, $H$, and $C_{ww}$, it becomes a function of $f$: in that sense it's a likelihood.. – whuber. Jun 25, 2014 at 17:55. • We have to do some careful interpreting because the notation is sloppy. Apparently the model is $Y=Hf+W$ where $W$ is a random vector-valued variable. This makes $Y$ a random variable, too. Given any value of $f$, any realization of $Y$, which is written $y$, corresponds to a realization $y-Hf$ of $W$. The probability density of that realization is given by the equation. The right hand side $\Lambda$ is a function of $(y,H,C_{ww},f)$. If you assume values for $H$ and $C_{ww}$, and are given the data $y$, $f$ remains the only variable and you can study how $\Lambda$ depends on $f$.. – whuber. Jun 25, 2014 at 18:10. • Your edits support my suppositions about how to read the slide. The $f$ you are trying to recover plays the role of unknown parameters; everything else is either known or assumed. Thus the likelihood will be considered a function of $f$ and you will later find values of $f$ that make the likelihood as large as possible. You might go even further and deduce confidence limits for your estimates of $f$ by studying how the likelihood varies as you vary $f$ around its maximizing value. You might possibly even adopt a "prior distribution" for $f$, but that would not alter the present interpretation.. – whuber.
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Jun 25, 2014 at 18:44. $p(w)=p(y|f)$:. It is because of what is annotated in red on the slide, you have $w$ and $y$ linked as:. $w=y-Hf$. ,so $p(w)=p(y-Hf)$ as well.. If $H$ and $f$ are held constant, $y$ is the only random variable which determines the probability:. $p(w)=p(y-Hf|f,H)=p(y|f)$.. I assume he omits $H$ because it is defined as constant anyways, so the probability is no longer dependent on $f$ neither on $H$.. He then correctly substitutes $'w=y-Hf'$ into the Gaussian density of $w$.. • One point worth addressing is the potential confusion of data with random variables. If the right hand side is called a "likelihood," then "$y$" must refer to data (that is, a realization), not to a random variable. Furthermore, if we accept that calling the RHS a likelihood was intentional, then we must emphasize its dependence on $f$ rather than dismiss it.. – whuber. Jun 25, 2014 at 18:24. • Thank you Emcor. FYI, I edited my question to give more details on the background setup. That said, unfortunately I still find myself somewhat puzzled by why, exactly, p(w) = p(y|f). Specifically, I am not sure why you are saying that f is held constant, when we are in fact trying to find it... Jun 25, 2014 at 18:26. • "Likelihood" is the density evaluated at a datapoint, which we have in p(w) as already clear. As we also have p(w)=p(y-Hf), it is a notational convention to write the parameter to be maximized in the Maximum Likelihood Function as $f(x|\theta)$. This convention might be confusing here. Jun 25, 2014 at 18:32. • Creatron, I believe you may be overthinking this. In a formula like $2-x^2$, $x$ has some definite but unknown value. Your circumstance is no different conceptually. You will estimate $f$ by maximizing the likelihood, just as you would consider varying the unknown $x$ to maximize $2-x^2$, even though the quantity $x$ refers to is whatever it is and that doesn't vary at all. For more about how the likelihood works I will refer you once more to the link I gave in an earlier comment: please read the thread at stats.stackexchange.com/questions/2641.. – whuber. Jun 25, 2014 at 18:35. • @whuber Thanks whuber, let me study that link and digest. Jun 25, 2014 at 19:55.
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