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what is 1,199,679 in expanded notation

what is 1,199,679 in expanded notation

Research, Knowledge and Information :


What is Expanded Form in Math? - Definition & Examples ...


What is Expanded Notation? ... it means one one, or 1. Expanded Form. ... What is Expanded Form in Math? - Definition & Examples Related Study Materials.
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7-1 Scientific Notation - Math Is Fun


Enter a number and see it in Scientific Notation: Now try to use Scientific Notation yourself: Other Ways of Writing It. 3.1 × 10^8.
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Quia - Mathematics


Quia Web allows users to create and share online educational activities in dozens of subjects, including Mathematics. Home FAQ About Log in ...
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Alabama Course of Study for Mathematics–Grade 6


176-179, 180-184, 190 (#40-43), 195 (#52-59, 62), 196, 197 (#1-8), 198, 199, 200 (#1-25), 201 ... 35-38), 679 (#25-26), 680 ... in expanded notation ...
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www.topmath.info


Write 8882 in expanded notation. 8000 + 800 + 80 + 2: 13) ... 199: 95) 2/11 + 1/3 = 17/33: 96) ... 679) How many pounds and ...
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30 CFR Ch. I (7-1-04 Edition) Mine Safety and Health Admin ...


... 2004 CODE OF FEDERAL REGULATIONS 30 Parts 1 to 199 Revised as of ... The Code of Federal Regulations is a codification of the general and permanent rules ...
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14 CFR Ch. I (1-1-00 Edition) Federal Aviation Administration ...


CODE OF FEDERAL REGULATIONS 14 Parts 1 to 59 Revised as of ... The first three volumes containing parts 1-199 are comprised of chapter I—Federal Aviation ...
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Decimals Worksheets - cnm.edu


Revised @2009 MLC Page 1 of 21 Decimals Worksheets Decimal Place Values ... 199.99 200 is the answer The place to round to The digit 2 is small. Do not round up.
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Exponents, Radicals and Scientific Notation - Industrial Wiki


exponents, radicals and scientific notation. print. ... 199. 207. 215. 223. 231. 540. 239. 247. 255. 263. 272. 280. 288. 296. 304. 312. 41. 320. ... 679. 687. 695 ...
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Suggested Questions And Answer :


binomial Expansion using Pascal's triangle?

If we have an equation such as (x+y)^n, then the expanded equation would be: (nP0)x^(n)*y^(0)+(nP1)x^(n-1)*y^(1)+(nP0)x^(n-2)*y(2)+(nP3)x^(n-3)*y(3) ... (nPn-1)x^(1)*y^(n-1)+(nPn)x^(0)*y^(n). Note the notation (aPb) here means "a choose b". Or in summation form: (x+y)^n=Σ((nPk)x^(n-k)*y^k, k, 0, n)
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how do you write 5087 in expanded notation

5 thousands, 0 hundreds, 8 tens, 7 ones 
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if n is a +ive Z, and..... (Calculus)

The first part is covered by Proof by Induction, and gives us the iteration formula I_(n-1) – I_n = 2π For brevity and clarity I am going to use the notation int[f(x)] to represent the integral, from 0 to 2π, of f(x) with respect to x. To prove: int[sin^2(nx/2) / sin^2(x/2)] = 2nπ sin^2(nx/2) / sin^2(x/2) = 2(1 – cos(nx)) / ({2(1 – cos(x))} = (1 – cos(nx)) / (1 – cos(x)) Therefore, int[sin^2(nx/2) / sin^2(x/2)] = int[(1 – cos(nx)) / (1 – cos(x))] int[] = int[(1) / (1 – cos(x))] –  int[(cos(nx)) / (1 – cos(x))] But, from part 1,  int[(cos(nx)) / (1 – cos(x))] = I_n, therefore, int[] = int[(1) / (1 – cos(x))] – I_n int[] = int[(1) / (1 – cos(x))] – I_(n-1) + 2π int[] = int[(1) / (1 – cos(x))] – int[(cos((n-1)x)) / (1 – cos(x))]  + 2π int[] = int[(1 – cos((n-1)x)) / (1 – cos(x))]  + 2π, i.e. int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-1)x)) / (1 – cos(x))]  + 2π The above is an iteration formula, with the nth term on the lhs and the (n-1)th term on the rhs. We can expand this iteration sequence as follows, int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-2)x)) / (1 – cos(x))]  + 2(2π) int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-3)x)) / (1 – cos(x))]  + 3(2π) ... int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos((n-(n-1))x)) / (1 – cos(x))]  + (n-1)(2π) int[(1 – cos(nx)) / (1 – cos(x))] = int[(1 – cos(x)) / (1 – cos(x))]  + (n-1)(2π) int[(1 – cos(nx)) / (1 – cos(x))] = int[ (1) ] + (n-1)(2π) int[(1 – cos(nx)) / (1 – cos(x))] = 2π + (n-1)(2π) int[(1 – cos(nx)) / (1 – cos(x))] = 2nπ int[sin^2(nx/2) / sin^2(x/2)] = 2nπ  
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how to find 4ײ-×=0

(y-4)²+7=0
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augmented matrix 2x-2y+2z=-6, 3x-5y+6z=3, 7y-4z=-7

The augmented matrix for the system: { 2 -2 2 | -6 } { 3 -5 6 | 3 } { 0 7 -4 | -7 } The curly brackets extending over three lines are meant to represent the matrix enclosure. Use the letter R to denote a matrix row and 1, 2 and 3 to define which row it is: R1, R2, R3. Work out R1/2: { 1 -1 1 -3 } that is R1→1/2R1. Work out 3R1-R2: { 0 2 -3 | -12 } that is, R2→3R1-R2. We now have two rows beginning with 0, this row and R3, so let's reduce the problem to a 3 by 2 matrix for brevity and to avoid unnecessary clutter and repetition. {  2 -3 | -12 } { 7 -4 | -7 } Call the rows r1 and r2 to distinguish from R1 and R2. Remember we're aiming for an identity matrix with zeroes and one 1 to the left of the vertical bar. Let's work out 3r2-4r1 in two steps: { 8 -12 | -48 } { 21 -12 | -21 } 3r2-4r1={ 13 0 | 27 } and divide this row by 13: { 1 0 | 27/13 } which can be expanded to { 0 1 0 | 27/13 } in the original matrix format, row 2. Similarly 2r2-7r1 becomes { 0 13 | 70 }, and dividing this by 13 gives { 0 1 | 70/13 } which can be expanded to { 0 0 1 | 70/13 in the original matrix format, row 3. This what we have: { 1 -1 1 | -3 } { 0 1 0 | 27/13 } { 0 0 1 | 70/13 } Finally, we can convert R1 by performing R1→R1+R2-R3: { 1 0 0 | -82/13 }, so the solution to the system is x=-82/13, y=27/13 and z=70/13.  
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sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13
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A standard Hyperbola has its vertices ar 5,7 and 3,-1.Find its equation and its asymptotes

this equation expands toThe vertices share no common coordinate so the hyperbola cannot be orientated in the usual x-y form. The hyperbola must therefore be skew (tilted axis) and the line joining the vertices is sloping. This line is given by the equation y=((7+1)/(5-3))x+c where c is to be found. y=4x+c, and -1=12+c, plugging in (3,-1). c=-13. Therefore the axis of the hyperbola is y=4x-13. The centre or origin of the hyperbola lies midway between them at ((1/2)(5+3),(1/2)(7-1))=(4,3) and the equation of the conjugate axis is given by y=-x/4+k where k is found by plugging in the midpoint: 3=-4/4+k, so k=4 and y=4-x/4 is the equation of the conjugate axis. Let's define the axes of the hyperbola as X and Y instead of x and y. We can place the centre of the hyperbola at (0,0) in the X-Y plane so that in this plane the equation of the hyperbola is Y^2/A^2-X^2/B^2=1. The vertices are found by finding Y when X=0, so (0,A) and (0,-A) are the vertices. The asymptotes are given by (Y/A+X/B)(Y/A-X/B)=0. So Y=AX/B and Y=-AX/B are their equations. The distance between the vertices is 2A so we can work this out from the x-y coordinates of the vertices using Pythagoras' Theorem: 2A=sqrt((7+1)^2+(5-3)^2)=sqrt(68). A=sqrt(68)/2, A^2=68/4=17. Since the hyperbola is rectangular, B=A, and the equation of the hyperbola is Y^2-X^2=17. The asymptotes then become Y=X and Y=-X. Now we have to transform X-Y to x-y, so that the equation can be expressed in the x-y plane. We use geometry and trigonometry  for this. Let O'(4,3) be the location of the origin of the X-Y plane. Let P(x,y) be any point. In the X-Y plane this is P'(X,Y). Join O'P'=O'P. This line can be represented in both reference frames. O'P^2=X^2+Y^2=(x-4)^2+(y-3)^2 by Pythagoras. The gradient of the Y axis with respect to the x-y frame is 4, and we can represent this as tan(w) so w=tan^-1(4) or tan(w)=4; sin(w)=4/sqrt(17); cos(w)=1/sqrt(17). The angle z that O'P makes with the line y=3 is given by tan(z)=(y-3)/(x-4). sin(z)=(y-3)/O'P; cos(z)=(x-4)/O'P. Let v=90-w. sin(v+z)=Y/O'P; cos(v+z)=X/O'P. v+z=90-w+z=90-(w-z), so sin(v+z)=cos(w-z)=cos(w)cos(z)+sin(w)sin(z)=(x-4)/(O'Psqrt(17))+4(y-3)/(O'Psqrt(17))=(x+4y-16)/(O'Psqrt(17)). cos(v+z)=sin(w-z)=sin(w)cos(z)-cos(w)sin(z)=4(x-4)/(O'Psqrt(17))-(y-3)/(O'Psqrt(17))=(4x-y-13)/(O'Psqrt(17)). Y=O'P*sin(v+z)=(x+4y-16)/sqrt(17); X=O'P*cos(v+z)=(4x-y-13)/sqrt(17). So P maps to the x-y plane using these transformations. Y^2-X^2=17 becomes (x+4y-16)^2-(4x-y-13)^2=289, the equation of the hyperbola. This expands to x^2+8x(y-4)+16y^2-128y+256-(16x^2-8x(y+13)+y^2+26y+169)=289; -15x^2+16xy+15y^2+72x-154y-202=0. The asymptotes are x+4y-16+4x-y-13=0; 5x+3y-29=0 and x+4y-16-4x+y+13=0; 5y-3x-3=0.
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compute the numbers of 1s needed to list in binary notation,the integers from 0 to 1023

the answer is 5120
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what is 654.321 in expanded notation using exponents

654.321  expanded notation is the same as scientific notation. 6.54321 * 10^(-2)  the exponent is negative due to the fact that we moved the decimal to the left and made the number smaller.
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interval notation for radical 9x/x^2-100?

9x/(x^2-100) >= 0 Something happens at x = 0, -10, and 10 Test:  x = -100, -5, 5, 100 x = -100 makes it - x = -5 makes it + x = 5 makes it - x = 100 makes it + We can't hit x = -10 because that would make the bottom of the equation 0, so at -10 there's a ( We can hit x = 0, so there's a ] at 0 We can't hit x = 10 because that would make the bottom of the equation 0, so at 10 there's a ( (-10,0], (10,Infinity)
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