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write expanded notatiob for 17,000,012,053

write expanded notation for 17,000,012,053

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Expanded Notation - Softschools.com

Expanded form or expanded notation is a helpful way to rewrite numbers in order to show case the place value of each digit. ... To write a number in the form, ...

Write Numbers in Expanded Form - MathCoach Interactive

NS 1.5 Use expanded notation to represent numbers ... Write Numbers in Expanded Form ... Write it in expanded form and in standard form. 1.

Expanded Form Calculator - Online Calculator Resource

Expanded form calculator to write numbers in expanded form given decimal place value. ... Expanded Notation Form: 20,000 + 3,000 + 900 + 50 + 8 . Expanded Factors Form:

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

What is Expanded Notation? ... we're writing that number in expanded form. Writing numbers in expanded form ... What is Expanded Form in Math? - Definition & Examples ...

Expanded Form - Math Dictionary - ICoachMath.com

Expanded Form is a way to break up a number ... Now, let's try to write the number in words and see how it helps us identify the expanded form. ...

Definition of Expanded Notation - Math Is Fun

Expanded Notation. Writing a number to show the value of each digit. ... See: Standard Notation. Search

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http://www.mathsisfun.com/definitions/expanded-notation.html

You will need 1.5 Expanded Form and Scientific Notation

scientific notation a way of writing a ... or expanded form. a) 1 300 000 b) 1.235 10 c) 1 103 2 102 3 10 5 1 5. ... 17. Answer each question ...

Math Skills - Scientific Notation

Scientific notation is the way that scientists easily handle very large numbers or very ... instead of writing 0.0000000056, we write 5.6 x 10-9 ... 0.053 = 5.3 x 10-2:

How to Write Numbers in Expanded Form | Sciencing

How to Write Numbers in Expanded Form ... For example, write 6,923.47 in expanded form using words by writing "6 thousands, 9 hundreds, 2 tens, 3 ones, ...

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.

The pythagorean theorem of this problem x^2+(2x+6)^2=(2x+4)^2

Pythagoras' theorem relates the lengths of the sides of a right-angled triangle: a^2+b^2=c^2 where c, the hypotenuse, is the longest side. In your question 2x+6 is the longest side, so there could be no solution, since the the sum of the two squares of the sides on the left-hand side must be bigger than the square of a side that's smaller than either one of them, and 2x+4 is smaller than 2x+6 (2x+6-(2x+4)=2). The equation x^2+(2x+4)^2=(2x+6)^2 can be solved. We can write it as: x^2=(2x+6)^2-(2x+4)^2. On the right-hand side we have the difference of two squares, which factorises to make the calculation simpler: (2x+6-(2x+4))(2x+6+2x+4)=2(4x+10). So now we have x^2=8x+20, by expanding the brackets. We can write this: x^2-8x=20 by subtracting 8x from each side. This is a quadratic that can be solved by completing the square. Halve the x coefficient then square it: 8/2=4, and 4^2=16. Add 16 to both sides: x^2-8x+16=36. The left-hand side is a perfect square of x-4: (x-4)^2=36. 36 is also a perfect square=6^2 so, by taking square roots of each side we have: x-4=6 or -6, because both 6 and -6 have a square of 36. We have two solutions: x=4+6=10 or 4-6=-2. Another way of solving for x is factorisation: x^2-8x=20 can also be written: x^2-8x-20=0 which factorises: (x-10)(x+2)=0, so either x-10=0 or x+2=0, and that also gives us x=10 and -2. The factors x-10 and x+2 are binomial expressions, and the equations arising from them use the zero factor idea: x-10=0 or x+2=0 to solve for x.

write the completed factored form of g(x)

Quadratic polynomial, so we should have something like this: (?x + ?)(?x + ?) f(-3) = 0 means (x+3) is a factor. f(2) = 0 means (x-2) is a factor (x+3)(x-2) = x^2 + x - 6 But we want the leading coefficient (on the x^2) to be 7, so Instead of (x+3)(x-2) = 0, we want 7(x+3)(x-2) = 0 Factored form:  7(x+3)(x-2) 7(x^2 + x - 6) Expanded form:  7x^2 + 7x - 42

If 2n+1 and 3n=1 are both perfect squares then show that n is divisible by 40

If we write 3n+1=(x+y)^2 and 2n+1=x^2, then we have the question reduced to algebraic quantities. If we expand the first equation we get 3n+1=x^2+2xy+y^2. To eliminate n and find a relationship between x and y, we need to multiply the first equation by 2 and the second by 3: 6n+2=2x^2+4xy+2y^2 and 6n+3=3x^2. If we subtract the former from the latter we get 1=x^2-4xy-2y^2, which we can write as a quadratic for y: 2y^2+4xy+1-x^2=0. From this we get: y=(-4x+/-sqrt(16x^2-8+8x^2))/4 = (-2x+/-sqrt(6x^2-2))/2. So x+y = +/-sqrt(6x^2-2)/2 or (x+y)^2=(3x^2-1)/2. We now have relationship between the perfect squares. If x =1 we can see that y=0 and n= 0. We can also appreciate that x must be odd because (3x^2-1) must be divisible by 2. When x=9, (3x^2-1)/2 = 121 = 11^2. 121-81=40, and 3n+1-(2n+1)=n must also be 40. So we have two values for n so far 0 and 40.

Genet multiplied a 3-digit number by 1002 and got AB007C, where A, B, and C stand for digits. What was Genet's original 3-digit number?

Write the three digit number as 100a+10b+c, where a, b and c are the digits. Write 1002 as 1000+2, now expand (100a+10b+c)(1000+2)= 100000a+10000b+1000c+200a+20b+2c=100000A+10000B+70+C. I found it easier to equate digits by comparison by arranging this sum as a layout in long multiplication: ............................................1 0 0 2 ...............................................a b c ...........................................c 0 0 2c ........................................b 0 0 2b 0 .....................................a 0 0 2a 0 0 .....................................A B 0 0 7 C We should be able to equate the digits by position by comparisons. C must be an even number 0, 2, 4, 6 or 8. But in the tens we have 7, so c is 5, 6, 7, 8 or 9, to produce a carryover converting 6 to 7 in the tens. That tells us that b must be 3 or 8 because 2*3+1=7 or 2*8+1=17. But in the hundreds we have 0, so b can't be 8 because there would be a carryover, so b=3. And we can see that 2a must be a number ending in zero, so a=0 or 5. But a can't be zero because we would have a 3-digit number starting with zero making it a 2-digit number so a=5, and 2a=10, which is a carryover to the thousands where c is. But the thousands digit is zero, so c=9. We have all the digits: a=5, b=3 and c=9 making the number 539. 539*1002=540078.

The value of 'a' for which one root of by equation (a^2-5a+3)x^2 +(3a-1)x +2=0 is twice as large as the other is

In a quadratic equation we can write the factors as (x-A)(x-B)=0 where A and B are the roots. In this case B=2A. When this is expanded we get x^2-(A+B)x+AB=0=x^2-3Ax+2A^2=0. Divide the given equation by a^2-5a+3: x^2+(3a-1)x/(a^2-5a+3)+2/(a^2-5a+3)=0. Equating terms we get: -3A=(3a-1)/(a^2-5a+3) and 2A^2=2/(a^2-5a+3), A^2=1/(a^2-5a+3)   So, 9A^2=(3a-1)^2/(a^2-5a+3)^2 and 9/(a^2-5a+3)=(3a-1)^2/(a^2-5a+3)^2. 9(a^2-5a+3)=(3a-1)^2=9a^2-6a+1. 9a^2-45a+27=9a^2-6a+1; -45a+27=-6a+1; 26=39a so a=26/39=2/3. (The original equation becomes: x^2/9+x+2=0 or x^2+9x+18=0=(x+6)(x+3) whose roots are -3 and -6, one root being twice the other. So a=2/3 is proven.)

how do you write 5087 in expanded notation

5 thousands, 0 hundreds, 8 tens, 7 ones

write 5706 in expanded form

5706 = 5(1000) + 7(100) + 0(10) + 6(1)

find a polynomial, P(x) , of degree 3 with zeros of 4, 1 and -1 if p(0) =8

Which question is to be answered? The polynomial or the complex number problem or both? p(x)=ax^3+bx^2+cx+d is the polynomial. When x=0 p(0)=8, so d=8. p(x)=0 when x=4, 1, -1, so these must be the roots of the polynomial: (x-4)(x-1)(x+1) which expands to (x-4)(x^2-1) becoming x^3-4x^2-x+4. This needs to be doubled so that p(x)=8 when x=0. The polynomial p(x) is 2x^3-8x^2-2x+8. More to come...