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# what is 250000000 divisable by?

if 25 is the divisible by 1,5,25. Then what is 250,000,000 divisable by?

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### N is a number which is divisible by 19. if (n+1) (n+8) is ...

N is a number which is divisible by 19. if (n+1) (n+8) is divided by 19, the remainder will be? Find answers now! No. 1 Questions & Answers ... What is 250000000 ...

### What is 250000000 divided by 4 - answers.com

What is 250000000 divided by 4? SAVE CANCEL. already exists. Would you like to merge this question into it? MERGE CANCEL. already exists as an ...

### What is 0.4 divided by 6? - weknowtheanswer.com

What is 250000000 divided by 4? ... Are all even numbers divisible by 10? What is the sum of the first 100 natural number? How many days in week-?

### Math Forum - Ask Dr. Math

You need to figure out how many times your number is divisible by 10. Any number divisible by 10 ends in a zero ... we get 250,000,000 which does end in seven zeros ...

### What is 14.7 percent of 250000000 - step by step solution

Simple and best practice solution for 14.7% of 250000000. Check how easy it is, and learn it for the future. ... What is 14.7 percent of 250000000 ...

### Integer Number 1000000000

Integer Number 1000000000 ... , 12500000, 15625000, 20000000, 25000000, 31250000, 40000000, 50000000, 62500000, 100000000, 125000000, 200000000, 250000000, ...

### What is the sum of 1000000000 divided by 4 - Answers.com

What is the sum of 1000000000 divided by 4? ... 250000000 Minor edit ... 64 is the only number divisible by 5 with a remainder of 4 ...

### Engineering notation - Wikipedia

Engineering notation or engineering form is a version of scientific notation in which the exponent of ten must be divisible by ... in engineering notation is ...

## Suggested Questions And Answer :

### 10 vedic maths rules for class IX

The divisions between 0 and 1 show how a fraction is made up. Let's take the case where the interval is divided up into 12 equal divisions. One division represents 1/12. Two divisions represent 2/12 which together represent 1/6 because 6*2=12 and the interval between 0 and 1 contains 6 sixths. Three divisions represent 3/12 which represent 1/4 because 4*3=12 and the interval between 0 and 1 contains 4 fourths or quarters, just like there are four quarters to \$1. Four divisions represent 4/12 or one third. Five divisions just represent 5/12 and 7 divisions represent 7/12; but 6 divisions represent 6/12 or one half and two lots of these make 1. So 2*6/12 is the same as 2*1/2. 11 divisions make 11/12. Eight divisions is 8/12. Since 4/12 is a third, 8/12 must be two thirds because 2*4=8 or 2*4/12=8/12. Nine divisions are 3*3/12=9/12, and since 3/12 is the same as a quarter, 9/12 must be three quarters. Ten divisions make 10/12, but since 5*2/12=10/12 and 2/12 is one sixth, 10/12 must be 5 sixths or 5/6. If the interval 0 to 1 is divided into 60 divisions we get more fractions. The divisions help us to add and subtract. The number of divisions is associated with the least common multiple (LCM). So the LCM of 2, 3, 4 and 6 is 12. For 60 as LCM we have 2, 3, 4, 5, 6, 10, 12, 15, 20, 30. These numbers of divisions give us the fractions 1/30, 1/20, 1/15, 1/12, 1/10, 1/6, 1/5, 1/4, 1/3, 1/2. So to add 1/5 and 1/6 we use the divisions 12/60+10/60=22/60=11/30 because 22=2*11 and 22/60=2*11/60=11/30. We can also see that 11/30-1/5 is the same as 22/60-12/60=10/60=1/6. This should give you some idea how dividing 0 to 1 into different divisions helps you see how fractions add and subtract.

### n^3-9n^2+20n prove that it is divisible by 6 for all integers n greater or equal to 1

Factor the given expression.   We have: n³-9n²+20n=n(n-4)(n-5) ··· Ex.1 If Ex.1 is divisible by 6, Ex.1 is also divisible by two prime factors of 6, 2 and 3. (6=2x3) A. If n is odd, (n-5) is even.   If n is even, (n-4) is also even.   Therefore, n(n-4)(n-5) is always a multiple of 2. B. If Exp.1 is divisible by 3, the remainder is 0,1,or 2. If n ≡ 0 (mod 3), n-4 ≡ -4 ≡ -1 (mod 3), and n-5 ≡ -5 ≡ -2 (mod 3), that is: n is a multiple of 3. If n ≡ 1 (mod 3), n-4 ≡ -3 ≡ 0 (mod 3), and n-5 ≡ -4 ≡ -1 (mod 3), that is: (n-4) is a multiple of 3. If n ≡ 2 (mod 3), n-4 ≡ -2 (mod 3), and n-5 ≡ -3 ≡ 0 (mod 3), that is: (n-5) is a multiple of 3. Therefore, n(n-4)(n-5) is always a multiple of 3. CK: If n=1, n(n-4)(n-5)=1(-3)(-4)=12, 12(n=2), 6(n=3), 0(n=4), 0(n=5), 12, 42, 96, 180 … CKD.  Therefore, n³-9n²+20n is divisible by 6 for all integers greater than or equal to 1.

### Divide x4 -2 by x+1

Well!   I think you meant (x^4 - 2) / (x+1)   (x^4 - 2) / (x+1) Writing it in polynomial form with co-efficients x^4 - 2 = 1 x^4 + 0 x^3 + 0 x^2 + 0 x - 2   Now, writing the co-efficients alone, for synthetic division Synthetic Division x c   x^4 x^3 x^2 x c   Value Poly 1 1   1 0 0 0 -2             1 1         1 x^3       0 -1 0 0 -2     REM         -1 -1       -1 x^2         0 +1 0 -2     REM           1 1     1 x           0 -1 -2     REM             -1 -1   -1 c             0 -1     REM Further Division is not Possible. Thus, (x^4 - 2) / (x+1) Quotient: x^3 - x^2 + x -1 Remainder: -1

### find HCF of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in front of them and do the arithmetic: 3-17+27-7-6=0. Did we get zero? Yes, so -1 is a zero and x+1 is a factor. Now use synthetic division: -1 | 3 17..27...7..-6 ......3 -3 -14 -13...6 ......3 14..13..-6 | 0 gives us 3x^3+14x^2+13x-6. Neither 1 nor -1 is a zero. The factors of the constant 6 include 2 and 3. So the zeroes may be 2 or -2, 3 or -3. We find that -3 is a zero, so x+3 is a factor: -3 | 3 14..13..-6 ......3..-9 -15...6  .....3...5...-2 | 0 gives us 3x^2+5x-2, which we can easily factorise (see above). 6x^4+7x^3-27x^2+17x-3=(x-1)(6x^3+13x^2-14x+3)=(x-1)(x+3)(6x^2-5x+1)= (x-1)(x+3)(3x-1)(2x-1) CLUE: Coefficients sum to zero, so 1 is a zero and x-1 a factor. 1 | 6...7 -27..17..-3 .....6...6..13 -14...3 .....6 13 -14....3 | 0 gives us 6x^3+13x^2-14x+3. Try -3 as a factor because the constant is 3 and +3 does not give us zero: -6*27+13*9+14*3+3=-162+117+42+3=0, so x+3 is a factor. -3 | 6..13 -14...3 ......6 -18..15..-3 ......6..-5.....1 | 0 gives us 6x^2-5x+1, which is easy to factorise (see above).   By inspecting the bracketed factors we can see those in common between the two polynomials. There are two, so we take both of them: HCF=(x+3)(3x-1)=3x^2+8x-3

### find HCF and LCM of3x^4+17x^3+27x^2+7x-6;6x^4+7x^3-27x^2+17x-3 by division method.

3x^4+17x^3+27x^2+7x-6=(x+1)(3x^3+14x^2+13x-6)=(x+1)(x+3)(3x^2+5x-2)= (x+1)(x+3)(3x-1)(x+2) CLUE: Write coefficients: 3 17 27 7 -6. If we add them together, do we get zero? No, so 1 is not a zero and x-1 is not a factor. Look at the odd powers of x (17x^3 and 7x). Change the sign in front of them and do the arithmetic: 3-17+27-7-6=0. Did we get zero? Yes, so -1 is a zero and x+1 is a factor. Now use synthetic division: -1 | 3 17..27...7..-6 ......3 -3 -14 -13...6 ......3 14..13..-6 | 0 gives us 3x^3+14x^2+13x-6. Neither 1 nor -1 is a zero. The factors of the constant 6 include 2 and 3. So the zeroes may be 2 or -2, 3 or -3. We find that -3 is a zero, so x+3 is a factor: -3 | 3 14..13..-6 ......3..-9 -15...6  .....3...5...-2 | 0 gives us 3x^2+5x-2, which we can easily factorise (see above). 6x^4+7x^3-27x^2+17x-3=(x-1)(6x^3+13x^2-14x+3)=(x-1)(x+3)(6x^2-5x+1)= (x-1)(x+3)(3x-1)(2x-1) CLUE: Coefficients sum to zero, so 1 is a zero and x-1 a factor. 1 | 6...7 -27..17..-3 .....6...6..13 -14...3 .....6 13 -14....3 | 0 gives us 6x^3+13x^2-14x+3. Try -3 as a factor because the constant is 3 and +3 does not give us zero: -6*27+13*9+14*3+3=-162+117+42+3=0, so x+3 is a factor. -3 | 6..13 -14...3 ......6 -18..15..-3 ......6..-5.....1 | 0 gives us 6x^2-5x+1, which is easy to factorise (see above).   LCM=(x-1)(x+1)(x+3)(3x-1)(x+2)(2x-1) contains a combination of the factors of both numbers. By inspecting the bracketed factors we can see those in common between the two polynomials. There are two, so we take both of them: HCF=(x+3)(3x-1)=3x^2+8x-3

First set of questions is all roots, or solutions, of the equations and second set is to find where the functions are zero, the rational roots only. The methods for roots and zeroes are basically the same. 1. 3x^2+10x+3=3x^2+9x+x+3=3x(x+3)+(x+3)=(3x+1)(x+3)=0; x=-1/3, -3 2. 5x^2+5x-x-1=5x(x+1)-(x+1)=(5x-1)(x+1)=0; x=1/5, -1 3. Cubic equation. Trial shows that x=-1 is a solution, so divide cubic by factor (x+1). Use synthetic division: -1 | 3 11  5 -3      | 3 -3 -8  3        3  8 -3 | 0 and  (x+1)(3x^2+8x-3)= (x+1)(3x^2+9x-x-3)=(x+1)(3x(x+3)-(x+3))=(x+1)(3x-1)(x+3)=0; x=-1, 1/3, -3 4. 3x^4-2x^2-5=(3x^2-5)(x^2+1)=0; 3x^2=5, so x=+/-sqrt(5/3) 5. Trial shows that 3x=-1 is a solution, i.e., x=-1/3. Use synthetic division: -1/3 | 3 10 30   9         | 3  -1 -3  -9           3   9 27 | 0      (3x+1)(x^2+3x+9)=0; x=-1/3 is the only real solution 6. (5x^2+7)(x^2-2)=0; x=+/-sqrt(2) 7. No real roots 8. (5x^2+2)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 9. (5x^2+9)(x^2-4)=(5x^2+9)(x+2)(x-2)=0; x=+/-2 10. (5x^2+3)(x^2-8)=0; x=+/-sqrt(8) or +/-2sqrt(2) 11. f(x)=5x(x^2-1)-(x^2-1)=(5x-1)(x-1)(x+1); x=1/5, 1, -1 12. Try x=1/2, f(x)=0. So f(x)=(2x-1)(x^2-11x-1) has no more rational roots 13. f(x)=3x(x^2-1)-(x^2-1)=(3x-1)(x-1)(x+1); zeroes at x=1/3, 1, -1 14. f(x)=(x+2)(x^2+7x+3); only rational zero at x=-2 15. (x-3)(2x^2+7x-3); rational zero at x=3 16. (x-1)^2(4x-1); x=1, 1/4 17. (x+1)(2x-1)(2x+1); x=-1, +/-1/2 18. (x+1)^2(3x-1); x=-1, 1/3 19. (x-5)(3x^2-7x-75); x=5, no other rational roots 20. (x-3)(2x^2+11x-4); x=3, other roots irrational

### .ABXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXQW x 4 = WQXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXBA

219999...9978*4=879999...9912, making A=2, B=1, Q=7, W=8, X=9. SOLUTION DETAILS  Let's find X first by taking an X within the row of Xs. 4X+c=10c+X, where c is the carryover in the result of multiplying the following X by 4. 3X=9c, X=3c, and c must be between 0 and 3. Therefore X=0, 3, 6, or 9. Now look at W. 4W=10p+A, where p is a carryover. And 4A+q=W where q is a carryover. Both p and q are less than 4. Multiply the last equation by 4: 16A+4q=4W=10p+A, so 15A=10p-4q. Therefore, we need to find values of p and q that are multiples of 15.   0 4 8 12 0 0 -4 -8 -12 10 10 6 2 -2 20 20 16 12 8 30 30 26 22 18 The first column contains 10p values and the first row 4q values. The other cells are the result of subtracting the column header from the row header. Only two cells are divisible by 15: 0 and 30, implying (p,q)=(0,0) or (3,0). We can reject (0,0) because it's likely to lead to the trivial solution where the numbers are zero. So we go for p=3, q=0, making A=2. This tells us that 4W=32 and W=8. And this means that there is no carryover for 4B, therefore B must be less than 3. We can also see that 4Q+3=10c+B, because 4W=32 making the carryover 3. We know B can only be 0, 1 or 2. So 4Q=10c-3, 10c-2, or 10c-1. This has to be divisible by 4, so we can eliminate 10c-3 and 10c-1 which are odd numbers. So B=1. That leaves 4Q=10c-2, and Q=2, c=1 or Q=7, c=3. X is therefore 3 or 9. To recap, we are left with 2133...328*4?=?8233...312 or 2199...978*4?=?8799...912. Only the latter satisfies the arithmetic.

### Use principle of mathematical induction to 3 prove that n3 —n is divisible by 3

Let p(n) = n3 - n i)If n=1 , p(1 )  = 1^3 - 1                  [ 1^3 means 1 cube ]                        = 1 - 1                        = 0                       [ 0 divisible by 3] If n =1 , p(1) is true--------------------( 1) ii)  Assume  that p( m ) is true.--------------------------(2 ) iii) we have to prove that p(m+ 1) is true.   p(m + 1) = (m + 1) ^3 - (m + 1)                      = (m + 1 ) [ (m + 1)^2  -1]                [ (m + 1) is the common factor ]                     = ( m + 1) [ m 2 + 2m + 1 - 1 ]         [ expansion of (m + 1 ) ^2 ]                     = ( m + 1 ) [ m2 + 2m ]                    = ( m + 1 ) m ( m + 2 )                      [ m is the common factor ]  {   divisibility rule of 3 is  If sum of the terms divisible by 3 then it is divisible by 3} sum = m + 1 + m + m + 2            = 3m + 3         = 3 ( m + 1)      it is divisible by 3. Therefore ( m+1) m (m+2 ) is divisible by 3 p(m+1) is divisible by 3 p(m+1) is true---------------------------(3) From (1), (2) and (3) for all values of n , p(n)  = n3 - 3 is divisible by 3

### using synthetic division what it x to the fourth power minus 1 divided x-1

x^4 - 1 = x^4 + 0 x^3 + 0 x^2 + 0 x - 1 1 . | . 1 . 0 . 0 . 0 . -1 .... | ...... 1 . 1 . 1 .. 1 .... ----------------------- ....... 1 . 1 . 1 . 1 .. 0 (x^4 - 1) / (x - 1) = (x^3 + x^2 + x + 1) ========================= NOTE: -1 is a zero of (x^3 + x^2 + x + 1), so you can use synthetic division again to find (x^3 + x^2 + x + 1) / (x + 1) -1 . | . 1 . 1 . 1 . 1 ..... | ..... -1 . 0 . -1 ..... ------------------ ....... 1 .. 0 .. 1 .. 0 (x^3 + x^2 + x + 1) / (x + 1) = x^2 + 1