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# which is greater, 2.6 or 2.60?

decimals questions

## Research, Knowledge and Information :

### What is greater 2/6 or 2/3? - Weknowtheanswer

What is greater 2/6 or 2/3? Find answers now! No. 1 Questions & Answers Place. More questions about Science & Mathematics, Mathematics, what

### What fraction is bigger? • Math Forum

... as you can see it by writing some multiplies of both 2 and 5 as shown below: 2 = 2, 4, 6, 8, ... 3 / 5 = 0.6 and 4 / 6 = 0.66667 so clearly 4 /6 is greater.

### Which is greater 2.6 or 2.36 - Answers.com

Which is greater 2.6 meters 26000 centimerters? 26,000cm > 2.6m 7 people found this useful Edit. Share to: Neodarwinian. 16,472 Contributions.

### Comparing fractions with > and < symbols (video ... - Khan ...

Comparing fractions with > and < symbols. ... So, once again, we could write 3/4 is greater than 2/4. And then finally, I encourage you to pause the video.

### book6 greater less - ALEX - alex.state.al.us

- 2 - Greater Than, Less Than, ... 6. 2 1 7. 6 8 8. 9 2 9. 3 3 10. 10 6 11. 7 10 ... 60 35 29 48 54 55 37 53 ...

### Compare Fractions: Equal Denominator | CoolMath4Kids

Compare Fractions: Equal Denominator. Page 1 of 3. Here's the problem: ... Which is greater: or ? Remember that the denominator tells us how many pieces something ...

### Greater than Calculator | [email protected]

Example problems on Greater than Comparison Back to Top. Is -5 greater than -2? Step 1 : On number line, -2 is allocated on right sideof -5, so -5 is not greater than -2.

## Suggested Questions And Answer :

### 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.

### how many ways can you have coins that total 18p using 1p , 2p, 5p, and 10p coins

in how many ways can you have coins that total exactly 18pence using 1p, 2p, 5p and 10p coins but you may wish to use as many of each sort as you wish. Start with the biggest coin, then the next biggest, etc. This is so that you always end up adding on just single coins of 1p. 10    we can only have 1*10 because 2*10 is greater than 18. 10 + 5   we can only add on 1*5 because adding on 2*5 will make the sum greater than 18 10 + 5 + 2   again we can only add on 1*2 10 + 5 + 2 + 1   our first arrangement  (1*10, 1*5, 1*2, 1*1) Now we modify the 2p intp 2*1p, the 5p into 2*2p + 1p and the 10 p into 2*5p (as well as 2p's and 1p). 10 + 5 + (1+1) + 1    our 2nd arrangement   (1*10, 1*5, 0*2, 3*1) Now modify the 5, in the 1st arrangement. 10 + (2+2+1) + 2 + 1    3rd arrangement    (1*10, 0*5, 3*2, 2*1) 10 + (2+2+1) + (1+1) + 1    etc.   (1*10, 0*5, 2*2, 4*1) 10 + (2+(1+1)+1) + (1+1) + 1    etc.  (1*10, 1*5, 1*2, 1*1) 10 + ((1+1)+(1+1)+1) + (1+1) + 1     (1*10, 0*5, 0*2, 8*1) Now modify the 10, in the 1st arrangement. (5+5) + 5 + 2 + 1   our 7th arrangement     (0*10, 3*5, 1*2, 1*1) (5+5) + 5 + (1+1) + 1               (0*10, 3*5, 0*2, 3*1) (5+5) + (2+2+1) + (1+1) + 1      (0*10, 2*5, 2*2, 4*1) (5+5) + (2+(1+1)+1) + (1+1) + 1    (0*10, 2*5, 1*2, 6*1) (5+5) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 2*5, 0*2, 8*1)  -- 11th arrangment (5+(2+2+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 1*5, 2*2, 9*1) (5+(2+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 1*5, 1*2, 11*1) (5+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 1*5, 0*2, 13*1)   --- 14th arrangememt ((2+2+1)+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 0*5, 2*2, 14*1) ((2+(1+1)+1)+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 0*5, 1*2, 16*1) (((1+1)+(1+1)+1)+((1+1)+(1+1)+1)) + ((1+1)+(1+1)+1) + (1+1) + 1    (0*10, 0*5, 0*2, 18*1) --17th arrangement So, in total there are 17 arrangements of 1p, 2p, 5p and 10p coins to give 18p

### Write an inequality and provide a value that may or may not be a solution to the inequality.

x^2-3x-4>0 is an example of an inequality. (x-4)(x+1)>0; for this to be true the two factors must be both positive or both negative. Both positive: x-4>0, x>4; x+1>0, x>-1. So x must be greater than 4 because 4>-1. Both negative: x-4<0, x<4; x+1<0, x<-1. So x must be less than -1 because -1<4. The solution is that x<-1 or x>4. Substitute x=0. This value is not expected to satisfy the inequality. With x=0 we have -4>0, which is false, as expected. Substitute x=-2: we have 4+6-4>0. True. Substitute x=5: we have 25-15-4>0. True. x=-1 expected to be false. We have 1+3-4=0. False because 0 is not greater than 0.

### how do you solve -x(x-5)(x+5)=0

-x(x-5)(x+5)=0 -x = 0 or x - 5 = 0 or x + 5 = 0 x = 0 or x = 5 or x = -5

### a, b, c are real numbers. such that a^2+b^2+c^2 = 1. prove that 1/2≤(ab+bc+ca)≤1

a^2 + b^2 + c^2 = 1 a, b, and c each have to be <=1 because if either was greater than 1, then a^2 + b^2 + c^2 > 1 a, b, and c each have to be >= -1 because if either was less than -1, then a^2 + b^2 + c^2 > 1 What we have now is:  -1 <= a, b, c <= 1 ab, bc, ca each have to be < 1 because if either was = 1, then both parts (a and b, b and c, or c and a) would have to be 1 and 1 or -1 and -1, which would make the squared parts (a^2 + b^2, b^2 + c^2, or c^2 + a^2) > 1 So the <=1 part of 1/2 <= ab + bc + ca <= 1 is incorrect. Now consider the case of a=1, b=0, c=0.  That satisfies a^2 + b^2 + c^2 = 1.  But does it satisfy 1/2 <= ab + bc + ca ? ab + bc + ca 1(0) + 0(0) + 0(0 = 0 0 is not greater than or equal to 1/2. So the 1/2 <= part of 1/2 <= ab + bc + ca <= 1 is incorrect. Answer:  a^2 + b^2 + c^2 = 1 does not mean 1/2 <= ab + bc + ca <= 1

### is 0.794 greater than or less than 0.05

You can put any zeroes to the begginning of the number before the dot, and you can put any zeroes after the end of the number after the dot. 0.05 = 0.050 = 0.05000000000000000000000 0.794 = 0000000000.794 =00.794000000000 Now you can place the numbers in a way that the dots will be on top of eachother 0000,05000000 0000.79400000 You don't need that many zeroes, you need only one zero here, so the two values will have the same number of digits. 0.050 0.794 Now compare the digits one-by-one from left to right. Whoever will have the first larger digit wins. 0 vs 0, they are equal, go to the next digit. 0 vs 7, 7 is greater than 0, 0.794 wins. 0.794 is greater than 0.05.

### inequalities with less than or greater than symbol

(x-3) / (x+2) >= 0 There are two points to consider: x - 3 = 0 >> x = 3 x + 2 = 0 >> x = -2 Number line: <-----(-2)-----(+3)-----> Let's try something to the left of -2: (x-3)/(x+2) when x = -3 (-3-3)/(-3+2) (-6)/(-1) +1 1 is >= 0, so that section is in. Let's try something between -2 and +3: (x-3)/(x+2) when x = 0 (-3)/(+2) -3/2 -3/2 is not >=0, so that section is out. Let's try something to the right of +3: (x-3)/(x+2) when x = 4 (4-3)/(4+2) 1/6 1/6 is >= 0, so that section is in. Answer:  x <= -2 or x >= +3

### ((5)/(x-1)) - ((2x)/(x+1)) is less than 1

((5)/(x-1)) - ((2x)/(x+1)) < 1 ( 5(x+1)-2x(x-1) ) / (x^2 - 1) < 1 ( 5x + 5 -2x^2 + 2x ) / (x^2 - 1) < 1 ( -2x^2 + 7x  + 5 ) / (x^2 - 1) < 1 -2x^2 + 7x + 5 = 0 x = (-7 +- sqrt(7^2 - 4(-2)(5)) / 2(-2) x = (-7 +- sqrt(49 + 40)) / -4 x = (7 +- sqrt(89)) / 4 x = (7-sqrt(89))/4, (7+sqrt(89))/4 x = about -0.6805, 4.1085 x^2 - 1 = 0 x = -1, 1 We have four places where things can change:  x = -1, -0.6805, 1, 4.1085 Let's test some values:  x = -10, -0.7, 0, 2, 5 f(x) = ((5)/(x-1)) - ((2x)/(x+1)) f(-10) = 5/-11 - (-20/-9) = -5/11 - 20/9 = negative = less than 1 f(-0.7) = 5/-1.7 - (-1.4/0.3) = -5/1.7 + 1.4/0.3 = about 1.7255 = greater than 1 f(0) = 5/-1 - 0 = -5 = less than 1 f(2) = 5/1 - 4/3 = 15/3 - 4/3 = 11/3 = greater than 1 f(5) = 5/4 - 10/6 = 15/12 - 20/12 = -5/12 = less than 1 So we don't want the regions including x = -0.7 and 2, and we do want the regions including x = -10, 0, and 5. Answer: (x can be any of these) x < -1 (7-sqrt(89))/4 < x < 1 (7+sqrt(89))/4 < x

### Find the solution set for each rational inequation. Graph the solution set on a number line.

(x+3)(x-2)/((x+2)(x-1))≥0. We need to look at various intervals marked by -3, -2, 1, 2 on the number line. x≤-3 satisfies the inequality; -3-3. (x-2)/(x^2-3x-10)<0. This is (x-2)/((x-5)(x+2))<0. Disallowed values are 5 and -2. Significant values are -2, 2, 5; x<-2 satisfies; -25 fails.