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Is -0.19 a integer

Is -19/100 (a fraction) a integer

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Whole Numbers and Integers - Math Is Fun


Whole Numbers and Integers ... {0, 1, 2, 3, ...}, depending on the subject. Integers. ... And some people say that zero is NOT a whole number.
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Integer - Wikipedia


An integer (from the Latin integer meaning "whole") is a number that can be written without a fractional component. For example, 21, 4, 0, ...
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0 - Wikipedia


The number 0 may or may not be considered a natural number, but it is a whole number and hence a rational number and a real number ...
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What is integer? - Definition from WhatIs.com


This definition explains what an integer is: a whole number (not a ... For example, it is intuitive from the list {..., -3, -2, -1, 0, 1, 2, 3, ...} that ...
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Is 0.8 real number, rational number, whole number, Integer ...


0.8 is a rational number, ... Is 0.8 real number, rational number, whole number, Integer, Irrational number? Algebra Properties of Real Numbers Order of Real Numbers.
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Is 0.1 a integer - Answers.com


Answers.com ® WikiAnswers ® Categories Science Math and Arithmetic Numbers Is 0.1 a integer? What would you like to ... 19 people found this useful Anna Rino ...
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Is 0 a rational, irrational, natural, whole, integer or real ...


Is 0 a rational, irrational, natural, whole, ... 19 minutes ago. ... irrational, natural, whole, integer or real number?
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sql - Is "NUMBER" and "NUMBER(*,0)" the same in Oracle ...


Is “NUMBER” and “NUMBER(*,0)” the same in Oracle? Ask Question. up vote 2 down vote favorite. 2. ... So, for example, NUMBER(19) is equivalent to NUMBER(19,0).
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Is 0.19 an irrational number - Answers.com


Is 0.19 an irrational number? SAVE CANCEL. already exists. Would ... An irrational number is a real number that cannot be expressed as a ratio of two integers.
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Integers - Math Goodies


Learn integers with lessons from Math Goodies. ... 0: 282 feet below sea ... Whole numbers less than zero are called negative integers. The integer zero is neither ...
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Suggested Questions And Answer :


relationships between sets of real numbers with visual representation

We can define some sets within the superset of all real numbers: A. Integers  A1. Positive integers or natural numbers, including zero   A1.1. Positive integers with integer exponents greater than zero.   A1.2. Negative integers with even integer exponents greater than zero  A2. Negative integers   A2.1. Negative integers with odd integer exponents greater than zero B. Fractions  B1. Proper fractions   B1.1. Positive proper fractions    B1.1.1. Positive integers with integer exponents less than zero   B1.2. Negative proper fractions    B1.2.1. Negative integers with integer exponents less than zero B2. Improper fractions (including mixed numbers)   B2.1. Positive improper fractions   B2.2. Negative improper fractions C. Irrational numbers  C1. Transcendental numbers (cannot be defined as the root of a fraction or integer)  C2. Integer root of a positive integer for integers>1  C3. Integer root of a positive fraction for integers>1  C4. Fractional root of a positive integer  C5. Fractional root of a positive fraction These are arbitrary sets that can be represented visually as circles. Some circles may be completely isolated from other circles; some may completely contain other circles; some may intersect other circles. Indentation above implies that the circle associated with the lesser indentation contains the whole of the circle with the greater indentation. For example, B1 is completely contained by B, B2.2 is contained in B2, which in turn is contained in B. The separate sets of odd and even numbers could be included. We could add the set of prime numbers, positive integers>1 with no other factors than 1 and the number itself. We could add the set of perfect numbers, integers>1 in which the factors, including 1 but excluding the number itself, add up to the number. We could add factorials (the product of consecutive integers up to the factorial integer itself). These would be included in the superset of positive integers. The visual representation is that the subset is totally enclosed by the superset as a circle inside another circle. An example of interlocking circles is the set of factorials with the set of perfect numbers, where 6 is contained in the overlap. 6 belongs to both sets.  
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From integers 1 up to 10, how many possible sets of integers are to be made, if each set has six integers? And from 10 up to 15.

The number of different sets created by 6 integers from a set of 10 is calculated as follows. Choice of first integer is all 10 integers: 10 Now there are 9 integers left, so there are 10*9=90 sets of 2 integers. Moving on to the third integer, we have 8 and so on. Continuing until we have 6 integers: 10*9*8*7*6*5. But in a set the order isn't important, just the combination so we need to divide by the number of arrangements  of 6 integers = 6*5*4*3*2*1=720. Therefore the number of sets is 10*9*...*5/(720)=210. There are only six integers between 10 and 15 so only one set of 6 integers is possible.
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Derrick is thinking of a negative integer. When he multiplies the integer by itself and then adds three times the integer to the product , he gets 180. What is Derek's integer?

Derrick is thinking of a negative integer. When he multiplies the integer by itself and then adds three times the integer to the product , he gets 180. What is Derek's integer? Our formula is x^2 + 3x = 180 Use the method called "completing the square." In order to factor the left side of the equation, we need to add the square of (1/2) * (b/a) a is the coefficient of x^2; b is the coefficient of x (1/2 * b/a)^2 = (1/2 * 3)^2 = 1.5^2 = 2.25 x^2 + 3x + 2.25 = 180 + 2.25 Factoring the left side of the equation gives (x + 1.5)*(x + 1.5) = 182.25 Notice that x * 1.5 (1.5 is 1/2 of our b/a) plus x * 1.5 gives us 3x. Also, 1.5 * 1.5 gives us 2.25 Take the square root of both sides. x + 1.5 = sqrt(182.25) The square root of 182.25 is +13.5; it is also -13.5 x + 1.5 = 13.5   and  x + 1.5 = -13.5 x = 13.5 - 1.5 = 12, not a negative number or x = -13.5 - 1.5 = -15, the number we want (-15)^2 + 3(-15) = 180 225 - 45 = 180
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x^3+y^3=z^3?

  x^3+y^3=z^3⇒(x+y)(x^2-xy+y^2)=z^3. If x+y=z, then x^2-xy+y^2=z^2; but that implies x^2-xy+y^2=x^2+2xy+y^2, and  3xy=0, which is the trivial solution and is disallowed by the theorem. So x+y needs to be an integer factor of z, x+y=az and x^2-xy+y^2=z^2/a; or x+y=az^2 and the quadratic x^2-xy+y^2=z/a. This respectively implies that (1) x^2-xy+y^2=(x^2+2xy+y^2)/a^2; or (2) x^2-xy+y^2=sqrt((x+y)/a)/a. (1) The equation becomes x^2(a^2-1)-xy(a^2+2)+y^2(a^2-1)=0 for a>1, and, dividing through by a^2-1 we get: x^2-xy(a^2+2)/(a^2-1)+y^2=0. Solve this by completing the square: (x-y(a^2+2)/(2(a^2-1)))^2+y^2-y^2((a^2+2)/(2(a^2-1))^2=0. x-y(a^2+2)/(2(a^2-1))=+ysqrt((a^2+2)^2-4(a^2-1)^2)/(2(a^2-1))= +ysqrt((a^2+2+2a^2-2)(a^2+2-2a^2+2))/(2(a^2-1))=+ysqrt(3a^2(4-a^2))/(2(a^2-1))= +aysqrt(3(4-a^2))/(2(a^2-1)). The square root cannot be negative and a>1, so a=2 and x=y(a^2+2)/(2(a^2-1))=6y/6=y. But this solution is based on the initial assumption that the quadratic could be equal to z/a. Clearly x^3+y^3=2x^3 cannot be z^3 where z is an integer. So the assumption (supposed condition) is false and there are no solutions under (1). (2) If x^2-xy+y^2=z/a and we write z/a=k, where k is a positive integer, then we can rewrite the quadratic: (x-y/2)^2-y^2/4+y^2=k, so (x-y/2)^2=k-(3/4)y^2. If y is an even integer y=2w where w is an integer. Then (x-w)^2=k-3w^2. So (x-w)=sqrt(k-3w^2). The left-hand side is an integer so the right-hand side must also be an integer, and this can only happen if k-3w^2=p^2 where p is an integer >0. So k=p^2+3w^2. x=w+p; y=2w; z=a(p^2+3w^2). We have the three variables in terms of three parameters, a, p and w. For the special case of p=0, we can see the effect of substituting the parametrised values into the quadratic: w^2-2w^2+4w^2=3w^2 and into x+y=3w, the product of which is 9w^3=x^3+y^3. This sum cannot be a perfect cube because 9 is not a perfect cube, therefore there are no solutions to the cubic equation when p=0. When p>0, x+y=3w+p and the quadratic is w^2+2pw+p^2-2w^2T2wp+4w^2=3w^2+p^2. The product is (3w+p)(3w^2+p^2)=9w^3+3wp^2+3pw^2+p^3=(2w)^3+(p+w)^3=y^3+x^3. If y is odd y=2v-1 where integer v>0 and (x-v+(1/2))^2=k-(3/4)(4v^2-4v+1)=k-3v(v-1)-3/4; (x-v)^2+x-v+1=k-3v(v-1) More...
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Divide the integers between 1 and 100 into four sets

multipels av 2...2, 4, 6, 8, 10....100####number integers=(100/2)+1 multipels av 3...3, 6, 9, 12, 15....99##### number integers=round-down(100/3) +1...=34 multipels av 5...5, 10, 15, 20, 25...100#### number integers=(100/5)+1=21
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the product of...

The product of two consecutive integers is 306. Find the integers. The hardest part is understanding what the question is asking and setting up the problem.  Once you have this, it is a piece of cake.  Here's what we do: First what its asking: (Break the question down into segments) The product: Ok this means Multiply. Two consecutive integers: This means that the second number comes right after the first number in a number line. Ex. 1,2,3,4,5... The integers belong to the number 306: Factor 306.                    306                      ^                     2*153                          ^                         9*17                         ^                        3*3 Now look 3 * 3 * 2 = 18 The other factor 306 = 17 These two are consecutive numbers so these are your integers since 18 * 17 = 306. A more Algebraic approach would be to say: two consecutive intergers are multiplied together to make 306 x(x + 1) = 306 x^2 + x = 306 x^2 + x - 306 = 0 (x - 17)(x+18) = 0 x - 17 = 0; x = 17 x - 18 = 0; x = 18 There are your two consecutive integers. 
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Pete challenges his friend Jill to find two consecutive odd integers that have the following relationship.

Two consecutive odd integers can be represented by 2x-1 and 2x+1. Their product is 4x^2-1 and their sum is 4x. So, 4x^2-1=3(4x+6)=12x+18. This is the quadratic: 4x^2-12x-19=0. This has no rational solutions. Or, reading the question slightly differently: 4x^2-1=12x+6, so 4x^2-12x-7=0. In the second case, we can factorise: (2x-7)(2x+1)=0, so x=3.5 or -0.5. The two integers are: 6 and 8, or -2 and 0. None of these are odd, so it isn't possible to find two consecutive odd integers. Why did I think there were two interpretations of the question? "3 times the sum of the integers plus 6" is ambiguous:  is it 3(sum+6) or 3*sum+6? Judging by the result, it was the latter.
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What are the integers?

The greatest of four consecutive even integer is 14 less than twice the smallest integer. What are the integers? Let teh four consectutive even integers be, I1 = 2n, I2 = 2n + 2, I3 = 2n + 4, I4 = 2n + 6 We have, The greatest is 14 less than twice the smallest i.e. I4 = 2*I1 - 14 So, 2n + 6 = 2(2n) - 14 2n = 4n - 20 20 = 2n n = 10 The numbers are: I1 = 20, I2 = 22, I3 = 24, I4 = 26
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State where the least integer function is continuous, left or right continuous, discontinuous

The least integer function is discontinuous because its results are only integers, that is, there are no values other than integers. Its input (domain) is continuous but its range is limited to integers making a graph of the function look like an infinitely long flight of stairs as the function leaps from integer to integer. For one output value there are an infinite number of different inputs. For example, 0.1, 0.1234, sqrt(0.5), 0.5, 1/5, 4/5, etc., have a least integer value of 1, which is the ceiling for all values of x where 0 Read More: ...

3x+2y=4z-5

We can write x=2n+1, where n is an integer. y=2n+3 and z=2n+5, because consecutive odd numbers are separated by 2. 3(2n+1)+2(2n+3)=4(2n+5)-5; 6n+3+4n+6=8n+20-5; 2n=6, n=3. x=2n+1=7, y=9, z=11, assuming x is the first integer (and the smallest), y the second and z the third and the largest). Other solutions, because the question doesn't relate x, y and z to the order of the numbers (e.g., is x the smallest or largest number? Is y the first integer? etc.): 3(2n+3)+2(2n+5)=4(2n+1)-5; 6n+9+4n+10=8n+4-5; 2n=-20, n=-10, so x=-17, y=-15, z=-19. 3(2n+5)+2(2n+1)=4(2n+3)-5; 2n=12-5-2-15=-10, n=-5, so x=-5, y=-9, z=-7. 3(2n+5)+2(2n+3)=4(2n+1)-5; 2n=4-5-15-6=-22, n=-11, so x=-17, y=-19, z=-21. 3(2n+1)+2(2n+5)=4(2n+3)-5; 2n=12-5-10-3=-6, n=-3, so x=-5, y=-1, z=-3. 3(2n+3)+2(2n+1)=4(2n+5)-5; 2n=20-5-9-2=4, n=2, so x=7, y=5, z=9.    
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