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what is the definition of algebraic sum

definition of algebraic sum

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Algebraic Sum | Definition of Algebraic Sum by Merriam-Webster


Define algebraic sum: ... (as + or −) according to the rules of addition in algebra compare arithmetical sum See the full definition ...
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Algebraic sum - definition of Algebraic sum by The Free ...


Define Algebraic sum. Algebraic sum synonyms, ... English dictionary definition of Algebraic sum. as distinguished from arithmetical ... Algebraic sum; Algebraic surface;
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Summation - Wikipedia


When it is necessary to clarify that numbers are added with their signs, the term algebraic sum is used. For example, ...
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algebraic sum definition | English definition dictionary ...


algebraic sum definition, algebraic sum meaning ... You can complete the definition of algebraic sum given by the English Definition dictionary with other English ...
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Algebraic Equation | Definition of Algebraic Equation by ...


Define algebraic equation: an equation obtained by equating to zero a sum of a finite number of terms each one of which is a ... Definition of algebraic equation: ...
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Algebraic sum | Definition of Algebraic sum by Webster's ...


Looking for definition of Algebraic sum? Algebraic sum explanation. Define Algebraic sum by Webster's Dictionary, ... Algebraic Algebraic curve ...
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algebraic sum definition | English dictionary for learners ...


algebraic sum meaning, definition, English dictionary, ... You can complete the definition of algebraic sum given by the English Cobuild dictionary with other ...
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Algebraic | Define Algebraic at Dictionary.com


Algebraic definition, of, occurring in, or utilizing algebra. See more. Dictionary.com; Word of the Day; Translate; ... algebraic investigation required writing: ...
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Algebraic sum - definition, etymology and usage, examples and ...


Definition of Algebraic sum in the Fine Dictionary. ... algebraid sum, algebraif sum, algebraiv sum, algebraic aum, algebraic wum, algebraic dum, algebraic xum ...
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Algebraic fraction - Wikipedia


... an algebraic fraction is a fraction whose numerator and denominator are algebraic ... The sum of two proper rational fractions is a proper rational fraction ...
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Suggested Questions And Answer :


If a monic polynomial of degree n has n roots in Z, then..... (elementary algebra)

True. Consider the polynomial factorised into its roots: (x-a)(x-b)...(x-z) where a, b, z, etc. are all in Z, then the expansion of the product of the factors will contain the products, sums of products, of combinations of a, b, ...z. But these quantities are, by definition, contained in Z. The products and sums of products form the coefficients in the expansion, therefore the coefficients are all in Z.
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where do polynomial functions originate from

A polynomial simply means many (poly-) terms (nom-) and is just a made-up word for a description of a sum of terms that are individually expressed as a variable (unknown) usually with a positive whole number exponent and a multiplying constant. So when such terms are strung together in a sum, mathematicians called the sum a polynomial. The simplest polynomial is ax+b because we have an unknown x and constants (numbers) a and b. Such a simple algebraic sum isn't usually called a polynomial because the word is normally reserved for greater powers or exponents of the unknown variable. So the word polynomial was invented as a name to cover all such algebraic sums. Just as the word elephant is used to describe the animal without going into a long description involving size, colour and long nose or trunk. The elephant existed before anyone invented a word for it. The same is true for polynomials. As for when polynomials (as we call them) were first introduced as a mathematical concept, well the date is probably around 300BC, but at that time the word polynomial hadn't been invented even though a descriptive word could have been used. But it seems the ancient Egyptians and Babylonians were solving linear and quadratic equations (which are polynomials) without "knowing" they were working with polynomials. They just called them something else in their own language. Polynomials date back to a time not long after algebra was introduced as a technique for representing unknown quantities. The word algebra is likely to have been derived from the Arabic word al-jabru, meaning restoration and balancing, the process of solving an equation.
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for any n belongs N define Pn as the set containing all primes that are the sum of n consecutive primes

The first few primes are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (P<50). The sum of consecutive primes that are themselves are primes can be identified: p1=1; p2=1+2=3; p4=1+2+3+5=11; p6=1+2+3+5+7+11=29; p8=1+2+3+5+7+11+13+17=59; p10=1+2+3+...+17+19+23+101, ... My understanding of the questions is that P1={ p1 }; P2={ p1 p2 }; P3={ p1 p2 p4 }; P4={ p1 p2 p4 p6 }; P5={ p1 p2 p4 p6 p8 }; P6={ p1 p2 p4 p6 p8 p10 }. P2 and P4 are defined in this list. In Pn (P sub n), n defines the number of elements, and since n is a finite number (natural number belonging to N), the sets are finite by definition. In the sums p sub k it follows that k will be even (apart from p1) because, after 2, all primes are odd, adding another prime will automatically result in an even number, so a further prime must be added to get an odd number which may be prime. However, not all odd numbers obtained this way will necessarily be prime because not all odd numbers are prime. What I'm not sure of is whether P3={ p1 p2 p4 } should in fact be defined as P4 because the highest pk is p4. If this is indeed the case then the list given earlier should define the sets for P1, P2, P4, P6, P8. P10 instead of P1, P2, P3,... Under this assumption, P4={ p1 p2 p4} while P2 is unaffected. However, part 4 suggests that P3 and P5 are definable. This takes us back to the original definition. P3 has only three elements under this definition: p1, p2, p4. If p1, p2 and p4 are themselves considered as sets of the primes comprising the sums, then, assuming duplicates are allowed, we have P3={ {1 } { 1 2 } { 1 2 3 5 } }. Five elements from these are, for example, { 1 1 1 2 2 } which also applies to P5.
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prove that P1 = P?

The first few primes are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (P<50). The sum of consecutive primes that are themselves are primes can be identified: p1=1; p2=1+2=3; p4=1+2+3+5=11; p6=1+2+3+5+7+11=29; p8=1+2+3+5+7+11+13+17=59; p10=1+2+3+...+17+19+23+101, ... My understanding of the questions is that P1={ p1 }; P2={ p1 p2 }; P3={ p1 p2 p4 }; P4={ p1 p2 p4 p6 }; P5={ p1 p2 p4 p6 p8 }; P6={ p1 p2 p4 p6 p8 p10 }. P2 and P4 are defined in this list. In Pn (P sub n), n defines the number of elements, and since n is a finite number (natural number belonging to N), the sets are finite by definition. In the sums p sub k it follows that k will be even (apart from p1) because, after 2, all primes are odd, adding another prime will automatically result in an even number, so a further prime must be added to get an odd number which may be prime. However, not all odd numbers obtained this way will necessarily be prime because not all odd numbers are prime. What I'm not sure of is whether P3={ p1 p2 p4 } should in fact be defined as P4 because the highest pk is p4. If this is indeed the case then the list given earlier should define the sets for P1, P2, P4, P6, P8. P10 instead of P1, P2, P3,... Under this assumption, P4={ p1 p2 p4} while P2 is unaffected. However, the question suggests that P3 and P5 are definable. This takes us back to the original definition. P3 has only three elements under this definition: p1, p2, p4. If p1, p2 and p4 are themselves considered as sets of the primes comprising the sums, then, assuming duplicates are allowed, we have P3={ {1 } { 1 2 } { 1 2 3 5 } }. Five elements from these are, for example, { 1 1 1 2 2 } which also applies to P5.  
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Find solution

The first few primes are 1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (P<50). The sum of consecutive primes that are themselves are primes can be identified: p1=1; p2=1+2=3; p4=1+2+3+5=11; p6=1+2+3+5+7+11=29; p8=1+2+3+5+7+11+13+17=59; p10=1+2+3+...+17+19+23+101, ... My understanding of the questions is that P1={ p1 }; P2={ p1 p2 }; P3={ p1 p2 p4 }; P4={ p1 p2 p4 p6 }; P5={ p1 p2 p4 p6 p8 }; P6={ p1 p2 p4 p6 p8 p10 }. P2 and P4 are defined in this list. In Pn (P sub n), n defines the number of elements, and since n is a finite number (natural number belonging to N), the sets are finite by definition. In the sums p sub k it follows that k will be even (apart from p1) because, after 2, all primes are odd, adding another prime will automatically result in an even number, so a further prime must be added to get an odd number which may be prime. However, not all odd numbers obtained this way will necessarily be prime because not all odd numbers are prime. What I'm not sure of is whether P3={ p1 p2 p4 } should in fact be defined as P4 because the highest pk is p4. If this is indeed the case then the list given earlier should define the sets for P1, P2, P4, P6, P8. P10 instead of P1, P2, P3,... Under this assumption, P4={ p1 p2 p4} while P2 is unaffected. However, (iv) suggests that P3 and P5 are definable. This takes us back to the original definition. P3 has only three elements under this definition: p1, p2, p4. If p1, p2 and p4 are themselves considered as sets of the primes comprising the sums, then, assuming duplicates are allowed, we have P3={ {1 } { 1 2 } { 1 2 3 5 } }. Five elements from these are, for example, { 1 1 1 2 2 } which also applies to P5. You have presented similar questions recently and I can only hope that this well-meant answer helps you to resolve the problem as you intended.
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How do you factor the sum of terms as a product of the GCF and a sum?

It took me a while to interpret your question, and I came up with a polynomial that had a GCF, but it doesn't have to be a polynomial: 36x^2+45x+63 simplifies when we note that the GCF of the numbers 36, 45 and 63 is 9. That means 9 is the largest integer to divide into the numbers: 36=9*4, 45=9*5 and 63=9*7. This means we can rewrite the polynomial as the product of a GCF and a sum: 9(4x^2+5x+7). It utilises the distributive property of numbers. Another example: abc+ace-ca^2. In this case the GCF is algebraic: ac, because all the terms contain ac, so we can factorise: ac(b+e-a). This is the product of the GCF ac and the sum b+e-a. Another example: 4x^2+2xy+6x=2x(2x+y+3). This time the GCF is 2x because the GCF of the numbers is 2 and of the algebraic quantities is x, so we just combine the two GCFs to make 2x. Does this help you in time? If you still don't understand send me a private message explaining your difficulties and providing examples.
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word problem


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Which statements are true? Select all that apply.

True True False True False
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algebraic expression- seven times the sum of a number n and 3

7(n+3) ...............
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how do you write 5 times the sum of a number and 3 is 10 as a algebraic expression

5(x+3)=10 5x+15=10 5x=-5 x=-1
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