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what is the algebraic expression for twice the product of -1 and 6, subtracted from -4

what is the algebraic expression for twice the product of -1 and 6, subtracted from -4    

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Writing Algebraic Expressions


Writing Algebraic Expressions is presnted ... five more than twice a number: 2n + 5: the product of a ... An algebraic expression is a mathematical expression ...
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Algebraic expressions - A complete course in algebra


When numbers are added or subtracted, ... 3x + 1 . Now an algebraic expression is not a sentence, ... Twice the product of two numbers is twenty.
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Algebraic Translations - MathBitsNotebook(A1 - CCSS Math)


... algebraic expressions do not contain an ... to properly group terms in your algebraic expression. ... rectangle whose length is twice its width decreased by 6.
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1-6 - mathppt.com


5 ×4 product =20 of 5 and 4 is 20. ... subtracted from less twice ()×2 ... Lesson 1-6 (cont.) Math 6 Slide Show: ...
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Translating Words into Algebra - Leeward Community College


Translating Words into Algebraic Expressions Operation Word Expression Algebraic Expression Addition Add, ... the product of,
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Translating Verbal Phrases to Algebraic Expressions


Translating Verbal Phrases to Algebraic ... (the first number subtracted from the ... Expression Division Phrases Expression Twice a number The product of 2 ...
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Translating Algebraic Expressions - OU Campus Login


Translating Algebraic Expressions M110.1 . ... product, multiply by, twice, double, ... 4 subtracted from 8 d) The quotient of 6 and 3 e) ...
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Name: Unit3 #2 Writing Expressions and Equations


Writing Expressions and Equations ... phrase as an algebraic expression. 1. ... Expression Division Phrases Expression Twice a number The product of 2 ...
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Suggested Questions And Answer :


what is the algebraic expression for twice the product of -1 and 6, subtracted from -4

product of -1 and 6: -1 * 6 twice the product of -1 and 6: 2(-1 * 6) twice the product of -1 and 6, subtracted from 4: 4 - 2(-1 * 6)
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With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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parimeter of a rectangle

L = 2W a).  P (peremeter) = 2(L + W) P = 2(2W + W) P = 2(3W) P = 6W b). A (area) = LW A = 2W(W) A = 2W^2 c).  L = 20 inches, W = 12 inches P = 2(L + W) P = 2(20 + 12) P = 2(32) = 64 inches d).  L = 20 inches, W = 12 inches A = LW A = 20*12 = 240 sq inches
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if a,b,c,d are four non-negative real numbers and a+b+c+d=1,then the maximum value of ab+bc+cd is = ?

Consider the expression ab+bc+cd. If a and d were zero then b+c=1, so let b=c=1/2. Then the expression=1/4. Note that b and c appear twice: b in ab and bc, and c in bc and cd, while a and d appear only once each. Therefore b and c contribute more to the sum than a and d. Although none of the quantities can be zero, a and d can be sufficiently close to zero so that the expression approaches 1/4. If b=1/2-x and c=1/2+x then bc=1/4-x^2 so the product is always <1/4. If b=1/2-x and a=x, and c=1/2-x and d=x, where x is very small and positive, a+b+c+d=1. The expression becomes: 2x(1/2-x)+1/4-x+x^2=x-2x^2+1/4-x+x^2=1/4-x^2. Therefore the maximum value is <1/4, but x can be made infinitesimally small. This makes the maximum 0.25. Example: x=1/100: a=d=0.01; b=c=0.49; expression=0.0049+0.2401+0.0049=0.2499. Example: x=1/1000: expression=0.249001+2*0.000499=0.249001+0.000998=0.249999.  
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Which statements can be used to write an algebraic expression to represent the phrase?

1 3 4 5
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A student translated the phrase below into an algebraic expression, and then evaluated if for y = 24. Is the student's work correct?

1st answer. Student should have subtracted, y-6; 24-6=18.
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how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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How do I write the algebraic expression for the verbal expression subtract 31 from a number?

x - 31
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What are the 4 conditions considering algebraic expressions as polynomial

A polynomial is really an algebraic expression that consists of two or more monomials. A monomial is a number, a variable, or a product of a number with non-negative exponents. Some examples of polynomials are:  2x+3y-1;3bc+7cd-y;7mny-my+n  To be exact, polynomials is any expression with the lowest power of unknown as 0 or more. 6z+5, 5x4+9x-1 are some examples of polynomials. However, if the unknown is in the denominator, it is not a polynomial. Also, if the expression has a negative power like 3-2, it will become a fraction when you remove the negative sign (3-2 = 1/32).-42 is a polynomial because it is also -42x0 except that people do not need to write the x0 since x0 = 1 and when you multiply -42x0, it still ends up becoming -42.
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Translate both into algebraic expressions: A. 5 more than a number C. B. 12 minus the product of a number P and 4.

A. C+5 B. 12-4P
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