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how do you write the product of 8 more than a number and 5 less than the number as an expression

write as a mathematial expression

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Writing Expressions (solutions, examples, videos)


Expression . 6 more than 5 times a number . ... 4 times the sum of a number and 7 . 4(y + 7) 5 less than the product of 3 and a number . ... How Do You Write ...
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writing one-operation equations, expressions, statements


writing one-operation equations, expressions, statements ... two less than a number: x ... I. Write the expression/equation then check the answer.
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Writing Algebraic Expressions


Writing Algebraic Expressions is presnted by ... five more than twice a number: 2n + 5: the product of a ... Write an algebraic expression to represent his ...
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What is the algebraic expression for 5 less than a number?


What is the algebraic expression for 5 less than a number? ... How do you write 24 less than three times a number in algebraic expression? ... More . Home New ...
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Writing Expressions and Equations


A number increased by 5 x 5 b. 7 less than a number x 7 c. 3 more than ... Lesson 7.1 Writing Expressions and Equations 319 Write the ... 7 8. The product of a number ...
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How To Write 8 More Than A Number X - answers.com


How To Write 8 More Than A Number X? ... How do you write 5 less than the number X? ... than 3x+8 is the algebraic expression for 8 more than triple a number.
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Name: Unit3 #2 Writing Expressions and Equations


Writing Expressions and Equations ... Subtraction Phrases Expression 8 more than a number ... is 7. __2n – 5 = 7____ 8. One less than the product of four and a ...
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Translating Verbal Expressions – Terms - Del Mar College


Translating Verbal Expressions – Terms . ... The product of four times a number and negative two is five 4x ... = 5 8. Three less than a number is greater than two ...
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Numerical Expression - Math Dictionary - ICoachMath.com


... the numerical expression, ... More About Numerical Expression. A numerical expression represents a particular number. For example, the numerical expression 4 ...
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Suggested Questions And Answer :


how do I write an algebraic expression for fourth grade pre-algebra?

1.) The sum lets you know its addition, a number could be anything we don't know and 8.   we will let x represent any number ( you can use any expression)  x+8 2.)  5 < x     3.)  product lets you know its multiplication, again one is unknown and the other is 9. you can write this like 9x or 9(x) or 9 x X 4.) this is the same as the one above, 4x or 4 x X (IF YOU DONT LIKE THE LETTER X YOU CAN USE ANY LETTER IT DOESNT MATTER like 4 x b) 5.) quotient means division: 18 divided by x or 18/x
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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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the product of five less than a number and seven. translate into a variable expression and simplify

Five less than a number means x - 5 Product of x - 5 and 7 means (x - 5) * 7 or just: 7(x-5)
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how do you write the product of 8 more than a number and 5 less than the number as an expression

(x+8)*(x-5) = x^2-5x+8x-40 =x^2 +3x -40
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please help me

Question: find 5 consecutive integers such that 3 times the product of the 3rd and 4th, minus the the product of the 2nd and 4 less than the first, plus 3 times the 5th is 232. Variables are x, x +1, x + 2, x + 3, x + 4 ? Let us rewrite your problem statement. find 5 consecutive integers such that 3 times the product of the 3rd and 4th, minus the product of the 2nd and 4 less than the first, plus 3 times the 5th is 232. Variables are x, x +1, x + 2, x + 3, x + 4, or v1, v2, v3, v4, v5 Turning those statements above into mathematical expressions, 3 times the product of the 3rd and 4th,                            becomes   3*v3*v4 minus the product of the 2nd and (4 less than the first),  becomes   - v2*(v1-4) plus 3 times the 5th                                                          becomes   + 3*v5 is 232.                                                                               becomes   = 232 As a single expression, this is 3*v3*v4 - v2*(v1 - 4) + 3*v5 = 232 3(x+2)(x+3) - (x+1)(x - 4) + 3(x+4) = 232 3(x^2 + 5x + 6) - (x^2 - 3x - 4) + (3x + 12) = 232 2x^2 + 21x + 34 = 232 2x^2 + 21x - 198 = 0 (2x + 33)(x - 6) = 0 x = -33/2, x = 6 But x is an integer, so the only solution is x = 6 The numbers then are : 6, 7, 8, 9, 10
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Shawn took in $69.15 in two hours work on Saturday selling Belts for $8.05 and earrings for $4.50. How many of each did Shawn sell?

Shawn took in $69.15 in two hours work on Saturday selling Belts for $8.05 and earrings for $4.50. How many of each did Shawn sell? I'm not too sure how ​you are expected to solve this problem. You will end up with a Diophantine Equation (one involving integer coefficients and integer solutions only) If you have never heard of Diophantus, then the precise mathematical solution later on should be ignored. Instead ...   The (Diophatine) equation is 161B + 90E = 1363 You should be able to show that Shawn can only sell a maximum of 8 belts (i.e. if he sells no earrings), and a maximum of 15 earrings (i.e. if he sells no belts). So start with the smaller number (8). Shawn sells up to 8 belts. So put B = 1 in the (Diophantine ) equation and see if B is an integer. If not, then B=1 cannot be a solution, So now try again with B = 2, and so on ..     My precise mathematical solution now follows. Let B be the number of belts sold @ $8.05 per belt. Let E be the number of earrings sold @ $4.50 per pair. Total money made is P = $69.15 Total money made is: B*$8.05 + E*$4.50 Equating the two expressions, 8.05B + 4.50E = 69.15 Making the coefficients as integers we end up with a Diophantine equation. 161B + 90E = 1383 We now manipulate the coefficients 161 and 90. We should end up with a relation between the two that should produce the value 1. This manipulation is in two parts. In the first line, we are writing the larger coefft, 161, as a multiple of the smaller coefft plus a remainder. 1st Part 161 = 1x90 + 71  90 = 1x71 + 19     now we write the 1st multiple as a multiple of the 1st remainder + remainder 71 = 3x19 + 14    we write the 2nd multiple as a multiple of the 2nd remainder + remainder 19 = 1x14 + 5 14 = 2x5 + 4 5 = 1x4 + 1 ============ 2nd Part We rewrite that last equation so that 1 is on the left hand side 1 = 5 – 1x4 We now use the remainder from the next equation up. 1 = 5 – 1x(14 – 2x5) 1 = 3x5 – 1x14 And again we use the remainder in the next equation up 1 = 3(19 – 1x14) – 1x14 1 = 3x19 – 4x14 1 = 3x19 – 4(71 – 3x19) 1 = 15x19 – 4x71 1 = 15(90 – 1x71) – 4x71 1 = 15x90 – 19x71 1 = 15x90 – 19(161 – 1x90) 1 = 34x90 – 19x161 ================ Therefore, 1383 = (-19x1383).161 + (34x1383).90 1383 = -26,277*161 + 57,022*90 Which implies B = -26,277 E = 47,022 which is obviously incorrect! But all is not lost. These two values for B and E are simply two values that happen to satisfy the original Diophantine equation. What we need is the general solution, which now follows. In general, B = -26,277 + 90k    (using the coefft of E) E = 47,022 – 161k    (using the coefft of B) If you substitute these expressions into the original Diophantine equation, the k-terms will cancel out, leaving you with the original eqn. We know that B and E must be smallish numbers and that neither can be negative. For example, if no earrings were sold, Shawn would need to sell over 8 belts to clear $69, and if no belts were sold, then Shawn would need to sell over 15 earrings. So Shawn needs to sell between 0 and 8 belts and between 0 and 15 earrings. Setting k = 292, B = -26,277 + 90*292 = 3 E = 47,022 – 161*292 = 10 B = 3, E = 10 If k is greater than or less than 292 by 1, or more, then either B will be negative or E will be negative. So this is the answer. Check 3*8.05 + 10*4.50 = 25.15 + 45.00 = 69.15 – Correct! Answer: Shawn sells 3 belts and 10 earrings
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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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i need help with finding the answer for the square root of 34

The only step by step way of finding the square root of a number which doesn't radicalise is to use an arithmetic technique as follows: We know that the answer is close to 6 because 6 squared is 36, so there will be decimals. We write digits in pairs thus: | 34 |•00 | 00 | 00 | 00 The answer is in single digits for each pair, so we will be writing the answer over the top. The decimal point is inserted in front of the digit that will go over the first pair of zeroes. I've put in 4 pairs of zeroes so we'll be working out 4 decimal places only. We know the first digit in the answer is 5, because the answer is a little less than 6, so we write 5 over 34. We write the square of 5, i.e., 25, under 34 and subtract it: ..........5...• 8.....3.....0.....9 ...... | 34 |•00 | 00 | 00 | 00 .........25 108 |...900 ...........864 1163 |...3600 .............3489 ................11100 116609 |..1110000 ................1049481 ....................60519 I'm using dots to space out and align the working so that you can see what I'm doing. Now we double 5 to make 10 and we need to guess what digit should follow 10 so that when we multiply by the digit we get a number which is smaller than or equal to the remainder so far, 900. The digit we need to add is 8, so we write 108 and multiply by 8=864, which is a little less than 900. We subtract 864 from 900 leaving 36 and we pull down another pair of zeroes. We write 8 next to the decimal point, so now we have 5.8. The next step is to double the answer we have so far ignoring the decimal point, so we have 116, and again we need to put a digit on the end of this and multiply by the same digit so that the product is as close to 3600 as we can get without exceeding it. Let's guess 3, so we have 1163*3=3489 (note that 4 would be too big). The remainder is 111 and we bring down the next pair of zeroes to make 11100. We carry on. Double the answer so far=1166. What digit are we going to add this time? The smallest is 1, but 11661 is bigger than the remainder 11100, so we have to put zero in the answer and bring down the next pair of zeroes. Double the answer 11660. Now we can add a digit, which is going to be high, so we'll pick 9 and multiply it to give 116609*9=1049481. We could carry on repeating the same process, bringing down pairs of zeroes as necessary. There are other ways of finding square roots, but I find this one fairly straightforward and it doesn't require a calculator. I think this method is based on Newton's method.
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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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the product of 15 and number, multiplied by itself four times(expression)

If the number is n, then when multiplied by itself four times we write n^4, which is usually printed as n with a small number 4 as a superscript (a number placed to the right of n but positioned a little higher than it). The product of a number and a symbol like x, n, y, etc., is not usually shown with a multiply sign (*) but is written immediately before the symbol, so 15n means 15*n. When we combine as a product 15 and n^4 we get 15n^4.
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