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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 ...

### 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.

### 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 ...

### 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 ...

### 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 ...

### 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.

### 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 ...

### 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 ...

### 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 ...

## 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

### 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)

### 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

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

### 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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### 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.