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the product of 12 and the difference between 4 and -8

the product of 12 and the difference between 4 and -8      

Research, Knowledge and Information :

What is the difference between the product of 12 and 8 and ...

What is the difference between the product of 12 and 8 and the sum of 12 and 8? ... What Is the difference for between 8 and 12 equals 4?
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4 × 3 = 12. In the product below, ... Since there were an odd number of integers, the product is the opposite of 1320, which is -1320, so 4 × (-2) × 3 × ...
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IXL - Find two numbers based on sum, difference, product, and ...

Improve your skills with free problems in 'Find two numbers based on sum, difference, product, and quotient' and thousands of other practice lessons. Sign in Remember.
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Write an algebraic expression to represent - Harmony

Write an algebraic expression to ... the product of 12 and the sum of a number and negative 3 $16:(5 the difference between the product of 4 and a number and ...
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Mathematical Phrase Translation - Cerritos College

Two-thirds of the product of 8 and x is the same as 12. ... Mathematical Phrase Translation ... 2 x + 7 or 7 2 x 4. The difference of a number and 8.
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Name: Unit3 #2 Writing Expressions and Equations

Writing Expressions and Equations ... The product of 4 and the difference of x and 3 The product of ... The product of 2 and the sum of 5 and t is 8. ___2(5 + t) = 8 ...
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Factors, Multiples, and Divisors

Factors, Multiples, and Divisors. ... the possible natural number products that can be formed where the product is twelve. That is, 1 × 12, ... 2, 3, 4, 6, 8, 12, 24 ...
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Fractions - Math League

The fractions 1/2, 2/4, ... 3/4 + 1/6 = 9/12 + 2 ... the result is a fraction with a numerator that is the product of the fractions' numerators and a denominator ...
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Translating Verbal Expressions – Terms

Translating Verbal Expressions – Terms . ADDITION: ... the difference between . x. and . y. and. ... The product of four times a number and negative two is five 4x ...
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Translating Words into Mathematical Symbols

Translating Words into Mathematical Symbols ... The sum of 4 times a number n and 7 4n + 7 The product of n ... The absolute value of the difference between x and ...
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Suggested Questions And Answer :

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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solve by factorisation the following equation (1).5x^2+2x-15=0 (2).21x^2-25x=4 (3).10x^2+3x-4=0

  (1) 5x^2+2x-15=0 I suspect that this should be 5x^2+22x-15=0, because the equation as written does not factorise. Write down the factors of the squared term coefficient and the constant term. We write these as ordered pairs (a,b) squared term (c,d): (1,5) and constant: (1,15), (15,1), (3,5), (5,3). The sign of the constant (minus) tells us that we are going to subtract the cross-products of  factors. If it had been plus we would be adding the cross-products. Now we create a little table: Quadratic factor table a b c d ad bc  |ad-bc| 1 5 15 1 1 75 74 1 5 1 15 15 5 10 1 5 3 5 5 15 10 1 5 5 3 3 25 22 The column |ad-bc| just means the difference between ad and bc, regardless of it being positive or negative, just take the smaller product away from the larger. If the coefficient of the x term is in the last column, then that row contains the factors you need. If it isn't in the last column, then the quadratic doesn't factorise, or you've missed some factors. If 22x is the middle term, then the factors are shown in the last row and (a,b,c,d)=(1,5,5,3). Now we look at the sign of the middle term. Whatever the sign is we put it in front of the number c or d for the larger cross-product. The cross-products are bc and ad. In this case, bc is bigger than ad so the + sign goes in front of c. The sign in front of the constant tells us whether the sign in front of d is different or the same. If the sign is plus it's the same, otherwise it's the opposite sign. So in this case, it's minus, so the minus sign goes in front of d and we have (ax+c)(bx-d) (note the order of the letters!) or (x+5)(5x-3), putting in the values for a, b, c and d. If (x+5)(5x-3)=0, then x+5=0 or 5x-3=0. So in the first case x=-5 and in the second case 5x=3 so x=3/5. (2) 21x^2-25x-4=0 (moving 4 over to the left to put the equation into standard form). Quadratic factor table a b c d ad bc  |ad-bc| 1 21 1 4 4 21 17 1 21 4 1 1 84 83 1 21 2 2 2 42 40 3 7 1 4 12 7 5 3 7 4 1 3 28 25 3 7 2 2 6 14 8 The row in bold print applies. The sign in front of 4 is minus so the signs in the brackets will be different. 28 is the larger product, so since we have -25x, the minus sign goes in front of c (4) and plus in front of d (1): (3x-4)(7x+1)=0. So the solution is x=4/3 or -1/7. (3) 10x^2+3x-4=0. Quadratic factor table a b c d ad bc  |ad-bc| 1 10 1 4 4 10 6 1 10 4 1 1 40 39 1 10 2 2 2 20 18 2 5 1 4 8 5 3 2 5 4 1 2 20 18 2 5 2 2 4 10 6 (2x-1)(5x+4)=0, so x=1/2 or -4/5.  
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Give two decimals whose product is 18.354 and whose difference is o.17?

Give two decimals whose product is 18.354 and whose difference is o.17? Give two decimals whose product is 18.354 and whose difference is o.17. y = x + 0.17 x * y = 18.345 x * (x + 0.17) = 18.345 There  are two sets of two numbers that satisfy the requirements, because x will have two values, one positive and one negative. x^2 + 0.17x = 18.345 x^2 + 0.17x + 0.085^2 = 18.345 + 0.085^2 x^2 + 0.17x + 0.085^2 = 18.345 + 0.085^2 x^2 + 0.17x + 0.085^2 = 18.352225 (x + 0.085)^2 = 18.352225 x + 0.085 = 4.2839 x = 4.2839 - 0.085    x = -4.2839 - 0.085 x = 4.1989            x = -4.3689 y = 4.369              y = -4.201 Due to rounding errors, the multiplication and subtraction will not produce exact results, but they are within reason. x = 4.1989, y = 4.369 x * y = 18.34499 y - x = 4.369 - 4.1989 y - x = 0.1701 x = -4.3689, y = -4.201 x * y = -4.3689 * (-4.201) = 18.3537 y - x = -4.201 - (-4.3689) y - x = 0.1679  
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A math word problem using the Quadratic Equation formula (URGENT)

A. The Golden Ratio I have a line divided into two unequal sections so that the ratio of the shorter length to the longer length is the same as the ratio of the longer length to the whole length of the line. What is the ratio? Solution Call the lengths of the two sections A and B. A/B=B/(A+B). Cross-multiply: A^2+AB=B^2, so A^2+AB-B^2=0 and, using the quadratic formula, A=(-B+sqrt(B^2+4B^2)/2=(-B+Bsqrt(5))/2=B(-1+sqrt(5))/2. Therefore the fraction A/B=(sqrt(5)-1)/2=0.618 or -1.618 approx. The positive root applies because A and B are considered positive lengths. An alternative solution is to let r=A/B, which is the ratio, then r=1/(r+1); r^2+r-1=0 and r=(sqrt(5)-1)/2.  This Golden Ratio keeps popping up in different contexts and seems to be aesthetically pleasing. B. The A sizes A rectangular piece of paper is folded in half so that the ratio of length to width remains the same after folding as before folding. What is the ratio? Solution If A and B are the length and width, the area is AB, and after folding in half the area becomes 1/2(AB). In the process of folding, only one side (choose the length A) is halved so the new dimensions of the rectangle are A/2 and B, but the ratio of the sides is still A/B, and the length and width are B and A/2 (the length has become the width and the width has become the length). So A/B=length/width=B/(A/2)=2B/A. If A/B=2B/A then A^2=2B^2. The ratio of the sides is A/B which we'll call r. So r^2=2 and r=sqrt(2)=1.414 approx. So the length is 1.414 times the width. The paper we use today is largely based on this ratio. A4 is the standard size used for printers. A3 is larger and tends to be used for drawings and posters. A5 is smaller and tends to be used for booklets and leaflets. All use the square root of 2 as the ratio of the sides. C. Identity Prove that the difference between the cubes of two consecutive integers is one more than three times their product. Solution The difference between the cubes of consecutive integers (x+1)^3-x^3 where x is an integer. The product of consecutive integers is x(x+1). The difference between the cubes is x^3+3x^2+3x+1-x^3=3x^2+3x+1=3x(x+1)+1 which is three times the product of the integers plus 1. These questions use in different ways a quadratic expression. Although this reply is later than you wished for, what goes round comes round, and you may have the opportunity of using the ideas.
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write in mathematical expression the difference between n and the product of 5 and n

The product of 5 and n = 5n The difference between n and the product of 5 and n:  n - 5n = -4n
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among all pairs of numbers with a sum of 224, find the pair whose product is maximum.

The pairs of numbers with a sum of 224 are (1,223), (2,222), (3,221), etc. The product is maximum when x(224-x) is maximum. This happens when x=112, so the product is 112^2=12544. If the numbers have to be different then one is 111 and the other 113, with a product of 12543. There is no need to introduce fractions because the maximum product is 12544 made up of 112*112 which are not fractions. If the numbers are to be different then one number would be 112-a and the other 112+a, where a is the fraction which can be as small as you like.
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Factor each expression 3 x^2 - 2x - 8

The factors of 3x^2 are 3x and x and the factors of 8 are 8 and 1, or 4 and 2. -8 tells us that there must be a plus sign in one bracket (one factor) and minus in the other. It also tells us that we need to look at the difference between cross products, and not their sum, and the minus sign tells us that the larger of the two cross products must be negative. So we have (3x 1)(x 8) or (3x 8)(x 1) or (3x 2)(x 4) or (3x 4)(x 2) where the sign is yet to be determined. The correspondence no cross product differences are: 23, 5, 10, 2. One of these matches the x coefficient, so since this coefficient is 2 we know that the last in the list of possible factors is the right one, and the larger cross product is 3*2=6 so the minus sign goes in front of 2: (3x+4)(x-2) is the answer.  
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the difference between the sum of 9 and 9 and the product of 9 and 9

(9 + 9) - ( 9 * 9) 18 - 81 = 63  take the absolute value for the difference.
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A plane which is perpendicular to the two planes 2x-2y+z=0 and x-y+2z=4, passes through (1,-2,1). The distance of the plane from the point (1,2,2) is

Let P1 and P2 be the planes 2x-2y+z=0 and x-y+2z=4 respectively. The normal vector to P1 is (2,-2,1) and to P2 is (1,-1,2). The perpendicular plane P3  has to contain point (1,-2,1). The equation of P3 can be written a(x-1)+b(y+2)+c(z-1)=0 being the scalar dot product of its normal vector (a,b,c) and the difference of the general point vector on P3 (x,y,z) and vector of point (1,-2,1). The scalar product of P1 normal with P3 normal must be zero because they are perpendicular, since P1 and P3 are perpendicular. That is, (2,-2,1)•(a,b,c)=0. That is, 2a-2b+c=0. Similarly P2 and P3 normals: (1,-1,2)•(a,b,c)=0, so a-b+2c=0. Doubling this: 2a-2b+4c=0, therefore, since 2a-2b=-c from the above equation c=0, and a=b, and a(x-1)+a(y+2)=0⇒x-1+y+2=0, x+y+1=0 is the equation of the plane P3, with the normal (1,1,0). The magnitude of the normal (1,1,0) of P3 is √(1^2+1^2+0^2)=√2. If we take any other point (x,y,z) on the plane P3, we can work out the difference vector between this and the required point (1,2,2). We know that x+y=-1 for all points on P3, so n • r=-1. The dot product n • (r - r0) where r=(x,y,z) and r0=(1,2,2), gives the projection on to the normal: n • (r - r0) = n • r - n • r0=-1-(1*1+1*2)=-4. We now have to divide this by the magnitude of the normal, √2, to give -4/√2=-4√2/2=-2√2. We need this as a magnitude, or absolute value,=2√2 or 2*2^0.5 or 2(2^0.5).
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Identify the factors of 25 and 42. The idea is to find a combination that leads to a difference of 5, the coefficient of the v term. 25 has factors 1 and 25, and 5 and 5; 42 has 1 and 42, 2 and 21, 3 and 14, 6 and 7. The coefficient 5 is a small number so it's unlikely we'll need large numbers like 25, 42 and 21. Let's go for numbers of similar size: 5 and 5, and 6 and 7 might work. Let's imagine a game between two teams. The first team is 5 and 5 and the other team is 6 and 7. So 5 plays against 6 and its partner 5 plays against 7. Play consists of the product of the opponents, so we have 5*6=30 and 5*7=35. The difference is 5. So we nearly have the answer. By the way, the reason we subtract these products is because the sign on the constant term is negative or minus, which means subtract. If it had been + we would have added the products. So we have (5v 6)(5v 7), but what signs go in the brackets? We look at the sign in front of the middle term. It's +, so this sign must be in front of the larger of 6 and 7. If it had been minus, we put minus in front of the larger number. Answer is (5v-6)(5v+7). Note that each brackets contain opponents in the game. (When doing the algebra note that 5v in the first bracket multiplies (plays) the 7 in the second bracket. Imagine a game where the players play diagonally across the playing court.)
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