Guide :

what 2 numbers can be multiplied to get the number 8 and the sum of negative 4

What is the procedure to find the answer

Research, Knowledge and Information :


What two numbers multiply to 8? - Research Maniacs


What 2 numbers do you multiply to get 8? In other words, what number can you multiply with another number to get 8? The formula to solve this would be:
Read More At : researchmaniacs.com...

Multiply Two Numbers - WebMath


Multiply Two Numbers - powered by WebMath. ... because it can show you how to multiply together very, very large numbers that can have up to 10 digits each!
Read More At : www.webmath.com...

How to multiply 4 numbers by two numbers - Quora


How do I multiply 4 numbers by two numbers? ... Why does a negative number multiplied with ... Why do we always count things as the sum of two numbers less than 4?
Read More At : www.quora.com...

Multiplication - Wikipedia


Any number multiplied by zero is zero. ... is the sum of M copies of N when N and M are positive whole numbers. ... Generalization to negative numbers can be done by ...
Read More At : en.wikipedia.org...

What two numbers multiply to 41? - Research Maniacs


What 2 numbers do you multiply to get 41? In other words, what number can you multiply with another number to get 41? The formula to solve this would be:
Read More At : researchmaniacs.com...

php - Find what 2 numbers add to something and multiply to ...


... efficient way to find what two numbers multiple to a specified number, ... Negative numbers must ... $a + $b == $sum is always true, and multiply $a and $b ...
Read More At : stackoverflow.com...

Multiplying Negatives Makes A Positive - Math Is Fun


Multiplying Negatives When We Multiply: ... "−" is the negative sign. When a number has no ... What About Multiplying 3 or More Numbers Together? Multiply two at a ...
Read More At : www.mathsisfun.com...

How to get two negative values multiplying in negative - Quora


How do I get two negative values multiplying in negative? ... What is the sum of two negative numbers? ... Why does a negative number multiplied with another negative ...
Read More At : www.quora.com...

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  
Read More: ...

how far apart are the fractions 8/11 and -4/5 on a number line

On a number line you can put zero in the middle, so that positive numbers are to the right of zero and negative numbers to the left. The distance from zero of the negative number is 4/5 and the distance from zero to the positive number is 8/11, so the distance between the numbers is the sum of the distances on either side of zero is 4/5+8/11. We can work this out by finding a common denominator of 5 and 11. They both go into 55. 5 goes into 55 11 times and we have 4 fifths so we need to multiply 11 by 4 to get 44/55. 11 goes into 55 5 times and we have 8 elevenths so we multiply 8 by 5 to get 40/55. Now we have both fractions with the same denominator. We add these together: 44+40 giving us 84/55. We can write this as 1 (because we can divide 55 into 84 just once) and 84-55=29 remainder, that's 29/55 so the answer can be written 1 29/55. This means that if we make a number line divided into 110 equal divisions with zero right in the middle, we count left from zero 44 divisions and right from zero 40 divisions to represent the two numbers. We have 84 divisions in between, which is 84/55. That's 55+29 or 1 whole plus 29/55.
Read More: ...

(2)(2)(-3)(4) - (3)(2)(-3)

Note the minus signs in the brackets, but come back to them later. Multiply the numbers in the first group of brackets: 48. Now multiply the numbers in the second set: 18. We had one minus sign in the first group, so we need to make the first number negative: -48. And there is a minus sign in the second group: -18. The minus sign between the groups negates the minus sign in the second group changing subtract to add. The sum becomes: -48 + 18. This is like saying what is 18 degrees warmer than 48 degrees below zero? The answer is 30 degrees below zero, or -30. That's your answer!
Read More: ...

10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
Read More: ...

0,2,4,4,6,8,9,11,13,15 what are the five numbers?

Call the 5 numbers A, B, C, D and E. We assume these to be integers, since there's no indication otherwise. The sums are A+B, A+C, A+D, A+E, B+C, B+D, B+E, C+D, C+E, D+E. Let's look at the sum zero. To get the sum of two number to be zero, they're either both zero or one is the negative of the other. Let's assume there are no negative numbers, so two numbers must be zero. So A=B=0. 6 of the remaining numbers involve A or B, so the sums would be C, D, E, C, D, E, i.e., three pairs of identical sums. But we only have one pair, so the assumption of no negative numbers is wrong. Therefore, there is at least one negative number in the set and A=-B. The sums involving A and B we can write C+A, C-A, D+A, D-A, E+A, E-A. Do we have 6 sums that could fit these? Yes, we do. The 6 sums would enable us to find A, since, for example, (C+A)-(C-A)=2A=(D+A)-(D-A)=(E+A)-(E-A). The difference between the pairs of sums is 2 for each pair, implying that A=1 and therefore B=-1. The first two pairs are 2 and 4 and 4 and 6, because we have two 4's. If we add the sums in each pair we get 6 and 10. These are respectively 2C and 2D, so C=3 and D=5, making C+D=8, another of the sums. We're left with sums 9, 11, 13, 15, and we haven't found E yet. We need sums of E with the other discovered numbers, E+1, E-1, E+3, E+5 and we have four remaining sums to fit these if possible. The lowest number is 9 which must correspond to E-1. So E=10 and the other sums become 11, 13 and 15, which matches the three remaining sums. So the answer is: the numbers are -1, 1, 3, 5, 10.
Read More: ...

can you get a negative answer when adding a positive number and several negative numbers?

Yes, you can. And no, you can't. It all depends on how many negatives there are in the equation. Like multiplying a positive with a negative will equal a negative. Or multiplying a positive times a negative times a negative times a negative will equal a negative, because a plus times a negative equals negative. Multiply the negative with a negative you get a positive. Multiply that positive with the last negative and you get a negative.
Read More: ...

how do you resolve this : n+(n)2=42

n + n^2 = 42 (i'm presuming that you want to find the possible values of n?) So, subtract 42 from both sides of the equation: n + n^2 - 42 = 0 So now you want to factorise this equation, this means you think of the possible numbers that you could multiply together to get 42. There are many different options for this. You could chose: 42x1, 21x2, 14x3, 7x6. But we need a value that is going to give us our equation and in this example only 7x6 will work (don't forget that the 42 is negative, and therefore one of the numbers that you place in the brackets has to be negative too): So,  n + n^2 - 42 = (n + 7)(n - 6) = 0 (n+ 7)(n - 6) = 0 at two values of n, when n=-7 and when n=6.
Read More: ...

what is 7m+108/30 is greater than-67/30 or 319/66m+408/66 is less than -1291/462

What horrible numbers! Never mind, let's see what we've got. I think what you're looking for is a range of values for m between two limits. 7m+(108/30)>-67/30. We can solve this for m as if the greater than sign were an equals. So we take 108/30 (positive) over to the right where it becomes -108/30. We already have a negative quantity so we make it more negative by adding the two fractions and making them negative. -(108/30-67/30) is -175/30, which cancels down when we divide top and bottom by 5 to -35/6. So 7m>-35/6 and m>-5/6. (319/66)m+408/66<1291/462. We can multiply both sides of the inequality by 66 to get rid of the fractions on the left: 319m+408<1291*66/462. As it happens, 66 divides into 462 exactly 7 times, so 319m+408<1291/7. Take 408 over to the right where it becomes negative. So we have 319m<1291/7 - 408. We need to get the right side over a common denominator, so we multiply 408 by 7 so that we get (1291-2856)/7=-1565/7. The inequality becomes 319m<-1565/7, so dividing both sides by 319 we get m<-(1565/7)/319. So m<-1565/2233. We can write the range for m as -5/6 Read More: ...

Derrick is thinking of a negative integer. When he multiplies the integer by itself and then adds three times the integer to the product , he gets 180. What is Derek's integer?

Derrick is thinking of a negative integer. When he multiplies the integer by itself and then adds three times the integer to the product , he gets 180. What is Derek's integer? Our formula is x^2 + 3x = 180 Use the method called "completing the square." In order to factor the left side of the equation, we need to add the square of (1/2) * (b/a) a is the coefficient of x^2; b is the coefficient of x (1/2 * b/a)^2 = (1/2 * 3)^2 = 1.5^2 = 2.25 x^2 + 3x + 2.25 = 180 + 2.25 Factoring the left side of the equation gives (x + 1.5)*(x + 1.5) = 182.25 Notice that x * 1.5 (1.5 is 1/2 of our b/a) plus x * 1.5 gives us 3x. Also, 1.5 * 1.5 gives us 2.25 Take the square root of both sides. x + 1.5 = sqrt(182.25) The square root of 182.25 is +13.5; it is also -13.5 x + 1.5 = 13.5   and  x + 1.5 = -13.5 x = 13.5 - 1.5 = 12, not a negative number or x = -13.5 - 1.5 = -15, the number we want (-15)^2 + 3(-15) = 180 225 - 45 = 180
Read More: ...

what 2 numbers can be multiplied to get the number 8 and the sum of negative 4

1....x*y=8 2....x+y=-4, so y=-x-4 1...x*(-x-4)=8 -x^2 -4x-8=0....or x^2 +4x +8=0 quadratik equashun giv roots=-2+-2i (komplex number)
Read More: ...

Tips for a great answer:

- Provide details, support with references or personal experience .
- If you need clarification, ask it in the comment box .
- It's 100% free, no registration required.
next Question || Previos Question
  • Start your question with What, Why, How, When, etc. and end with a "?"
  • Be clear and specific
  • Use proper spelling and grammar
all rights reserved to the respective owners || www.math-problems-solved.com || Terms of Use || Contact || Privacy Policy
Load time: 0.1388 seconds