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:

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### 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?

### 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 ...

### 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:

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### 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 ...

### 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 ...

## Suggested Questions And Answer :

### 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.

### 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.

### 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.

### 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.

### 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

### 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)