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negative 4 plus 7

negative 4 plus 7

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What is negative 7 plus negative 4 - Answers.com


A negative plus a negative is a negative number. It is a larger negative number, that is, it is a smaller number - IOW, it is farther to the left on the number line.
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Adding Positive and Negative Numbers | Wyzant Resources


Adding Positive and Negative Numbers. Once you understand the basics of positive and negative numbers, you can start to add them together. Sometimes this seems ...
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What is negative 4 plus 7 - Answers.com


What is negative 4 plus 6?-6+4=-2 If you find it hard to work negatives out, just switch the numbers around, 4-6=-2 2 people found this useful Edit. Share to: ...
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Subtracting Positive and Negative Integers


Subtracting Negative and Positive Integers. To subtract integers, ... If both signs are negative, the answer will be negative. Example: -8 - (+4) = -8 - 4 = -12;
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Negative 3 minus Negative 7 = POSITIVE FOUR?


Negative 3 minus Negative 7 = POSITIVE FOUR? ... If you want to make a number more negative, you add a negative number to it. by Anonymous: reply 1: 08/31/2015:
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Adding and Subtracting Positive and Negative Numbers


Positive and Negative Numbers ... "Positive 2 plus Positive 3 equals Positive 5" ... Now let's see what adding and subtracting negative numbers looks like:
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MathSteps: Grade 6: Negative Numbers: What Is It?


Operations with Negative Numbers. Commonly, plus and minus signs are used to indicate addition and subtraction, respectively. Those signs can also be used to indicate ...
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Subtracting Positive and Negative Numbers - Wyzant


Subtracting Positive and Negative Numbers. Subtracting positive and negative numbers can also be tricky because there are several rules to remember and follow.
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Plus and minus signs - Wikipedia


The plus and minus signs (+ and −) are mathematical symbols used to represent the notions of positive and negative as well as the operations of addition and subtraction
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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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negative eight minus negative eight equals

-8 - -8= -16 the answer is negative 16 because a negative plus a negative is a positive. so u add. so 8 plus 8 is 16 and then its a negative so negative 16.
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-8/5 + 2/-5

-8/5+2/-5 is the same as -8/5 - 2/5, because the minus sign applies to the whole fraction not to just the top or just the bottom part. The fractions have the same denominator so we can write (-8-2)/5, that is we can add together the two negatives. How do you add two negative numbers? Think of the sign as indicating direction. Plus means go to the right or forwards and minus means go to the left or backwards. In this case take eight steps backwards then go back another two steps. How many steps backwards have you gone? 10, of course, which is -10. -10/5, if there was no minus, would be 2, so you do the division first then call the result negative, so the answer is -2. The rule is: if you multiply or divide two numbers with different signs (no sign at all means plus) the result is always negative; if the signs are the same, two positives or two negatives, the result is always positive.
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five times a number increased by 2 is equal to twice the number decreased by 4.

If the number is x, 5x+2=2x-4, so putting x's on the left and numbers on the right we get 3x=-6, so x is -2. The equals sign changes positive to negative and negative to positive when transfers are made across it. Put x=-2 and check out the question: 5 times the number is -10 plus 2 is -8, and twice the number is -4 less 4 is also -8. Treat negatives as direction backwards and positives as direction forward: negative is steps back and positive is steps forward, so the number -2 is two steps back. Multiply by 5 means 10 steps back; plus 2 is 2 steps forward, making 8 steps back. Twice 2 steps back is 4 steps back; less 4 is 4 steps further back making 8 steps back.
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what is the product of (-3) (-6)

Simple Algebra skills... (Minus X Minus) = positive number. eg. -2x-10 = 20 (Minus X Plus) = negative number. eg. -2x10 = -20 (Plus X Minus) = negative number. (Plus X Plus) = positive number. eg. 2x10 = 20 To answer your question. (-3)(-6) is 18. Brackets or parenthesises in maths, placed next to each other means Times
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Whats 2-(-4)

When you have "minus a negative" back to back... that becomes "plus a positive" 2 - (-4) = 2 + 4 {minus a negative becomes plus a positive} = 6 {added} www.algebrahouse.com        
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View this ans answer this for me please.

Before, below, loss are examples of negatives, while after, above, gain or profit are examples of positives. All all the things mentioned in the question must happen before the launch, so we add the times together: 2hr 15m plus 20m plus 40m=2hr 15m plus 1hr (40+20 minutes=1hr)=3hr 15m. These must take place BEFORE launch, so this time is negative: -3hr 15m, answer 1.
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what is -t+5=t-19

People have a hang-up over negatives, but really there's no need to panic. After all if the temperature goes down to -2 it just means 2 degrees below zero. You can see this on a thermometer and everyone understands what it means, no problem. Just because negatives show up in mathematical and algebraic expressions is no big deal either. Back to your problem. To solve the equation and find t we need to gather together the unknowns (in this case the variable t) and the knowns, which are just numbers. We need to have the knowns on one side of the equals and the unknowns on the other side for this type of equation. It doesn't matter which side is used for what, although most people feel happier with the unknown(s) on the left and plain numbers on the right. We don't want loads of negatives so what we'll do is take the t's over to the right. When you cross from side of equals to the other, plus changes to minus and minus to plus. Divide changes to multiply and multiply changes to divide. Bring the -t from the left to right and it becomes +t or just t, added to the t already there makes 2t. Now bring the -19 from right to left where it becomes +19, added to the 5 already there makes 24. So we have 24=2t or put another way 2t=24. Take 2 from the left where it's multiplying over to the right where it divides into 24, making 12. So t=12. Bingo!
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integral from 0 to infinity of (cos x * cos(x^2)) dx

The behaviour of this function f(x)=cos(x)cos(x^2) is interesting. The integral is the area between the curve and the x axis. If the functions cos(x) and -cos(x) are plotted on the same graph, the latter form an envelope for f(x). Between x=(pi)/2 and 3(pi)/2, the curve has 4 maxima and 3 minima; between x=3(pi)/2 and 5(pi)/2 there are 7 maxima and 6 minima; between x=(2n-1)(pi)/2 and (2n+1)(pi)/2 there are 3n+1 maxima and 3n minima (integer n>0), a total of 6n+1. (These are based purely on observation, and need to be supported by sound mathematical deduction.) As n becomes large the envelope appears to fill as the extrema become closer together. As x tends to infinity n also tends to infinity. The envelope apparently has as much area above (positive) as below (negative) the x axis so the total area will be zero as the positive and negative areas cancel out. The question is: do the areas cancel out exactly? As x gets larger, the curve starts to develop irregularities and patterns, but it stays within the envelope, and positive irregularities appear to be balanced by negative irregularities, so the overall symmetry appears to be preserved. f(x)=0 when cos(x)=0 or cos(x^2)=0, which means that x=(2n-1)(pi)/2 or sqrt((2n-1)(pi)/2), where n>0. Between 3(pi)/2 and 5(pi)/2, for example, we have sqrt(3(pi)/2), sqrt(5(pi)/2), ..., sqrt(13(pi)/2), because sqrt(13(pi)/2)<3(pi)/20, and this lies between (2n-1) and (2n+1); so n is defined by 2n-1<(2m+1)^2(pi)/2<2n+1.  For m-1 we have 2(n-z)-1<(2m-1)^2(pi)/2<2(n-z)+1, where z is related to the number of zeroes in the current "batch". For example, take m=3: 2n-1<49(pi)/2<2n+1; 49(pi)/2=76.97 approx., so 2n-1=75, and n=38. Also 2(n-z)-1<25(pi)/2<2(n-z)+1 so, because 25(pi)/2=39.27 approx., 2(n-z)+1=41, n-z=20, and z=18. When m=2, 2n-1=39, n=20; 2(20-z)-1<9(pi)/2<2(20-z)+1; 2(20-z)-1=13, 20-z=7, z=13. The actual number of zeroes, Z, including the end points is 2 more than this: Z=z+2. Now we have an exact way to calculate the number of zeroes in each batch. So Z and n are both related to m. The number of extrema, E=Z-1=z+1. In fact, E=int(2(pi)(m-1)+1), where int(a) means the integer part of a, so as m increases, there are proportionately more extrema over the range (2m-1)(pi)/2 to (2m+1)(pi)/2. The figure of 6n+1 deduced earlier by observation approximates to the mathematical findings, because 2(pi) is approximately equal to 6. But we still need to show, or disprove, that the areas above and below the x axis are equal and therefore cancel out. Unfortunately, if we consider the area under the first maximum (between x=(pi)/2 and sqrt(3(pi)/2)), and the area above the first minimum (between x=sqrt(3(pi)/2) and sqrt(5(pi)/2)), they are not the same, so do not cancel out. More...
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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.
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