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i need help with adding surds

what is 4√7 + 4√5

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adding surds, please help! - The Student Room


Hi everyone, I am having a bit of trouble with surds. I don't know what to do when adding them, when there is a number in front and they are different.
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Adding and Subtracting Surds - YouTube


Dec 20, 2011 · This video will introduce the rules you need to follow when adding and subtracting surds

Operations with Surds - mathsteacher.com.au


Radicand, like surds, unlike surds, addition and subtraction of surds.
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Surds - Mathematics GCSE Revision


Addition and Subtraction of Surds. Adding and subtracting surds are simple- however we need the numbers being square rooted ...
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Mathspace



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How to Add and Subtract Surds


How to Add and Subtract Surds. ... The same thing happens to \sqrt{a} - \sqrt{b} as subtraction can be thought of as adding a negative number.
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BBC - GCSE Bitesize: Basic rules


A secondary school revision resource for GCSE Maths about higher level factors, powers, roots and surds
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Surds - Math Is Fun


Surds. When we can't simplify a number to remove a square root (or cube root etc) then it is a surd.
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add surds - The Student Room


Gcse maths surds question; Desperately need maths help, Surds; Surds; help on surds; Surds are like turds - please help! TSR Support Team. ... The Student Room, ...
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Addition and Subtraction of Surds - Sunshine Maths


Addition and subtraction of surds involve a few simple rules: we can add or subtract surds only when they are in the simplest form; ... adding like terms. ...
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i need help with adding surds

?????? "surds" ?????? it look like 4*sqrt(7) +4*sqrt(5)or 4*[sqrt(7)+sqrt(5)] sqrt(5)=2.23606797749979 sqrt(7)=2.6457513110645907 sum=4.88181928856438 4*=19.52727715425752
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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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what is 13/56 plus 5/7 equals

When adding or subtracting fractions we need to work out the LCD (lowest common denominator). That means we need to find the smallest number that the denominators of the fractions all divide into without a remainder. Sometimes this is easy to do by factorising each denominator so we can see what numbers make it up. In your case,  we see that 56 is one denominator that can be split into 8*7, and it just so happens that the other fraction has a denominator of 7. How many 56ths do we need to make 1/7? We need 8. So for 5/7 we need 5 times as many, so we need 40. Now that we have both fractions sharing a common basis, we can do the sum. 13/56 + 5/7 is the same as 13/56 + 40/56. That makes 53/56 just by adding the numerators. It's like adding 13 apples to 40 apples, where an apple is represented by a 56th. To help you with this sort of problem when the denominators have no common factors just multiply the denominators together to create the LCD. Also when you've finished the sum, check to see if the result cancels down to a simpler fraction because numerator and denominator share a common factor.
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8-3[25-(7+25)]

8-3[25-(7+25)] i need help evaluating the expression 8-3[25-(7+25)] 8 - 3[25 - (7 + 25)] Ordinarily, I would say perform the addition inside the parentheses, but in this problem we are adding 25 then subtracting 25. 8 - 3[-7] Now, perform the multiplication. 8 - (-21) Last, the subtraction. Subtracting a negative is adding a positive. 29
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Use the following clues to find the ages of the three children.

Welcome to Math Homework Answers - Math Homework Answers is a math help site where students, teachers, and math lovers can ask and answer math homework questions to help create a useful fun math community. Ask or answer math questions in basic math, pre-algebra, algebra, trigonometry, geometry, pre-calculus, calculus, unit conversion or any other math related topic. Register and play the math game against others to see how many points you can earn. All categories Pre-Algebra Answers (518) Algebra 1 Answers (1,109) Algebra 2 Answers (450) Geometry Answers (245) Trigonometry Answers (71) Calculus Answers (238) Statistics Answers (71) Word Problem Answers (372) Site News (11) Use the following clues to find the ages of the three children. Use the following clues to find the ages of the three children. The product of their ages is 96. The middle child is two years older then the youngest.   A = Eldest B = Middle child C = Youngest   AxBxC = 96 <= eq1 B = 2 + C <= eq2   You need to supply one more clue to solve the problem    
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Explain the results of the following options: Option 1: 6% compound interest quarterly for 5 years. Option 2: 8% compound interest annually for 5 years. Option 3: 14.5% simple interest for 10 years.

MEMO TO CLIENT Simple interest applies the interest rate proportionately, so the amount of interest on a particular investment is directly proportional to the length of time invested. This means that, for example, if the investment period is 5 years, the interest is 5 times the interest earned in one year; for 10 years it is 10 times that earned in a year. It is also easy to calculate because of this simple proportion. Compound interest is more rewarding to the investor. After a period of time, for example, a year, the interest earned in the year is added to the original amount invested. So at this point, it is the same as simple interest. But what happens next is different. The investment plus the interest becomes the invested amount for the next period, the next year, for example. At the end of this period the process continues, and the interest is again added and becomes the investment amount for the next period. So it is clear that over a period of time more and more interest is earned. An important feature that investors need to be aware of is: how regularly is compound interest added? The shorter the period, the bigger the interest earned. Interest can be compounded annually, quarterly, monthly, daily or continuously. So, if the investor is quoted a particular annual rate of interest, then the largest amount of interest gained will correspond to the shortest compound interest period. As an example, take 6% per annum, or annual interest rate. After a year with interest compounded annually, 6% interest will be earned. If interest is compounded quarterly, then each quarter the interest will be added at a rate of 1.5% each quarter, but by the end of a year, the effective interest will be more than 6.1%. If interest is compounded monthly, the monthly rate would be 0.5% and after a year would be effectively closer to 6.2%. Interest compounded daily would be even closer to 6.2% and continuously would be slightly more. Growth is a convenient way of expressing the factor by which an investment increases over time, and companies will often publish tables to simplify calculations of expected returns on investments at fixed rates. The time periods will be typically 5, 10, 15, 25 years for a range of annual rates. So investors can quickly calculate the returns on varying amounts of money. As an example, take 15 years. The growth rates at 6% per annum would be: 1.9 (simple interest); 2.40 (compounded annually); 2.44 (compounded quarterly); 2.45 (compounded monthly); 2.46 (compounded daily or continuously). Option 1 6% annually is 1.5% quarterly, so growth is 1.015^20=1.3469, where 20 is 20*(1/4)=5 years. $500000*1.3469=$673,427.50 to best accuracy. Option 2 8% compounded annually: growth=1.08^5=1.4693 and amount is $500000*1.4693=$734,664.04. Option 3 14.5% simple interest for 10 years: 145% interest=1.45*500000=$725,000 interest+500000=$1.225 million. The first two options have the same investment time period, and option 2 is better. Option 3 has double the time period. If option 3 were to be applied over 5 years instead of 10, the interest would have been $362,500 (half of $725,000) and the total amount would have been $862,500. However, the investment is over ten years so the investor would need to wait 10 years before taking full advantage of the investment. Take option 2 over 10 years and we get a growth rate of 2.1589 making the investment worth $1,079,462.50, which is still smaller than option 3, which had a growth rate of 2.45 (1.45+1) because of the higher interest rate.  
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what is the perimeter of 115 1/2, 66 1/4, 106 1/8, and 110 1/4?

Problem: what is the perimeter of 115 1/2, 66 1/4, 106 1/8, and 110 1/4? i need help with this one. i dont like fractions! When adding fractions, you need to have a common denominator. All of these are multiples of 1/8. 115 1/2 + 66 1/4 + 106 1/8 + 110 1/4 115 4/8 + 66 2/8 + 106 1/8 + 110 2/8 397 9/8 398 1/8
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help solving 2c-5=c+4

first you need to get all like term on one side of the equal sign, so you would do that by adding 5 to both sides of the equal sign which would give you 2c-5+5=c+4+5 the (-5) and +5 cancel out so now your equation should look like this 2c=c+4+5 which = 2c=c+9 now you need to get c on the other side of the equal sign so since c is positive we do the opposite again and subtract c from both sides which gives you an equation that looks like this 2c-c=c-c+9 which = 1c=9 or you want to get c by itself so you do the opposite again so instead of multiplying c by 1 you'd divide it by 1 but you have to divide both sides by 1 which gives you 1c/1=9/1 = c=9 so your answer is c=9
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3x+2y=17 2x+5y=26

3x+2y=17 2x+5y=26 Elimination i need help solving it We need to eliminate either the x or the y by adding or subtracting the equations. In order for that to work, the coefficients must be equal. Multiply the first equation by 5. 5 * (3x + 2y) = 5 * 17 15x + 10y = 85 Multiply the second equation by 2. 2 * (2x + 5y) = 2 * 26 4x + 10y = 52 Subtract the re-worked second equation from the re-worked first equation. 15x + 10y = 85 -(4x + 10y = 52) ----------------------- 11x           = 33 x = 3 Substitute that into one of the original equations. 2x + 5y = 26 2(3) + 5y = 26 6 + 5y = 26 5y = 26 - 6 5y = 20 y = 4 x = 3, y = 4
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Please help

You need to see through the problem and apply whatever is necessary to reduce the number of variables till eventually there's only one to find. Remember a simple fact: if you have two variables you always need two independent equations to find them; for three variables, three equations; four variables, four equations. You use the multiplication property if it helps you to eliminate a variable between two equations. Take some examples: x+y=10, x-y=3; simply adding these two equations will eliminate y and help you find x. 2x=13 so x=6.5 and y=3.5; 2x+y=10, x-2y=10; we could double one equation or the other so as to match the coefficients of one or other of the variables; but since it's easier to add two equations rather than to subtract them, where we have a minus in one equation and a plus in the other, we would prefer to use the multiplier for the relevant variable. So we double the first equation and add to the second: 4x+2y=20 PLUS x-2y=10: 5x=30, making x=6 and y=-2. The last pair of equations could have been written: 2x+y=x-2y=10, but it's still two equations. There is no one way to solve equations, and you can save yourself a lot of stress by not assuming you have to remember a rigid technique or formula as “The Way to do it”. You'll find mathematics is more fun when you intelligently try different methods and use your natural creativity to guide you. And here's another interesting thing. Those questions about finding a missing number in a series can be tackled in many cases as solving simultaneous equations. You only need n equations to find n variables, and a series can be seen as a set of terms generated by a function y=f(x) for different values of x (the position in the series), giving different values of y (the terms in the series). f(x) is a polynomial of the type ax^n+bx^(n-1)+cx^(n-2)+... If there are four given terms n=3 and the variables are a, b, c and d; if there are 3 given terms, n=2 and the variables are a, b and c. There is always a solution, we just have to work through and find it!
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