Guide :

# i need help with adding surds

what is 4√7 + 4√5

## Research, Knowledge and Information :

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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## Suggested Questions And Answer :

### 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

### 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.

### 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

### 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.

### 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

### 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

### 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

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!