Guide :

Are multiples of 8 also multiples of 2 and 4?

multiples of 8,4,2

Research, Knowledge and Information :


Multiplication: Multiples of 8 are also multiples of 2 and 4 ...


Mar 02, 2012 · http://www.mathincolor.com All multiples of 8 are also multiples of 2 and 4 using Math in Color charts. Buy posters and handouts at www.mathincolor.com.

Are multiples of 8 also multiples of 4 - Answers.com


Why are multiples of 8 also multiples of 4? Multiples of 8 are also multiples of 4 because 8 is a multiple 4. In any number that is a multiple of 8, ...
Read More At : www.answers.com...

Are multiples of 8 also multiples of 2 - Answers.com


Yes, multiples of 8 are also multiples of 2 because 8 is a multipleof 2 itself. Go. Log In Sign Up. ... Multiples of 8 are also multiples of 4 because 8 is a multiple 4.
Read More At : www.answers.com...

are multiples of 8 also multiples of 4 - Brainly.com


are multiples of 8 also multiples of 4 - 2182610. 1. Log in Join now Katie; a few seconds ago; ... Multiples of 8 are also multiples of 4 because 8 is a multiple 4.
Read More At : brainly.com...

Q jose said all of the multiples of 8 are also multiples of 2


jose said all of the multiples of 8 are also multiples of 2. jamila ... both 2 and 4 are multiples of 8 so any multiple of 8 ... multiples of 157; multiples;
Read More At : homework.boodom.com...

Multiplicaiton: Multiples of 4 are also multiples of 2 - Math ...


Mar 02, 2012 · http://www.mathincolor.com Multiples of 4 are multiples of 2. ... Multiplication Multiples Factors ... Multiples of 8 are also multiples of 2 and 4 ...

Multiples Calculator


Multiples of 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34 ...
Read More At : www.calculatorsoup.com...

Why is multiple of 8 also multiple of 2 and 4 - Brainly.com


Why is multiple of 8 also multiple of 2 and 4 2. Ask for ... Log in to add a comment Answers StevenLopez1999; 2016-11-22T17:57:57-05:00. 4 x 2 = 8 ok, 2x4=8, 4,8 did ...
Read More At : brainly.com...

Multiples and Least - Math Goodies


Multiples and Least Common Multiples: Unit 3 > Lesson 2 of 9: Problem: ... We have also found the Least Common Multiple (LCM) of 4 and 5, which is 20.
Read More At : www.mathgoodies.com...

Suggested Questions And Answer :


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: ...

solve equation 6x+5y=0, 3x-2y = 19

Exponents Exponents are supported on variables using the ^ (caret) symbol. For example, to express x2, enter x^2. Note: exponents must be positive integers, no negatives, decimals, or variables. Exponents may not currently be placed on numbers, brackets, or parentheses. Parentheses and Brackets Parentheses ( ) and brackets [ ] may be used to group terms as in a standard equation or expression. Multiplication, Addition, and Subtraction For addition and subtraction, use the standard + and - symbols respectively. For multiplication, use the * symbol. A * symbol is not necessiary when multiplying a number by a variable. For instance: 2 * x can also be entered as 2x. Similarly, 2 * (x + 5) can also be entered as 2(x + 5); 2x * (5) can be entered as 2x(5). The * is also optional when multiplying with parentheses, example: (x + 1)(x - 1). Order of Operations The calculator follows the standard order of operations taught by most algebra books - Parentheses, Exponents, Multiplication and Division, Addition and Subtraction. The only exception is that division is not currently supported; attempts to use the / symbol will result in an error. Division, Square Root, Radicals, Fractions The above features are not supported at this time. A future release will add this functionality.
Read More: ...

n^3-9n^2+20n prove that it is divisible by 6 for all integers n greater or equal to 1

Factor the given expression.   We have: n³-9n²+20n=n(n-4)(n-5) ··· Ex.1 If Ex.1 is divisible by 6, Ex.1 is also divisible by two prime factors of 6, 2 and 3. (6=2x3) A. If n is odd, (n-5) is even.   If n is even, (n-4) is also even.   Therefore, n(n-4)(n-5) is always a multiple of 2. B. If Exp.1 is divisible by 3, the remainder is 0,1,or 2. If n ≡ 0 (mod 3), n-4 ≡ -4 ≡ -1 (mod 3), and n-5 ≡ -5 ≡ -2 (mod 3), that is: n is a multiple of 3. If n ≡ 1 (mod 3), n-4 ≡ -3 ≡ 0 (mod 3), and n-5 ≡ -4 ≡ -1 (mod 3), that is: (n-4) is a multiple of 3. If n ≡ 2 (mod 3), n-4 ≡ -2 (mod 3), and n-5 ≡ -3 ≡ 0 (mod 3), that is: (n-5) is a multiple of 3. Therefore, n(n-4)(n-5) is always a multiple of 3. CK: If n=1, n(n-4)(n-5)=1(-3)(-4)=12, 12(n=2), 6(n=3), 0(n=4), 0(n=5), 12, 42, 96, 180 … CKD.  Therefore, n³-9n²+20n is divisible by 6 for all integers greater than or equal to 1.  
Read More: ...

how many ways are there to add and get the sum of 180

There are an infinite number of ways to get 180 from two numbers, if we count decimals and fractions as well as other real numbers; but if we are limited to positive integers greater than zero and just the sum of two of them, we are limited to x and 180-x. If we also exclude 90+90 because the numbers are the same, then we have 1 to 89 combined with 179 to 91, which is 89 pairs. Moving on to the sum of three different numbers, let's make 1 plus another two different numbers adding up to 179. So we have 2+177, 3+176, ..., 87+92, 88+91, 89+90, which is 88 groups combined with 1. Move on to 2 plus another two different numbers adding up to 178: 3+175, ..., 87+91, 88+90, which is 86 groups. Then we move on to 3 plus 177: 4+173, ..., 86+91, 87+90, 88+89, 85 groups. And so on, with reducing numbers, until we get to 59, 60 and 61. Let's divide the numbers into two groups A and B. In A we start with 1 and in B we put 2 and (180-A-B)=177 as a pair (2,177). Then we put the next pair in group B: (3,176), then (4,175) and keep going till we have used up all the numbers, ending up with (88,90). Then we count how many pairs there are in group B and pair it up with the number in group A, so we start with (1,88) which covers all the combinations of numbers in group B. Now we move to 2 in group A, put all the pairs adding up to 178 in group B, and finally put the count of these pairs with 2 in group A: (2,86). We then move on to 3, and so on, putting in the counts to make up the number pair in group A. When we've finished by putting the last count in group A, which is (59,1), we can forget about group B and look at the pattern in group A. What we see is this: (1,88), (2,86), (3,85), (4,83), (5,82), (6,80), (7,79), ... See how the counts come in pairs with a gap? All the multiples of 3 are missing in the counts sequence (e.g., 87, 84, 81). We find there are 29 pairs and one odd count, 88, which is unpaired. Number the pairs 0 to 28 and refer to the pair number as N. Add the counts in the pairs together so we start with pair 0 as 86+85=171, pair 1 as 165, pair 2 as 159, and so on. The sequence 171, 165, 159, ..., 3 is an arithmetic sequence with a start of 171 and a difference of 6 between each term in the sequence. [Note also that the terms in the series are all multiples of 3: 3*57, 3*55, 3*55, ...] The rule for the Nth term is 171-6N. When N=0 we have the first term 171 and when N=28 the last term is 3. There is one more term at the end which is unpaired made up of the numbers 59, 60 and 61. We can combine this with the unpaired (1,88). We can find the sum of the terms in the series, which will tell us how many ways there are of adding three different integers so that their sum is 180 (like the sum of the angles of a triangle).  To find the sum of the terms of the series we note that there are 29 terms (0 to 28) and they all contain 171, so that's 171*29=4959. We also have to subtract 6(0+1+2+3+...+28)=6*28*29/2=2436. So 4959-2436=2523. [The sum of the series is also 3(57+55+53+...+5+3+1)=2523.] To this we add the "odd couple" 88+1=89 and 2523+89=2612. Add also the 89 which is the number of pairs of integers adding up to 180 we calculated at the beginning. The total so far is 2612+89=2701 ways of adding 2 or 3 positive integers so that their sum is 180. If you want to go further, please feel free to do so!
Read More: ...

are multiples of 3 also multiples of 63

other wae round...multipel av 63 gotta be multipel av 3
Read More: ...

how do i add fractions

To add or subtract fractions, obtain a least common denominator. Subtract the numerators in the correct order and retain the same least common denominator for your answer. Simplify. To multiply fractions, multiple the numerators. The product will be the numerator of your answer. Repeat with denominators. Simplify. To divide fractions, take the reciprocal of what you are dividing by. Multiply the reciprocal with the initial number (see above for multiplication process). Simplify. Evaluate means to solve. You can solve fraction problems using the above processes. You can only simplify if both the numerator and denominator are divisible by the same number. If the denominator is odd, you can only simplify it if the numerator also is divisible by a same number. Ex. 88/33. Although the denominator is odd, both the numerator and denominator are divisible by 11 resulting in 8/3 as the simplified answer. To pace yourself during a test do the following. Find out how long you have for the test. Divide this by the total number of problems on the test. Example. 1 hour for 20 problems on your test. This means you have 3 minutes per problem. If you spend more than 3 minutes on a problem, skip it. Continue until you attempt all the problems. Go back with the remainder of the time to retry these problems you skipped. Most likely they are the most difficult, hence why you spent alot of time on them. This method of pacing allows you to skip the hard problems at first, attempt all problems, and finish the easier problems for sure.
Read More: ...

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 do you know that 6 to 14 is equivalent to 15 to 35 using a multiplication table?

Rows 3 and 7 and columns 2 and 5 are significant. Row 3 and column 2 represent 3*2=2*3=6; row 7 and column 2 represent 2*7=7*2=14, making up the fraction 6/14=2*3/(2*7)=3/7, because the common factor 2 cancels out. The ratio 6:14 is the same as 6/14 so 3:7 is the same as 3/7. Similarly, row 3 and column 5 represent 3*5=5*3=15; row 7 and column 5 represent 7*5=5*7=35, making up 15/35 or ratio 15:35=3*5/(7*5)=3/7, because the common factor 5 cancels out.  So the table shows by the common rows 3/7 that both fractions reduce to the same fraction or ratio 3/7 or 3:7. Take column 9, for example. Here we have 27 and 63 so 27/63 is also 3/7. This is how the multiplication table can be used to show which fractions or ratios are equivalent.   In the table, you can also replace the word column for row and row for column, and the same rule applies.
Read More: ...

? divided by 9=6 then

A number divided by 9 gives 6. If you write out your 9 times table you have: 1*9=9 2*9=18 3*9=27 4*9=36 5*9=45 6*9=54 7*9=63 8*9=72 9*9=81 10*9=90 Right? See anything with 6 and 9? 6*9=54. This tells you, for example, that if you gave $6 to each of 9 people you would give out $54 altogether. But it also tells you that if you had $54 you could divide it up so that each receives $6. It doesn't have to be money. It could be M&Ms. And you could divide $54 between 6 people so that they had $9 each, or M&Ms or whatever. The same rule applies. Division is the reverse process of multiplication. 6*9=54 can also be written 6=54/9 or 9=54/6. Know your multiplication tables and you'll be OK. (You could also have written out your 6 times table, because 9*6=54.)
Read More: ...

If m, element of N, is a multiple of 4 then Fm is a multiple of 3.

The series is 1 1 2 3 5 8 13 21 etc. We can see that every 4th term is a multiple of 3: 3 21, etc so we need to look at the series closely to find out why. F2=F1=1. F4=F3+F2=F2+F1+F2=3F1 F8=F7+F6=F6+F5+F5+F4=F6+2F5+F4=F5+F4+2F5+F4=3F5+2F4, But we know F4 is a multiple of 3 (F4=3F1), so F8=3F5+6F1=3(F5+2F1), Therefore F8 is also a multiple of 3. By induction, every 4th term is a multiple of 3, so Fm is a multiple of 3 when m is a natural number which is a multiple of 4.
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: 0.0469 seconds