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# 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, ...

### 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.

### 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.

### 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;

### 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 ...

### 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 ...

### 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.

## Suggested Questions And Answer :

### 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.

### 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.

### are multiples of 3 also multiples of 63

other wae round...multipel av 63 gotta be multipel av 3

### 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.

### 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.

### ? 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.)

### 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.