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Q: If a and b are even then G(ab) must be 2?
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How do you play 'Gently Sleep' on the recorder?

very easy : GAB AGA GAB ABG there and if you dont know the holes are : for g the top 3 for a the top 2 and for b the first hole now you try it good luck


Why is an even number minus an even number an even number?

In math, even is just another way of saying divisible by 2.If you add or subtract two numbers that share a divisor, the resulting number will also share that divisor. So the sum or difference of a pair of even numbers will (always and also) be divisible by two.In algebraic terms:If a/2 results in a whole number, then a must be even. Likewise, b/2.a/2 - b/2 = (a-b)/2.So if both a/2 and b/2 are whole numbers, then their difference must be, as well.


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on the flute its the best day ever the words is GAB NOB BAG GANO BAG BABAG GAB NOB BAG GAB BAG B


Is the cube root of 2 irrational?

The cube root of 2 is irrational. The proof that the square root of 2 is irrational may be used, with only slight modification. Assume the cube root of 2 is rational. Then, it may be written as a/b, where a and b are integers with no common factors. (This is possible for all nonzero rational numbers). Since a/b is the cube root of 2, its cube must equal 2. That is, (a/b)3 = 2 a3/b3 = 2 a3 = 2b3. The right side is even, so the left side must be even also, that is, a3 is even. Since a3 is even, a is also even (because the cube of an odd number is always odd). Since a is even, there exists an integer c such that a = 2c. Now, (2c)3 = 2b3 8c3 = 2b3 4c3 = b3. The left side is now even, so the right side must be even too. The product of two odd numbers is always odd, so b3 cannot be odd; it must be even. Therefore b is even as well. Since a and b are both even, the fraction a/b is not in lowest terms, thus contradicting our initial assumption. Since the initial assumption cannot have been true, it must be false, and the cube root of 2 is irrational.


If the equation x2 minus ax plus 2b equals 0 has prime roots where a and b are positive integers then a minus b is equal to?

I think you mean the roots are prime numbers. Let the two roots be primes p and q Then the equation factorises to (x - p)(x - q) = 0 which can be expanded to give: x² - (p + q)x + pq = 0 Which comparing coefficients of the original gives: a = p + q 2b = pq as b is an integer, pq must be even, → at least one of p or q must be even → as they are both primes and at least one is even, it MUST be 2 as 2 is the ONLY even prime Assume p is an even prime, ie p = 2 → a = 2 + q 2b = 2q → b = q → a - b = (2 + q) - q = 2 (It doesn't matter if the other prime is even (2) or not as it cancels out from a - b.)


Why is the sum of 2 even numbers always an even number?

If a is even then there is an integer x such that a = 2x If b is even then there is an integer y such that b = 2y Then a+b = 2x+2y = 2(x+y) ie 2 is a factor of a+b, that is, a+b is even.


How do you play saria's song on recorder?

FAB FAB FAB High D B BCBGE Low D EGE FAB FAB FAB High D B BCBGE Low D EGE DEF GAB CBE DEF GAB CAE DEF GAB CBE FEAGBAD (x2) Hope this helps =D


How do you play songs on recorder?

FAB FAB FAB High D B BCBGE Low D EGE FAB FAB FAB High D B BCBGE Low D EGE DEF GAB CBE DEF GAB CAE DEF GAB CBE FEAGBAD (x2) Hope this helps =D


What are the recorder notes for ode to joy?

aab^ccb^ag ffgaagg aab^ccb^ag ffgagff ggaf gab^af gab^agfgc aab^ccb^ab^g ffgagff b^ is b flat c is middle C, all other notes are higher


Is even times odd always even?

Yes. Suppose you are multiplying two even numbers, A and B. Then make B odd by adding 1. You can represent any even number as 2*x. So let's rewrite A*B as (2*x) * (2*y +1) Which is 2*x*2*y + 2*x which is 2*(x*2*y+x) Which is of the form 2*(something) which has to be even.


Scrabble words using the letters a b g I i k w?

big gab bag wag wig


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seize hull beady mill= Cecil B. DeMille