For any number n>1, is

Mathematics

1 answer:

Simora [160]3 years ago

7 0

The right answer is A ! :)

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Divide. Reduce the answer to lowest terms 2 1/9 ÷ 3 1/3

aleksandrvk [35]

Answer:

I am pretty sure the answer is 19/30

Step-by-step explanation:

If u turn 2 1/9 ÷ 3 1/3 into improper fractions u will get 19/9 Divided by 10/3

Then u do the butterfly. So this would be 9 x 10 =90 and 19 x 3 =57 so this is what it looks like 57/90 and that can be simpflyed into 19/30. Becasue 3 x 19 = 57 and 3 x 30 = 90

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Show that if x and y may both be written as the sum of the squares of two rational integers, then their product xy may also be w

Georgia [21]

Answer:

See proof below

Step-by-step explanation:

One way to solve this problem is to "add a zero" to complete the required squares in the expression of xy.

Let $x=m^2+n^2$ and $y=l^2+k^2$ with $m,n,l,k\in \mathbb{Z}$ . Multiplying the two equations with the distributive law and reordering the result with the commutative law, we get $xy=(m^2+n^2)(l^2+k^2)=m^2l^2+m^2k^2+n^2l^2+n^2k^2=n^2l^2+m^2k^2+m^2l^2+n^2k^2$

Now, note that $0=2lkmn-2lkmn=2nkml-2nlmk$ by the commutativity of rational integers. Add this convenient zero the the previous equation to obtain $xy=n^2l^2-2nlmk+m^2k^2+m^2l^2+2nkml+n^2k^2=(nl-mk)^2+(ml+nk)^2$ , thus xy is the sum of the squares of $nl-mk,ml+nk\in \mathbb{Z}$ .