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Nakerria
18.12.2020 •
Mathematics
Evaluate: 5.12 ÷ 102.
Solved
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Ответ:
0.050196
Step-by-step explanation:
normally, 2 place values is the answer so 0.05
Ответ:
Step-by-step explanation:
a). Let![x^{2}+y^{2}=4](/tpl/images/1297/4360/f3db0.png)
Now since
is a square of an integer, its value is
0.
Therefore,
≥ 0, since x is an integer
where
= 1,2,3,...
and x = 1
But as x =
,... cannot be integer
∴![y^{2}=4-x^{2}](/tpl/images/1297/4360/0f3c2.png)
= 1.732 which is not an integer
Thus, any positive integer cannot be written as the sum of the squares of the two integers.
b). Let n be an integer
∴![3(n^{2}+2n+3)-2n^{2}](/tpl/images/1297/4360/3bf2d.png)
On solving we get,
which is a perfect square
Hence proved.