384|2
192|2
96|2
48|2
24|2
12|2
6|2
3|3
1
![384=2^7\cdot3=2^6\cdot2\cdot3=(2^2)^3\cdot6=4^3\cdot6\\\\ \sqrt[3]{384}=\\ \sqrt[3]{4^3\cdot6}=\\ \sqrt[3]{4^3}\cdot\sqrt[3]{6}=\\ 4\sqrt[3]6\Rightarrow C](https://tex.z-dn.net/?f=384%3D2%5E7%5Ccdot3%3D2%5E6%5Ccdot2%5Ccdot3%3D%282%5E2%29%5E3%5Ccdot6%3D4%5E3%5Ccdot6%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B384%7D%3D%5C%5C%0A%5Csqrt%5B3%5D%7B4%5E3%5Ccdot6%7D%3D%5C%5C%0A%5Csqrt%5B3%5D%7B4%5E3%7D%5Ccdot%5Csqrt%5B3%5D%7B6%7D%3D%5C%5C%0A4%5Csqrt%5B3%5D6%5CRightarrow%20C)
192|2
96|2
48|2
24|2
12|2
6|2
3|3
1
2/9 converted into decimal rounded to the nearest hundreth
2 answers:
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Answer:
If then
and
a | b | a + b (answer)
0 | 0 | 0
0 | 1 | 1
0 | 2 | 2
1 | 0 | 1
2 | 0 | 2
1 | 1 | 2
2 | 1 | 3
Step-by-step explanation:
Considering the following conditions for the real numbers:
Following the rules of these in-equations, it is possible to deduce:
Then, if the proposed statement is:
The conditions above shall comply the requirements established, but first, analyzing the statement:
If and
then
,
and
.
If and b a non negative real number, then
, but because to
, then
. Due to the commutative property of sums, the same behavior will be presented if
and a a non negative real number.
According to that, if , then
and
.