I honestly can't tell if you're mocking lookagain.
Is it proper to write f(-lookagain) = f(greg1313) or is more appropriate to write
f(-lookagain) = -f(greg1313)?
If u>0 then the RHS of the original question is larger
If u<0 then the LHS of the original question is larger
Cube both sides ...
4+u=(3−2)2(3−2)
4+u=(11−62)(3−2)
4+u=45−292
u=41−292
Approximate by knowing root 2 to 3dp...
u≈41−29×1.414
=41−(30×1.414−1.414)
=41−(42.42−1.414)
=41−41.006
The above approximation of root 2 is rounded down from the real value, therefore the "41.006" would only become larger if more decimal places are used. Therefore u<0 which implies the LHS of the original question is larger.
If u>0 then the RHS of the original question is larger
If u<0 then the LHS of the original question is larger
Cube both sides ...
4+u=(3−2)2(3−2)
4+u=(11−62)(3−2)
4+u=45−292
u=41−292
Approximate by knowing root 2 to 3dp...
u≈41−29×1.414
=41−(30×1.414−1.414)
=41−(42.42−1.414)
=41−41.006
The above approximation of root 2 is rounded down from the real value, therefore the "41.006" would only become larger if more decimal places are used. Therefore u<0 which implies the LHS of the original question is larger.
This was posted in another forum where my response used the fact that sqrt(2) ~ 1.414 and I was quickly told that I could not use any calculations. Then how the hay can I add up the three values to compare to 5?
This was posted in another forum where my response used the fact that sqrt(2) ~ 1.414 and I was quickly told that I could not use any calculations. Then how the hay can I add up the three values to compare to 5?
No, the partial phrasing is "I don't recommend doing it your way." That means you do it at your risk, or you would have to be extra careful. For example, how would you know you are getting enough correct decimals?
So, after correcting your typo, you have (approximately 3.414) + 4^(1/3) versus 5, or something similar to that.
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