In the equation 10n=120, is n 10?
2 answers:
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Since
The square root of 11 is somewhere between 3 and 4. In order to round it to the nearest tenth, we have to try all numbers between 3 and 4 with one decimal digit, and see which is closest to 11 when squared. We have
So, the square root of 11 is somewhere between 3.3 and 3.4.
Given the following question:
First expression:
Second expression:
Answer:
see explanation
Step-by-step explanation:
Note that (x² + 5)(x² - 5) are the factors of a difference of squares
a² - b² = (a + b)(a - b)
with a = x² and b = 5, hence
(x² + 5)(x² - 5) = (x²)² - 5² = - 25, thus
3( - 25) = 3
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