Assuming the order required is as n-> inf.

As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.

We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).

So yes, both are o(log(n)).

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8 0

3 years ago

Serenity has 2\frac{1}{2} 2 1 cups of dough to make dumplings. If she uses \frac{3}{4} 4 3 cup of dough for each dumpling, h

BARSIC [14]

3 dumplings can be made by Serenity

Solution:

Given that,

Serenity has $2\frac{1}{2}$ cups of dough to make dumplings

She uses $\frac{3}{4}$ cup of dough for each dumpling

To find: Number of whole dumplings made

From given information,

$\text{Total cups of dough } = 2\frac{1}{2} = \frac{2 \times 2+1}{2} =\frac{5}{2}$

$\text{Cups of dough for 1 dumpling } = \frac{3}{4}$

To find the number of whole dumplings made, we can divide the total cups of dough by cups of dough for 1 dumpling

$\text{Number of whole dumplings made } = \frac{\text{total cups of dough}}{\text{cups of dough for 1 dumpling}}\\\\\text{Number of whole dumplings made } = \frac{\frac{5}{2}}{\frac{3}{4}} = \frac{5}{2} \times \frac{4}{3}\\\\\text{Number of whole dumplings made } = \frac{10}{3} = 3.33 \approx 3$

Thus 3 dumplings can be made

6 0

3 years ago