How to approximate to 3 significant figures

Mathematics

1 answer:

kondaur [170]2 years ago

3 0

Answer:

We round a number to three significant figures in the same way that we would round to three decimal places. We count from the first non-zero digit for three digits. We then round the last digit. We fill in any remaining places to the right of the decimal point with zeros.

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How many elements does A contains if it has 64 subsets

Amanda [17]

<h3>Answer: 6</h3>

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Explanation:

Rule: If a set has n elements in it, then it will have 2^n subsets.

For example, there are n = 3 elements in the set {a,b,c}. This means there are 2^n = 2^3 = 8 subsets. The eight subsets are listed below.

{a,b,c} .... any set is a subset of itself
{a,b}
{a,c}
{b,c}
{a}
{b}
{c}
{ } ..... the empty set

Subsets 2 through 4 are subsets with exactly 2 elements. Subsets 5 through 7 are singletons (aka sets with 1 element). The last subset is the empty set which is a subset of any set. You could use the special symbol $\varnothing$ to indicate the empty set.

For more information, check out concepts relating to the power set.

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The problem is asking what value of n will make 2^n = 64 true.

You could guess-and-check your way to see that 2^n = 64 has the solution n = 6.

Another approach is to follow these steps.

$2^n = 64\\\\2^n = 2^6\\\\n = 6$

Which is fairly trivial.

Or you can use logarithms to solve for the exponent.

$2^n = 64\\\\\text{Log}\left(2^n\right)=\text{Log}\left(64\right)\\\\n*\text{Log}\left(2\right)=\text{Log}\left(64\right)\\\\n=\frac{\text{Log}\left(64\right)}{\text{Log}\left(2\right)}\\\\n\approx\frac{1.80617997398389} {0.30102999566399} \ \text{ ... using base 10 logs}\\\\n\approx5.99999999999983\\\\$