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VARVARA [1.3K]
2 years ago
5

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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Help asap pls and ty
PolarNik [594]

Answer:

B

Step-by-step explanation: i know

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3 years ago
What is the range of the function f(x) = 3x^2 + 6x - 8?
Airida [17]

R; all quadratic functions are going to have ranges and domains of *ALL REAL NUMBERS*.

3 0
3 years ago
How many elements does A contains if it has 64 subsets​
Amanda [17]
<h3>Answer:  6</h3>

===========================================================

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.

  1. {a,b,c} .... any set is a subset of itself
  2. {a,b}
  3. {a,c}
  4. {b,c}
  5. {a}
  6. {b}
  7. {c}
  8. { } ..... 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.

-------------------

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\\\\

Due to rounding error, we don't land exactly on 6 even though we should.

5 0
2 years ago
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Reil [10]
B) 1:3 

The ratio is 1 to 3


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3 years ago
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