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3 years ago

Which expression has the fewest significant figures?

Mathematics

2 answers:

pickupchik [31]3 years ago

5 0

Answer:

Step-by-step explanation:

yanalaym [24]3 years ago

4 0

Hello!

Here are some rules to determine the number of significant figures.

Numbers that are not zero are significant (45 - all are sigfigs)
Zeros between non-zero digits are significant (3006 → all are sigfigs)
Trailing zeros are not significant (0.067 → the first two zeros are not sigfigs)
Trailing zeros after a decimal point are always significant (1.000 → all are sigfigs)
Trailing zeros in a whole number are not significant (7800 → the last two zeros are not sigfigs)
In scientific notation, the exponential digits are not significant, known as place holders (6.02 x 10² → 10² is not a sigfig)

Now, let's find the number of significant figures in each given number.

A). 296.54

Since these digits are all <em>non-zero</em>, there are 5 significant figures.

B). 5003.1

Since the two <em>zeros are between non-zero digits</em>, they are significant figures. Thus, there are 5 significant figures.

C). 360.01

Again, the two zeros are between non-zero digits. There are 5 significant figures.

D). 18.3

All of these digits are non-zero, hence, there are 3 significant figures.

Therefore, expression D has the fewest number of significant figures being 3.

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What is the intermediate step in the form (x+a)2 = b as a result of completing the square for the following

skelet666 [1.2K]

For this case we have the following expression:

$x ^ 2-4x + 10 = 4x + 10$

We follow the steps below:

We subtract 4x on both sides of the equation:

$x ^ 2-4x + 10-4x = 4x + 10-4x\\x ^ 2-8x + 10 = 10$

We subtract 10 from both sides of the equation:

$x ^ 2-8x + 10-10 = 10-10\\x ^ 2-8x = 0$

Now, we must complete squares.

When we have an equation of the form:

$ax ^ 2 + bx + c = 0$ , if we want to complete squares we must subtract c on both sides of the equation obtaining:

$ax ^ 2 + bx + c-c = -c\\ax ^ 2 + bx = -c$

The square is completed by adding to both sides of the equation: $(\frac{b}{2})^{2}$

So, we have left:

$ax^2+bx + (\frac {b} {2})^{2} = - c + (\frac {b} {2}) ^ {2}$

In the given expression we have:

$a = 1\\b = -8\\c = 0$

And to complete the square we have:

$(\frac {b} {2}) ^ {2} = (\frac {-8} {2}) ^ {2} = 16$

Rewriting we have:

$x ^ 2-8x + 16 = 16$

We factor the left side of the equation, that is, we look for two numbers that when added together result in -8 and when multiplied as a result 16. We have:

$-4-4 = -8\\-4 * -4 = 16$