Finite Math Examples

$\frac{0.3 (m cg)}{\frac{kg}{\min}} \cdot 1.8 kg$

Step 1

Multiply the numerator by the reciprocal of the denominator.

$0.3 m cg (\frac{\min}{kg}) \cdot 1.8 kg$

Step 2

Combine $0.3 m cg$ and $\frac{\min}{kg}$ .

$0.3 m (\frac{cg \min}{kg}) \cdot 1.8 kg$

Step 3

Convert $cg$ to $kg$ .

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Step 3.1

Convert $cg$ to $kg$ .

$0.3 m (\frac{cg \min}{kg} \cdot \frac{1 g}{100 cg} \cdot \frac{1 kg}{1000 g}) \cdot 1.8 kg$

Step 3.2

Cancel the common units and simplify.

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Step 3.2.1

Cross cancel units.

$0.3 m (\frac{cg \min}{kg} \cdot \frac{1 g}{100 cg} \cdot \frac{1 kg}{1000 g}) \cdot 1.8 kg$

Step 3.2.2

Remove the canceled units.

$0.3 m (\frac{\min}{kg} \cdot \frac{1}{100} \cdot \frac{1 kg}{1000}) \cdot 1.8 kg$

Step 3.2.3

Cross cancel units.

$0.3 m (\frac{\min}{kg} \cdot \frac{1}{100} \cdot \frac{1 kg}{1000}) \cdot 1.8 kg$

Step 3.2.4

Simplify the expression.

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Step 3.2.4.1

Remove the canceled units.

$0.3 m (\frac{\min}{1} \cdot \frac{1}{100} \cdot \frac{1}{1000}) \cdot 1.8 kg$

Step 3.2.4.2

Combine fractions.

$0.3 m (\frac{1}{1000 \cdot 100} \min) \cdot 1.8 kg$

Step 3.2.4.3

Multiply $1000$ by $100$ .

$0.3 m (\frac{1}{100000} \min) \cdot 1.8 kg$

Step 3.2.5

Convert to decimal.

$0.3 m (0.00001 \min) \cdot 1.8 kg$

Step 4

Multiply $0.00001$ by $0.3$ .

$0.000003 m \min \cdot 1.8 kg$

Step 5

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, 1$

Step 6

Since $1, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1$ then find LCM for the variable part .

Step 7

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 8

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 9

The LCM of $1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$