algebra division
- Thread starter soprano
- Start date
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,962
Is this what you're trying to describe?
\(\displaystyle \frac{38 m^{-15} n^{19}}{40 m^{11} n^5}\)
If so, here's how we type it, using a keyboard.
[38 m^(-15) n^19]/[40 m^11 n^5]
Otherwise, I cannot figure out what you've been given.
If this is the algebraic ratio to simplify, then we can first factor it into a product of ratios.
\(\displaystyle \frac{38}{40} \cdot \frac{m^{-15}}{m^{11}} \cdot \frac{n^{19}}{n^5}\)
Next, reduce the Rational number 38/40 to lowest terms, and also rewrite each of the other two factors as single powers by using the following property of exponents.
\(\displaystyle \frac{m^a}{m^b} = m^{a - b}\)
In other words, to simplify a ratio of two powers with equal bases, we subtract the lower exponent from the upper exponent.
EG:
\(\displaystyle \frac{m^7}{m^3} = m^{7 - 3} = m^4\)
\(\displaystyle \frac{38 m^{-15} n^{19}}{40 m^{11} n^5}\)
If so, here's how we type it, using a keyboard.
[38 m^(-15) n^19]/[40 m^11 n^5]
Otherwise, I cannot figure out what you've been given.
If this is the algebraic ratio to simplify, then we can first factor it into a product of ratios.
\(\displaystyle \frac{38}{40} \cdot \frac{m^{-15}}{m^{11}} \cdot \frac{n^{19}}{n^5}\)
Next, reduce the Rational number 38/40 to lowest terms, and also rewrite each of the other two factors as single powers by using the following property of exponents.
\(\displaystyle \frac{m^a}{m^b} = m^{a - b}\)
In other words, to simplify a ratio of two powers with equal bases, we subtract the lower exponent from the upper exponent.
EG:
\(\displaystyle \frac{m^7}{m^3} = m^{7 - 3} = m^4\)