Mathematics
OpenStudy (anonymous):

I need to calculate the scientific notation 8.3 x 10e6 kg/2.5 x 10e9 cme3. I can't convert the kg into cm. Anyone know how to do this problem?

OpenStudy (anonymous):

kg doesnt convert to cm, kg is a weight where as cm is a distance does that make sense? is the problem given to you in fraction form? if so you can have your final answer be kg/cm

OpenStudy (anonymous):

Looking at it it will be $\frac{kg}{cm^3}$

OpenStudy (anonymous):

The problem can be viewed as: $\frac{8.3}{2.5}*\frac{10^6}{10^9}*\frac{kg}{cm^3}$

OpenStudy (anonymous):

cm^3 is a volume so that can be done and $1cm ^{3} = 1ml$

OpenStudy (anonymous):

wait a thats a volume not a weight. okay either way there not going to cancel out so you'll multiply the top across and the bottom across and then your lable will just be kg/$cm ^{3}$

OpenStudy (anonymous):

That's a density problem (how much a known volume weighs) and kg*cm^-3 is considered a derived unit. $\left(\begin{matrix}8.3*10^6 \\ 2.5*10^9\end{matrix}\right)kg*cm ^{-3}=3.32*10^{-3} kg*cm ^{-3}$
The text passage contains a mathematical question about calculating a scientific notation expression and determining the units of the result. The question is answered in the text, including the final calculated value.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Mathematics
OpenStudy (anonymous):

I need to calculate the scientific notation 8.3 x 10e6 kg/2.5 x 10e9 cme3. I can't convert the kg into cm. Anyone know how to do this problem?

OpenStudy (anonymous):

kg doesnt convert to cm, kg is a weight where as cm is a distance does that make sense? is the problem given to you in fraction form? if so you can have your final answer be kg/cm

OpenStudy (anonymous):

Looking at it it will be $\frac{kg}{cm^3}$

OpenStudy (anonymous):

The problem can be viewed as: $\frac{8.3}{2.5}*\frac{10^6}{10^9}*\frac{kg}{cm^3}$

OpenStudy (anonymous):

cm^3 is a volume so that can be done and $1cm ^{3} = 1ml$

OpenStudy (anonymous):

wait a thats a volume not a weight. okay either way there not going to cancel out so you'll multiply the top across and the bottom across and then your lable will just be kg/$cm ^{3}$

OpenStudy (anonymous):

That's a density problem (how much a known volume weighs) and kg*cm^-3 is considered a derived unit. $\left(\begin{matrix}8.3*10^6 \\ 2.5*10^9\end{matrix}\right)kg*cm ^{-3}=3.32*10^{-3} kg*cm ^{-3}$
user
The text passage contains a mathematical question about calculating a scientific notation expression and determining the units of the result. The question is answered in the text, including the final calculated value.

TYPE: SINGLE_QUESTION_WITH_ANSWER