simplify the mathematical expression to determine the appropriate india product or quotient in a scientific notation round round do the 1st fractor goes to the 10th place 3.3x10^3 7.8 3.7 2.1x10^-3

From the power rules, the matching of each expression with its result is:
[tex](8.6*10^7)(9.1*10^{-8})=7.8[/tex]
[tex](6.9*10^5)(4.8*10^{-3})=3.3*10^3[/tex]
[tex]\frac{(3.7*10^2)*(4.6*10^{-3})}{(1.4*10^{-6})*(5.7*10^{8})} =2.1*10^{-3}[/tex]
[tex]\frac{(3.1*10^5)*(5.3*10^{-9})}{(7.3*10^{2})*(6.1*10^{-7})}=3.7[/tex]
There are different power rules, see some them:
1. Multiplication with the same base: you should repeat the base and add the exponents.
2. Division with the same base: you should repeat the base and subctract the exponents.
3.Power. For this rule, you should repeat the base and multiply the exponents.
4. Zero Exponent. When you have an exponent equals to zero, the result must be 1.
From this for your question, you have:
[tex]\frac{(3.7*10^2)*(4.6*10^{-3})}{(1.4*10^{-6})*(5.7*10^{8})} =\frac{(3.7*4.6)*(10^2*10^{-3})}{(1.4*5.7)*(10^{-6}*10^{8})} =\frac{17.02*10^{-1}}{7.98*10^2} =2.13*10^{-3}=2.1*10^{-3}[/tex]
[tex]\frac{(3.1*10^5)*(5.3*10^{-9})}{(7.3*10^{2})*(6.1*10^{-7})}=\frac{(3.1*5.3)*(10^{5}*10^{-9})}{(7.3*6.1)*(10^{2}*10^{-7})} =\frac{16.43*10^{-4}}{44.53*10^{-5}}=0.36896*10^1=3.7[/tex]
Read more about power rules here:
brainly.com/question/12140519
$SPJ1