Answered

Join the IDNLearn.com community and get your questions answered by knowledgeable individuals. Find the information you need quickly and easily with our comprehensive and accurate Q&A platform.

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

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 class=

Sagot :

Lhannearaujoidn

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]

Power Rules

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](8.6*10^7)(9.1*10^{-8})[/tex] -  From the distributive property of multiplication  and power rules you can do:[tex](8.6*10^7)(9.1*10^{-8})=(8.6*9.1)*(10^7*10^{-8})=78.26*10^{-1}=7.8[/tex]

  • [tex](6.9*10^5)(4.8*10^{-3})[/tex] -   From the distributive property of multiplication and power rules you can do:[tex](6.9*10^5)(4.8*10^{-3})=(6.9*4.8)*(10^5*10{-3})=33.12*10^2=3.3*10^3[/tex]

  • [tex]\frac{(3.7*10^2)*(4.6*10^{-3})}{(1.4*10^{-6})*(5.7*10^{8})}[/tex] -   From the distributive property of multiplication and power rules you can do:

[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})}[/tex] -   From the distributive property of multiplication and power rules you can do:

[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

We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.