how do i solve 8E+05 * 3E+13
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okay, thank you for the help. can you help me solve this math problem?
Is your question \[\huge 8^5 \times 3^{13}\]
yes, we are currently dealing with scientific notation, so that makes sense. how do i solve? i have a few other questions just like it and i'm super confused :/
Are you there?
Yes I'm here
Hold on.
Okay, so what is 8 times 3?
It should be this right? \[\huge(8 \times 10^5) \times( 3 \times 10^{13})\]
hmm. im not sure. that looks right too D:
Lawl how are we supposed to do then? ;) Wait you've said you were in a scientific notation class right now?
Okay so its the second one
well, industry and trades. starting with scientifc notation so yes
Yeah so, first multiply 8 and 3
24
Okay, now, as for exponents, you'll have to add them, what is 5 plus 13?
18
Okay now you get \[\huge 24 \times 10^{18}\] This is not your answer, you'll have to put it in decimal coz there's 'tens', you need 'ones' So when you decrease the number 24 to 2.4, you add 1 to the exponent, so became 19 \[\huge 2.4 \times 10^{19}\] Done
thank you so much! can i ask for help on one more, a little more challenging?
Okay :) you can hit me with anything, I'm up till university level :p
haha must be nice! the next problem is 2.3E-05 / 1.15E-10
Okay, so now, you'll have to divide 2.3 with 1.15
so 2
Then, you've notice that its dividing, so exponents are to subtract, what is (-5)-(-10)?
yeah i have no idea. haha
Well, a minus and a minus gives a plus, so -5-(-10) turns into -5+10 and that you get 5
The whole thing turns into \[\huge 2 \times 10^5\]
okay. 8th grade is comin back to me now
This might be useful \[(+)(+)=(+) \\ (+)(-)=(-) \\ (-)(+)=(-) \\ (-)(-)=(+)\]
yes it is. i remember that now. so will i then just do 20 times itself five times?
No don't do that, that is \[2^5=2 \times 2 \times 2 \times 2 \times 2=32\] If its \[2 \times 10^5 \] means \[2 \times 10 \times 10\times 10\times 10\times 10=200000\]
oh my lord. alright thank you so much for all your help. if i have questions again how do i request you? or can i?
Of course c:
and you're welcome :)
Thank you! expect to hear from me soon haha! have a good day
good luck and have fun :)
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