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Mathematics 17 Online
OpenStudy (anonymous):

How would I put this decimial into scientific notation? Please help... 0.00000004049 Is this correct? 4.049 x 10^8

OpenStudy (lgbasallote):

close..it's -8 because the decimal is on the left ;)

Parth (parthkohli):

Hmm but the exponent shouldve been negative

Parth (parthkohli):

Yeah, when it's 1 over something, we have a negative exponent.

OpenStudy (campbell_st):

just count the number of places you move the decimal point to the right to get a number between 1 and 10. that will be the power... just needs the negative for a decimal

OpenStudy (anonymous):

So it's 4.049 x 10^-8 ...That explains why got it wrong, but how do I know if it should be negative or positive... these always confuse me... @campbell_st @ParthKohli @lgbasallote

Parth (parthkohli):

Well it's easy. First convert this into fraction.

OpenStudy (lgbasallote):

i remember it this way 2 x 10 = 20 2x x 10^2 = 2 x 100 = 200 therefore i can conclude that if it's positive exponent then the zeroes to the right increase ;)

OpenStudy (lgbasallote):

negative exponent is opposite of course

Parth (parthkohli):

\( \color{Black}{\Rightarrow \Large {1 \over 4.049 \times 10^8} }\) Now, as 1/(x^y) = x^(-y), we'll have 4.049 * 10^-8

OpenStudy (anonymous):

I never get fractions :( I still don't get why it is negative ? @ParthKohli @lgbasallote

OpenStudy (lgbasallote):

1 DIVIDED by 10 (10 is 10^1) = 0.1 1 DIVIDED by 100 (100 is 10^2) = 0.01 1 DIVIDED by 1000 (1000 is 10^3) = 0.001 is that making sense?

OpenStudy (lgbasallote):

\[\large \frac{1}{10} = 10^{-1}\] \[\large \frac{1}{100} = 10^{-2}\] \[\large \frac{1}{1000} = 10^{-3}\]

Parth (parthkohli):

\( \color{Black}{\Rightarrow \Large {1 \over x^y} = x^{-y} }\) This is the identity. Do you understand now?

OpenStudy (anonymous):

Yesss! :) Thanks everyone! :) @ParthKohli @lgbasallote @campbell_st

Parth (parthkohli):

yw! ^_^

OpenStudy (lgbasallote):

most welcome ^_^ \[\huge \color{maroon}{\mathtt{\text{<tips hat>}}}\]

Parth (parthkohli):

OpenStudy (anonymous):

haha Thats cute!!! :-* @ParthKohli @lgbasallote

Parth (parthkohli):

lol

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