Question

I need help with the problem shown at http://pbckt.com/pK.uBYzsG .

The link is not spam.

Thanks.

Answer 1

@saifoo.khan

Answer 2

An easy way to do these is to remember that for each time you move the decimal point the right, you subtract 1 from the exponent on the 10.

Each time you move the decimal point to the left, you add 1 to the exponent on the 10.

Answer 3

The exponent is NEGATIVE. So this means the value will be very very small.

Answer 4

ah. okay.

Answer 5

so. what would the answer be?

Answer 6

j?

Answer 7

Well how many times do you need to move the decimal point? Try it like this:

if we move the decimal point once to the left, the number becomes:
\[0.25 \times 10^{-8}\]

Answer 8

oh.

Answer 9

h then.

Answer 10

saifoo said the amount would be small, so f and g are big therefore not the answers,no?

Answer 11

Yep, since it's 10^-9, it's a small number. We need to move the decimal point 9 places to the left.

Answer 12

okedoke:) i get it now:D

Answer 13

Which answer did you get?

Answer 14

h,no?

Answer 15

Try it to check. Write 2.5 and move the decimal point 9 places to the left, and then see how many zeroes there'll be.

Answer 16

okedoke, i am doing so right now

Answer 17

oh. 9

Answer 18

:0 its a miracle! lol

Answer 19

So it should look like this after moving it 9 times:
0.0000000025
which is answer...

Answer 20

h.

Answer 21

Count how many zeroes are between the decimal point and the 2

Answer 22

okay

Answer 23

And then compare that to how many zeroes are between the decimal and the 2 in:
0.0000000025

Answer 24

eight?

Answer 25

Yep, so which of h or j has eight zeroes between the decimal and the 2?

Answer 26

so. it Is j,

Answer 27

Yep.

Answer 28

thank you:)

Answer 29

No prob. And remember that for each time you move the decimal point the right, you subtract 1 from the exponent on the 10.

Each time you move the decimal point to the left, you add 1 to the exponent on the 10.

Answer 30

okedoke:)

Answer 31

It helps to remember it like this:



This way you can easily remember "okay, i need to move the decimal to the left, so i have to add 1 to the exponent for each time i move it" and vice versa if you're moving the decimal to the right, you subtract 1 from the exponent each time.

Answer 32

thanks.:)