Answer:
The right answer is "5.9". A further solving is provided below.
Step-by-step explanation:
The given values are:
Initial Magnitude,
M₁ = 3.9
Let Second magnitude be M₂.
Now,
⇒ 
On substituting the values, we get
⇒ 
On putting the value of "log 120", we get
⇒ 
On adding "3.9" both sides, we get
⇒ 
⇒ 
That will depend, you have to ask yourself first how many kilobytes one picture is. Let's just say that the size of one picture is 100 kb (which is the average size of a picture) Then first you multiply 100 kb with the number of pictures which is 43. Now you have a total used up memory of 4300 kb. After that, you minus the used up memory which is 4300 kb, to the total available space which is 32,834.5 and you will get an available space of 28534.5 kb. After that, you divide the remaining available space with the size of each picture. So this will be 28534.5 divided by 100. You will get 285. You can still take 285 pictures.
17000
This time its thousands not hundreds, so you do 17 plus 000 wich would be thousands, like this
one thousand
1
thousand
000
:)
Answer:
Step-by-step explanation:
25
