Klang

New member
Well this is embarrassing but I'm struggling with reading the bellow equation.

So t' is adjusted time with new filtration and t = current exposure time
The dots signify multiplication and the line is division but in what order would the sum be done as I cannot seem to get any where near the answer in the example?

Thank you

Subhotosh Khan

Super Moderator
Staff member
You said, "I cannot seem to get any where near the answer..."

Klang

New member

Hi,

Well if I take the top row for example and multiply them i.e. 1.17x2.01x1.00 and then do the same for the bottom 1.14 etc, then divide the totals of each. But that doesn't take into account the time etc

Or is it t i.e. 10 x 1.17 divided by 1.14

Subhotosh Khan

Super Moderator
Staff member
Hi,

Well if I take the top row for example and multiply them i.e. 1.17x2.01x1.00 and then do the same for the bottom 1.14 etc, then divide the totals of each. But that doesn't take into account the time etc

Or is it t i.e. 10 x 1.17 divided by 1.14
Please rewrite the complete equation with which you are working and then show your work.

firemath

Full Member
Well this is embarrassing......
Don't be embarrassed. That will not help you win the struggle. Everyone needs help sometimes.

HallsofIvy

Elite Member
Hi,

Well if I take the top row for example and multiply them i.e. 1.17x2.01x1.00 and then do the same for the bottom 1.14 etc, then divide the totals of each. But that doesn't take into account the time etc

Or is it t i.e. 10 x 1.17 divided by 1.14
Yes, of course- your equation clearly says t times the fraction $$\displaystyle \frac{km'\cdot ky'\cdot kc'}{km\cdot ky \cdot kc}$$ and each example, with t= 10, shows 10 times the fraction.

Klang

New member
Yes, of course- your equation clearly says t times the fraction $$\displaystyle \frac{km'\cdot ky'\cdot kc'}{km\cdot ky \cdot kc}$$ and each example, with t= 10, shows 10 times the fraction.
Thank you got it.

10x(1.17x2.01x1.00)÷(1.14x1.69x1.00)=12.20 Seconds

I guess to expand on that so the above gives me the correct answer when inputted into a scientific calculator but out of interest in what order would this equation be performed when using a non scientific calculator i.e. written by hand or with a calculator without the brackets function. Does that question make sense?

Subhotosh Khan

Super Moderator
Staff member
Thank you got it.

10x(1.17x2.01x1.00)÷(1.14x1.69x1.00)=12.20 Seconds

I guess to expand on that so the above gives me the correct answer when inputted into a scientific calculator but out of interest in what order would this equation be performed when using a non scientific calculator i.e. written by hand or with a calculator without the brackets function. Does that question make sense?
You may do it as follows:

1.17*2.01 = 2.3517

2.2517*1 = 2.3517

2.3517 / 1.14 = 2.0628947368421

2.06289 / 1.69 = 1.2206449704142

and so on....

HallsofIvy

Elite Member
Thank you got it.

10x(1.17x2.01x1.00)÷(1.14x1.69x1.00)=12.20 Seconds

I guess to expand on that so the above gives me the correct answer when inputted into a scientific calculator but out of interest in what order would this equation be performed when using a non scientific calculator i.e. written by hand or with a calculator without the brackets function. Does that question make sense?
Since only multiplication and division are involved, the order doesn't matter!a

If there were multiplications mixed with additions, then using "Please Excuse My Dear Aunt Sally" or "PEMDAS", 'parentheses", "exponents", "multiplication", "division", "addition", "subtraction", you would do all multiplications before the additions.

Subhotosh Khan

Super Moderator
Staff member
Since only multiplication and division are involved, the order doesn't matter!a

If there were multiplications mixed with additions, then using "Please Excuse My Dear Aunt Sally" or "PEMDAS", 'parentheses", "exponents", "multiplication", "division", "addition", "subtraction", you would do all multiplications before the additions.
Just be careful with those divisions - without parentheses.