Addition and Subtraction Calculations 2

Jules Bruno
Was this helpful?
using the method discussed above. Determine the answer to the following question. All right, so remember, we're gonna go with the largest value. The largest value we have here is 10 to the negative 11. That would mean my other two values have to be converted. So their exponents also become 10 to the negative 11. So this stays the same. So we're gonna bring it over now for the next number, it's 10 to the negative 12. So I needed needed to increase by one so that it matches 10 to the negative if I want to increase it by one. That means I need to make my coefficient smaller by one decimal place so I can move this decimal over here so it's gonna become negative 0.117 times 10 to the negative 11. This one is to the negative 13. I also needed to get it to 10 to the negative 11. But now I'm gonna have to move it to decimal places, so it's gonna go 12 so it's gonna come 0.335 times 10 to the negative 11. Now that all the exponents or 10 to the negative 11 that just comes down and stays constant. So it's gonna be this minus this minus this to give me 8.9295 But remember, when it comes toe adding and subtracting, we want the least number of decimal places. So here, this one here as two decimal places, this one here as three decimal places. This one here has four decimal places. So you want least number of decimal places. So we need to. So it's gonna become 8.93 times 10 to the negative 11 as my final answer. So that will be my answer when I'm converting all of my scientific notation values to the same exponents and then subtracting them from one another.