Addition and Subtraction Operations - Video Tutorials & Practice Problems
Addition and subtraction of values written in scientific notation.
Addition and Subtraction
1
example
Addition and Subtraction
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1m
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So we're told here when you add or subtract values in scientific notation they must have the same exponents. Now we're gonna say the coefficients add or subtract but the coefficients remain constant. So if we're taking a look at these two examples here, we have a times 10 to the x minus b. Times 10 to the x. The exponent portion must remain constant or answer at the end will be still times 10 to the X. And here we subtract our two coefficients, it would be a minus B when we're adding them, it would be a plus B Times 10 to the X. Now realize here that um you could easily put this into your calculator, get an answer at the end but if your professor is asking um to do that without the use of a calculator. This is the method that we have to use to get the right answer. Now we're gonna see if the exponents are not the same, then we transform the smaller value so that they do. And we're gonna say remember when adding or subtracting values that the final answer must have the least decimal places. Okay, so we'll have the least decimal places when we're doing either a minus B or a plus B. Using this logic will attempt to do the examples below. Once we've gotten a handle of it, you'll attempt to do some on your own. Okay, so come back and look at the next video and see how I approach the example problem listed right below us
If the exponents are not the same then we transform the smaller value so that they do.
2
example
Addition and Subtraction
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2m
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So here it says using the method discussed above determine the answer to the following question. So here we have 8.17 times 10 to the eight plus 1.25 times 10 to the nine. Remember the smaller value? We have to manipulate its exponent so that it matches up with the larger value here. 10 to the nine is larger. So I need to change this 10 to the eight also to 10 to the nine. So we need this to increase by one in order for that to increase by one. That means that our coefficient this portion has to decrease by one. That means I'm gonna take the decimal point here and move it over by one, decreasing that by one increases my exponent by one. Remember there's an inverse relationship between my coefficient and my exponent. So plugging what we have we're gonna have .817 times 10 to the nine And that's gonna be plus 1.25 times 10 to the nine Because they're both 10-9. Now we can just bring that down where it's constant. And then we're gonna add this value to this value here. When we do that, that's gonna give me 2.067. But remember when you're adding or subtracting its least number of decimal points that we have to have for our coefficient here. This has three decimal points or decimal places .817. This one has two decimal places .25. So my answer at the end has to have two decimal places. So we're gonna keep this portion here because this is a seven here, we'll have to round this one up At the end, that's gonna give me 2.07 times 10 to the nine as my final answer. So just remember this is the method that we have to use in order to get the correct answer. Now example to attempt to do example to on your own if you get stuck, don't worry, just come back and see how I approach that same exact question. So go ahead, try it out, come back and see what I do.
Remember when adding or subtracting values that the final answer must have the least number of decimal places.
3
example
Addition and Subtraction
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2m
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So remember the largest expo and we're gonna hold constant and we have to convert the smaller ones to match it here. The largest expo that we have is 10 to the negative 11. So we're gonna have to convert the other two to match it. So this is gonna stay still so we're gonna keep it as is this one we need to bring it down by one so it matches 10 to the negative 11. So it has to increase by going from negative 12 to negative 11 by one. If it's increasing by one that means that the coefficient has to decrease by one. We're gonna move the decimal 10.1 over. So that's gonna be minus 0.117 times 10 to the negative 11. This one it's negative 13. We're gonna have to increase it by two so that it goes and becomes a negative 11. That means I'm gonna have to decrease this one by two. So 12. So this is gonna be .0335 times 10 to the -11. So you got 10 to the negative 11 for all of them. That part's gonna remain constant here. We're gonna subtract all of these from one another. So by subtracting from one another we have 8.92 95 times 10 to the -11. Here. This one has two decimal places. This one here has three decimal places and this one here has four decimal places we want two decimal places at the end. And because there's a nine here we're gonna have to round up To 8.93 times 10 to the -11. So that'll be our final answer there. Now. We we all know in terms of rounding this is the approach we have to take. But just for future references, let's say we got a point 9 to 5 flat Here because this is five, were traditionally traditionally thought taught to increase it by one. But when it comes to analytical chemistry, if it ends with just five flat with nothing afterwards or like just 50 with a bunch of Zeros, we're gonna keep this number the same. It'd be 8.92 4 to be rounded up. It has to be like 51 or 52 or 53, something larger than 50 here in order for us to round up to the next highest value for this too. But again, if it's just five or five trailed by a bunch of Zeros, we went round up, we just keep it the same exact value, so it stay as 8.92. Okay, we haven't seen that example um as of yet, but just remember if it did pop up this is the approach that we have to take in terms of rounding up or holding the number of the same
