EMAT 4680

MIDTERM

a) Find a question for which the answer is particularly good.
Provide a link to it from your page. Then explain what it is

b) Find a question for which the answer is particularly poor.
Provide a link to it from your page. Then explain what it is
question yourself (naturally you will want to answer it in a
way you would consider 'good'), as if you were responding
to the original e-mail (only you are placing your response
on your web page--don't actually e-mail the person!).

This student asks two questions, first he asks what the next two terms in the sequence are, and second he asks what the formula for the nth term would be.

I think that Doctor Marko answers these questions very efficiently. He first tells Chris that he was right in guessing the next two terms, then he heads into the discussion of the more difficult question.

He first says what the formula would actually do, then he goes on to pose a problem. He says that yes, we can find the next term by dividing the previos one in half, but what about if we didn't know the previous term, or what if we didn't know a few of the preceding terms, what do we do then? This is a good question to pose because he leads Chris in the direction of learning the proper way to think about a problem like this.

He starts to walk through the calculations from the 1st term on step by step, in order to show the calculations. However, he makes an observation along the way, not only are we dividing each term by two, but we are dividing each term by a power of two, i.e.

" the nth term is equal to 32/(2)^n."

Right there, would have been enough. He could have stopped and Chris would have had a perfectly good formula for the nth term, but to give him a better formula he takes it down to saying that 32 is also 2 to the 5th power, so he could rewrite this problem as (2)^5/(2)^n.

Again, this would have been enough to give Chris a correct formula, however to give him a review of some exponents he tells him that this is actually equal to (2)^5-n.

In conclusion, Doctor Marko gave Chris a good direct answer to his question. He made the answer as clear as he could. He also tried to help lead Chris in the right direction when dealing with problems like this in the future. He also gave Chris a little review of exponents by taking this problem down to it's simplest form.

I would first like to note that this question is only an elementary school question. However, I think that the question was answered very poorly, and I really wanted to point out the flaws with this response.

To start off with, Doctor Andrew has gone through his steps incorrectly. In the body of the multiplication, he has put his carrying digits in the wrong place. This seems like a simple mistake to you and I, but when you think that an elementary student asked this question, he was probably confused by this.

The Doctor also fails to explain why he executed the steps that he did. Again, older students would probably understand this, but what about an elementary student who asked what is 697*56? He probably asked this because he gets confused with the steps of multiplication. All the Doctor says in the way of explanation is: <-- Carries from the 6. What does this mean? The "4" actually carries over from the fact that 7*6 is 42, "Carries from the 6" does not come close to explaining this to an elementary school student who is having trouble multiplying.

This, coupled with the fact that he put all of this in one step instead of splitting the problem up, probably didn't help Chris get what he wanted. Sure, he now knows that 697*56 = 39032, but does he really know why?

I pose the following as my solution to Chris' question:

Hi Chris,

In order to answer your question I will show you two ways to multiply numbers.

```1) a) 697 --> b)   54  --> c)   43  -->  d)   697
* 56          697          697          * 56
* 56         * 56          4182
4182         4182        +34850
+34850         39032```

Step a) this is how we set up a multiplication problem

Step b) 7*6 is 42. The 2 goes in the ones column, and the 4 gets carried to the next multiplication, 6*9 is 54 +4 is 58. The 8 goes in the next column, and the 5 gets carried over. 6*6 is 36 + 5 is 41, both numbers get dropped down since there are no more numbers to multiply.

Step c) We drop the 0 into the ones column and repeat step b, using the 5 this time.

Step d) here we add the resulting numbers together to get our answer of 39,032.

This takes you through multiplication step by step.

2) Here is another way to think about multiplication. It might make it easier to think about how we multiply numbers.

We have 697*56. It might be easier to break this multiplication up into parts.

697*56 means that we want to add 697, 56 times. So lets break this up:

We know that 600 + 97 = 697. So let's make the multiplication be (600*56) plus (97*56).

I know that 697 is more complicated to multiply than 600 , so this might be easier.

Let's also change the 97 to 90 + 7. So now we have (600*56) + (90*56) + (7*56).

It will probably be easier for you to multiply these three numbers and add them up, then it might be easier to see what 697*56 is asking for.

If you are starting to see how this works, you can tell that we can break up the 56 as well. I will give you a more broken up version of the original problem, and you can multiply the numbers out to see if you get the same answer. Try this:

(600*50) + (600*6) + (90*50) + (90*6) + (7*50) + (7*6) = 39,032 ?

I hope this has answered your question, and helped you to make tough numbers easier to multiply.

P.S. Why does this method work as well?

What are the similarities between this and the first method?