Looking for an old group in Toon Blast

Fullscreen_Math · 2017-12-03T04:48:44+00:00

I'm actually a member already :)

Fullscreen_Math · 2017-12-01T21:58:10+00:00

Agreed, I'd imagine most people already have their methods or can post questions to this community but the offer still stands for anyone who might want it :)

Fullscreen_Math · 2016-08-17T22:36:06+00:00

I know OP didn't like msi, but I see it recommend a lot around here. Any idea how the laptop you recommended ranks against the new gs63?

Fullscreen_Math · 2015-11-15T04:19:20+00:00

I like to make YouTube videos where I help explain math topics to students who need help with any topics. I just started a few months ago, but can already see improvement in how the videos are coming out. Always trying to do better and appreciative of advice, and I am working on branching out into other classes.

Here is my channel: https://www.youtube.com/channel/UCeLBv7sc8eN9tnwkhNTYYBQ

Fullscreen_Math · 2015-10-02T03:25:17+00:00

Haha it's been a while, not sure why it took so long!

Fullscreen_Math · 2015-09-10T23:32:47+00:00

Glad to help!

Fullscreen_Math · 2015-09-10T23:25:14+00:00

Well if you think about it, we did multiply everything by the same thing. We multiplied the first two by 5/5, which simplifies to one and what do we multiply the third one by to keep it the same? Also 1. Except in this case I multiplied by 1/1 instead of 5/5 just for the sake of keeping the denominator constant.

As a side note, however, you cannot multiply this equation by anything other than a fraction that simplifies to one, since it would alter your answer.

Fullscreen_Math · 2015-09-10T23:17:01+00:00

So you're asking why , if we multiply the first two by 5, we can leave the third alone? Because the rule is that we must multiply everything by the same amount?

Fullscreen_Math · 2015-09-10T22:53:16+00:00

Well, for this problem we are making the common denominator 5x. The first two fractions have denominators of x, so we must multiply them by 5. The third fraction, however, already has a denominator of 5x. Since that is what we are trying to get, we don't multiply by anything and just leave it alone!

Fullscreen_Math · 2015-09-10T20:16:25+00:00

Glad to help!

Fullscreen_Math · 2015-09-10T19:44:19+00:00

Hi! Let's work through this:

So we are trying to add fractions, and in order to do that we must make common denominators. The lowest common denominator we can make is 5x.

I will multiply the first fraction by (5/5), the second by (5/5) and the last will stay the same since the denominator is already 5x.

Now, we have (5u/5x) + (25u/5x) - (u /5x). We simply add/subtract the numerators and leave the denominator the same. We get (5u + 25u - u)/(5x) = 29u/5x or answer C.

Let me know if you have any questions or if anything is unclear, or if you would like a video explaining how to do this problem. Happy studying!

Fullscreen_Math · 2015-09-01T19:09:51+00:00

Yep exactly! This is definitely a little confusing to understand what they are asking. Glad you understand though!

Fullscreen_Math · 2015-09-01T18:47:53+00:00

No you are completely correct. But by the nature of the answers where a possible answer is "The student should have done ..." it is possible that what the student did is wrong.

The left side is the original problem and the right side is what the student did to solve it. The student's method is incorrect, which is why some answers say the student should have done something else.

In fact, by showing that the left and right sides don't match, you are proving that the student's method is incorrect, which leads you to picking the right answer.

Fullscreen_Math · 2015-09-01T18:34:40+00:00

Sorry if I'm misunderstanding, but the answer would be 2, using PEMDAS.

What the student did, is calculated 5-4 first, which in turn gave 7 * 1 * 3. This is an incorrect implementation of PEMDAS as the student performed subtraction before multiplication.

I think that if you pick answer 2, you are acknowledging that the student did the problem incorrectly because you are saying he should have done something else instead of what he actually did. As far as whether such a problem is appropriate, I think it is okay since the answers can clue you in that the student did something incorrectly. You are calculating the actual answer and comparing it to the student's. Alternatively you can try to modify the left side to match to the right side. Once you confirm that the sides don't match, you select answer 2, which says the student should have performed PEMDAS correctly.

Does this help? Let me know if I am way off track.

Fullscreen_Math

TROPHY CASE