The Math Thread!

Lol, Don. Just lol.

Yoshidude10 said:
Me being a non-calculus person, I figured that since you rode 30 MPH for the first lap, you would need to find the number that has a mean of 60 when combined. Since 30 + 90 = 120, and 120 / 2 = 60, 90 "would" be the correct answer in my mind.
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Unfortunately, that's not how things work. To solve this problem, you would need a very important formula:
Code: 
Average Speed = Distance / Time
Let us see what the total average speed is when you go 90 MPH in the second lap. For the first lap we have:
Code: 
30 MPH = 1 mile / Time
(30 MPH) * Time = 1 Mile
Time = 1 Mile / 30MPH = 1/30 Hour = 2 minutes
So it takes two minutes to complete the first lap. If we go at 90 MPH at the second lap, which is three times as fast, we would finish the second lap in one third of the time it takes to complete the second lap which would be (1/30 Hour) / 3 = 1/90 of an hour. So this is what we have:
Code: 
Total Distance = 2 Miles
Total Time = 1/30 + 1/90 Hour = 2/45 Hour
Average speed = Distance / Time = (2 Miles) / (2/45 Hour) = 45 MPH
I omitted a few steps so if I went too fast, please let me know.
 
I haven't gotten that far. But now I know stuff some people in my class don't know! Thanks!~
 
No, I've just skimmed it (haven't actually gone through with paper to confirm), but everything seems to make sense.
 
Did you finish Algebra I, P99? I think you should be able to do the problem.
 
Right now,I'm in seventh grade,and I'm in algebra II.No joke.It's also no joke that at the moment,IT'S FREAKING HARD(eccept for the whole "order of operations" thing.Then it's a complete synch)!
 
The closest I got was 120, so when you plug it in, the MPH would be 48. I need some help ; would I need to increase my value, or decrease it? Sorry, I'm still in Algebra I.
 
If you want to increase the total average speed, you want to go faster. It's easier if you go backwards, actually. You know that you want a total average speed of 60 MPH and a total distance of two miles. You need to figure out how much total time you have to complete the two miles overall.

Once you know how much time you have to complete the two laps, subtract the total time with how much time it took to complete the first lap (which I uncovered was 1/30 hours or 2 minutes). The remaining time is how long you get to complete the second lap.

That's as much as I'll say for now. See if you can tell me this "remaining" time.
 
Yoshidude10 said:
I believe 90 MPH, but I'm probably wrong.

I'm taking Algebra I this year (for 8th grade), and then I take Geometry. Is Geometry harder than Algebra?
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I think they are both not that hard. I finished Algebra is all A's and Geometry isn't to hard, besides the Vocabulary. There is a lot. And I don't like Truth Tables.
 
Okay, I know that 60 MPH will mean 2 miles will be 2 MPH. Then I have the formula T = D / S, so T = 2 / 2, which is 1, which equals 60 minutes. If you subtract 2 from 60, you get 58 minutes.....? I know I'm doing something wrong.
 
Yoshidude10 said:
Okay, I know that 60 MPH will mean 2 miles will be 2 MPH. Then I have the formula T = D / S, so T = 2 / 2, which is 1, which equals 60 minutes. If you subtract 2 from 60, you get 58 minutes....? I know I'm doing something wrong.
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I'm not sure what that is, lol. Isn't S = 60 MPH as already givien?
 
I get it now, but I was confused about translating from P | Q to ~P and ~Q, because I didnt know what to put into the spaces between, but now I know ~p |~q.
 
Well, I simplified 60 / 1 to....oh wait, I'm stupid. But then it doesn't make sense, because T = D / S, which would then be T = 2 / 60 = 1/30 or 2 minutes. Oh well, I'm not in Calculus, so maybe this isn't the thing for me.
 
But then subtracting 2 from 2 would get you 0, so you have 0 minutes to complete the second lap....

Just to clarify, 2 minutes is the total amount of time you have to complete both laps, correct?
 
Yup! No matter how fast you go, you can't get an average speed of 60 MPH (unless you can complete the second lap instantly)! =D
 
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