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Old 09-11-2009, 10:51 PM   #1
PCSkiBum_21
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Default The Official Homework Help Thread!!

Woot. I don't know why I haven't thought of this before.

Post up homework questions you might need help with. If you are bored and know how to solve or help out please chime in!

Well, here we go!
(i feel dumb but i can't figure this out)

Math:
"Two planes leave simultaneously from Salt Lake City International, one flying due north and one due east. The northbound plane is flying 60 miles per hour faster than the east bound plane. After 4 hours the planes are 1987 miles apart. The speed of the slower plane is ______ miles per hour. Round your answer to the nearest mile per hour. Assume the earth is flat, this is wildly inaccurate over a distance of 1987 miles!
-Hint: Use the Pythagorean Theorem."
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Last edited by PCSkiBum_21; 09-11-2009 at 10:56 PM.
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Old 09-11-2009, 10:55 PM   #2
PCSkiBum_21
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^ my attempts...

(pythagorean theorem says.. a^2 + b^2 = c^2 )

(60+x) + (x) = 1987 miles
2x + 60 = 1987
2x = 1947
x = 963.5

check: 963.5 + (963.5+60) = 963.5 + 1023.5 = 1987

but it doesn't accept my answer what am i missing?
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Old 09-11-2009, 10:57 PM   #3
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Way over my head.
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Old 09-11-2009, 11:22 PM   #4
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unimportant, what was here is.

Last edited by Himynameisgreg; 09-11-2009 at 11:46 PM.
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Old 09-11-2009, 11:33 PM   #5
PCSkiBum_21
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i wasn't factoring the "4 hours" into the equation.. i think that's part of why i'm off.
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Old 09-11-2009, 11:44 PM   #6
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winner winner chicken dinner!

319.97mph for the slower plane. i was way off the first attempts lol.

solution:
((4)(60+x))^2 + ((4)(x))^2 = 1987^2
16x^2 + 16x^2 + 960x + 960x + 57600 = 3948169
32x^2 + 1920x + 57600 = 3948169
32x^2 + 1920x - 3890569 = 0
(put into quadratic formula, which is hard to show by typing, but it worked)

Last edited by PCSkiBum_21; 09-11-2009 at 11:52 PM.
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Old 09-11-2009, 11:46 PM   #7
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What math are you in?
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Old 09-11-2009, 11:55 PM   #8
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^ just pre-calc haha. i'm dumb.
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Old 09-11-2009, 11:56 PM   #9
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Damn NVM.
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Old 09-12-2009, 01:16 AM   #10
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I won't even say what math i'm in. hahahahaha
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Old 09-12-2009, 02:28 AM   #11
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Totally just need all your kinematics equations for this one. All the ones I should know but for got.
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Old 09-12-2009, 02:38 AM   #12
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Reading this thread made me just realize how young a lot of you guys are... I feel old.
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Old 09-12-2009, 02:40 AM   #13
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If you ever need more math help I've taken 5 calculus classes.
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Old 09-12-2009, 02:46 AM   #14
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Quote:
Originally Posted by Not-EWRX View Post
Totally just need all your kinematics equations for this one. All the ones I should know but for got.
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I assume that when you say kinematic equations (physics) you mean algebra?
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Old 09-12-2009, 04:16 AM   #15
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Use the distance Formula:

D=RT
Distance=Rate x Time

and set up two equations trying to solve for one variable, such as Rate. Solve for rate and plug it back into one of the original equations to solve the other variables if needs be...or something like that. I'd do it but it's late, I'm tired and I'm going to bed
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Old 09-12-2009, 05:58 AM   #16
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Quote:
Originally Posted by Dragicon View Post
Reading this thread made me just realize how young a lot of you guys are... I feel old.
Hah. I'm not even that much older than them, but if it's any consolation it makes me feel old too.
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Old 09-12-2009, 10:29 AM   #17
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Default Re: The Official Homework Help Thread!!

Quote:
Originally Posted by Not-EWRX
If you ever need more math help I've taken 5 calculus classes.
I've taken five also. But I still can't pass it.

I'll be chiming in later with some questions.
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Old 09-12-2009, 11:42 AM   #18
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d=rt is basically my entire job...
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Old 09-12-2009, 11:58 AM   #19
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Quote:
Originally Posted by Dragicon View Post
I assume that when you say kinematic equations (physics) you mean algebra?
Go nuts.
http://en.wikipedia.org/wiki/Kinematics
Kinematics is basically the study motion. There are like four fundamental equations that anyone in the physics/ engineering world should know. With out them, these problems are impossible.

Or you can just remember how to intigrate things with respect to time.
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Old 09-12-2009, 02:16 PM   #20
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ummmm... i'm horrible at math. But if anyone needs help in a graphic design class, I'm your man!
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Old 09-12-2009, 07:24 PM   #21
Dragicon
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Quote:
Originally Posted by Not-EWRX View Post
Go nuts.
http://en.wikipedia.org/wiki/Kinematics
Kinematics is basically the study motion. There are like four fundamental equations that anyone in the physics/ engineering world should know. With out them, these problems are impossible.

Or you can just remember how to intigrate things with respect to time.
I know what the kinematic equations are. Why would you waste all kinds of time and effort with the kinematic equations when you can simply solve it with algebra like the OP did? I guess algebra can defy the impossible.... This is a 3 minute problem with pythagorean & algebra.

If somebody asks, "If I drive 3 hours @ 40 MPH, how far have I driven?", are you really going to do all kinds of BS work with kinematics, or are you going to do 40x3?
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Old 09-12-2009, 10:32 PM   #22
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fluent in Spanish, though my grammar has been a bit off, i'm no Latin but i'll help where i can
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Old 10-03-2009, 12:03 AM   #23
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question..

"You and your friend part at an intersection. Your friend drives away north at a constant speed. You linger at the intersection for 1 hour, then drive off due east at a speed that is 10mph faster than your friend's speed. 2 hours after your friend's departure the distance between the two of you is 130 miles. 5 hours after your friend's departure your distance from your friend is _____ miles."

At the first distance, I found that "your" speed is 56mph.
(2x)^2 + (x+10)^2 = 130^2

But for some reason I can't figure out what the distance is at 5 hours. I'm trying this...
(5x)^2 + (4x+10)^2

5 being total hours of driving for your friend, and 4 being total hours driving for you.. but it doesn't work. I is lost.
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Old 10-03-2009, 12:16 AM   #24
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Quote:
Originally Posted by bhhamblin View Post
Way over my head.
anything school related is over your head.
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Old 10-03-2009, 12:32 AM   #25
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Quote:
Originally Posted by PCSkiBum_21 View Post
question..

"You and your friend part at an intersection. Your friend drives away north at a constant speed. You linger at the intersection for 1 hour, then drive off due east at a speed that is 10mph faster than your friend's speed. 2 hours after your friend's departure the distance between the two of you is 130 miles. 5 hours after your friend's departure your distance from your friend is _____ miles."

At the first distance, I found that "your" speed is 56mph.
(2x)^2 + (x+10)^2 = 130^2

But for some reason I can't figure out what the distance is at 5 hours. I'm trying this...
(5x)^2 + (4x+10)^2

5 being total hours of driving for your friend, and 4 being total hours driving for you.. but it doesn't work. I is lost.
if your 56 mph is correct (I dont have a calculator other than this windows one ), i believe your logic for the second part is correct. I got 398.84 miles.
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