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**5 members have voted**

July 25th, 2020 at 3:21:49 PM
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One morning it starts to snow at a constant rate. Later, at 6:00am, a snowplow sets out to clear a straight street. The plow can remove a fixed volume of snow per unit time, in other words its speed it inversely proportional to the depth of the snow. If the plow covered 10 miles from 6 to 7am and 5 miles from 7 to 8am, what time did it start snowing?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

July 25th, 2020 at 3:37:56 PM
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Let m be the number of minutes before 6 AM when it started snowing

The amount of snow cleared in the first 10 miles = 10m + 5 x 60

The amount of snow cleared in the next 5 miles = 5 (m + 60) + 5/2 x 60

10m + 300 = 5m + 300 + 150

m = 30, so it started snowing at 5:30 AM

July 25th, 2020 at 4:41:38 PM
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Quote:Wizard...what time did it start snowing?...

that it started snowing at approximately 5:23 AM.

I see that the answer has something to do with the golden ratio--the snow started before 6:00 AM by one hour divided by the golden ratio.

I see that the answer has something to do with the golden ratio--the snow started before 6:00 AM by one hour divided by the golden ratio.

July 25th, 2020 at 4:46:19 PM
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Forgot way to much to solve it the intended way tonight...

But my excel-solution seems to converge towards (phi-1) hours before 6

So i guess it started 06:22:55

6:30 is for sure not correct

But my excel-solution seems to converge towards (phi-1) hours before 6

So i guess it started 06:22:55

6:30 is for sure not correct

July 25th, 2020 at 5:19:40 PM
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Quote:ChesterDogthat it started snowing at approximately 5:23 AM.

I see that the answer has something to do with the golden ratio--the snow started before 6:00 AM by one hour divided by the golden ratio.

I agree. Remember, for full credit you must show your work.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

July 25th, 2020 at 5:57:16 PM
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A try..

I assume it started to snow at time 0:

The plow have a speed at 1/X

so f(X) = 1/x

We have that the definite integral from x to (x+1) = 2 times the definite integral from (x+1) to (x+2)

The integral for 1/x = ln(x)

That gives:

ln(x+1) - ln(x) = 2*( ln(x+2) - ln(x+1) )

3*ln(x+1) -ln(x) = 2*ln(x+2)

3*ln(x+1) = 2*ln(x+2)+ln(x)

ln((x+1)^3) = ln(x) + ln((x+2)^2)

ln((x+1)^3) = ln((x+2)^2 * x)

ln(x^3+3x^2+3x+1)= ln(x^3+4x^2+4x)

x^3+3x^2+3x+1= x^3+4x^2+4x

x^2 +x-1 = 0

X = -1,618 and 0,618

As it started to snow before the plow started, the solution have to be positive and the answer is 0,618.

=it started to snow at 06:22:55

Sorry my poor english and formating.

July 25th, 2020 at 7:34:30 PM
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Quote:TorghattenForgot way to much to solve it the intended way tonight...

But my excel-solution seems to converge towards (phi-1) hours before 6

So i guess it started 06:22:55

6:30 is for sure not correct

Quote:Torghatten

A try..

I assume it started to snow at time 0:

The plow have a speed at 1/X

so f(X) = 1/x

We have that the definite integral from x to (x+1) = 2 times the definite integral from (x+1) to (x+2)

The integral for 1/x = ln(x)

That gives:

ln(x+1) - ln(x) = 2*( ln(x+2) - ln(x+1) )

3*ln(x+1) -ln(x) = 2*ln(x+2)

3*ln(x+1) = 2*ln(x+2)+ln(x)

ln((x+1)^3) = ln(x) + ln((x+2)^2)

ln((x+1)^3) = ln((x+2)^2 * x)

ln(x^3+3x^2+3x+1)= ln(x^3+4x^2+4x)

x^3+3x^2+3x+1= x^3+4x^2+4x

x^2 +x-1 = 0

X = -1,618 and 0,618

As it started to snow before the plow started, the solution have to be positive and the answer is 0,618.

=it started to snow at 06:22:55

Sorry my poor english and formating.

Really? Twice you have him plowing BEFORE it starts snowing??? I think you missed a digit or something.

This math is was beyond my abilities, but I know that’s wrong.

I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁

July 25th, 2020 at 8:19:16 PM
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Quote:Torghatten

A try..

I assume it started to snow at time 0:

The plow have a speed at 1/X

so f(X) = 1/x

We have that the definite integral from x to (x+1) = 2 times the definite integral from (x+1) to (x+2)

The integral for 1/x = ln(x)

That gives:

ln(x+1) - ln(x) = 2*( ln(x+2) - ln(x+1) )

3*ln(x+1) -ln(x) = 2*ln(x+2)

3*ln(x+1) = 2*ln(x+2)+ln(x)

ln((x+1)^3) = ln(x) + ln((x+2)^2)

ln((x+1)^3) = ln((x+2)^2 * x)

ln(x^3+3x^2+3x+1)= ln(x^3+4x^2+4x)

x^3+3x^2+3x+1= x^3+4x^2+4x

x^2 +x-1 = 0

X = -1,618 and 0,618

As it started to snow before the plow started, the solution have to be positive and the answer is 0,618.

=it started to snow at 06:22:55

Sorry my poor english and formating.

Great answer and solution Tor! I don't recall you being a member of the Beer Club yet, in which case welcome. This means I owe you a beer upon your next visit to Vegas.

To be perfectly honest, I asked this problem because I made a video going over the solution. This is my first math video in this kind of format so please go easy on me with your reviews.

Direct: https://www.youtube.com/watch?v=33KYd7OSKjI

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

July 25th, 2020 at 10:02:24 PM
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July 26th, 2020 at 6:12:47 AM
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Quote:DJTeddyBear

Really? Twice you have him plowing BEFORE it starts snowing??? I think you missed a digit or something.

This math is was beyond my abilities, but I know that’s wrong.

Sorry, misread and used 07-09 instead of 06-08.

Quote:Wizard

Great answer and solution Tor! I don't recall you being a member of the Beer Club yet, in which case welcome. This means I owe you a beer upon your next visit to Vegas.

To be perfectly honest, I asked this problem because I made a video going over the solution. This is my first math video in this kind of format so please go easy on me with your reviews.

Direct: https://www.youtube.com/watch?v=33KYd7OSKjI

Thx :)

Idk when im able/allowed to go to Vegas again, but most likely it will be next summer :p