What's his score? (Answer Page 4)
02-19-2009, 07:34 PM
#64
Nontypical Buck
Join Date: Oct 2005
Location:
Posts: 1,424
RE: What's his score? (Answer known - will post tomorrow)
This site is getting rediculous. Is everybody that gets on here just out to get everybody else? I think it's a nice looking buck regardless of score. I believe your measurements, you took them, what reason do you have to lie since I don't know you? I'm sick of everyone on this site bitching everyone else out because of something they don't know. IT'S A PICTURE PEOPLE!, unless you held the sheds, how in the holy hell do you know? I'm just pissed people are so damn arrogant, and rediculous towards each other.
He's gonna be even better this fall. 000hhhhhhh
02-26-2009, 05:09 PM
#66
Fork Horn
Join Date: Jan 2008
Location:
Posts: 343
RE: What's his score? (Answer known - will post tomorrow)
I just want to know what I won. 3 days on the property hunting this deer, a new sentenal, a trip down south to suck up some warm weather hog hunting, shopping spree at cabelas, etc. The anxioty is killing me. By the way, I cheated; pulled out my calculus book and solved the simple triple integral to calculate the volume and then solved for the circumference followed by the length. Everyone knows that you can solve for any equation if you have only 1 unknown. Here is the answer 
The volume of the three-dimensional region E is given by the integral ∫∫∫dv=∫∫∫dxdydz. We need to set the boundaries of each
variable. Let's take a look at our plane 2x+y+z=4, it can be written
as z=4-2x-y. So, we can claim that z varies from 0 to 4-2x-y. Now
we need to find the variety of x&y. To do so, let's find out what
shade or region this plane cover from the x-y plain :
In the x-y plane, z=0, meaning : 2x+y=4 --> y=4-2x. This is the equation of the line which represent the limit of the region. (note that the other limits are the lines x=0 & y=0). Now we know that y
varies from 0 to 4-2x. Good, but what about the boundaries of x ??
Well, when y=0 & z=0, x=2 (because 2x+0+0=4) So we can say that x
varies from 0 to 2. Let's sum up our calculations :
1. dz : 0 --> 4-2x-y
2. dy : 0 --> 4-2x
3. dx : 0 --> 3
Now we will move to calculating the triple integral :
∫∫∫dxdydz==∫∫[4-2x-y]dxdy=∫dx[4y-2xy-(y^2)/2]{y:0-->4-2x}=
=∫[4(4-2x)-2x(4-2x)-((4-2x)^2)/2]dx { x goes from 0 to 2 }
I'll leave it to you as an exercise to calculate this immediate
integral.
The volume of the three-dimensional region E is given by the integral ∫∫∫dv=∫∫∫dxdydz. We need to set the boundaries of each
variable. Let's take a look at our plane 2x+y+z=4, it can be written
as z=4-2x-y. So, we can claim that z varies from 0 to 4-2x-y. Now
we need to find the variety of x&y. To do so, let's find out what
shade or region this plane cover from the x-y plain :
In the x-y plane, z=0, meaning : 2x+y=4 --> y=4-2x. This is the equation of the line which represent the limit of the region. (note that the other limits are the lines x=0 & y=0). Now we know that y
varies from 0 to 4-2x. Good, but what about the boundaries of x ??
Well, when y=0 & z=0, x=2 (because 2x+0+0=4) So we can say that x
varies from 0 to 2. Let's sum up our calculations :
1. dz : 0 --> 4-2x-y
2. dy : 0 --> 4-2x
3. dx : 0 --> 3
Now we will move to calculating the triple integral :
∫∫∫dxdydz==∫∫[4-2x-y]dxdy=∫dx[4y-2xy-(y^2)/2]{y:0-->4-2x}=
=∫[4(4-2x)-2x(4-2x)-((4-2x)^2)/2]dx { x goes from 0 to 2 }
I'll leave it to you as an exercise to calculate this immediate
integral.
02-28-2009, 01:32 PM
#67
Fork Horn
Thread Starter
Join Date: Apr 2007
Location: West Virginia
Posts: 329
RE: What's his score? (Answer known - will post tomorrow)
ORIGINAL: niehenke
I just want to know what I won. 3 days on the property hunting this deer, a new sentenal, a trip down south to suck up some warm weather hog hunting, shopping spree at cabelas, etc. The anxioty is killing me. By the way, I cheated; pulled out my calculus book and solved the simple triple integral to calculate the volume and then solved for the circumference followed by the length. Everyone knows that you can solve for any equation if you have only 1 unknown. Here is the answer
The volume of the three-dimensional region E is given by the integral ∫∫∫dv=∫∫∫dxdydz. We need to set the boundaries of each
variable. Let's take a look at our plane 2x+y+z=4, it can be written
as z=4-2x-y. So, we can claim that z varies from 0 to 4-2x-y. Now
we need to find the variety of x&y. To do so, let's find out what
shade or region this plane cover from the x-y plain :
In the x-y plane, z=0, meaning : 2x+y=4 --> y=4-2x. This is the equation of the line which represent the limit of the region. (note that the other limits are the lines x=0 & y=0). Now we know that y
varies from 0 to 4-2x. Good, but what about the boundaries of x ??
Well, when y=0 & z=0, x=2 (because 2x+0+0=4) So we can say that x
varies from 0 to 2. Let's sum up our calculations :
1. dz : 0 --> 4-2x-y
2. dy : 0 --> 4-2x
3. dx : 0 --> 3
Now we will move to calculating the triple integral :
∫∫∫dxdydz==∫∫[4-2x-y]dxdy=∫dx[4y-2xy-(y^2)/2]{y:0-->4-2x}=
=∫[4(4-2x)-2x(4-2x)-((4-2x)^2)/2]dx { x goes from 0 to 2 }
I'll leave it to you as an exercise to calculate this immediate
integral.
I just want to know what I won. 3 days on the property hunting this deer, a new sentenal, a trip down south to suck up some warm weather hog hunting, shopping spree at cabelas, etc. The anxioty is killing me. By the way, I cheated; pulled out my calculus book and solved the simple triple integral to calculate the volume and then solved for the circumference followed by the length. Everyone knows that you can solve for any equation if you have only 1 unknown. Here is the answer
The volume of the three-dimensional region E is given by the integral ∫∫∫dv=∫∫∫dxdydz. We need to set the boundaries of each
variable. Let's take a look at our plane 2x+y+z=4, it can be written
as z=4-2x-y. So, we can claim that z varies from 0 to 4-2x-y. Now
we need to find the variety of x&y. To do so, let's find out what
shade or region this plane cover from the x-y plain :
In the x-y plane, z=0, meaning : 2x+y=4 --> y=4-2x. This is the equation of the line which represent the limit of the region. (note that the other limits are the lines x=0 & y=0). Now we know that y
varies from 0 to 4-2x. Good, but what about the boundaries of x ??
Well, when y=0 & z=0, x=2 (because 2x+0+0=4) So we can say that x
varies from 0 to 2. Let's sum up our calculations :
1. dz : 0 --> 4-2x-y
2. dy : 0 --> 4-2x
3. dx : 0 --> 3
Now we will move to calculating the triple integral :
∫∫∫dxdydz==∫∫[4-2x-y]dxdy=∫dx[4y-2xy-(y^2)/2]{y:0-->4-2x}=
=∫[4(4-2x)-2x(4-2x)-((4-2x)^2)/2]dx { x goes from 0 to 2 }
I'll leave it to you as an exercise to calculate this immediate
integral.
02-28-2009, 01:43 PM
#68
Giant Nontypical
Join Date: Sep 2005
Location: Mont County, Pa
Posts: 7,368
RE: What's his score? (Answer known - will post tomorrow)
ORIGINAL: niehenke
I just want to know what I won. 3 days on the property hunting this deer, a new sentenal, a trip down south to suck up some warm weather hog hunting, shopping spree at cabelas, etc. The anxioty is killing me. By the way, I cheated; pulled out my calculus book and solved the simple triple integral to calculate the volume and then solved for the circumference followed by the length. Everyone knows that you can solve for any equation if you have only 1 unknown. Here is the answer
The volume of the three-dimensional region E is given by the integral ∫∫∫dv=∫∫∫dxdydz. We need to set the boundaries of each
variable. Let's take a look at our plane 2x+y+z=4, it can be written
as z=4-2x-y. So, we can claim that z varies from 0 to 4-2x-y. Now
we need to find the variety of x&y. To do so, let's find out what
shade or region this plane cover from the x-y plain :
In the x-y plane, z=0, meaning : 2x+y=4 --> y=4-2x. This is the equation of the line which represent the limit of the region. (note that the other limits are the lines x=0 & y=0). Now we know that y
varies from 0 to 4-2x. Good, but what about the boundaries of x ??
Well, when y=0 & z=0, x=2 (because 2x+0+0=4) So we can say that x
varies from 0 to 2. Let's sum up our calculations :
1. dz : 0 --> 4-2x-y
2. dy : 0 --> 4-2x
3. dx : 0 --> 3
Now we will move to calculating the triple integral :
∫∫∫dxdydz==∫∫[4-2x-y]dxdy=∫dx[4y-2xy-(y^2)/2]{y:0-->4-2x}=
=∫[4(4-2x)-2x(4-2x)-((4-2x)^2)/2]dx { x goes from 0 to 2 }
I'll leave it to you as an exercise to calculate this immediate
integral.
I just want to know what I won. 3 days on the property hunting this deer, a new sentenal, a trip down south to suck up some warm weather hog hunting, shopping spree at cabelas, etc. The anxioty is killing me. By the way, I cheated; pulled out my calculus book and solved the simple triple integral to calculate the volume and then solved for the circumference followed by the length. Everyone knows that you can solve for any equation if you have only 1 unknown. Here is the answer
The volume of the three-dimensional region E is given by the integral ∫∫∫dv=∫∫∫dxdydz. We need to set the boundaries of each
variable. Let's take a look at our plane 2x+y+z=4, it can be written
as z=4-2x-y. So, we can claim that z varies from 0 to 4-2x-y. Now
we need to find the variety of x&y. To do so, let's find out what
shade or region this plane cover from the x-y plain :
In the x-y plane, z=0, meaning : 2x+y=4 --> y=4-2x. This is the equation of the line which represent the limit of the region. (note that the other limits are the lines x=0 & y=0). Now we know that y
varies from 0 to 4-2x. Good, but what about the boundaries of x ??
Well, when y=0 & z=0, x=2 (because 2x+0+0=4) So we can say that x
varies from 0 to 2. Let's sum up our calculations :
1. dz : 0 --> 4-2x-y
2. dy : 0 --> 4-2x
3. dx : 0 --> 3
Now we will move to calculating the triple integral :
∫∫∫dxdydz==∫∫[4-2x-y]dxdy=∫dx[4y-2xy-(y^2)/2]{y:0-->4-2x}=
=∫[4(4-2x)-2x(4-2x)-((4-2x)^2)/2]dx { x goes from 0 to 2 }
I'll leave it to you as an exercise to calculate this immediate
integral.
Damn, I've got a friggen head ache now.
02-28-2009, 09:42 PM
#69
Nontypical Buck
Join Date: Sep 2006
Location: Illinois
Posts: 3,558
RE: What's his score? (Answer known - will post tomorrow)
Before I even got to page 4 I said 130" in my head. That buck is definitely mature and that is why it is so deceiving in the trail cam pics. Nice deer and a good find getting both sides. How far apart did you find each side?
